3.403 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\)

Optimal. Leaf size=325 \[ \frac{2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(a (A-B)+b (A+B)) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(a (A-B)+b (A+B)) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)} \]

[Out]

-(((b*(A - B) - a*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((b*(A - B) - a*
(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) + (2*b^(5/2)*(A*b - a*B)*ArcTan[(Sqrt
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(a^(5/2)*(a^2 + b^2)*d) + ((a*(A - B) + b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[
c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) - ((a*(A - B) + b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) - (2*A)/(3*a*d*Tan[c + d*x]^(3/2)) + (2*(A*b - a*B))/(a^2*d*Sq
rt[Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.974389, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.394, Rules used = {3609, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d \left (a^2+b^2\right )}-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )}+\frac{(a (A-B)+b (A+B)) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}-\frac{(a (A-B)+b (A+B)) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]

[Out]

-(((b*(A - B) - a*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((b*(A - B) - a*
(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) + (2*b^(5/2)*(A*b - a*B)*ArcTan[(Sqrt
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(a^(5/2)*(a^2 + b^2)*d) + ((a*(A - B) + b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[
c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) - ((a*(A - B) + b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) - (2*A)/(3*a*d*Tan[c + d*x]^(3/2)) + (2*(A*b - a*B))/(a^2*d*Sq
rt[Tan[c + d*x]])

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n
+ 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e +
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ
[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

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]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \int \frac{\frac{3}{2} (A b-a B)+\frac{3}{2} a A \tan (c+d x)+\frac{3}{2} A b \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx}{3 a}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}+\frac{4 \int \frac{-\frac{3}{4} \left (a^2 A-A b^2+a b B\right )-\frac{3}{4} a^2 B \tan (c+d x)+\frac{3}{4} b (A b-a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{3 a^2}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}+\frac{4 \int \frac{-\frac{3}{4} a^2 (a A+b B)+\frac{3}{4} a^2 (A b-a B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{3 a^2 \left (a^2+b^2\right )}+\frac{\left (b^3 (A b-a B)\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}+\frac{8 \operatorname{Subst}\left (\int \frac{-\frac{3}{4} a^2 (a A+b B)+\frac{3}{4} a^2 (A b-a B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{3 a^2 \left (a^2+b^2\right ) d}+\frac{\left (b^3 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}+\frac{\left (2 b^3 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^2 \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{(a (A-B)+b (A+B)) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}+\frac{(b (A-B)-a (A+B)) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac{(a (A-B)+b (A+B)) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}+\frac{(a (A-B)+b (A+B)) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}\\ &=\frac{2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}+\frac{(a (A-B)+b (A+B)) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{(a (A-B)+b (A+B)) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}+\frac{(b (A-B)-a (A+B)) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}-\frac{(b (A-B)-a (A+B)) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}\\ &=-\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}+\frac{(b (A-B)-a (A+B)) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right ) d}+\frac{2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}+\frac{(a (A-B)+b (A+B)) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{(a (A-B)+b (A+B)) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right ) d}-\frac{2 A}{3 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b-a B)}{a^2 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.21279, size = 174, normalized size = 0.54 \[ \frac{\frac{6 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} \left (a^2+b^2\right )}-\frac{2 ((3 a B-3 A b) \tan (c+d x)+a A)}{a^2 \tan ^{\frac{3}{2}}(c+d x)}+\frac{3 \sqrt [4]{-1} (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{a-i b}+\frac{3 \sqrt [4]{-1} (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{a+i b}}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]

[Out]

((3*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/(a - I*b) + (6*b^(5/2)*(A*b - a*B)*ArcTan[(Sqr
t[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(a^(5/2)*(a^2 + b^2)) + (3*(-1)^(1/4)*(A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan
[c + d*x]]])/(a + I*b) - (2*(a*A + (-3*A*b + 3*a*B)*Tan[c + d*x]))/(a^2*Tan[c + d*x]^(3/2)))/(3*d)

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Maple [B]  time = 0.052, size = 666, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x)

[Out]

2/d/a^2*b^4/(a^2+b^2)/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A-2/d/a*b^3/(a^2+b^2)/(a*b)^(1/2)*arc
tan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B-2/3*A/a/d/tan(d*x+c)^(3/2)+2/d/a^2/tan(d*x+c)^(1/2)*A*b-2*B/a/d/tan(d*x+
c)^(1/2)-1/2/d/(a^2+b^2)*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a-1/2/d/(a^2+b^2)*A*2^(1/2)*arctan(-1+2^
(1/2)*tan(d*x+c)^(1/2))*a-1/4/d/(a^2+b^2)*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(
d*x+c)^(1/2)+tan(d*x+c)))*a-1/2/d/(a^2+b^2)*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b-1/2/d/(a^2+b^2)*B*2
^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b-1/4/d/(a^2+b^2)*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+
c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*b+1/4/d/(a^2+b^2)*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*
x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*b+1/2/d/(a^2+b^2)*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*
b+1/2/d/(a^2+b^2)*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b-1/4/d/(a^2+b^2)*B*2^(1/2)*ln((1-2^(1/2)*tan(
d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a-1/2/d/(a^2+b^2)*B*2^(1/2)*arctan(1+2^(1/2)
*tan(d*x+c)^(1/2))*a-1/2/d/(a^2+b^2)*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a

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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((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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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((A+B*tan(d*x+c))/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.83159, size = 459, normalized size = 1.41 \begin{align*} -\frac{{\left (\sqrt{2} A a + \sqrt{2} B a - \sqrt{2} A b + \sqrt{2} B b\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{{\left (\sqrt{2} A a + \sqrt{2} B a - \sqrt{2} A b + \sqrt{2} B b\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{{\left (\sqrt{2} A a - \sqrt{2} B a + \sqrt{2} A b + \sqrt{2} B b\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{2} d + b^{2} d\right )}} + \frac{{\left (\sqrt{2} A a - \sqrt{2} B a + \sqrt{2} A b + \sqrt{2} B b\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{2 \,{\left (B a b^{3} - A b^{4}\right )} \arctan \left (\frac{b \sqrt{\tan \left (d x + c\right )}}{\sqrt{a b}}\right )}{{\left (a^{4} d + a^{2} b^{2} d\right )} \sqrt{a b}} - \frac{2 \,{\left (3 \, B a \tan \left (d x + c\right ) - 3 \, A b \tan \left (d x + c\right ) + A a\right )}}{3 \, a^{2} d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(sqrt(2)*A*a + sqrt(2)*B*a - sqrt(2)*A*b + sqrt(2)*B*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)
)))/(a^2*d + b^2*d) - 1/2*(sqrt(2)*A*a + sqrt(2)*B*a - sqrt(2)*A*b + sqrt(2)*B*b)*arctan(-1/2*sqrt(2)*(sqrt(2)
- 2*sqrt(tan(d*x + c))))/(a^2*d + b^2*d) - 1/4*(sqrt(2)*A*a - sqrt(2)*B*a + sqrt(2)*A*b + sqrt(2)*B*b)*log(sq
rt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/(a^2*d + b^2*d) + 1/4*(sqrt(2)*A*a - sqrt(2)*B*a + sqrt(2)*A*b +
sqrt(2)*B*b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/(a^2*d + b^2*d) - 2*(B*a*b^3 - A*b^4)*arctan(
b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^4*d + a^2*b^2*d)*sqrt(a*b)) - 2/3*(3*B*a*tan(d*x + c) - 3*A*b*tan(d*x + c)
+ A*a)/(a^2*d*tan(d*x + c)^(3/2))