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

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

[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

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Rubi [A]  time = 0.240723, antiderivative size = 147, normalized size of antiderivative = 1., 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 = {3494, 3493} \[ \frac{8 i a^2 \sec ^5(c+d x)}{33 d \sqrt{a+i a \tan (c+d x)}}+\frac{64 i a^3 \sec ^5(c+d x)}{231 d (a+i a \tan (c+d x))^{3/2}}+\frac{256 i a^4 \sec ^5(c+d x)}{1155 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 i a \sec ^5(c+d x) \sqrt{a+i a \tan (c+d x)}}{11 d} \]

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 3494

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]

Rule 3493

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]

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*}

Mathematica [A]  time = 0.954824, size = 109, normalized size = 0.74 \[ \frac{2 a \sec ^4(c+d x) (\cos (d x)-i \sin (d x)) \sqrt{a+i a \tan (c+d x)} (\sin (3 c+2 d x)+i \cos (3 c+2 d x)) (494 \cos (2 (c+d x))+110 i \tan (c+d x)+215 i \sin (3 (c+d x)) \sec (c+d x)+39)}{1155 d} \]

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 = 0.321, size = 125, normalized size = 0.9 \begin{align*}{\frac{2\,a \left ( 512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+512\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -64\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+192\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -20\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+140\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +105\,i \right ) }{1155\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

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)

[Out]

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

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Maxima [B]  time = 69.6067, size = 1345, normalized size = 9.15 \begin{align*} \text{result too large to display} \end{align*}

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]

-(-17075520*I*sqrt(2)*a*cos(6*d*x + 6*c) - 14636160*I*sqrt(2)*a*cos(4*d*x + 4*c) - 6504960*I*sqrt(2)*a*cos(2*d
*x + 2*c) + 17075520*sqrt(2)*a*sin(6*d*x + 6*c) + 14636160*sqrt(2)*a*sin(4*d*x + 4*c) + 6504960*sqrt(2)*a*sin(
2*d*x + 2*c) - 1182720*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)*s
qrt(a)/(((5336100*cos(2*d*x + 2*c)^3 + 1334025*(4*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 5336100*I*sin(2*d
*x + 2*c)^3 + 1334025*(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) + 53
36100*(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) + 8004150*(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) + 12006225*cos(2*d*x + 2*c)^2 - (-13
34025*I*cos(2*d*x + 2*c)^2 - 1334025*I*sin(2*d*x + 2*c)^2 - 2668050*I*cos(2*d*x + 2*c) - 1334025*I)*sin(8*d*x
+ 8*c) - (-5336100*I*cos(2*d*x + 2*c)^2 - 5336100*I*sin(2*d*x + 2*c)^2 - 10672200*I*cos(2*d*x + 2*c) - 5336100
*I)*sin(6*d*x + 6*c) - (-8004150*I*cos(2*d*x + 2*c)^2 - 8004150*I*sin(2*d*x + 2*c)^2 - 16008300*I*cos(2*d*x +
2*c) - 8004150*I)*sin(4*d*x + 4*c) - (-5336100*I*cos(2*d*x + 2*c)^2 - 10672200*I*cos(2*d*x + 2*c) - 5336100*I)
*sin(2*d*x + 2*c) + 8004150*cos(2*d*x + 2*c) + 1334025)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1
)) - (-5336100*I*cos(2*d*x + 2*c)^3 + (-5336100*I*cos(2*d*x + 2*c) - 1334025*I)*sin(2*d*x + 2*c)^2 + 5336100*s
in(2*d*x + 2*c)^3 + (-1334025*I*cos(2*d*x + 2*c)^2 - 1334025*I*sin(2*d*x + 2*c)^2 - 2668050*I*cos(2*d*x + 2*c)
- 1334025*I)*cos(8*d*x + 8*c) + (-5336100*I*cos(2*d*x + 2*c)^2 - 5336100*I*sin(2*d*x + 2*c)^2 - 10672200*I*co
s(2*d*x + 2*c) - 5336100*I)*cos(6*d*x + 6*c) + (-8004150*I*cos(2*d*x + 2*c)^2 - 8004150*I*sin(2*d*x + 2*c)^2 -
16008300*I*cos(2*d*x + 2*c) - 8004150*I)*cos(4*d*x + 4*c) - 12006225*I*cos(2*d*x + 2*c)^2 + 1334025*(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) + 5336100*(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) + 8004150*(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) + 5336100*(cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x +
2*c) - 8004150*I*cos(2*d*x + 2*c) - 1334025*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 = 2.33687, size = 437, normalized size = 2.97 \begin{align*} \frac{\sqrt{2}{\left (14784 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 12672 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 5632 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 1024 i \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{1155 \,{\left (d e^{\left (11 i \, d x + 11 i \, c\right )} + 5 \, d e^{\left (9 i \, d x + 9 i \, c\right )} + 10 \, d e^{\left (7 i \, d x + 7 i \, c\right )} + 10 \, d e^{\left (5 i \, d x + 5 i \, c\right )} + 5 \, d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

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]

1/1155*sqrt(2)*(14784*I*a*e^(6*I*d*x + 6*I*c) + 12672*I*a*e^(4*I*d*x + 4*I*c) + 5632*I*a*e^(2*I*d*x + 2*I*c) +
1024*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e^(11*I*d*x + 11*I*c) + 5*d*e^(9*I*d*x + 9*I*c
) + 10*d*e^(7*I*d*x + 7*I*c) + 10*d*e^(5*I*d*x + 5*I*c) + 5*d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))

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

[Out]

Timed out

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Giac [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}} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

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)