3.356 \(\int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((8*I)/63)*a^2*Sec[c + d*x]^7)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I)/9)*a*Sec[c + d*x]^7)/(d*(a + I*a*Ta
n[c + d*x])^(5/2))

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Rubi [A]  time = 0.128931, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac{8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((8*I)/63)*a^2*Sec[c + d*x]^7)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I)/9)*a*Sec[c + d*x]^7)/(d*(a + I*a*Ta
n[c + d*x])^(5/2))

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 \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac{1}{9} (4 a) \int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac{2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.275391, size = 80, normalized size = 1.1 \[ \frac{2 (7 \tan (c+d x)-11 i) \sec ^5(c+d x) (\sin (2 (c+d x))+i \cos (2 (c+d x)))}{63 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*Sec[c + d*x]^5*(I*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*(-11*I + 7*Tan[c + d*x]))/(63*a*d*(-I + Tan[c + d*x]
)*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.295, size = 117, normalized size = 1.6 \begin{align*}{\frac{64\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+64\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}-8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+24\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -34\,i\cos \left ( dx+c \right ) -14\,\sin \left ( dx+c \right ) }{63\,{a}^{2}d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}\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)^7/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

2/63/d/a^2*(32*I*cos(d*x+c)^5+32*sin(d*x+c)*cos(d*x+c)^4-4*I*cos(d*x+c)^3+12*cos(d*x+c)^2*sin(d*x+c)-17*I*cos(
d*x+c)-7*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4

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Maxima [B]  time = 1.72257, size = 659, normalized size = 9.03 \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)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/63*(-11*I*sqrt(a) - 30*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 12*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 - 86*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 9*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 10
8*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 108*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 9*I*sqrt(a)*
sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 86*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 12*I*sqrt(a)*sin(d*x +
c)^10/(cos(d*x + c) + 1)^10 - 30*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 11*I*sqrt(a)*sin(d*x + c)^12/
(cos(d*x + c) + 1)^12)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(3/2)
/((a^2 - 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 20*a^2*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*a^2*sin(d*x + c)^10/(cos(d*x +
c) + 1)^10 + a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 - 1)^(3/2))

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Fricas [B]  time = 2.08432, size = 325, normalized size = 4.45 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (288 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i\right )} e^{\left (i \, d x + i \, c\right )}}{63 \,{\left (a^{2} d e^{\left (9 i \, d x + 9 i \, c\right )} + 4 \, a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} + 6 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} 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)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/63*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(288*I*e^(2*I*d*x + 2*I*c) + 64*I)*e^(I*d*x + I*c)/(a^2*d*e^(9*
I*d*x + 9*I*c) + 4*a^2*d*e^(7*I*d*x + 7*I*c) + 6*a^2*d*e^(5*I*d*x + 5*I*c) + 4*a^2*d*e^(3*I*d*x + 3*I*c) + a^2
*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)**7/(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 \frac{\sec \left (d x + c\right )^{7}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^7/(I*a*tan(d*x + c) + a)^(3/2), x)