3.143 \(\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=32 \[ \frac {i \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^3} \]

[Out]

1/3*I*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^3

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Rubi [A]  time = 0.04, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {3488} \[ \frac {i \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=\frac {i \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 32, normalized size = 1.00 \[ \frac {i \sec ^3(c+d x)}{3 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 17, normalized size = 0.53 \[ \frac {i \, e^{\left (-3 i \, d x - 3 i \, c\right )}}{3 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I*e^(-3*I*d*x - 3*I*c)/(a^3*d)

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giac [A]  time = 1.70, size = 36, normalized size = 1.12 \[ \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{3 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*tan(1/2*d*x + 1/2*c)^2 - 1)/(a^3*d*(tan(1/2*d*x + 1/2*c) - I)^3)

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maple [A]  time = 0.45, size = 57, normalized size = 1.78 \[ \frac {\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x)

[Out]

2/d/a^3*(1/(tan(1/2*d*x+1/2*c)-I)-4/3/(tan(1/2*d*x+1/2*c)-I)^3+2*I/(tan(1/2*d*x+1/2*c)-I)^2)

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maxima [A]  time = 0.37, size = 29, normalized size = 0.91 \[ \frac {i \, \cos \left (3 \, d x + 3 \, c\right ) + \sin \left (3 \, d x + 3 \, c\right )}{3 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(I*cos(3*d*x + 3*c) + sin(3*d*x + 3*c))/(a^3*d)

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mupad [B]  time = 3.44, size = 68, normalized size = 2.12 \[ -\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}-\mathrm {i}\right )}{3\,a^3\,d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,3{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(tan(c/2 + (d*x)/2)^2*3i - 1i))/(3*a^3*d*(tan(c/2 + (d*x)/2)*3i - 3*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)
/2)^3*1i + 1))

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sympy [A]  time = 2.09, size = 80, normalized size = 2.50 \[ \begin {cases} - \frac {\sec ^{3}{\left (c + d x \right )}}{3 a^{3} d \tan ^{3}{\left (c + d x \right )} - 9 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 9 a^{3} d \tan {\left (c + d x \right )} + 3 i a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{3}{\relax (c )}}{\left (i a \tan {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-sec(c + d*x)**3/(3*a**3*d*tan(c + d*x)**3 - 9*I*a**3*d*tan(c + d*x)**2 - 9*a**3*d*tan(c + d*x) + 3
*I*a**3*d), Ne(d, 0)), (x*sec(c)**3/(I*a*tan(c) + a)**3, True))

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