3.386 \(\int \frac{1}{(a+b \tan ^4(c+d x))^2} \, dx\)

Optimal. Leaf size=648 \[ \frac{\sqrt [4]{b} \left (\sqrt{a}-3 \sqrt{b}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} d (a+b)}+\frac{\sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} d (a+b)^2}-\frac{\sqrt [4]{b} \left (\sqrt{a}-3 \sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} d (a+b)}-\frac{\sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt{2} a^{3/4} d (a+b)^2}-\frac{\sqrt [4]{b} \left (\sqrt{a}+3 \sqrt{b}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} d (a+b)}-\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} d (a+b)^2}+\frac{\sqrt [4]{b} \left (\sqrt{a}+3 \sqrt{b}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} d (a+b)}+\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} d (a+b)^2}+\frac{b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a d (a+b) \left (a+b \tan ^4(c+d x)\right )}+\frac{x}{(a+b)^2} \]

[Out]

x/(a + b)^2 + ((Sqrt[a] - Sqrt[b])*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[2]*a^(3
/4)*(a + b)^2*d) + ((Sqrt[a] - 3*Sqrt[b])*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(8*Sqrt[
2]*a^(7/4)*(a + b)*d) - ((Sqrt[a] - Sqrt[b])*b^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(2*Sq
rt[2]*a^(3/4)*(a + b)^2*d) - ((Sqrt[a] - 3*Sqrt[b])*b^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)]
)/(8*Sqrt[2]*a^(7/4)*(a + b)*d) - ((Sqrt[a] + Sqrt[b])*b^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d
*x] + Sqrt[b]*Tan[c + d*x]^2])/(4*Sqrt[2]*a^(3/4)*(a + b)^2*d) - ((Sqrt[a] + 3*Sqrt[b])*b^(1/4)*Log[Sqrt[a] -
Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(16*Sqrt[2]*a^(7/4)*(a + b)*d) + ((Sqrt[a] + S
qrt[b])*b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(4*Sqrt[2]*a^(3/
4)*(a + b)^2*d) + ((Sqrt[a] + 3*Sqrt[b])*b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*
Tan[c + d*x]^2])/(16*Sqrt[2]*a^(7/4)*(a + b)*d) + (b*Tan[c + d*x]*(1 - Tan[c + d*x]^2))/(4*a*(a + b)*d*(a + b*
Tan[c + d*x]^4))

________________________________________________________________________________________

Rubi [A]  time = 0.661279, antiderivative size = 648, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {3661, 1239, 203, 1179, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\sqrt [4]{b} \left (\sqrt{a}-3 \sqrt{b}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} d (a+b)}+\frac{\sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} d (a+b)^2}-\frac{\sqrt [4]{b} \left (\sqrt{a}-3 \sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} d (a+b)}-\frac{\sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt{2} a^{3/4} d (a+b)^2}-\frac{\sqrt [4]{b} \left (\sqrt{a}+3 \sqrt{b}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} d (a+b)}-\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b}\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} d (a+b)^2}+\frac{\sqrt [4]{b} \left (\sqrt{a}+3 \sqrt{b}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} d (a+b)}+\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b}\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} d (a+b)^2}+\frac{b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a d (a+b) \left (a+b \tan ^4(c+d x)\right )}+\frac{x}{(a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^4)^(-2),x]

[Out]

x/(a + b)^2 + ((Sqrt[a] - Sqrt[b])*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[2]*a^(3
/4)*(a + b)^2*d) + ((Sqrt[a] - 3*Sqrt[b])*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(8*Sqrt[
2]*a^(7/4)*(a + b)*d) - ((Sqrt[a] - Sqrt[b])*b^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(2*Sq
rt[2]*a^(3/4)*(a + b)^2*d) - ((Sqrt[a] - 3*Sqrt[b])*b^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)]
)/(8*Sqrt[2]*a^(7/4)*(a + b)*d) - ((Sqrt[a] + Sqrt[b])*b^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d
*x] + Sqrt[b]*Tan[c + d*x]^2])/(4*Sqrt[2]*a^(3/4)*(a + b)^2*d) - ((Sqrt[a] + 3*Sqrt[b])*b^(1/4)*Log[Sqrt[a] -
Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(16*Sqrt[2]*a^(7/4)*(a + b)*d) + ((Sqrt[a] + S
qrt[b])*b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(4*Sqrt[2]*a^(3/
4)*(a + b)^2*d) + ((Sqrt[a] + 3*Sqrt[b])*b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*
Tan[c + d*x]^2])/(16*Sqrt[2]*a^(7/4)*(a + b)*d) + (b*Tan[c + d*x]*(1 - Tan[c + d*x]^2))/(4*a*(a + b)*d*(a + b*
Tan[c + d*x]^4))

