3.179 \(\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=68 \[ \frac {i \sec ^7(c+d x)}{63 a d (a+i a \tan (c+d x))^7}+\frac {i \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^8} \]

[Out]

1/9*I*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^8+1/63*I*sec(d*x+c)^7/a/d/(a+I*a*tan(d*x+c))^7

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Rubi [A]  time = 0.08, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3502, 3488} \[ \frac {i \sec ^7(c+d x)}{63 a d (a+i a \tan (c+d x))^7}+\frac {i \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^8} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^8,x]

[Out]

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

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]

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]

Rubi steps

\begin {align*} \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=\frac {i \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^8}+\frac {\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^7} \, dx}{9 a}\\ &=\frac {i \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^8}+\frac {i \sec ^7(c+d x)}{63 a d (a+i a \tan (c+d x))^7}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 40, normalized size = 0.59 \[ -\frac {(\tan (c+d x)-8 i) \sec ^7(c+d x)}{63 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^8,x]

[Out]

-1/63*(Sec[c + d*x]^7*(-8*I + Tan[c + d*x]))/(a^8*d*(-I + Tan[c + d*x])^8)

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fricas [A]  time = 0.47, size = 30, normalized size = 0.44 \[ \frac {{\left (9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-9 i \, d x - 9 i \, c\right )}}{126 \, a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/126*(9*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(-9*I*d*x - 9*I*c)/(a^8*d)

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giac [B]  time = 6.55, size = 125, normalized size = 1.84 \[ \frac {2 \, {\left (63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 63 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 483 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 315 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 693 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 189 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 225 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8\right )}}{63 \, a^{8} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

2/63*(63*tan(1/2*d*x + 1/2*c)^8 - 63*I*tan(1/2*d*x + 1/2*c)^7 - 483*tan(1/2*d*x + 1/2*c)^6 + 315*I*tan(1/2*d*x
+ 1/2*c)^5 + 693*tan(1/2*d*x + 1/2*c)^4 - 189*I*tan(1/2*d*x + 1/2*c)^3 - 225*tan(1/2*d*x + 1/2*c)^2 + 9*I*tan
(1/2*d*x + 1/2*c) + 8)/(a^8*d*(tan(1/2*d*x + 1/2*c) - I)^9)

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maple [B]  time = 0.52, size = 156, normalized size = 2.29 \[ \frac {-\frac {172}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {256}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{9}}-\frac {128 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{8}}+\frac {272}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {152 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {14 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {992 i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{6}}-\frac {1856}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{7}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}}{a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^8,x)

[Out]

2/d/a^8*(-86/3/(tan(1/2*d*x+1/2*c)-I)^3+128/9/(tan(1/2*d*x+1/2*c)-I)^9-64*I/(tan(1/2*d*x+1/2*c)-I)^8+136/(tan(
1/2*d*x+1/2*c)-I)^5-76*I/(tan(1/2*d*x+1/2*c)-I)^4+7*I/(tan(1/2*d*x+1/2*c)-I)^2+496/3*I/(tan(1/2*d*x+1/2*c)-I)^
6-928/7/(tan(1/2*d*x+1/2*c)-I)^7+1/(tan(1/2*d*x+1/2*c)-I))

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maxima [A]  time = 0.68, size = 53, normalized size = 0.78 \[ \frac {7 i \, \cos \left (9 \, d x + 9 \, c\right ) + 9 i \, \cos \left (7 \, d x + 7 \, c\right ) + 7 \, \sin \left (9 \, d x + 9 \, c\right ) + 9 \, \sin \left (7 \, d x + 7 \, c\right )}{126 \, a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/126*(7*I*cos(9*d*x + 9*c) + 9*I*cos(7*d*x + 7*c) + 7*sin(9*d*x + 9*c) + 9*sin(7*d*x + 7*c))/(a^8*d)

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mupad [B]  time = 3.74, size = 37, normalized size = 0.54 \[ \frac {2\,\left (\frac {{\mathrm {e}}^{-c\,7{}\mathrm {i}-d\,x\,7{}\mathrm {i}}\,9{}\mathrm {i}}{4}+\frac {{\mathrm {e}}^{-c\,9{}\mathrm {i}-d\,x\,9{}\mathrm {i}}\,7{}\mathrm {i}}{4}\right )}{63\,a^8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

(2*((exp(- c*7i - d*x*7i)*9i)/4 + (exp(- c*9i - d*x*9i)*7i)/4))/(63*a^8*d)

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sympy [A]  time = 35.77, size = 311, normalized size = 4.57 \[ \begin {cases} - \frac {\tan {\left (c + d x \right )} \sec ^{7}{\left (c + d x \right )}}{63 a^{8} d \tan ^{8}{\left (c + d x \right )} - 504 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 1764 a^{8} d \tan ^{6}{\left (c + d x \right )} + 3528 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 4410 a^{8} d \tan ^{4}{\left (c + d x \right )} - 3528 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 1764 a^{8} d \tan ^{2}{\left (c + d x \right )} + 504 i a^{8} d \tan {\left (c + d x \right )} + 63 a^{8} d} + \frac {8 i \sec ^{7}{\left (c + d x \right )}}{63 a^{8} d \tan ^{8}{\left (c + d x \right )} - 504 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 1764 a^{8} d \tan ^{6}{\left (c + d x \right )} + 3528 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 4410 a^{8} d \tan ^{4}{\left (c + d x \right )} - 3528 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 1764 a^{8} d \tan ^{2}{\left (c + d x \right )} + 504 i a^{8} d \tan {\left (c + d x \right )} + 63 a^{8} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{7}{\relax (c )}}{\left (i a \tan {\relax (c )} + a\right )^{8}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7/(a+I*a*tan(d*x+c))**8,x)

[Out]

Piecewise((-tan(c + d*x)*sec(c + d*x)**7/(63*a**8*d*tan(c + d*x)**8 - 504*I*a**8*d*tan(c + d*x)**7 - 1764*a**8
*d*tan(c + d*x)**6 + 3528*I*a**8*d*tan(c + d*x)**5 + 4410*a**8*d*tan(c + d*x)**4 - 3528*I*a**8*d*tan(c + d*x)*
*3 - 1764*a**8*d*tan(c + d*x)**2 + 504*I*a**8*d*tan(c + d*x) + 63*a**8*d) + 8*I*sec(c + d*x)**7/(63*a**8*d*tan
(c + d*x)**8 - 504*I*a**8*d*tan(c + d*x)**7 - 1764*a**8*d*tan(c + d*x)**6 + 3528*I*a**8*d*tan(c + d*x)**5 + 44
10*a**8*d*tan(c + d*x)**4 - 3528*I*a**8*d*tan(c + d*x)**3 - 1764*a**8*d*tan(c + d*x)**2 + 504*I*a**8*d*tan(c +
d*x) + 63*a**8*d), Ne(d, 0)), (x*sec(c)**7/(I*a*tan(c) + a)**8, True))

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