3.7 \(\int \cot ^4(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=89 \[ -\frac {a (B+i A) \cot ^2(c+d x)}{2 d}+\frac {a (A-i B) \cot (c+d x)}{d}-\frac {a (B+i A) \log (\sin (c+d x))}{d}+a x (A-i B)-\frac {a A \cot ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3591, 3529, 3531, 3475} \[ -\frac {a (B+i A) \cot ^2(c+d x)}{2 d}+\frac {a (A-i B) \cot (c+d x)}{d}-\frac {a (B+i A) \log (\sin (c+d x))}{d}+a x (A-i B)-\frac {a A \cot ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

a*(A - I*B)*x + (a*(A - I*B)*Cot[c + d*x])/d - (a*(I*A + B)*Cot[c + d*x]^2)/(2*d) - (a*A*Cot[c + d*x]^3)/(3*d)
- (a*(I*A + B)*Log[Sin[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 3529

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

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 3591

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.72, size = 102, normalized size = 1.15 \[ -\frac {a \left (3 (B+i A) \left (\cot ^2(c+d x)+2 (\log (\tan (c+d x))+\log (\cos (c+d x)))\right )+2 A \cot ^3(c+d x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\tan ^2(c+d x)\right )+6 i B \cot (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(c+d x)\right )\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

-1/6*(a*(2*A*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2] + (6*I)*B*Cot[c + d*x]*Hypergeom
etric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] + 3*(I*A + B)*(Cot[c + d*x]^2 + 2*(Log[Cos[c + d*x]] + Log[Tan[c + d*x
]]))))/d

________________________________________________________________________________________

fricas [B]  time = 0.41, size = 166, normalized size = 1.87 \[ \frac {{\left (18 i \, A + 12 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-18 i \, A - 18 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (8 i \, A + 6 \, B\right )} a + {\left ({\left (-3 i \, A - 3 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (9 i \, A + 9 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-9 i \, A - 9 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (3 i \, A + 3 \, B\right )} a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*((18*I*A + 12*B)*a*e^(4*I*d*x + 4*I*c) + (-18*I*A - 18*B)*a*e^(2*I*d*x + 2*I*c) + (8*I*A + 6*B)*a + ((-3*I
*A - 3*B)*a*e^(6*I*d*x + 6*I*c) + (9*I*A + 9*B)*a*e^(4*I*d*x + 4*I*c) + (-9*I*A - 9*B)*a*e^(2*I*d*x + 2*I*c) +
(3*I*A + 3*B)*a)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*
d*x + 2*I*c) - d)

________________________________________________________________________________________

giac [B]  time = 0.96, size = 221, normalized size = 2.48 \[ \frac {A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 i \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 i \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, {\left (i \, A a + B a\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 24 \, {\left (i \, A a + B a\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {-44 i \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 i \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 i \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/24*(A*a*tan(1/2*d*x + 1/2*c)^3 - 3*I*A*a*tan(1/2*d*x + 1/2*c)^2 - 3*B*a*tan(1/2*d*x + 1/2*c)^2 - 15*A*a*tan(
1/2*d*x + 1/2*c) + 12*I*B*a*tan(1/2*d*x + 1/2*c) + 48*(I*A*a + B*a)*log(tan(1/2*d*x + 1/2*c) + I) - 24*(I*A*a
+ B*a)*log(tan(1/2*d*x + 1/2*c)) - (-44*I*A*a*tan(1/2*d*x + 1/2*c)^3 - 44*B*a*tan(1/2*d*x + 1/2*c)^3 - 15*A*a*
tan(1/2*d*x + 1/2*c)^2 + 12*I*B*a*tan(1/2*d*x + 1/2*c)^2 + 3*I*A*a*tan(1/2*d*x + 1/2*c) + 3*B*a*tan(1/2*d*x +
1/2*c) + A*a)/tan(1/2*d*x + 1/2*c)^3)/d

________________________________________________________________________________________

maple [A]  time = 0.39, size = 129, normalized size = 1.45 \[ -\frac {i A a \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {i A a \ln \left (\sin \left (d x +c \right )\right )}{d}-i B a x -\frac {i B \cot \left (d x +c \right ) a}{d}-\frac {i B a c}{d}-\frac {a A \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a A \cot \left (d x +c \right )}{d}+a A x +\frac {A a c}{d}-\frac {a B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a B \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.91, size = 103, normalized size = 1.16 \[ \frac {6 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a - 3 \, {\left (-i \, A - B\right )} a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \, {\left (-i \, A - B\right )} a \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )^{2} - {\left (3 i \, A + 3 \, B\right )} a \tan \left (d x + c\right ) - 2 \, A a}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.25, size = 80, normalized size = 0.90 \[ -\frac {\left (-A\,a+B\,a\,1{}\mathrm {i}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+\left (\frac {B\,a}{2}+\frac {A\,a\,1{}\mathrm {i}}{2}\right )\,\mathrm {tan}\left (c+d\,x\right )+\frac {A\,a}{3}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^3}-\frac {a\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i),x)

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 0.78, size = 168, normalized size = 1.89 \[ - \frac {i a \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 8 i A a - 6 B a + \left (18 i A a e^{2 i c} + 18 B a e^{2 i c}\right ) e^{2 i d x} + \left (- 18 i A a e^{4 i c} - 12 B a e^{4 i c}\right ) e^{4 i d x}}{- 3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} - 9 d e^{2 i c} e^{2 i d x} + 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

-I*a*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-8*I*A*a - 6*B*a + (18*I*A*a*exp(2*I*c) + 18*B*a*exp(2*I*c
))*exp(2*I*d*x) + (-18*I*A*a*exp(4*I*c) - 12*B*a*exp(4*I*c))*exp(4*I*d*x))/(-3*d*exp(6*I*c)*exp(6*I*d*x) + 9*d
*exp(4*I*c)*exp(4*I*d*x) - 9*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

________________________________________________________________________________________