3.180 \(\int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=138 \[ \frac{2 i \sec ^5(c+d x)}{1155 a^3 d (a+i a \tan (c+d x))^5}+\frac{2 i \sec ^5(c+d x)}{231 a^2 d (a+i a \tan (c+d x))^6}+\frac{i \sec ^5(c+d x)}{33 a d (a+i a \tan (c+d x))^7}+\frac{i \sec ^5(c+d x)}{11 d (a+i a \tan (c+d x))^8} \]

[Out]

((I/11)*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^8) + ((I/33)*Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^7)
+ (((2*I)/231)*Sec[c + d*x]^5)/(a^2*d*(a + I*a*Tan[c + d*x])^6) + (((2*I)/1155)*Sec[c + d*x]^5)/(a^3*d*(a + I*
a*Tan[c + d*x])^5)

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Rubi [A]  time = 0.175295, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3502, 3488} \[ \frac{2 i \sec ^5(c+d x)}{1155 a^3 d (a+i a \tan (c+d x))^5}+\frac{2 i \sec ^5(c+d x)}{231 a^2 d (a+i a \tan (c+d x))^6}+\frac{i \sec ^5(c+d x)}{33 a d (a+i a \tan (c+d x))^7}+\frac{i \sec ^5(c+d x)}{11 d (a+i a \tan (c+d x))^8} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((I/11)*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^8) + ((I/33)*Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^7)
+ (((2*I)/231)*Sec[c + d*x]^5)/(a^2*d*(a + I*a*Tan[c + d*x])^6) + (((2*I)/1155)*Sec[c + d*x]^5)/(a^3*d*(a + I*
a*Tan[c + d*x])^5)

Rule 3502

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

Rule 3488

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)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=\frac{i \sec ^5(c+d x)}{11 d (a+i a \tan (c+d x))^8}+\frac{3 \int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^7} \, dx}{11 a}\\ &=\frac{i \sec ^5(c+d x)}{11 d (a+i a \tan (c+d x))^8}+\frac{i \sec ^5(c+d x)}{33 a d (a+i a \tan (c+d x))^7}+\frac{2 \int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^6} \, dx}{33 a^2}\\ &=\frac{i \sec ^5(c+d x)}{11 d (a+i a \tan (c+d x))^8}+\frac{i \sec ^5(c+d x)}{33 a d (a+i a \tan (c+d x))^7}+\frac{2 i \sec ^5(c+d x)}{231 a^2 d (a+i a \tan (c+d x))^6}+\frac{2 \int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^5} \, dx}{231 a^3}\\ &=\frac{i \sec ^5(c+d x)}{11 d (a+i a \tan (c+d x))^8}+\frac{i \sec ^5(c+d x)}{33 a d (a+i a \tan (c+d x))^7}+\frac{2 i \sec ^5(c+d x)}{231 a^2 d (a+i a \tan (c+d x))^6}+\frac{2 i \sec ^5(c+d x)}{1155 a^3 d (a+i a \tan (c+d x))^5}\\ \end{align*}

Mathematica [A]  time = 0.208569, size = 73, normalized size = 0.53 \[ \frac{i \sec ^8(c+d x) (55 i \sin (c+d x)+63 i \sin (3 (c+d x))+440 \cos (c+d x)+168 \cos (3 (c+d x)))}{4620 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((I/4620)*Sec[c + d*x]^8*(440*Cos[c + d*x] + 168*Cos[3*(c + d*x)] + (55*I)*Sin[c + d*x] + (63*I)*Sin[3*(c + d*
x)]))/(a^8*d*(-I + Tan[c + d*x])^8)

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Maple [A]  time = 0.109, size = 189, normalized size = 1.4 \begin{align*} 2\,{\frac{1}{d{a}^{8}} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1}+{\frac{512}{3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{9}}}-{\frac{88\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+{\frac{932}{5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{5}}}-{\frac{128}{11\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{11}}}-{\frac{2376}{7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{7}}}-{\frac{288\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{8}}}+{\frac{7\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}+{\frac{64\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{10}}}+{\frac{292\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{6}}}-30\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-3} \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))^8,x)

