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

Optimal. Leaf size=74 \[ -\frac {\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {3 i \tan (c+d x)}{2 a d}-\frac {\log (\cos (c+d x))}{a d}+\frac {3 i x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3550, 3525, 3475} \[ -\frac {\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {3 i \tan (c+d x)}{2 a d}-\frac {\log (\cos (c+d x))}{a d}+\frac {3 i x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3550

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a^2), Int[(c + d*Tan[e +
f*x])^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac {\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan (c+d x) (2 a-3 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac {3 i x}{2 a}-\frac {3 i \tan (c+d x)}{2 a d}-\frac {\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan (c+d x) \, dx}{a}\\ &=\frac {3 i x}{2 a}-\frac {\log (\cos (c+d x))}{a d}-\frac {3 i \tan (c+d x)}{2 a d}-\frac {\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 1.14, size = 174, normalized size = 2.35 \[ -\frac {i \cos (c) \sec (c+d x) (\cos (d x)+i \sin (d x)) \left ((-4-4 i \tan (c)) \tan ^{-1}(\tan (d x))-4 d x \tan ^2(c)-2 i d x \tan (c)+4 d x \sec ^2(c)-i \tan (c) \sin (2 d x)-2 i \log \left (\cos ^2(c+d x)\right )+(\tan (c)+i) \cos (2 d x)+4 \sec (c) \sin (d x) \sec (c+d x)+2 \tan (c) \log \left (\cos ^2(c+d x)\right )+4 i \tan (c) \sec (c) \sin (d x) \sec (c+d x)-6 d x+\sin (2 d x)\right )}{4 d (a+i a \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 93, normalized size = 1.26 \[ \frac {10 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (10 i \, d x + 9\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, {\left (e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 1}{4 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(10*I*d*x*e^(4*I*d*x + 4*I*c) + (10*I*d*x + 9)*e^(2*I*d*x + 2*I*c) - 4*(e^(4*I*d*x + 4*I*c) + e^(2*I*d*x +
2*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) + 1)/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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giac [A]  time = 2.42, size = 70, normalized size = 0.95 \[ -\frac {\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{a} - \frac {5 \, \log \left (-i \, \tan \left (d x + c\right ) - 1\right )}{a} + \frac {4 i \, \tan \left (d x + c\right )}{a} + \frac {5 \, \tan \left (d x + c\right ) - 3 i}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(log(tan(d*x + c) + I)/a - 5*log(-I*tan(d*x + c) - 1)/a + 4*I*tan(d*x + c)/a + (5*tan(d*x + c) - 3*I)/(a*
(tan(d*x + c) - I)))/d

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maple [A]  time = 0.15, size = 73, normalized size = 0.99 \[ -\frac {i \tan \left (d x +c \right )}{d a}-\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {i}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {5 \ln \left (\tan \left (d x +c \right )-i\right )}{4 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I/d/a*tan(d*x+c)-1/4/d/a*ln(tan(d*x+c)+I)-1/2*I/d/a/(tan(d*x+c)-I)+5/4/d/a*ln(tan(d*x+c)-I)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 4.00, size = 73, normalized size = 0.99 \[ \frac {5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{4\,a\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{a\,d}+\frac {1}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*log(tan(c + d*x) - 1i))/(4*a*d) - log(tan(c + d*x) + 1i)/(4*a*d) - (tan(c + d*x)*1i)/(a*d) + 1/(2*a*d*(tan(
c + d*x)*1i + 1))

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sympy [A]  time = 0.40, size = 119, normalized size = 1.61 \[ \begin {cases} \frac {e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 4 a d e^{2 i c} \neq 0 \\x \left (- \frac {i \left (1 - 5 e^{2 i c}\right ) e^{- 2 i c}}{2 a} - \frac {5 i}{2 a}\right ) & \text {otherwise} \end {cases} - \frac {2}{- a d e^{2 i c} e^{2 i d x} - a d} + \frac {5 i x}{2 a} - \frac {\log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+I*a*tan(d*x+c)),x)

[Out]

Piecewise((exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(4*a*d*exp(2*I*c), 0)), (x*(-I*(1 - 5*exp(2*I*c))*exp(-2*I*c)/
(2*a) - 5*I/(2*a)), True)) - 2/(-a*d*exp(2*I*c)*exp(2*I*d*x) - a*d) + 5*I*x/(2*a) - log(exp(2*I*d*x) + exp(-2*
I*c))/(a*d)

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