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

Optimal. Leaf size=50 \[ \frac{2 i}{d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{i \log (\cos (c+d x))}{a^3 d}-\frac{x}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.0486496, antiderivative size = 50, 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} \[ \frac{2 i}{d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{i \log (\cos (c+d x))}{a^3 d}-\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{a-x}{(a+x)^2} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\frac{1}{-a-x}+\frac{2 a}{(a+x)^2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{x}{a^3}-\frac{i \log (\cos (c+d x))}{a^3 d}+\frac{2 i}{d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.214589, size = 88, normalized size = 1.76 \[ \frac{\sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x))) (\log (\cos (c+d x))+\tan (c+d x) (i \log (\cos (c+d x))+d x+i)-i d x-1)}{a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*(-1 - I*d*x + Log[Cos[c + d*x]] + (I + d*x + I*Log[Cos
[c + d*x]])*Tan[c + d*x]))/(a^3*d*(-I + Tan[c + d*x])^3)

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Maple [A]  time = 0.08, size = 40, normalized size = 0.8 \begin{align*}{\frac{i\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{3}}}+2\,{\frac{1}{d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 0.98214, size = 89, normalized size = 1.78 \begin{align*} -\frac{\frac{4 \,{\left (-i \, \tan \left (d x + c\right ) - 1\right )}}{2 i \, a^{3} \tan \left (d x + c\right )^{2} + 4 \, a^{3} \tan \left (d x + c\right ) - 2 i \, a^{3}} - \frac{i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.26739, size = 157, normalized size = 3.14 \begin{align*} -\frac{{\left (2 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i\right )} 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)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.20099, size = 138, normalized size = 2.76 \begin{align*} -\frac{-\frac{2 i \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}{a^{3}} + \frac{i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac{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 )^{2} + 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 i}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-2*I*log(tan(1/2*d*x + 1/2*c) - I)/a^3 + I*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + I*log(abs(tan(1/2*d*x +
1/2*c) - 1))/a^3 + (3*I*tan(1/2*d*x + 1/2*c)^2 + 10*tan(1/2*d*x + 1/2*c) - 3*I)/(a^3*(tan(1/2*d*x + 1/2*c) - I
)^2))/d