3.3.100 \(\int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [300]

Optimal. Leaf size=147 \[ \frac {256 i a^4 \sec ^5(c+d x)}{1155 d (a+i a \tan (c+d x))^{5/2}}+\frac {64 i a^3 \sec ^5(c+d x)}{231 d (a+i a \tan (c+d x))^{3/2}}+\frac {8 i a^2 \sec ^5(c+d x)}{33 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d} \]

[Out]

8/33*I*a^2*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(1/2)+2/11*I*a*sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2)/d+256/1155*I
*a^4*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(5/2)+64/231*I*a^3*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.19, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \begin {gather*} \frac {256 i a^4 \sec ^5(c+d x)}{1155 d (a+i a \tan (c+d x))^{5/2}}+\frac {64 i a^3 \sec ^5(c+d x)}{231 d (a+i a \tan (c+d x))^{3/2}}+\frac {8 i a^2 \sec ^5(c+d x)}{33 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((256*I)/1155)*a^4*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((64*I)/231)*a^3*Sec[c + d*x]^5)/(d*(a
+ I*a*Tan[c + d*x])^(3/2)) + (((8*I)/33)*a^2*Sec[c + d*x]^5)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((2*I)/11)*a*S
ec[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 3574

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(
d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3575

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}+\frac {1}{11} (12 a) \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {8 i a^2 \sec ^5(c+d x)}{33 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}+\frac {1}{33} \left (32 a^2\right ) \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {64 i a^3 \sec ^5(c+d x)}{231 d (a+i a \tan (c+d x))^{3/2}}+\frac {8 i a^2 \sec ^5(c+d x)}{33 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}+\frac {1}{231} \left (128 a^3\right ) \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\\ &=\frac {256 i a^4 \sec ^5(c+d x)}{1155 d (a+i a \tan (c+d x))^{5/2}}+\frac {64 i a^3 \sec ^5(c+d x)}{231 d (a+i a \tan (c+d x))^{3/2}}+\frac {8 i a^2 \sec ^5(c+d x)}{33 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}\\ \end {align*}

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Mathematica [A]
time = 1.13, size = 109, normalized size = 0.74 \begin {gather*} \frac {2 a \sec ^4(c+d x) (\cos (d x)-i \sin (d x)) (i \cos (3 c+2 d x)+\sin (3 c+2 d x)) (39+494 \cos (2 (c+d x))+215 i \sec (c+d x) \sin (3 (c+d x))+110 i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{1155 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sec[c + d*x]^4*(Cos[d*x] - I*Sin[d*x])*(I*Cos[3*c + 2*d*x] + Sin[3*c + 2*d*x])*(39 + 494*Cos[2*(c + d*x)]
+ (215*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (110*I)*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(1155*d)

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Maple [A]
time = 0.76, size = 125, normalized size = 0.85

method	result	size
default	\(\frac {2 \left (512 i \left (\cos ^{6}\left (d x +c \right )\right )+512 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )-64 i \left (\cos ^{4}\left (d x +c \right )\right )+192 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-20 i \left (\cos ^{2}\left (d x +c \right )\right )+140 \sin \left (d x +c \right ) \cos \left (d x +c \right )+105 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{1155 d \cos \left (d x +c \right )^{5}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155/d*(512*I*cos(d*x+c)^6+512*sin(d*x+c)*cos(d*x+c)^5-64*I*cos(d*x+c)^4+192*sin(d*x+c)*cos(d*x+c)^3-20*I*co
s(d*x+c)^2+140*sin(d*x+c)*cos(d*x+c)+105*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^5*a

