3.365 \(\int \frac{(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=279 \[ -\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}-\frac{d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 f}+\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{2 \sqrt{2} a^2 f}+\frac{d^{3/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 f}-\frac{d^{3/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2 \tan (e+f x)+a^2\right )} \]

[Out]

-(d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*f) - (d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])
/Sqrt[d]])/(2*Sqrt[2]*a^2*f) + (d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(2*Sqrt[2]*a^2*f)
+ (d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) - (d^(3/2)*Lo
g[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) + (d*Sqrt[d*Tan[e + f*x]])
/(2*f*(a^2 + a^2*Tan[e + f*x]))

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Rubi [A]  time = 0.508814, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3567, 3653, 12, 16, 3476, 329, 297, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}-\frac{d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 f}+\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{2 \sqrt{2} a^2 f}+\frac{d^{3/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 f}-\frac{d^{3/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2 \tan (e+f x)+a^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(3/2)/(a + a*Tan[e + f*x])^2,x]

[Out]

-(d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*f) - (d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])
/Sqrt[d]])/(2*Sqrt[2]*a^2*f) + (d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(2*Sqrt[2]*a^2*f)
+ (d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) - (d^(3/2)*Lo
g[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) + (d*Sqrt[d*Tan[e + f*x]])
/(2*f*(a^2 + a^2*Tan[e + f*x]))

Rule 3567

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

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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{(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx &=\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\int \frac{\frac{a d^2}{2}-a d^2 \tan (e+f x)-\frac{1}{2} a d^2 \tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{2 a^2}\\ &=\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\int -\frac{2 a^2 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{4 a^4}-\frac{d^2 \int \frac{1+\tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a}\\ &=\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac{d^2 \int \frac{\tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{2 a^2}-\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac{d \int \sqrt{d \tan (e+f x)} \, dx}{2 a^2}-\frac{d \operatorname{Subst}\left (\int \frac{1}{a+\frac{a x^2}{d}} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 a f}\\ &=-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac{d^2 \operatorname{Subst}\left (\int \frac{\sqrt{x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{2 a^2 f}\\ &=-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac{d^2 \operatorname{Subst}\left (\int \frac{x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 f}\\ &=-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac{d^2 \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 a^2 f}+\frac{d^2 \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 a^2 f}\\ &=-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac{d^{3/2} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}+2 x}{-d-\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 f}+\frac{d^{3/2} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}-2 x}{-d+\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 f}+\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^2 f}+\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^2 f}\\ &=-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}+\frac{d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 f}-\frac{d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac{d^{3/2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 f}-\frac{d^{3/2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 f}\\ &=-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 f}-\frac{d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 f}+\frac{d^{3/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 f}+\frac{d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 f}-\frac{d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 f}+\frac{d \sqrt{d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.92393, size = 229, normalized size = 0.82 \[ \frac{\sec (e+f x) (d \tan (e+f x))^{3/2} (\sin (e+f x)+\cos (e+f x))^2 \left (\frac{2 \cot (e+f x)}{\sin (e+f x)+\cos (e+f x)}-\frac{\csc (e+f x) \left (2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )-2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )+4 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )-\sqrt{2} \log \left (-\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}-1\right )+\sqrt{2} \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )}{2 \sqrt{\tan (e+f x)}}\right )}{4 a^2 f (\tan (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(3/2)/(a + a*Tan[e + f*x])^2,x]

[Out]

(Sec[e + f*x]*(Cos[e + f*x] + Sin[e + f*x])^2*((2*Cot[e + f*x])/(Cos[e + f*x] + Sin[e + f*x]) - (Csc[e + f*x]*
(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + 4*ArcTa
n[Sqrt[Tan[e + f*x]]] - Sqrt[2]*Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] + Sqrt[2]*Log[1 + Sqrt[2]*
Sqrt[Tan[e + f*x]] + Tan[e + f*x]]))/(2*Sqrt[Tan[e + f*x]]))*(d*Tan[e + f*x])^(3/2))/(4*a^2*f*(1 + Tan[e + f*x
])^2)

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Maple [A]  time = 0.03, size = 234, normalized size = 0.8 \begin{align*}{\frac{{d}^{2}\sqrt{2}}{8\,f{a}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{d}^{2}\sqrt{2}}{4\,f{a}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{d}^{2}\sqrt{2}}{4\,f{a}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{d}^{2}}{2\,f{a}^{2} \left ( d\tan \left ( fx+e \right ) +d \right ) }\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{1}{2\,f{a}^{2}}{d}^{{\frac{3}{2}}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x)

[Out]

1/8/f/a^2*d^2/(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*ta
n(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/4/f/a^2*d^2/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/4/f/a^2*d^2/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*
x+e))^(1/2)+1)+1/2/f/a^2*d^2*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)-1/2*d^(3/2)*arctan((d*tan(f*x+e))^(1/2)/d^(
1/2))/a^2/f

