3.29 \(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=69 \[ \frac {3 i a^3 \log (\sin (c+d x))}{d}+\frac {i a^3 \log (\cos (c+d x))}{d}-\frac {\cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-4 a^3 x \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3553, 3589, 3475, 3531} \[ \frac {3 i a^3 \log (\sin (c+d x))}{d}+\frac {i a^3 \log (\cos (c+d x))}{d}-\frac {\cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-4 a^3 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3,x]

[Out]

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

Rule 3475

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

Rule 3531

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

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{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[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3589

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

Rubi steps

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

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Mathematica [B]  time = 1.60, size = 144, normalized size = 2.09 \[ \frac {a^3 \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \csc (c+d x) \left (14 d x \cos (2 c+d x)+12 \sin (c) \sin (c+d x) \tan ^{-1}(\tan (4 c+d x))-i \cos (2 c+d x) \log \left (\cos ^2(c+d x)\right )+\cos (d x) \left (3 i \log \left (\sin ^2(c+d x)\right )+i \log \left (\cos ^2(c+d x)\right )-14 d x\right )-3 i \cos (2 c+d x) \log \left (\sin ^2(c+d x)\right )+4 \sin (d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Csc[c/2]*Csc[c + d*x]*Sec[c/2]*(14*d*x*Cos[2*c + d*x] - I*Cos[2*c + d*x]*Log[Cos[c + d*x]^2] + Cos[d*x]*(
-14*d*x + I*Log[Cos[c + d*x]^2] + (3*I)*Log[Sin[c + d*x]^2]) - (3*I)*Cos[2*c + d*x]*Log[Sin[c + d*x]^2] + 4*Si
n[d*x] + 12*ArcTan[Tan[4*c + d*x]]*Sin[c]*Sin[c + d*x]))/(8*d)

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fricas [A]  time = 0.45, size = 90, normalized size = 1.30 \[ \frac {-2 i \, a^{3} + {\left (i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (3 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.90, size = 119, normalized size = 1.72 \[ -\frac {-2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 16 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - 6 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {-6 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(-2*I*a^3*log(tan(1/2*d*x + 1/2*c) + 1) + 16*I*a^3*log(tan(1/2*d*x + 1/2*c) + I) - 2*I*a^3*log(tan(1/2*d*
x + 1/2*c) - 1) - 6*I*a^3*log(tan(1/2*d*x + 1/2*c)) - a^3*tan(1/2*d*x + 1/2*c) - (-6*I*a^3*tan(1/2*d*x + 1/2*c
) - a^3)/tan(1/2*d*x + 1/2*c))/d

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maple [A]  time = 0.32, size = 63, normalized size = 0.91 \[ \frac {i a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 i a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-4 a^{3} x -\frac {a^{3} \cot \left (d x +c \right )}{d}-\frac {4 a^{3} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x)

[Out]

I*a^3*ln(cos(d*x+c))/d+3*I*a^3*ln(sin(d*x+c))/d-4*a^3*x-a^3*cot(d*x+c)/d-4/d*a^3*c

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maxima [A]  time = 0.67, size = 56, normalized size = 0.81 \[ -\frac {4 \, {\left (d x + c\right )} a^{3} + 2 i \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 3 i \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {a^{3}}{\tan \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.81, size = 38, normalized size = 0.55 \[ -\frac {a^3\,\left (\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}+\mathrm {cot}\left (c+d\,x\right )-\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,3{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: NotInvertible

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