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 1239

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a +
c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, p, q}, x] && ((IntegerQ[p] && IntegerQ[q]) || IGtQ[p, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1179

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)*(a + c*x^4)^(p + 1))/(
4*a*(p + 1)), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b x^4\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(a+b)^2 \left (1+x^2\right )}+\frac{b-b x^2}{(a+b) \left (a+b x^4\right )^2}+\frac{b-b x^2}{(a+b)^2 \left (a+b x^4\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{(a+b)^2 d}+\frac{\operatorname{Subst}\left (\int \frac{b-b x^2}{a+b x^4} \, dx,x,\tan (c+d x)\right )}{(a+b)^2 d}+\frac{\operatorname{Subst}\left (\int \frac{b-b x^2}{\left (a+b x^4\right )^2} \, dx,x,\tan (c+d x)\right )}{(a+b) d}\\ &=\frac{x}{(a+b)^2}+\frac{b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a (a+b) d \left (a+b \tan ^4(c+d x)\right )}-\frac{\left (1-\frac{\sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a} \sqrt{b}+b x^2}{a+b x^4} \, dx,x,\tan (c+d x)\right )}{2 (a+b)^2 d}+\frac{\left (1+\frac{\sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a} \sqrt{b}-b x^2}{a+b x^4} \, dx,x,\tan (c+d x)\right )}{2 (a+b)^2 d}-\frac{\operatorname{Subst}\left (\int \frac{-3 b+b x^2}{a+b x^4} \, dx,x,\tan (c+d x)\right )}{4 a (a+b) d}\\ &=\frac{x}{(a+b)^2}+\frac{b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a (a+b) d \left (a+b \tan ^4(c+d x)\right )}-\frac{\left (1-\frac{\sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\tan (c+d x)\right )}{4 (a+b)^2 d}-\frac{\left (1-\frac{\sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\tan (c+d x)\right )}{4 (a+b)^2 d}-\frac{\left (\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\tan (c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\tan (c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (1-\frac{3 \sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a} \sqrt{b}+b x^2}{a+b x^4} \, dx,x,\tan (c+d x)\right )}{8 a (a+b) d}+\frac{\left (1+\frac{3 \sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a} \sqrt{b}-b x^2}{a+b x^4} \, dx,x,\tan (c+d x)\right )}{8 a (a+b) d}\\ &=\frac{x}{(a+b)^2}-\frac{\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}+\frac{b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a (a+b) d \left (a+b \tan ^4(c+d x)\right )}-\frac{\left (\left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} (a+b)^2 d}+\frac{\left (\left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (1-\frac{3 \sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\tan (c+d x)\right )}{16 a (a+b) d}-\frac{\left (1-\frac{3 \sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\tan (c+d x)\right )}{16 a (a+b) d}-\frac{\left (\left (\sqrt{a}+3 \sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\tan (c+d x)\right )}{16 \sqrt{2} a^{7/4} (a+b) d}-\frac{\left (\left (\sqrt{a}+3 \sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\tan (c+d x)\right )}{16 \sqrt{2} a^{7/4} (a+b) d}\\ &=\frac{x}{(a+b)^2}+\frac{\left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (\sqrt{a}+3 \sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} (a+b) d}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}+\frac{\left (\sqrt{a}+3 \sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} (a+b) d}+\frac{b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a (a+b) d \left (a+b \tan ^4(c+d x)\right )}-\frac{\left (\left (\sqrt{a}-3 \sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} (a+b) d}+\frac{\left (\left (\sqrt{a}-3 \sqrt{b}\right ) \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} (a+b) d}\\ &=\frac{x}{(a+b)^2}+\frac{\left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} (a+b)^2 d}+\frac{\left (\sqrt{a}-3 \sqrt{b}\right ) \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} (a+b) d}-\frac{\left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (\sqrt{a}-3 \sqrt{b}\right ) \sqrt [4]{b} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} (a+b) d}-\frac{\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}-\frac{\left (\sqrt{a}+3 \sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} (a+b) d}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{4 \sqrt{2} a^{3/4} (a+b)^2 d}+\frac{\left (\sqrt{a}+3 \sqrt{b}\right ) \sqrt [4]{b} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{b} \tan ^2(c+d x)\right )}{16 \sqrt{2} a^{7/4} (a+b) d}+\frac{b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a (a+b) d \left (a+b \tan ^4(c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.27356, size = 598, normalized size = 0.92 \[ -\frac{b \tan ^3(c+d x) \text{Hypergeometric2F1}\left (\frac{3}{4},2,\frac{7}{4},-\frac{b \tan ^4(c+d x)}{a}\right )}{3 a^2 d (a+b)}-\frac{3 \left (\frac{2 \left (\frac{\sqrt{2} b^{3/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{\sqrt [4]{a}}-\frac{\sqrt{2} b^{3/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{a}}\right )}{\sqrt{a}}+\frac{\frac{\sqrt{2} b^{3/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{\sqrt [4]{a}}-\frac{\sqrt{2} b^{3/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{\sqrt [4]{a}}}{\sqrt{a}}\right )}{32 a d (a+b)}+\frac{b \tan (c+d x)}{4 a d (a+b) \left (a+b \tan ^4(c+d x)\right )}+\frac{\tan ^{-1}(\tan (c+d x))}{d (a+b)^2}+\frac{\left (\sqrt{a}-\sqrt{b}\right ) \left (\frac{\sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{\sqrt [4]{a}}-\frac{\sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{a}}\right )}{4 \sqrt{a} d (a+b)^2}-\frac{\left (\sqrt{a}+\sqrt{b}\right ) \left (\frac{\sqrt{2} \sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{\sqrt [4]{a}}-\frac{\sqrt{2} \sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}{\sqrt [4]{a}}\right )}{8 \sqrt{a} d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^4)^(-2),x]

