3.183 \(\int \frac{\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=75 \[ \frac{i \tan ^2(c+d x)}{2 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 i \log (\sin (c+d x))}{a^3 d}-\frac{4 i \log (\tan (c+d x))}{a^3 d}+\frac{4 x}{a^3} \]

[Out]

(4*x)/a^3 + ((4*I)*Log[Sin[c + d*x]])/(a^3*d) - ((4*I)*Log[Tan[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 = 0.0805796, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3088, 848, 88} \[ \frac{i \tan ^2(c+d x)}{2 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 i \log (\sin (c+d x))}{a^3 d}-\frac{4 i \log (\tan (c+d x))}{a^3 d}+\frac{4 x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x)/a^3 + ((4*I)*Log[Sin[c + d*x]])/(a^3*d) - ((4*I)*Log[Tan[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 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^3 (i a+a x)^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\left (-\frac{i}{a}+\frac{x}{a}\right )^2}{x^3 (i a+a x)} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{i}{a^3 x^3}-\frac{3}{a^3 x^2}-\frac{4 i}{a^3 x}+\frac{4 i}{a^3 (i+x)}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{4 x}{a^3}+\frac{4 i \log (\sin (c+d x))}{a^3 d}-\frac{4 i \log (\tan (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 = 0.599273, size = 110, normalized size = 1.47 \[ \frac{i \sec (c) \sec ^2(c+d x) (\cos (c) (4 \log (\cos (c+d x))-4 i d x+1)-i (2 \cos (c+2 d x) (d x+i \log (\cos (c+d x)))+2 \cos (3 c+2 d x) (d x+i \log (\cos (c+d x)))-6 \sin (d x) \cos (c+d x)))}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*Sec[c]*Sec[c + d*x]^2*(Cos[c]*(1 - (4*I)*d*x + 4*Log[Cos[c + d*x]]) - I*(2*Cos[c + 2*d*x]*(d*x + I*Log[
Cos[c + d*x]]) + 2*Cos[3*c + 2*d*x]*(d*x + I*Log[Cos[c + d*x]]) - 6*Cos[c + d*x]*Sin[d*x])))/(a^3*d)

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Maple [A]  time = 0.196, size = 52, normalized size = 0.7 \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)^3/(a*cos(d*x+c)+I*a*sin(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/d/a^3*ln(tan(d*x+c)-I)

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Maxima [B]  time = 1.64775, size = 404, normalized size = 5.39 \begin{align*} \frac{-8 i \, d x +{\left (4 i \, \cos \left (4 \, d x + 4 \, c\right ) + 8 i \, \cos \left (2 \, d x + 2 \, c\right ) - 4 \, \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right ) + 4 i\right )} \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) +{\left (-8 i \, d x - 8 i \, c\right )} \cos \left (4 \, d x + 4 \, c\right ) +{\left (-16 i \, d x - 16 i \, c - 4\right )} \cos \left (2 \, d x + 2 \, c\right ) +{\left (2 \, \cos \left (4 \, d x + 4 \, c\right ) + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, \sin \left (4 \, d x + 4 \, c\right ) + 4 i \, \sin \left (2 \, d x + 2 \, c\right ) + 2\right )} \log \left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) + 8 \,{\left (d x + c\right )} \sin \left (4 \, d x + 4 \, c\right ) +{\left (16 \, d x + 16 \, c - 4 i\right )} \sin \left (2 \, d x + 2 \, c\right ) - 8 i \, c - 6}{{\left (-i \, a^{3} \cos \left (4 \, d x + 4 \, c\right ) - 2 i \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, a^{3} \sin \left (2 \, d x + 2 \, c\right ) - i \, a^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-8*I*d*x + (4*I*cos(4*d*x + 4*c) + 8*I*cos(2*d*x + 2*c) - 4*sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c) + 4*I)*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) + (-8*I*d*x - 8*I*c)*cos(4*d*x + 4*c) + (-16*I*d*x - 16*I*c - 4)*c
os(2*d*x + 2*c) + (2*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 2*I*sin(4*d*x + 4*c) + 4*I*sin(2*d*x + 2*c) + 2)*
log(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) + 8*(d*x + c)*sin(4*d*x + 4*c) + (16*d*x
+ 16*c - 4*I)*sin(2*d*x + 2*c) - 8*I*c - 6)/((-I*a^3*cos(4*d*x + 4*c) - 2*I*a^3*cos(2*d*x + 2*c) + a^3*sin(4*
d*x + 4*c) + 2*a^3*sin(2*d*x + 2*c) - I*a^3)*d)

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Fricas [A]  time = 0.484404, size = 317, normalized size = 4.23 \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)^3/(a*cos(d*x+c)+I*a*sin(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)**3/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.18891, size = 176, normalized size = 2.35 \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)^3/(a*cos(d*x+c)+I*a*sin(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