[Out]

2/d/a^8*(1/(tan(1/2*d*x+1/2*c)-I)+512/3/(tan(1/2*d*x+1/2*c)-I)^9-88*I/(tan(1/2*d*x+1/2*c)-I)^4+932/5/(tan(1/2*
d*x+1/2*c)-I)^5-128/11/(tan(1/2*d*x+1/2*c)-I)^11-2376/7/(tan(1/2*d*x+1/2*c)-I)^7-288*I/(tan(1/2*d*x+1/2*c)-I)^
8+7*I/(tan(1/2*d*x+1/2*c)-I)^2+64*I/(tan(1/2*d*x+1/2*c)-I)^10+292*I/(tan(1/2*d*x+1/2*c)-I)^6-30/(tan(1/2*d*x+1
/2*c)-I)^3)

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Maxima [A]  time = 1.23661, size = 131, normalized size = 0.95 \begin{align*} \frac{105 i \, \cos \left (11 \, d x + 11 \, c\right ) + 385 i \, \cos \left (9 \, d x + 9 \, c\right ) + 495 i \, \cos \left (7 \, d x + 7 \, c\right ) + 231 i \, \cos \left (5 \, d x + 5 \, c\right ) + 105 \, \sin \left (11 \, d x + 11 \, c\right ) + 385 \, \sin \left (9 \, d x + 9 \, c\right ) + 495 \, \sin \left (7 \, d x + 7 \, c\right ) + 231 \, \sin \left (5 \, d x + 5 \, c\right )}{9240 \, a^{8} d} \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))^8,x, algorithm="maxima")

[Out]

1/9240*(105*I*cos(11*d*x + 11*c) + 385*I*cos(9*d*x + 9*c) + 495*I*cos(7*d*x + 7*c) + 231*I*cos(5*d*x + 5*c) +
105*sin(11*d*x + 11*c) + 385*sin(9*d*x + 9*c) + 495*sin(7*d*x + 7*c) + 231*sin(5*d*x + 5*c))/(a^8*d)

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Fricas [A]  time = 2.4299, size = 177, normalized size = 1.28 \begin{align*} \frac{{\left (231 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 495 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 385 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 105 i\right )} e^{\left (-11 i \, d x - 11 i \, c\right )}}{9240 \, a^{8} d} \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))^8,x, algorithm="fricas")

[Out]

1/9240*(231*I*e^(6*I*d*x + 6*I*c) + 495*I*e^(4*I*d*x + 4*I*c) + 385*I*e^(2*I*d*x + 2*I*c) + 105*I)*e^(-11*I*d*
x - 11*I*c)/(a^8*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))**8,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.25228, size = 204, normalized size = 1.48 \begin{align*} \frac{2 \,{\left (1155 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} - 3465 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 13860 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 23100 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 37422 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 32802 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 27060 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 11220 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4895 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 517 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 152\right )}}{1155 \, a^{8} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{11}} \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))^8,x, algorithm="giac")

[Out]

2/1155*(1155*tan(1/2*d*x + 1/2*c)^10 - 3465*I*tan(1/2*d*x + 1/2*c)^9 - 13860*tan(1/2*d*x + 1/2*c)^8 + 23100*I*
tan(1/2*d*x + 1/2*c)^7 + 37422*tan(1/2*d*x + 1/2*c)^6 - 32802*I*tan(1/2*d*x + 1/2*c)^5 - 27060*tan(1/2*d*x + 1
/2*c)^4 + 11220*I*tan(1/2*d*x + 1/2*c)^3 + 4895*tan(1/2*d*x + 1/2*c)^2 - 517*I*tan(1/2*d*x + 1/2*c) - 152)/(a^
8*d*(tan(1/2*d*x + 1/2*c) - I)^11)