3.1.99 \(\int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [99]

Optimal. Leaf size=204 \[ \frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {368 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}-\frac {472 a (a+i a \tan (c+d x))^{3/2}}{315 d} \]

[Out]

4*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-368/105*a^2*(a+I*a*tan(d*x+c))^(1/2)
/d+92/105*a^2*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d+38/63*I*a^2*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3/d-2/9*
a^2*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^4/d-472/315*a*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]
time = 0.33, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3637, 3678, 3673, 3608, 3561, 212} \begin {gather*} \frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}+\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {368 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d}-\frac {472 a (a+i a \tan (c+d x))^{3/2}}{315 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (368*a^2*Sqrt[a + I*a*Tan[c + d*
x]])/(105*d) + (92*a^2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(105*d) + (((38*I)/63)*a^2*Tan[c + d*x]^3*Sq
rt[a + I*a*Tan[c + d*x]])/d - (2*a^2*Tan[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]])/(9*d) - (472*a*(a + I*a*Tan[c
+ d*x])^(3/2))/(315*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3561

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

Rule 3608

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

Rule 3637

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 3673

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

Rule 3678

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]

Rubi steps

\begin {align*} \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}+\frac {1}{9} (2 a) \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {17 a}{2}+\frac {19}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}+\frac {4}{63} \int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {57 i a^2}{2}+\frac {69}{2} a^2 \tan (c+d x)\right ) \, dx\\ &=\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}+\frac {8 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-69 a^3-\frac {177}{2} i a^3 \tan (c+d x)\right ) \, dx}{315 a}\\ &=\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}-\frac {472 a (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {8 \int \sqrt {a+i a \tan (c+d x)} \left (\frac {177 i a^3}{2}-69 a^3 \tan (c+d x)\right ) \, dx}{315 a}\\ &=-\frac {368 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}-\frac {472 a (a+i a \tan (c+d x))^{3/2}}{315 d}+\left (4 i a^2\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {368 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}-\frac {472 a (a+i a \tan (c+d x))^{3/2}}{315 d}+\frac {\left (8 a^3\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=\frac {4 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {368 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}-\frac {472 a (a+i a \tan (c+d x))^{3/2}}{315 d}\\ \end {align*}

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Mathematica [A]
time = 2.36, size = 176, normalized size = 0.86 \begin {gather*} -\frac {a^2 e^{-i (c+2 d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} (\cos (d x)+i \sin (d x)) \left (-10080 \sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt {1+e^{2 i (c+d x)}} \sec ^5(c+d x) (2331+3012 \cos (2 (c+d x))+961 \cos (4 (c+d x))+282 i \sin (2 (c+d x))+331 i \sin (4 (c+d x)))\right )}{1260 \sqrt {2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1260*(a^2*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(Cos[d*x] +
I*Sin[d*x])*(-10080*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^5*(2331 + 3012*Cos[
2*(c + d*x)] + 961*Cos[4*(c + d*x)] + (282*I)*Sin[2*(c + d*x)] + (331*I)*Sin[4*(c + d*x)])))/(Sqrt[2]*d*E^(I*(
c + 2*d*x)))

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Maple [A]
time = 0.19, size = 131, normalized size = 0.64

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {9}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d \,a^{2}}\)	\(131\)
default	\(-\frac {2 \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {9}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d \,a^{2}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/a^2*(1/9*(a+I*a*tan(d*x+c))^(9/2)-1/7*a*(a+I*a*tan(d*x+c))^(7/2)+1/5*a^2*(a+I*a*tan(d*x+c))^(5/2)+1/3*a^3
*(a+I*a*tan(d*x+c))^(3/2)+2*a^4*(a+I*a*tan(d*x+c))^(1/2)-2*a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2
)*2^(1/2)/a^(1/2)))

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Maxima [A]
time = 0.49, size = 156, normalized size = 0.76 \begin {gather*} -\frac {2 \, {\left (315 \, \sqrt {2} a^{\frac {13}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 45 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 63 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} + 105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 630 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{6}\right )}}{315 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(315*sqrt(2)*a^(13/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*t
an(d*x + c) + a))) + 35*(I*a*tan(d*x + c) + a)^(9/2)*a^2 - 45*(I*a*tan(d*x + c) + a)^(7/2)*a^3 + 63*(I*a*tan(d
*x + c) + a)^(5/2)*a^4 + 105*(I*a*tan(d*x + c) + a)^(3/2)*a^5 + 630*sqrt(I*a*tan(d*x + c) + a)*a^6)/(a^4*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (161) = 322\).
time = 0.50, size = 412, normalized size = 2.02 \begin {gather*} \frac {2 \, {\left (315 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 315 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 2 \, \sqrt {2} {\left (646 \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} + 1647 \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 2331 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 1365 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 315 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(315*sqrt(2)*sqrt(a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) +
4*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^
(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 315*sqrt(2)*sqrt(a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I
*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I*d*x + I*c) - sqrt(a^5/d
^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 2*sqrt(2)*(646*a^2*
e^(9*I*d*x + 9*I*c) + 1647*a^2*e^(7*I*d*x + 7*I*c) + 2331*a^2*e^(5*I*d*x + 5*I*c) + 1365*a^2*e^(3*I*d*x + 3*I*
c) + 315*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I
*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \tan ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 4.29, size = 142, normalized size = 0.70 \begin {gather*} -\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}-\frac {4\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{7\,a\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{9/2}}{9\,a^2\,d}-\frac {2\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}-\frac {\sqrt {2}\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

(2*(a + a*tan(c + d*x)*1i)^(7/2))/(7*a*d) - (4*a^2*(a + a*tan(c + d*x)*1i)^(1/2))/d - (2*(a + a*tan(c + d*x)*1
i)^(5/2))/(5*d) - (2*(a + a*tan(c + d*x)*1i)^(9/2))/(9*a^2*d) - (2*a*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d) - (2
^(1/2)*a^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*4i)/d

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