[Out]

ArcTan[Tan[c + d*x]]/((a + b)^2*d) + ((Sqrt[a] - Sqrt[b])*((Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Tan[c
+ d*x])/a^(1/4)])/a^(1/4) - (Sqrt[2]*b^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/a^(1/4)))/(4*
Sqrt[a]*(a + b)^2*d) - ((Sqrt[a] + Sqrt[b])*((Sqrt[2]*b^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*
x] + Sqrt[b]*Tan[c + d*x]^2])/a^(1/4) - (Sqrt[2]*b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] +
Sqrt[b]*Tan[c + d*x]^2])/a^(1/4)))/(8*Sqrt[a]*(a + b)^2*d) - (3*((2*((Sqrt[2]*b^(3/4)*ArcTan[1 - (Sqrt[2]*b^(1
/4)*Tan[c + d*x])/a^(1/4)])/a^(1/4) - (Sqrt[2]*b^(3/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/a^(
1/4)))/Sqrt[a] + ((Sqrt[2]*b^(3/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2
])/a^(1/4) - (Sqrt[2]*b^(3/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/a^
(1/4))/Sqrt[a]))/(32*a*(a + b)*d) - (b*Hypergeometric2F1[3/4, 2, 7/4, -((b*Tan[c + d*x]^4)/a)]*Tan[c + d*x]^3)
/(3*a^2*(a + b)*d) + (b*Tan[c + d*x])/(4*a*(a + b)*d*(a + b*Tan[c + d*x]^4))

________________________________________________________________________________________

Maple [A]  time = 0.037, size = 886, normalized size = 1.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c)^4)^2,x)

[Out]

1/d/(a+b)^2*arctan(tan(d*x+c))-1/4/d/(a+b)^2*b/(a+b*tan(d*x+c)^4)*tan(d*x+c)^3-1/4/d/(a+b)^2*b^2/(a+b*tan(d*x+
c)^4)/a*tan(d*x+c)^3+1/4/d/(a+b)^2*b/(a+b*tan(d*x+c)^4)*tan(d*x+c)+1/4/d/(a+b)^2*b^2/(a+b*tan(d*x+c)^4)/a*tan(
d*x+c)+7/16/d/(a+b)^2*b/a*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1)+3/16/d/(a+b)^2*b^2/a^2*
(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1)-7/16/d/(a+b)^2*b/a*(a/b)^(1/4)*2^(1/2)*arctan(-2^
(1/2)/(a/b)^(1/4)*tan(d*x+c)+1)-3/16/d/(a+b)^2*b^2/a^2*(a/b)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(a/b)^(1/4)*tan(d*x
+c)+1)+7/32/d/(a+b)^2*b/a*(a/b)^(1/4)*2^(1/2)*ln((tan(d*x+c)^2+(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2))/(ta
n(d*x+c)^2-(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2)))+3/32/d/(a+b)^2*b^2/a^2*(a/b)^(1/4)*2^(1/2)*ln((tan(d*x
+c)^2+(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2))/(tan(d*x+c)^2-(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2)))-5
/32/d/(a+b)^2/(a/b)^(1/4)*2^(1/2)*ln((tan(d*x+c)^2-(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2))/(tan(d*x+c)^2+(
a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2)))-1/32/d/(a+b)^2*b/a/(a/b)^(1/4)*2^(1/2)*ln((tan(d*x+c)^2-(a/b)^(1/4
)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2))/(tan(d*x+c)^2+(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2)))-5/16/d/(a+b)^2/(a
/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1)-1/16/d/(a+b)^2*b/a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(a/b)^(1/4)*tan(d*x+c)+1)+5/16/d/(a+b)^2/(a/b)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1)+1/16
/d/(a+b)^2*b/a/(a/b)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+tan(d*x+c)^4*b)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 3.67707, size = 9750, normalized size = 15.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+tan(d*x+c)^4*b)^2,x, algorithm="fricas")

