3.152 \(\int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=63 \[ \frac{\tan (c+d x)}{a^4 d}+\frac{4 i}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{4 i \log (\cos (c+d x))}{a^4 d}-\frac{4 x}{a^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.053086, antiderivative size = 63, 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} \[ \frac{\tan (c+d x)}{a^4 d}+\frac{4 i}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{4 i \log (\cos (c+d x))}{a^4 d}-\frac{4 x}{a^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-4*x)/a^4 - ((4*I)*Log[Cos[c + d*x]])/(a^4*d) + Tan[c + d*x]/(a^4*d) + (4*I)/(d*(a^4 + I*a^4*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 ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{(a-x)^2}{(a+x)^2} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (1+\frac{4 a^2}{(a+x)^2}-\frac{4 a}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{4 x}{a^4}-\frac{4 i \log (\cos (c+d x))}{a^4 d}+\frac{\tan (c+d x)}{a^4 d}+\frac{4 i}{d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.571658, size = 214, normalized size = 3.4 \[ \frac{\sec (c) \sec (c+d x) (-\cos (c+d x)+i \sin (c+d x)) (2 i d x \sin (c+2 d x)-2 \sin (c+2 d x)+2 i d x \sin (3 c+2 d x)-\sin (3 c+2 d x)+2 d x \cos (3 c+2 d x)-i \cos (3 c+2 d x)+2 i \cos (3 c+2 d x) \log (\cos (c+d x))+2 \cos (c+2 d x) (d x+i \log (\cos (c+d x)))+\cos (c) (4 i \log (\cos (c+d x))+4 d x-3 i)-2 \sin (c+2 d x) \log (\cos (c+d x))-2 \sin (3 c+2 d x) \log (\cos (c+d x))+\sin (c))}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.073, size = 53, normalized size = 0.8 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{{a}^{4}d}}+4\,{\frac{1}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{4\,i\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{{a}^{4}d}} \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))^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.997645, size = 130, normalized size = 2.06 \begin{align*} \frac{\frac{12 \,{\left (\tan \left (d x + c\right )^{2} - 2 i \, \tan \left (d x + c\right ) - 1\right )}}{3 \, a^{4} \tan \left (d x + c\right )^{3} - 9 i \, a^{4} \tan \left (d x + c\right )^{2} - 9 \, a^{4} \tan \left (d x + c\right ) + 3 i \, a^{4}} + \frac{4 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{4}} + \frac{\tan \left (d x + c\right )}{a^{4}}}{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))^4,x, algorithm="maxima")

[Out]

(12*(tan(d*x + c)^2 - 2*I*tan(d*x + c) - 1)/(3*a^4*tan(d*x + c)^3 - 9*I*a^4*tan(d*x + c)^2 - 9*a^4*tan(d*x + c
) + 3*I*a^4) + 4*I*log(I*tan(d*x + c) + 1)/a^4 + tan(d*x + c)/a^4)/d

________________________________________________________________________________________

Fricas [A]  time = 2.42734, size = 286, normalized size = 4.54 \begin{align*} -\frac{8 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (8 \, 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 )} - 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 2 i}{a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}} \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))^4,x, algorithm="fricas")

[Out]

-(8*d*x*e^(4*I*d*x + 4*I*c) + (8*d*x - 4*I)*e^(2*I*d*x + 2*I*c) - (-4*I*e^(4*I*d*x + 4*I*c) - 4*I*e^(2*I*d*x +
2*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 2*I)/(a^4*d*e^(4*I*d*x + 4*I*c) + a^4*d*e^(2*I*d*x + 2*I*c))

________________________________________________________________________________________

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))**4,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [B]  time = 1.19539, size = 198, normalized size = 3.14 \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^{4}} - \frac{2 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{2 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{2 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 i}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{4}} + \frac{-6 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 i}{a^{4}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{2}}\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))^4,x, algorithm="giac")

[Out]

2*(4*I*log(tan(1/2*d*x + 1/2*c) - I)/a^4 - 2*I*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 2*I*log(abs(tan(1/2*d*
x + 1/2*c) - 1))/a^4 + (2*I*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) - 2*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)
*a^4) + (-6*I*tan(1/2*d*x + 1/2*c)^2 - 16*tan(1/2*d*x + 1/2*c) + 6*I)/(a^4*(tan(1/2*d*x + 1/2*c) - I)^2))/d