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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((d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.67936, size = 4344, normalized size = 15.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/8*((2*d*cos(f*x + e)*sin(f*x + e) + d)*sqrt(-d)*log(-(6*d^2*cos(f*x + e)*sin(f*x + e) - d^2 - 4*(d*cos(f*x
+ e)^2 - d*cos(f*x + e)*sin(f*x + e))*sqrt(-d)*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(2*cos(f*x + e)*sin(f*x + e)
+ 1)) - 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*
a^2*d^4*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(a^8*f^4))^(1/4) + d^6 - sqrt(2)*a^2*f*sqrt((sqrt(2)*a^6*d^4*
f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(a^8*f^4))^(3/4)*cos(f*x + e) + a^4*d^6*f^2*sqrt(d^6/(a^8*f^4))*cos
(f*x + e) + d^9*sin(f*x + e))/cos(f*x + e))*(d^6/(a^8*f^4))^(1/4))/d^6) - 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(
f*x + e) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*a^2*d^4*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(
d^6/(a^8*f^4))^(1/4) - d^6 - sqrt(2)*a^2*f*sqrt(-(sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(
a^8*f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt(d^6/(a^8*f^4))*cos(f*x + e) - d^9*sin(f*x + e))/cos(f*x + e))*
(d^6/(a^8*f^4))^(1/4))/d^6) - (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))^(1/4
)*log((sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(a^8*f^4))^(3/4)*cos(f*x + e) + a^4*d^6*f^2*
sqrt(d^6/(a^8*f^4))*cos(f*x + e) + d^9*sin(f*x + e))/cos(f*x + e)) + (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e
) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))^(1/4)*log(-(sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(a^8
*f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt(d^6/(a^8*f^4))*cos(f*x + e) - d^9*sin(f*x + e))/cos(f*x + e)) + 4
*(d*cos(f*x + e)^2 + d*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(2*a^2*f*cos(f*x + e)*sin
(f*x + e) + a^2*f), -1/8*(4*(2*d*cos(f*x + e)*sin(f*x + e) + d)*sqrt(d)*arctan(sqrt(d*sin(f*x + e)/cos(f*x + e
))/sqrt(d)) + 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))^(1/4)*arctan(-(sqr
t(2)*a^2*d^4*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(a^8*f^4))^(1/4) + d^6 - sqrt(2)*a^2*f*sqrt((sqrt(2)*a^6
*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(a^8*f^4))^(3/4)*cos(f*x + e) + a^4*d^6*f^2*sqrt(d^6/(a^8*f^4)
)*cos(f*x + e) + d^9*sin(f*x + e))/cos(f*x + e))*(d^6/(a^8*f^4))^(1/4))/d^6) + 4*(2*sqrt(2)*a^2*f*cos(f*x + e)
*sin(f*x + e) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*a^2*d^4*f*sqrt(d*sin(f*x + e)/cos(f*x +
e))*(d^6/(a^8*f^4))^(1/4) - d^6 - sqrt(2)*a^2*f*sqrt(-(sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(
d^6/(a^8*f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt(d^6/(a^8*f^4))*cos(f*x + e) - d^9*sin(f*x + e))/cos(f*x +
e))*(d^6/(a^8*f^4))^(1/4))/d^6) + (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))
^(1/4)*log((sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6/(a^8*f^4))^(3/4)*cos(f*x + e) + a^4*d^6
*f^2*sqrt(d^6/(a^8*f^4))*cos(f*x + e) + d^9*sin(f*x + e))/cos(f*x + e)) - (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*
x + e) + sqrt(2)*a^2*f)*(d^6/(a^8*f^4))^(1/4)*log(-(sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^6
/(a^8*f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt(d^6/(a^8*f^4))*cos(f*x + e) - d^9*sin(f*x + e))/cos(f*x + e)
) - 4*(d*cos(f*x + e)^2 + d*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(2*a^2*f*cos(f*x + e
)*sin(f*x + e) + a^2*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}{\tan ^{2}{\left (e + f x \right )} + 2 \tan{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(3/2)/(a+a*tan(f*x+e))**2,x)

[Out]

Integral((d*tan(e + f*x))**(3/2)/(tan(e + f*x)**2 + 2*tan(e + f*x) + 1), x)/a**2

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Giac [A]  time = 1.19247, size = 362, normalized size = 1.3 \begin{align*} \frac{1}{8} \, d^{3}{\left (\frac{2 \, \sqrt{2}{\left | d \right |}^{\frac{3}{2}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} + 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{a^{2} d^{3} f} + \frac{2 \, \sqrt{2}{\left | d \right |}^{\frac{3}{2}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} - 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{a^{2} d^{3} f} - \frac{\sqrt{2}{\left | d \right |}^{\frac{3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{a^{2} d^{3} f} + \frac{\sqrt{2}{\left | d \right |}^{\frac{3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{a^{2} d^{3} f} - \frac{4 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{2} d^{\frac{3}{2}} f} + \frac{4 \, \sqrt{d \tan \left (f x + e\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )} a^{2} d f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/8*d^3*(2*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)
))/(a^2*d^3*f) + 2*sqrt(2)*abs(d)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sq
rt(abs(d)))/(a^2*d^3*f) - sqrt(2)*abs(d)^(3/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d))
+ abs(d))/(a^2*d^3*f) + sqrt(2)*abs(d)^(3/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) +
abs(d))/(a^2*d^3*f) - 4*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*d^(3/2)*f) + 4*sqrt(d*tan(f*x + e))/((d*tan(
f*x + e) + d)*a^2*d*f))