[Out]

1/32*(32*a*b*d*x*tan(d*x + c)^4 + 32*a^2*d*x - 8*(a*b + b^2)*tan(d*x + c)^3 + ((a^3*b + 2*a^2*b^2 + a*b^3)*d*t
an(d*x + c)^4 + (a^4 + 2*a^3*b + a^2*b^2)*d)*sqrt(((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2*sqrt(
-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b
+ 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)) + 70*a^2*b
+ 44*a*b^2 + 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2))*log((625*a^5 - 750*a^4*b - 1376*a
^3*b^2 - 594*a^2*b^3 - 81*a*b^4 + (625*a^4*b - 750*a^3*b^2 - 1376*a^2*b^3 - 594*a*b^4 - 81*b^5)*tan(d*x + c)^2
+ 2*(2*(a^11 + 5*a^10*b + 10*a^9*b^2 + 10*a^8*b^3 + 5*a^7*b^4 + a^6*b^5)*d^3*sqrt(-(625*a^6*b - 1950*a^5*b^2
- 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^
3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4))*tan(d*x + c) + (125*a^7 + 5*a^6*b - 44
2*a^5*b^2 - 490*a^4*b^3 - 195*a^3*b^4 - 27*a^2*b^5)*d*tan(d*x + c))*sqrt(((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b
^3 + a^3*b^4)*d^2*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81
*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7
*b^8)*d^4)) + 70*a^2*b + 44*a*b^2 + 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2)) + ((25*a^9
+ 109*a^8*b + 186*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2*tan(d*x + c)^2 - (25*a^9 + 109*a^8*b +
186*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2)*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*
a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*
a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)))/(tan(d*x + c)^2 + 1)) - ((a^3*b + 2*a^2*b^2 + a*b^3)*d*tan
(d*x + c)^4 + (a^4 + 2*a^3*b + a^2*b^2)*d)*sqrt(((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2*sqrt(-(
625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b +
28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)) + 70*a^2*b +
44*a*b^2 + 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2))*log((625*a^5 - 750*a^4*b - 1376*a^3
*b^2 - 594*a^2*b^3 - 81*a*b^4 + (625*a^4*b - 750*a^3*b^2 - 1376*a^2*b^3 - 594*a*b^4 - 81*b^5)*tan(d*x + c)^2 -
2*(2*(a^11 + 5*a^10*b + 10*a^9*b^2 + 10*a^8*b^3 + 5*a^7*b^4 + a^6*b^5)*d^3*sqrt(-(625*a^6*b - 1950*a^5*b^2 -
529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3
+ 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4))*tan(d*x + c) + (125*a^7 + 5*a^6*b - 442*
a^5*b^2 - 490*a^4*b^3 - 195*a^3*b^4 - 27*a^2*b^5)*d*tan(d*x + c))*sqrt(((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3
+ a^3*b^4)*d^2*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b
^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b
^8)*d^4)) + 70*a^2*b + 44*a*b^2 + 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2)) + ((25*a^9 +
109*a^8*b + 186*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2*tan(d*x + c)^2 - (25*a^9 + 109*a^8*b + 18
6*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2)*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^
3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^
10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)))/(tan(d*x + c)^2 + 1)) - ((a^3*b + 2*a^2*b^2 + a*b^3)*d*tan(d
*x + c)^4 + (a^4 + 2*a^3*b + a^2*b^2)*d)*sqrt(-((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2*sqrt(-(6
25*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b +
28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)) - 70*a^2*b - 4
4*a*b^2 - 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2))*log(-(625*a^5 - 750*a^4*b - 1376*a^3
*b^2 - 594*a^2*b^3 - 81*a*b^4 + (625*a^4*b - 750*a^3*b^2 - 1376*a^2*b^3 - 594*a*b^4 - 81*b^5)*tan(d*x + c)^2 +
2*(2*(a^11 + 5*a^10*b + 10*a^9*b^2 + 10*a^8*b^3 + 5*a^7*b^4 + a^6*b^5)*d^3*sqrt(-(625*a^6*b - 1950*a^5*b^2 -
529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3
+ 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4))*tan(d*x + c) - (125*a^7 + 5*a^6*b - 442*
a^5*b^2 - 490*a^4*b^3 - 195*a^3*b^4 - 27*a^2*b^5)*d*tan(d*x + c))*sqrt(-((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^
3 + a^3*b^4)*d^2*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*
b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*
b^8)*d^4)) - 70*a^2*b - 44*a*b^2 - 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2)) - ((25*a^9
+ 109*a^8*b + 186*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2*tan(d*x + c)^2 - (25*a^9 + 109*a^8*b + 1
86*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2)*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a
^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a
^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)))/(tan(d*x + c)^2 + 1)) + ((a^3*b + 2*a^2*b^2 + a*b^3)*d*tan(
d*x + c)^4 + (a^4 + 2*a^3*b + a^2*b^2)*d)*sqrt(-((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2*sqrt(-(
625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b +
28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)) - 70*a^2*b -
44*a*b^2 - 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2))*log(-(625*a^5 - 750*a^4*b - 1376*a^
3*b^2 - 594*a^2*b^3 - 81*a*b^4 + (625*a^4*b - 750*a^3*b^2 - 1376*a^2*b^3 - 594*a*b^4 - 81*b^5)*tan(d*x + c)^2
- 2*(2*(a^11 + 5*a^10*b + 10*a^9*b^2 + 10*a^8*b^3 + 5*a^7*b^4 + a^6*b^5)*d^3*sqrt(-(625*a^6*b - 1950*a^5*b^2 -
529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3
+ 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4))*tan(d*x + c) - (125*a^7 + 5*a^6*b - 442
*a^5*b^2 - 490*a^4*b^3 - 195*a^3*b^4 - 27*a^2*b^5)*d*tan(d*x + c))*sqrt(-((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b
^3 + a^3*b^4)*d^2*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81
*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7
*b^8)*d^4)) - 70*a^2*b - 44*a*b^2 - 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2)) - ((25*a^9
+ 109*a^8*b + 186*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2*tan(d*x + c)^2 - (25*a^9 + 109*a^8*b +
186*a^7*b^2 + 154*a^6*b^3 + 61*a^5*b^4 + 9*a^4*b^5)*d^2)*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*
a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b^4 + 56*
a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)))/(tan(d*x + c)^2 + 1)) + 8*(a*b + b^2)*tan(d*x + c))/((a^3*
b + 2*a^2*b^2 + a*b^3)*d*tan(d*x + c)^4 + (a^4 + 2*a^3*b + a^2*b^2)*d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+tan(d*x+c)**4*b)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.39658, size = 698, normalized size = 1.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+tan(d*x+c)^4*b)^2,x, algorithm="giac")

