3.161 \(\int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=68 \[ \frac{i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3}+\frac{i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4} \]

[Out]

((I/5)*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^4) + ((I/15)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3)

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Rubi [A]  time = 0.0793596, antiderivative size = 68, normalized size of antiderivative = 1., 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} \[ \frac{i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3}+\frac{i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/5)*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^4) + ((I/15)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3)

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 ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=\frac{i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}+\frac{\int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{5 a}\\ &=\frac{i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}+\frac{i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3}\\ \end{align*}

Mathematica [A]  time = 0.0796416, size = 40, normalized size = 0.59 \[ -\frac{(\tan (c+d x)-4 i) \sec ^3(c+d x)}{15 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-(Sec[c + d*x]^3*(-4*I + Tan[c + d*x]))/(15*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.089, size = 90, normalized size = 1.3 \begin{align*} 2\,{\frac{1}{{a}^{4}d} \left ( 8/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-5}+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1}-{\frac{4\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+{\frac{3\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}-14/3\, \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)^3/(a+I*a*tan(d*x+c))^4,x)

[Out]

2/d/a^4*(8/5/(tan(1/2*d*x+1/2*c)-I)^5+1/(tan(1/2*d*x+1/2*c)-I)-4*I/(tan(1/2*d*x+1/2*c)-I)^4+3*I/(tan(1/2*d*x+1
/2*c)-I)^2-14/3/(tan(1/2*d*x+1/2*c)-I)^3)

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Maxima [A]  time = 0.99218, size = 72, normalized size = 1.06 \begin{align*} \frac{3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 5 i \, \cos \left (3 \, d x + 3 \, c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 5 \, \sin \left (3 \, d x + 3 \, c\right )}{30 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/30*(3*I*cos(5*d*x + 5*c) + 5*I*cos(3*d*x + 3*c) + 3*sin(5*d*x + 5*c) + 5*sin(3*d*x + 3*c))/(a^4*d)

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Fricas [A]  time = 2.32098, size = 90, normalized size = 1.32 \begin{align*} \frac{{\left (5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{30 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/30*(5*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d*x - 5*I*c)/(a^4*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)**3/(a+I*a*tan(d*x+c))**4,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.17768, size = 99, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 15 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 25 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4\right )}}{15 \, a^{4} 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)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 15*I*tan(1/2*d*x + 1/2*c)^3 - 25*tan(1/2*d*x + 1/2*c)^2 + 5*I*tan(1/2*d*x +
1/2*c) + 4)/(a^4*d*(tan(1/2*d*x + 1/2*c) - I)^5)