\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [29]

Optimal result

Integrand size = 24, antiderivative size = 69 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-4 a^3 x+\frac {i a^3 \log (\cos (c+d x))}{d}+\frac {3 i a^3 \log (\sin (c+d x))}{d}-\frac {\cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d} \]

[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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3634, 3670, 3556, 3612} \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {3 i a^3 \log (\sin (c+d x))}{d}+\frac {i a^3 \log (\cos (c+d x))}{d}-\frac {\cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-4 a^3 x \]

[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 3556

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

Rule 3612

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 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] /;
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[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3670

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=a^3 \left (-\frac {\cot (c+d x)}{d}+\frac {3 i \log (\tan (c+d x))}{d}-\frac {4 i \log (i+\tan (c+d x))}{d}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(\frac {a^{3} \left (3 i \ln \left (\tan \left (d x +c \right )\right )-2 i \ln \left (\sec ^{2}\left (d x +c \right )\right )-4 d x -\cot \left (d x +c \right )\right )}{d}\)	\(43\)
derivativedivides	\(-\frac {a^{3} \left (-3 i \ln \left (\tan \left (d x +c \right )\right )+\frac {1}{\tan \left (d x +c \right )}+2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+4 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(51\)
default	\(-\frac {a^{3} \left (-3 i \ln \left (\tan \left (d x +c \right )\right )+\frac {1}{\tan \left (d x +c \right )}+2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+4 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(51\)
norman	\(\frac {-\frac {a^{3}}{d}-4 a^{3} x \tan \left (d x +c \right )}{\tan \left (d x +c \right )}+\frac {3 i a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(68\)
risch	\(\frac {8 a^{3} c}{d}-\frac {2 i a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {3 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(75\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {-2 i \, a^{3} + {\left (i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 3 \, {\left (-i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \]

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=- \frac {2 i a^{3}}{d e^{2 i c} e^{2 i d x} - d} + \frac {a^{3} \cdot \left (3 i \log {\left (e^{2 i d x} - e^{- 2 i c} \right )} + i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}\right )}{d} \]

[In]

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

[Out]

-2*I*a**3/(d*exp(2*I*c)*exp(2*I*d*x) - d) + a**3*(3*I*log(exp(2*I*d*x) - exp(-2*I*c)) + I*log(exp(2*I*d*x) + e
xp(-2*I*c)))/d

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {4 \, {\left (d x + c\right )} a^{3} + 2 i \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 3 i \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {a^{3}}{\tan \left (d x + c\right )}}{d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.71 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.72 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {-2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 16 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - 6 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {-6 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 4.64 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.55 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3\,\left (\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}+\mathrm {cot}\left (c+d\,x\right )-\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,3{}\mathrm {i}\right )}{d} \]

[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