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

Optimal. Leaf size=258 \[ -\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{363 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{64 d}-\frac{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

(363*(-1)^(3/4)*a^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(64*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 - (((149*I)/64)*
a^2*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + (107*a^2*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])
/(96*d) + (((17*I)/24)*a^2*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d - (a^2*Tan[c + d*x]^(7/2)*Sqrt[a +
I*a*Tan[c + d*x]])/(4*d)

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Rubi [A]  time = 0.841514, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ -\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{363 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{64 d}-\frac{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(363*(-1)^(3/4)*a^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(64*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 - (((149*I)/64)*
a^2*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + (107*a^2*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])
/(96*d) + (((17*I)/24)*a^2*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d - (a^2*Tan[c + d*x]^(7/2)*Sqrt[a +
I*a*Tan[c + d*x]])/(4*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{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{1}{4} a \int \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{15 a}{2}+\frac{17}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{1}{12} \int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{85 i a^2}{4}+\frac{107}{4} a^2 \tan (c+d x)\right ) \, dx\\ &=\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{\int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-\frac{321 a^3}{8}-\frac{447}{8} i a^3 \tan (c+d x)\right ) \, dx}{24 a}\\ &=-\frac{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{447 i a^4}{16}-\frac{1089}{16} a^4 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{24 a^2}\\ &=-\frac{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{1}{128} (363 i 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 i a^2\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\left (363 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{128 d}+\frac{\left (8 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{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\left (363 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{64 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{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\left (363 i 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 )}{64 d}\\ &=\frac{363 (-1)^{3/4} 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 )}{64 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{149 i a^2 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{107 a^2 \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{96 d}+\frac{17 i a^2 \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 d}-\frac{a^2 \tan ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}\\ \end{align*}

Mathematica [A]  time = 3.96806, size = 232, normalized size = 0.9 \[ \frac{a^2 \sqrt{a+i a \tan (c+d x)} \left (\frac{6 e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (512 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-363 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-i \sqrt{\tan (c+d x)} \sec ^3(c+d x) (70 i \sin (c+d x)+262 i \sin (3 (c+d x))+1205 \cos (c+d x)+583 \cos (3 (c+d x)))\right )}{768 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*((6*Sqrt[-1 + E^((2*I)*(c + d*x))]*(512*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] - 363*Sqr
t[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)))]) - I*Sec[c + d*x]^3*(1205*Cos[c + d*x] + 583*Cos[3*(c + d*x)] +
(70*I)*Sin[c + d*x] + (262*I)*Sin[3*(c + d*x)])*Sqrt[Tan[c + d*x]])*Sqrt[a + I*a*Tan[c + d*x]])/(768*d)

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Maple [B]  time = 0.035, size = 492, normalized size = 1.9 \begin{align*} -{\frac{{a}^{2}}{384\,d}\sqrt{\tan \left ( dx+c \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 96\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}\sqrt{-ia}-272\,i \left ( \tan \left ( dx+c \right ) \right ) ^{2}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}\sqrt{-ia}+384\,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+384\,\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+1089\,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+894\,i\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-428\,\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 ) +1536\,\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))^(5/2),x)

[Out]

-1/384/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(96*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2
)*(I*a)^(1/2)*(-I*a)^(1/2)-272*I*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+3
84*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+384*(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+1089*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+894*I*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-4
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)+1536*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(-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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.61159, size = 2326, normalized size = 9.02 \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))^(5/2),x, algorithm="fricas")

[Out]

1/192*(sqrt(2)*(-845*I*a^2*e^(6*I*d*x + 6*I*c) - 1275*I*a^2*e^(4*I*d*x + 4*I*c) - 1135*I*a^2*e^(2*I*d*x + 2*I*
c) - 321*I*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) + 96*sqrt(131769/4096*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*
I*d*x + 2*I*c) + d)*log(1/363*(363*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))*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) + 128*I*sqrt(131769/4096*I*a^5/d^2
)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - 96*sqrt(131769/4096*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3
*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(1/363*(363*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)
- 128*I*sqrt(131769/4096*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) - 96*sqrt(32*I*a^5/d^2)*(
d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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) + I*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) + 96*sqrt(32*I*a^
5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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) - I*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^(6*
I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.39251, size = 231, normalized size = 0.9 \begin{align*} -\frac{{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a +{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2}\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 (a \tan \left (d x + c\right ) - i \, a\right )} a^{2} + 2 i \, a^{3}} \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))^(5/2),x, algorithm="giac")

[Out]

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