\(\int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [20]

Optimal result

Integrand size = 32, antiderivative size = 107 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=4 a^3 (i A+B) x+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+\frac {a^3 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3675, 3670, 3556, 3612} \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+4 a^3 x (B+i A)+\frac {a^3 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d} \]

[In]

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

[Out]

4*a^3*(I*A + B)*x + (a^3*(3*A - (4*I)*B)*Log[Cos[c + d*x]])/d + (a^3*A*Log[Sin[c + d*x]])/d + ((I/2)*a*B*(a +
I*a*Tan[c + d*x])^2)/d - ((A - (2*I)*B)*(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 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]

Rule 3675

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

Rubi steps \begin{align*} \text {integral}& = \frac {i a B (a+i a \tan (c+d x))^2}{2 d}+\frac {1}{2} \int \cot (c+d x) (a+i a \tan (c+d x))^2 (2 a A+2 a (i A+2 B) \tan (c+d x)) \, dx \\ & = \frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (2 a^2 A+2 a^2 (3 i A+4 B) \tan (c+d x)\right ) \, dx \\ & = \frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) \left (2 a^3 A+8 a^3 (i A+B) \tan (c+d x)\right ) \, dx-\left (a^3 (3 A-4 i B)\right ) \int \tan (c+d x) \, dx \\ & = 4 a^3 (i A+B) x+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\left (a^3 A\right ) \int \cot (c+d x) \, dx \\ & = 4 a^3 (i A+B) x+\frac {a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+\frac {a^3 A \log (\sin (c+d x))}{d}+\frac {i a B (a+i a \tan (c+d x))^2}{2 d}-\frac {(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.65 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^3 \left (2 A \log (\tan (c+d x))-8 (A-i B) \log (i+\tan (c+d x))+(-2 i A-6 B) \tan (c+d x)-i B \tan ^2(c+d x)\right )}{2 d} \]

[In]

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

[Out]

(a^3*(2*A*Log[Tan[c + d*x]] - 8*(A - I*B)*Log[I + Tan[c + d*x]] + ((-2*I)*A - 6*B)*Tan[c + d*x] - I*B*Tan[c +
d*x]^2))/(2*d)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(-\frac {a^{3} \left (-8 i A x d +i B \left (\tan ^{2}\left (d x +c \right )\right )+2 i A \tan \left (d x +c \right )-4 i B \ln \left (\sec ^{2}\left (d x +c \right )\right )-8 B d x +4 A \ln \left (\sec ^{2}\left (d x +c \right )\right )-2 A \ln \left (\tan \left (d x +c \right )\right )+6 B \tan \left (d x +c \right )\right )}{2 d}\)	\(87\)
derivativedivides	\(\frac {a^{3} \left (\frac {\left (4 i B -4 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-4 i A -4 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )+\left (-4 i B +3 A \right ) \ln \left (\cot \left (d x +c \right )\right )-\frac {i A +3 B}{\cot \left (d x +c \right )}-\frac {i B}{2 \cot \left (d x +c \right )^{2}}\right )}{d}\)	\(98\)
default	\(\frac {a^{3} \left (\frac {\left (4 i B -4 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-4 i A -4 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )+\left (-4 i B +3 A \right ) \ln \left (\cot \left (d x +c \right )\right )-\frac {i A +3 B}{\cot \left (d x +c \right )}-\frac {i B}{2 \cot \left (d x +c \right )^{2}}\right )}{d}\)	\(98\)
norman	\(\left (4 i A \,a^{3}+4 B \,a^{3}\right ) x -\frac {\left (i A \,a^{3}+3 B \,a^{3}\right ) \tan \left (d x +c \right )}{d}-\frac {i B \,a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {A \,a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 \left (-i B \,a^{3}+A \,a^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(105\)
risch	\(-\frac {8 a^{3} B c}{d}-\frac {8 i a^{3} A c}{d}-\frac {2 i a^{3} \left (i A \,{\mathrm e}^{2 i \left (d x +c \right )}+4 B \,{\mathrm e}^{2 i \left (d x +c \right )}+i A +3 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(141\)

