3.202 \(\int \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=219 \[ -\frac{45 \sqrt [4]{-1} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

(-45*(-1)^(1/4)*a^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(8*d) - ((
4 - 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (19*a^2*Sqrt[Ta
n[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) + (((13*I)/12)*a^2*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]
)/d - (a^2*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d)

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Rubi [A]  time = 0.702864, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3556, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{45 \sqrt [4]{-1} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(-45*(-1)^(1/4)*a^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(8*d) - ((
4 - 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (19*a^2*Sqrt[Ta
n[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) + (((13*I)/12)*a^2*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]
)/d - (a^2*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3597

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*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]

Rule 3601

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

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

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]

Rule 3599

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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])

Rubi steps

\begin{align*} \int \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{1}{3} a \int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{11 a}{2}+\frac{13}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{1}{6} \int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-\frac{39 i a^2}{4}+\frac{57}{4} a^2 \tan (c+d x)\right ) \, dx\\ &=\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{57 a^3}{8}-\frac{135}{8} i a^3 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{6 a}\\ &=\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{1}{16} (45 a) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx-\left (4 a^2\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\left (45 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}+\frac{\left (8 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\left (45 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 d}\\ &=-\frac{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\left (45 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-i a x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}\\ &=-\frac{45 \sqrt [4]{-1} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}-\frac{(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{19 a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{13 i a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 2.92574, size = 258, normalized size = 1.18 \[ \frac{i a^2 e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (64 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-45 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{8 \sqrt{2} d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac{a^2 \sqrt{\tan (c+d x)} \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (26 i \sin (2 (c+d x))+65 \cos (2 (c+d x))+49)}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((I/8)*a^2*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(64*ArcTanh[
E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] - 45*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2
*I)*(c + d*x))]]))/(Sqrt[2]*d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))
]) + (a^2*Sec[c + d*x]^2*(49 + 65*Cos[2*(c + d*x)] + (26*I)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*
Tan[c + d*x]])/(48*d)

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Maple [B]  time = 0.035, size = 448, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}}{48\,d}\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 16\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}+48\,i\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) a-52\,i\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\tan \left ( dx+c \right ) -48\,\sqrt{ia}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) a+192\,i\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) \sqrt{-ia}a-135\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) a\sqrt{-ia}-114\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \right ){\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}{\frac{1}{\sqrt{ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-1/48/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(16*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)
*(I*a)^(1/2)*tan(d*x+c)^2+48*I*(I*a)^(1/2)*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-52*I*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2
)*tan(d*x+c)-48*(I*a)^(1/2)*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*t
an(d*x+c))/(tan(d*x+c)+I))*a+192*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2
)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a-135*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1
/2)+a)/(I*a)^(1/2))*a*(-I*a)^(1/2)-114*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(I*a)^(
1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \tan \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(3/2), x)

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Fricas [B]  time = 2.5315, size = 2060, normalized size = 9.41 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/24*(sqrt(2)*(91*a^2*e^(4*I*d*x + 4*I*c) + 98*a^2*e^(2*I*d*x + 2*I*c) + 39*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 12*sqrt(-2025/64*I*a^5/d^2
)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/45*(45*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)
+ 16*sqrt(-2025/64*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - 12*sqrt(-2025/64*I*a^5/d^2)*(
d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/45*(45*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 1
6*sqrt(-2025/64*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - 12*sqrt(-32*I*a^5/d^2)*(d*e^(4*I
*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I
*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + sqrt(-32*I*
a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + 12*sqrt(-32*I*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*
d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)
)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - sqrt(-32*I*a^5/d^2)*d*e^(2*I*
d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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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(tan(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.37927, size = 203, normalized size = 0.93 \begin{align*} \frac{{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} -{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a\right )} \sqrt{2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 2 \, a^{2}}{\left (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

((I*a*tan(d*x + c) + a)^3 - (I*a*tan(d*x + c) + a)^2*a)*sqrt(2*(I*a*tan(d*x + c) + a)*a - 2*a^2)*((-I*(I*a*tan
(d*x + c) + a)*a + I*a^2)/sqrt((I*a*tan(d*x + c) + a)^2*a^2 - 2*(I*a*tan(d*x + c) + a)*a^3 + a^4) + 1)*log(sqr
t(I*a*tan(d*x + c) + a))/((I*a*tan(d*x + c) + a)*a - 2*a^2)