3.134 \(\int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=58 \[ \frac{i \tan ^2(c+d x)}{2 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 i \log (\cos (c+d x))}{a^3 d}+\frac{4 x}{a^3} \]

[Out]

(4*x)/a^3 + ((4*I)*Log[Cos[c + d*x]])/(a^3*d) - (3*Tan[c + d*x])/(a^3*d) + ((I/2)*Tan[c + d*x]^2)/(a^3*d)

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Rubi [A]  time = 0.0472475, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3487, 43} \[ \frac{i \tan ^2(c+d x)}{2 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 i \log (\cos (c+d x))}{a^3 d}+\frac{4 x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(4*x)/a^3 + ((4*I)*Log[Cos[c + d*x]])/(a^3*d) - (3*Tan[c + d*x])/(a^3*d) + ((I/2)*Tan[c + d*x]^2)/(a^3*d)

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{(a-x)^2}{a+x} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (-3 a+x+\frac{4 a^2}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=\frac{4 x}{a^3}+\frac{4 i \log (\cos (c+d x))}{a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{i \tan ^2(c+d x)}{2 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.37478, size = 113, normalized size = 1.95 \[ \frac{\sec (c) \sec ^2(c+d x) (-3 \sin (c+2 d x)+2 d x \cos (3 c+2 d x)+2 i \cos (3 c+2 d x) \log (\cos (c+d x))+2 \cos (c+2 d x) (d x+i \log (\cos (c+d x)))+\cos (c) (4 i \log (\cos (c+d x))+4 d x+i)+3 \sin (c))}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c]*Sec[c + d*x]^2*(2*d*x*Cos[3*c + 2*d*x] + 2*Cos[c + 2*d*x]*(d*x + I*Log[Cos[c + d*x]]) + Cos[c]*(I + 4*
d*x + (4*I)*Log[Cos[c + d*x]]) + (2*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]] + 3*Sin[c] - 3*Sin[c + 2*d*x]))/(2*a
^3*d)

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Maple [A]  time = 0.079, size = 52, normalized size = 0.9 \begin{align*} -3\,{\frac{\tan \left ( dx+c \right ) }{d{a}^{3}}}+{\frac{{\frac{i}{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d{a}^{3}}}-{\frac{4\,i\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 0.968436, size = 61, normalized size = 1.05 \begin{align*} \frac{\frac{i \, \tan \left (d x + c\right )^{2} - 6 \, \tan \left (d x + c\right )}{a^{3}} - \frac{8 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*((I*tan(d*x + c)^2 - 6*tan(d*x + c))/a^3 - 8*I*log(I*tan(d*x + c) + 1)/a^3)/d

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Fricas [B]  time = 2.33946, size = 317, normalized size = 5.47 \begin{align*} \frac{8 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d x +{\left (16 \, d x - 4 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i}{a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

(8*d*x*e^(4*I*d*x + 4*I*c) + 8*d*x + (16*d*x - 4*I)*e^(2*I*d*x + 2*I*c) + (4*I*e^(4*I*d*x + 4*I*c) + 8*I*e^(2*
I*d*x + 2*I*c) + 4*I)*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I)/(a^3*d*e^(4*I*d*x + 4*I*c) + 2*a^3*d*e^(2*I*d*x + 2*
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)**6/(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.23457, size = 176, normalized size = 3.03 \begin{align*} \frac{2 \,{\left (-\frac{4 i \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}{a^{3}} + \frac{2 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac{2 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{-3 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 i}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

2*(-4*I*log(tan(1/2*d*x + 1/2*c) - I)/a^3 + 2*I*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 2*I*log(abs(tan(1/2*d
*x + 1/2*c) - 1))/a^3 + (-3*I*tan(1/2*d*x + 1/2*c)^4 + 3*tan(1/2*d*x + 1/2*c)^3 + 7*I*tan(1/2*d*x + 1/2*c)^2 -
3*tan(1/2*d*x + 1/2*c) - 3*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3))/d