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

Optimal. Leaf size=254 \[ -\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{23 (-1)^{3/4} a^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}+\frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d} \]

[Out]

(23*(-1)^(3/4)*a^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(8*d) + ((2
+ 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((I/3)*a^2*Tan[c
+ d*x]^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Tan[c + d*x]^(7/2))/(3*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((
(9*I)/8)*a*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + (7*a*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x
]])/(12*d)

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Rubi [A]  time = 0.846424, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3556, 3595, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{23 (-1)^{3/4} a^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}+\frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*(-1)^(3/4)*a^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(8*d) + ((2
+ 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((I/3)*a^2*Tan[c
+ d*x]^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Tan[c + d*x]^(7/2))/(3*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((
(9*I)/8)*a*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + (7*a*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x
]])/(12*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 3595

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

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{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{1}{3} a \int \frac{\tan ^{\frac{5}{2}}(c+d x) \left (\frac{13 a}{2}+\frac{11}{2} i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{5 i a^2}{2}-\frac{7}{2} a^2 \tan (c+d x)\right ) \, dx}{3 a}\\ &=\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{\int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (\frac{21 a^3}{4}+\frac{27}{4} i a^3 \tan (c+d x)\right ) \, dx}{6 a^2}\\ &=\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{27 i a^4}{8}+\frac{69}{8} a^4 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{6 a^3}\\ &=\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{23}{16} i \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx+(2 i a) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{\left (23 i a^2\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 (4 a^3\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{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{\left (23 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 d}\\ &=\frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}-\frac{\left (23 i a^2\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{23 (-1)^{3/4} a^{3/2} \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{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a^2 \tan ^{\frac{5}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x)}{3 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{7 a \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}\\ \end{align*}

Mathematica [A]  time = 2.79723, size = 213, normalized size = 0.84 \[ \frac{a \sqrt{a+i a \tan (c+d x)} \left (\frac{6 e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (32 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-23 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+2 \sqrt{\tan (c+d x)} \sec ^2(c+d x) (14 \sin (2 (c+d x))-35 i \cos (2 (c+d x))-19 i)\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((6*Sqrt[-1 + E^((2*I)*(c + d*x))]*(32*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] - 23*Sqrt[2]
*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I
)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) + 2*Sec[c + d*x]^2*(-19*I - (35*I)*Cos[2*(c + d*x)] + 14*Sin[2*(c +
d*x)])*Sqrt[Tan[c + d*x]])*Sqrt[a + I*a*Tan[c + d*x]])/(96*d)

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Maple [B]  time = 0.056, size = 447, normalized size = 1.8 \begin{align*} -{\frac{a}{48\,d}\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( -16\,i \left ( \tan \left ( dx+c \right ) \right ) ^{2}\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+24\,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+24\,\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+69\,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+54\,i\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-28\,\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 ) +96\,\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} \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)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-1/48/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(-16*I*tan(d*x+c)^2*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+
c)*(1+I*tan(d*x+c)))^(1/2)+24*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+24*(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+69*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+54*I*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1
+I*tan(d*x+c)))^(1/2)-28*(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)+96*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))/(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(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{3}{2}} \tan \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.46041, size = 2018, normalized size = 7.94 \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)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/24*(sqrt(2)*(-49*I*a*e^(4*I*d*x + 4*I*c) - 38*I*a*e^(2*I*d*x + 2*I*c) - 21*I*a)*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*(d*e^(4*I*d*x + 4*I*c)
+ 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(529/64*I*a^3/d^2)*log(1/23*(23*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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) + 16*
I*sqrt(529/64*I*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) - 12*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2
*I*d*x + 2*I*c) + d)*sqrt(529/64*I*a^3/d^2)*log(1/23*(23*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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) - 16*I*sqrt(529/
64*I*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) - 12*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*
I*c) + d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt((-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) + I*sqrt(8*I*a^3/d^2)*d*e^(2*I*d*x
+ 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) + 12*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(8*I*a^3/d^2
)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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) - I*sqrt(8*I*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2
*I*c)/a))/(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)**(5/2)*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.24757, size = 231, normalized size = 0.91 \begin{align*} \frac{{\left (-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} + 2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a +{\left (a \tan \left (d x + c\right ) - i \, a\right )} a^{2}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a{\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^{3} - 2 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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