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 994 vs. \(2 (115) = 230\).
time = 10.55, size = 994, normalized size = 6.76 \begin {gather*} \frac {64 \, {\left (231 i \, \sqrt {2} a \cos \left (6 \, d x + 6 \, c\right ) + 198 i \, \sqrt {2} a \cos \left (4 \, d x + 4 \, c\right ) + 88 i \, \sqrt {2} a \cos \left (2 \, d x + 2 \, c\right ) - 231 \, \sqrt {2} a \sin \left (6 \, d x + 6 \, c\right ) - 198 \, \sqrt {2} a \sin \left (4 \, d x + 4 \, c\right ) - 88 \, \sqrt {2} a \sin \left (2 \, d x + 2 \, c\right ) + 16 i \, \sqrt {2} a\right )} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sqrt {a}}{1155 \, {\left ({\left (4 \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 i \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (8 \, d x + 8 \, c\right ) + 4 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + 6 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (8 \, d x + 8 \, c\right ) + 4 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (6 \, d x + 6 \, c\right ) + 6 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (4 \, d x + 4 \, c\right ) + 4 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right ) + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (4 i \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (4 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} - 4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \cos \left (8 \, d x + 8 \, c\right ) + 4 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \cos \left (6 \, d x + 6 \, c\right ) + 6 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 i \, \cos \left (2 \, d x + 2 \, c\right )^{2} - {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (8 \, d x + 8 \, c\right ) - 4 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) - 6 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 4 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + 6 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

64/1155*(231*I*sqrt(2)*a*cos(6*d*x + 6*c) + 198*I*sqrt(2)*a*cos(4*d*x + 4*c) + 88*I*sqrt(2)*a*cos(2*d*x + 2*c)
- 231*sqrt(2)*a*sin(6*d*x + 6*c) - 198*sqrt(2)*a*sin(4*d*x + 4*c) - 88*sqrt(2)*a*sin(2*d*x + 2*c) + 16*I*sqrt
(2)*a)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a)/(((4*cos(2*d*x + 2*c)^
3 + (4*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 4*I*sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2
*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)*cos(6*d*x + 6*c) + 6*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x +
4*c) + 9*cos(2*d*x + 2*c)^2 + (I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(8*d
*x + 8*c) + 4*(I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(6*d*x + 6*c) + 6*(I
*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(4*d*x + 4*c) + 4*(I*cos(2*d*x + 2*c
)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(2*d*x + 2*c) + 6*cos(2*d*x + 2*c) + 1)*cos(3/2*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c) + 1)) + (4*I*cos(2*d*x + 2*c)^3 + (4*I*cos(2*d*x + 2*c) + I)*sin(2*d*x + 2*c)^2 - 4*sin(2*d*x
+ 2*c)^3 + (I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(8*d*x + 8*c) + 4*(I*co
s(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(6*d*x + 6*c) + 6*(I*cos(2*d*x + 2*c)^2
+ I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(4*d*x + 4*c) + 9*I*cos(2*d*x + 2*c)^2 - (cos(2*d*x + 2
*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(8*d*x + 8*c) - 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*
c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c) - 6*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2
*c) + 1)*sin(4*d*x + 4*c) - 4*(cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c) + 6*I*cos(2*d*x +
2*c) + I)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*d)

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Fricas [A]
time = 0.38, size = 125, normalized size = 0.85 \begin {gather*} -\frac {64 \, \sqrt {2} {\left (-231 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 198 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 88 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{1155 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-64/1155*sqrt(2)*(-231*I*a*e^(6*I*d*x + 6*I*c) - 198*I*a*e^(4*I*d*x + 4*I*c) - 88*I*a*e^(2*I*d*x + 2*I*c) - 16
*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x +
6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*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 {3}{2}} \sec ^{5}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 7.25, size = 293, normalized size = 1.99 \begin {gather*} \frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{5\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,192{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{11\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(5*d*(ex
p(c*2i + d*x*2i) + 1)^2) - (a*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i
) + 1))^(1/2)*192i)/(7*d*(exp(c*2i + d*x*2i) + 1)^3) + (a*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i
- 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(3*d*(exp(c*2i + d*x*2i) + 1)^4) - (a*exp(- c*1i - d*x*1i)*(a -
(a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(11*d*(exp(c*2i + d*x*2i) + 1)^5)

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