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

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Rubi [A]  time = 0.08, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 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]))

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 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])

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*}

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Mathematica [A]  time = 0.62, 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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fricas [A]  time = 1.08, size = 110, normalized size = 1.47 \[ \frac {8 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d x + {\left (16 \, d x - 4 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i}{a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d} \]

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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giac [A]  time = 0.30, size = 128, normalized size = 1.71 \[ \frac {2 \, {\left (\frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {4 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{3}} + \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} + \frac {-3 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}\right )}}{d} \]

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

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maple [A]  time = 0.29, size = 52, normalized size = 0.69 \[ -\frac {3 \tan \left (d x +c \right )}{a^{3} d}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2 a^{3} d}-\frac {4 i \ln \left (\tan \left (d x +c \right )-i\right )}{a^{3} d} \]

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

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maxima [B]  time = 0.50, size = 299, normalized size = 3.99 \[ \frac {-8 i \, d x + {\left (4 i \, \cos \left (4 \, d x + 4 \, c\right ) + 8 i \, \cos \left (2 \, d x + 2 \, c\right ) - 4 \, \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right ) + 4 i\right )} \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) + {\left (-8 i \, d x - 8 i \, c\right )} \cos \left (4 \, d x + 4 \, c\right ) + {\left (-16 i \, d x - 16 i \, c - 4\right )} \cos \left (2 \, d x + 2 \, c\right ) + {\left (2 \, \cos \left (4 \, d x + 4 \, c\right ) + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, \sin \left (4 \, d x + 4 \, c\right ) + 4 i \, \sin \left (2 \, d x + 2 \, c\right ) + 2\right )} \log \left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) + 8 \, {\left (d x + c\right )} \sin \left (4 \, d x + 4 \, c\right ) + {\left (16 \, d x + 16 \, c - 4 i\right )} \sin \left (2 \, d x + 2 \, c\right ) - 8 i \, c - 6}{{\left (-i \, a^{3} \cos \left (4 \, d x + 4 \, c\right ) - 2 i \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, a^{3} \sin \left (2 \, d x + 2 \, c\right ) - i \, a^{3}\right )} d} \]

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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mupad [B]  time = 0.92, size = 104, normalized size = 1.39 \[ -\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )\,8{}\mathrm {i}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,4{}\mathrm {i}}{a^3\,d}+\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,2{}\mathrm {i}-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a*cos(c + d*x) + a*sin(c + d*x)*1i)^3),x)

[Out]

(tan(c/2 + (d*x)/2)^2*2i - 6*tan(c/2 + (d*x)/2) + 6*tan(c/2 + (d*x)/2)^3)/(a^3*d*(tan(c/2 + (d*x)/2)^2 - 1)^2)
- (log(tan(c/2 + (d*x)/2) - 1i)*8i - log(tan(c/2 + (d*x)/2)^2 - 1)*4i)/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{- i \sin ^{3}{\left (c + d x \right )} - 3 \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} + 3 i \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + \cos ^{3}{\left (c + d x \right )}}\, dx}{a^{3}} \]

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]

Integral(sec(c + d*x)**3/(-I*sin(c + d*x)**3 - 3*sin(c + d*x)**2*cos(c + d*x) + 3*I*sin(c + d*x)*cos(c + d*x)*
*2 + cos(c + d*x)**3), x)/a**3

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