3.2.17 \(\int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [117]

Optimal. Leaf size=205 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {78 \sqrt {a+i a \tan (c+d x)}}{5 a^2 d}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}-\frac {151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d} \]

[Out]

-1/4*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)+78/5*(a+I*a*tan(d*x+c))^(1/2)/a^2
/d-39/10*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/a^2/d+19/6*I*tan(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^(1/2)-1/3*tan(
d*x+c)^4/d/(a+I*a*tan(d*x+c))^(3/2)-151/30*(a+I*a*tan(d*x+c))^(3/2)/a^3/d

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Rubi [A]
time = 0.32, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3639, 3676, 3678, 3673, 3608, 3561, 212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}+\frac {78 \sqrt {a+i a \tan (c+d x)}}{5 a^2 d}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(Sqrt[2]*a^(3/2)*d) - Tan[c + d*x]^4/(3*d*(a + I*a*
Tan[c + d*x])^(3/2)) + (((19*I)/6)*Tan[c + d*x]^3)/(a*d*Sqrt[a + I*a*Tan[c + d*x]]) + (78*Sqrt[a + I*a*Tan[c +
d*x]])/(5*a^2*d) - (39*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(10*a^2*d) - (151*(a + I*a*Tan[c + d*x])^(3
/2))/(30*a^3*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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(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 - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[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 3676

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 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 \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan ^3(c+d x) \left (-4 a+\frac {11}{2} i a \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {57 i a^2}{2}-\frac {117}{4} a^2 \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}+\frac {2 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {117 a^3}{2}-\frac {453}{8} i a^3 \tan (c+d x)\right ) \, dx}{15 a^5}\\ &=-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}-\frac {151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}+\frac {2 \int \sqrt {a+i a \tan (c+d x)} \left (\frac {453 i a^3}{8}+\frac {117}{2} a^3 \tan (c+d x)\right ) \, dx}{15 a^5}\\ &=-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {78 \sqrt {a+i a \tan (c+d x)}}{5 a^2 d}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}-\frac {151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}-\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {78 \sqrt {a+i a \tan (c+d x)}}{5 a^2 d}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}-\frac {151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}-\frac {\text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{2 a d}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {19 i \tan ^3(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {78 \sqrt {a+i a \tan (c+d x)}}{5 a^2 d}-\frac {39 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{10 a^2 d}-\frac {151 (a+i a \tan (c+d x))^{3/2}}{30 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 1.87, size = 149, normalized size = 0.73 \begin {gather*} \frac {e^{-2 i (c+d x)} \left (-5+115 e^{2 i (c+d x)}+855 e^{4 i (c+d x)}+1105 e^{6 i (c+d x)}+466 e^{8 i (c+d x)}-15 e^{3 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{30 a d \left (1+e^{2 i (c+d x)}\right )^3 \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5 + 115*E^((2*I)*(c + d*x)) + 855*E^((4*I)*(c + d*x)) + 1105*E^((6*I)*(c + d*x)) + 466*E^((8*I)*(c + d*x)) -
15*E^((3*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(5/2)*ArcSinh[E^(I*(c + d*x))])/(30*a*d*E^((2*I)*(c + d*x))*
(1 + E^((2*I)*(c + d*x)))^3*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 0.18, size = 131, normalized size = 0.64

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+8 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {9 a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{4}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}}{a^{4} d}\)	\(131\)
default	\(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+8 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {9 a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{4}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}}{a^{4} d}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^4*(1/5*(a+I*a*tan(d*x+c))^(5/2)-a*(a+I*a*tan(d*x+c))^(3/2)+4*a^2*(a+I*a*tan(d*x+c))^(1/2)+9/4*a^3/(a+I*a
*tan(d*x+c))^(1/2)-1/6*a^4/(a+I*a*tan(d*x+c))^(3/2)-1/8*a^(5/2)*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 = 0.51, size = 157, normalized size = 0.77 \begin {gather*} \frac {15 \, \sqrt {2} a^{\frac {9}{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 ) + 48 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 240 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 960 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4} + \frac {20 \, {\left (27 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} - 2 \, a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{120 \, a^{6} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a))) + 48*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 240*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 960*sqrt(I*a*ta
n(d*x + c) + a)*a^4 + 20*(27*(I*a*tan(d*x + c) + a)*a^5 - 2*a^6)/(I*a*tan(d*x + c) + a)^(3/2))/(a^6*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. 388 vs. \(2 (160) = 320\).
time = 0.48, size = 388, normalized size = 1.89 \begin {gather*} -\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (466 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1105 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 855 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 115 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 5\right )}}{60 \, {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt(1/2)*(a^2*d*e^(7*I*d*x + 7*I*c) + 2*a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt
(1/(a^3*d^2))*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 15*sqrt(1/2)*(a^2*d*e^(7*I*d*x + 7*I*c) + 2*a^2*d*e^
(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x +
2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) -
sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(466*e^(8*I*d*x + 8*I*c) + 1105*e^(6*I*d*x + 6*I*c) + 855*e^(4*I*d*x
+ 4*I*c) + 115*e^(2*I*d*x + 2*I*c) - 5))/(a^2*d*e^(7*I*d*x + 7*I*c) + 2*a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(
3*I*d*x + 3*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.20, size = 138, normalized size = 0.67 \begin {gather*} \frac {8\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{a^3\,d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^4\,d}+\frac {\sqrt {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 )\,1{}\mathrm {i}}{4\,a^{3/2}\,d}+\frac {\frac {25\,a}{6}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,9{}\mathrm {i}}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*(a + a*tan(c + d*x)*1i)^(1/2))/(a^2*d) - (2*(a + a*tan(c + d*x)*1i)^(3/2))/(a^3*d) + (2*(a + a*tan(c + d*x)
*1i)^(5/2))/(5*a^4*d) + (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(4*a^(3/2)*d
) + ((25*a)/6 + (a*tan(c + d*x)*9i)/2)/(a*d*(a + a*tan(c + d*x)*1i)^(3/2))

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