[Out]

-1/16*(2*(pi*floor((d*x + c)/pi + 1/2) + arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*tan(d*x + c))/(a/b)^(1/4)
))*((a*b^3)^(3/4)*(5*a + b) - (a*b^3)^(1/4)*(7*a*b^2 + 3*b^3))/(sqrt(2)*a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqrt(2)*
a^2*b^4) + 2*(pi*floor((d*x + c)/pi + 1/2) + arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*tan(d*x + c))/(a/b)^
(1/4)))*((a*b^3)^(3/4)*(5*a + b) - (a*b^3)^(1/4)*(7*a*b^2 + 3*b^3))/(sqrt(2)*a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqr
t(2)*a^2*b^4) - ((a*b^3)^(3/4)*(5*a + b) + (a*b^3)^(1/4)*(7*a*b^2 + 3*b^3))*log(tan(d*x + c)^2 + sqrt(2)*(a/b)
^(1/4)*tan(d*x + c) + sqrt(a/b))/(sqrt(2)*a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqrt(2)*a^2*b^4) + ((a*b^3)^(3/4)*(5*a
+ b) + (a*b^3)^(1/4)*(7*a*b^2 + 3*b^3))*log(tan(d*x + c)^2 - sqrt(2)*(a/b)^(1/4)*tan(d*x + c) + sqrt(a/b))/(s
qrt(2)*a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqrt(2)*a^2*b^4) - 16*(d*x + c)/(a^2 + 2*a*b + b^2) + 4*(b*tan(d*x + c)^3
- b*tan(d*x + c))/((b*tan(d*x + c)^4 + a)*(a^2 + a*b)))/d