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

Optimal. Leaf size=98 \[ \frac{2 i \sec (c+d x)}{15 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac{i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3} \]

[Out]

((I/5)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^3) + (((2*I)/15)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^2) +
(((2*I)/15)*Sec[c + d*x])/(d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.0766288, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3502, 3488} \[ \frac{2 i \sec (c+d x)}{15 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac{i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/5)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^3) + (((2*I)/15)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^2) +
(((2*I)/15)*Sec[c + d*x])/(d*(a^3 + I*a^3*Tan[c + d*x]))

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 (c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=\frac{i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac{2 \int \frac{\sec (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{5 a}\\ &=\frac{i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac{2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac{2 \int \frac{\sec (c+d x)}{a+i a \tan (c+d x)} \, dx}{15 a^2}\\ &=\frac{i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac{2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac{2 i \sec (c+d x)}{15 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.117521, size = 54, normalized size = 0.55 \[ -\frac{\sec ^3(c+d x) (6 i \sin (2 (c+d x))+9 \cos (2 (c+d x))+5)}{30 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3*(5 + 9*Cos[2*(c + d*x)] + (6*I)*Sin[2*(c + d*x)]))/(30*a^3*d*(-I + Tan[c + d*x])^3)

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Maple [A]  time = 0.049, size = 90, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{d{a}^{3}} \left ( -8/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-3}-{\frac{2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+4/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-5}+{\frac{2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^3*(-8/3/(tan(1/2*d*x+1/2*c)-I)^3-2*I/(tan(1/2*d*x+1/2*c)-I)^4+4/5/(tan(1/2*d*x+1/2*c)-I)^5+2*I/(tan(1/2*
d*x+1/2*c)-I)^2+1/(tan(1/2*d*x+1/2*c)-I))

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Maxima [A]  time = 0.991177, size = 93, normalized size = 0.95 \begin{align*} \frac{3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 10 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 10 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*I*cos(5*d*x + 5*c) + 10*I*cos(3*d*x + 3*c) + 15*I*cos(d*x + c) + 3*sin(5*d*x + 5*c) + 10*sin(3*d*x + 3
*c) + 15*sin(d*x + c))/(a^3*d)

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Fricas [A]  time = 2.02872, size = 128, normalized size = 1.31 \begin{align*} \frac{{\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*I*e^(4*I*d*x + 4*I*c) + 10*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d*x - 5*I*c)/(a^3*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)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.1774, size = 99, normalized size = 1.01 \begin{align*} \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 30*I*tan(1/2*d*x + 1/2*c)^3 - 40*tan(1/2*d*x + 1/2*c)^2 + 20*I*tan(1/2*d*x +
1/2*c) + 7)/(a^3*d*(tan(1/2*d*x + 1/2*c) - I)^5)