[In]

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

[Out]

-1/2*a^3*(-8*I*A*x*d+I*B*tan(d*x+c)^2+2*I*A*tan(d*x+c)-4*I*B*ln(sec(d*x+c)^2)-8*B*d*x+4*A*ln(sec(d*x+c)^2)-2*A
*ln(tan(d*x+c))+6*B*tan(d*x+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.61 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left (A - 4 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, {\left (A - 3 i \, B\right )} a^{3} + {\left ({\left (3 \, A - 4 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (3 \, A - 4 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (3 \, A - 4 i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (94) = 188\).

Time = 1.60 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.09 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {A a^{3} \log {\left (\frac {- A a^{3} + 2 i B a^{3}}{A a^{3} e^{2 i c} - 2 i B a^{3} e^{2 i c}} + e^{2 i d x} \right )}}{d} + \frac {a^{3} \cdot \left (3 A - 4 i B\right ) \log {\left (e^{2 i d x} + \frac {- 2 A a^{3} + 2 i B a^{3} + a^{3} \cdot \left (3 A - 4 i B\right )}{A a^{3} e^{2 i c} - 2 i B a^{3} e^{2 i c}} \right )}}{d} + \frac {2 A a^{3} - 6 i B a^{3} + \left (2 A a^{3} e^{2 i c} - 8 i B a^{3} e^{2 i c}\right ) e^{2 i d x}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} \]

[In]

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

[Out]

A*a**3*log((-A*a**3 + 2*I*B*a**3)/(A*a**3*exp(2*I*c) - 2*I*B*a**3*exp(2*I*c)) + exp(2*I*d*x))/d + a**3*(3*A -
4*I*B)*log(exp(2*I*d*x) + (-2*A*a**3 + 2*I*B*a**3 + a**3*(3*A - 4*I*B))/(A*a**3*exp(2*I*c) - 2*I*B*a**3*exp(2*
I*c)))/d + (2*A*a**3 - 6*I*B*a**3 + (2*A*a**3*exp(2*I*c) - 8*I*B*a**3*exp(2*I*c))*exp(2*I*d*x))/(d*exp(4*I*c)*
exp(4*I*d*x) + 2*d*exp(2*I*c)*exp(2*I*d*x) + d)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.83 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {i \, B a^{3} \tan \left (d x + c\right )^{2} + 8 \, {\left (d x + c\right )} {\left (-i \, A - B\right )} a^{3} + 4 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, A a^{3} \log \left (\tan \left (d x + c\right )\right ) + 2 \, {\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*(I*B*a^3*tan(d*x + c)^2 + 8*(d*x + c)*(-I*A - B)*a^3 + 4*(A - I*B)*a^3*log(tan(d*x + c)^2 + 1) - 2*A*a^3*
log(tan(d*x + c)) + 2*(I*A + 3*B)*a^3*tan(d*x + c))/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (93) = 186\).

Time = 0.86 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.47 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \, A a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, {\left (3 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 16 \, {\left (A a^{3} - i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 2 \, {\left (3 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 28 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, A a^{3} - 12 i \, B a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*A*a^3*log(tan(1/2*d*x + 1/2*c)) + 2*(3*A*a^3 - 4*I*B*a^3)*log(tan(1/2*d*x + 1/2*c) + 1) - 16*(A*a^3 - I
*B*a^3)*log(tan(1/2*d*x + 1/2*c) + I) + 2*(3*A*a^3 - 4*I*B*a^3)*log(tan(1/2*d*x + 1/2*c) - 1) - (9*A*a^3*tan(1
/2*d*x + 1/2*c)^4 - 12*I*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 4*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*B*a^3*tan(1/2*d*
x + 1/2*c)^3 - 18*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 28*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 4*I*A*a^3*tan(1/2*d*x + 1
/2*c) + 12*B*a^3*tan(1/2*d*x + 1/2*c) + 9*A*a^3 - 12*I*B*a^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 7.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {A\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^3+a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\right )}{d}-\frac {4\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,d} \]

[In]

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

[Out]

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