3.15 \(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=117 \[ -\frac{a^2 (3 B+4 i A) \cot ^2(c+d x)}{6 d}+\frac{2 a^2 (A-i B) \cot (c+d x)}{d}-\frac{2 a^2 (B+i A) \log (\sin (c+d x))}{d}+2 a^2 x (A-i B)-\frac{A \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

[Out]

2*a^2*(A - I*B)*x + (2*a^2*(A - I*B)*Cot[c + d*x])/d - (a^2*((4*I)*A + 3*B)*Cot[c + d*x]^2)/(6*d) - (2*a^2*(I*
A + B)*Log[Sin[c + d*x]])/d - (A*Cot[c + d*x]^3*(a^2 + I*a^2*Tan[c + d*x]))/(3*d)

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Rubi [A]  time = 0.255599, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3593, 3591, 3529, 3531, 3475} \[ -\frac{a^2 (3 B+4 i A) \cot ^2(c+d x)}{6 d}+\frac{2 a^2 (A-i B) \cot (c+d x)}{d}-\frac{2 a^2 (B+i A) \log (\sin (c+d x))}{d}+2 a^2 x (A-i B)-\frac{A \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

2*a^2*(A - I*B)*x + (2*a^2*(A - I*B)*Cot[c + d*x])/d - (a^2*((4*I)*A + 3*B)*Cot[c + d*x]^2)/(6*d) - (2*a^2*(I*
A + B)*Log[Sin[c + d*x]])/d - (A*Cot[c + d*x]^3*(a^2 + I*a^2*Tan[c + d*x]))/(3*d)

Rule 3593

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

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]

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 3475

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

Rubi steps

\begin{align*} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=-\frac{A \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{1}{3} \int \cot ^3(c+d x) (a+i a \tan (c+d x)) (a (4 i A+3 B)-a (2 A-3 i B) \tan (c+d x)) \, dx\\ &=-\frac{a^2 (4 i A+3 B) \cot ^2(c+d x)}{6 d}-\frac{A \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{1}{3} \int \cot ^2(c+d x) \left (-6 a^2 (A-i B)-6 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (A-i B) \cot (c+d x)}{d}-\frac{a^2 (4 i A+3 B) \cot ^2(c+d x)}{6 d}-\frac{A \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{1}{3} \int \cot (c+d x) \left (-6 a^2 (i A+B)+6 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=2 a^2 (A-i B) x+\frac{2 a^2 (A-i B) \cot (c+d x)}{d}-\frac{a^2 (4 i A+3 B) \cot ^2(c+d x)}{6 d}-\frac{A \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}-\left (2 a^2 (i A+B)\right ) \int \cot (c+d x) \, dx\\ &=2 a^2 (A-i B) x+\frac{2 a^2 (A-i B) \cot (c+d x)}{d}-\frac{a^2 (4 i A+3 B) \cot ^2(c+d x)}{6 d}-\frac{2 a^2 (i A+B) \log (\sin (c+d x))}{d}-\frac{A \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [B]  time = 3.30089, size = 435, normalized size = 3.72 \[ \frac{a^2 \csc (c) \csc ^3(c+d x) (\cos (2 d x)+i \sin (2 d x)) \left (-48 (A-i B) \sin (c) \sin ^3(c+d x) \tan ^{-1}(\tan (3 c+d x))+3 \cos (d x) \left ((-3 B-3 i A) \log \left (\sin ^2(c+d x)\right )+4 A (3 d x-i)+2 B (-1-6 i d x)\right )-18 A \sin (2 c+d x)+14 A \sin (2 c+3 d x)+12 i A \cos (2 c+d x)-36 A d x \cos (2 c+d x)-12 A d x \cos (2 c+3 d x)+12 A d x \cos (4 c+3 d x)+9 i A \cos (2 c+d x) \log \left (\sin ^2(c+d x)\right )+3 i A \cos (2 c+3 d x) \log \left (\sin ^2(c+d x)\right )-3 i A \cos (4 c+3 d x) \log \left (\sin ^2(c+d x)\right )-24 A \sin (d x)+12 i B \sin (2 c+d x)-12 i B \sin (2 c+3 d x)+6 B \cos (2 c+d x)+36 i B d x \cos (2 c+d x)+12 i B d x \cos (2 c+3 d x)-12 i B d x \cos (4 c+3 d x)+9 B \cos (2 c+d x) \log \left (\sin ^2(c+d x)\right )+3 B \cos (2 c+3 d x) \log \left (\sin ^2(c+d x)\right )-3 B \cos (4 c+3 d x) \log \left (\sin ^2(c+d x)\right )+24 i B \sin (d x)\right )}{24 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

(a^2*Csc[c]*Csc[c + d*x]^3*(Cos[2*d*x] + I*Sin[2*d*x])*((12*I)*A*Cos[2*c + d*x] + 6*B*Cos[2*c + d*x] - 36*A*d*
x*Cos[2*c + d*x] + (36*I)*B*d*x*Cos[2*c + d*x] - 12*A*d*x*Cos[2*c + 3*d*x] + (12*I)*B*d*x*Cos[2*c + 3*d*x] + 1
2*A*d*x*Cos[4*c + 3*d*x] - (12*I)*B*d*x*Cos[4*c + 3*d*x] + (9*I)*A*Cos[2*c + d*x]*Log[Sin[c + d*x]^2] + 9*B*Co
s[2*c + d*x]*Log[Sin[c + d*x]^2] + (3*I)*A*Cos[2*c + 3*d*x]*Log[Sin[c + d*x]^2] + 3*B*Cos[2*c + 3*d*x]*Log[Sin
[c + d*x]^2] - (3*I)*A*Cos[4*c + 3*d*x]*Log[Sin[c + d*x]^2] - 3*B*Cos[4*c + 3*d*x]*Log[Sin[c + d*x]^2] + 3*Cos
[d*x]*(2*B*(-1 - (6*I)*d*x) + 4*A*(-I + 3*d*x) + ((-3*I)*A - 3*B)*Log[Sin[c + d*x]^2]) - 24*A*Sin[d*x] + (24*I
)*B*Sin[d*x] - 48*(A - I*B)*ArcTan[Tan[3*c + d*x]]*Sin[c]*Sin[c + d*x]^3 - 18*A*Sin[2*c + d*x] + (12*I)*B*Sin[
2*c + d*x] + 14*A*Sin[2*c + 3*d*x] - (12*I)*B*Sin[2*c + 3*d*x]))/(24*d*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [A]  time = 0.066, size = 154, normalized size = 1.3 \begin{align*} 2\,{a}^{2}Ax+2\,{\frac{{a}^{2}A\cot \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{2}c}{d}}-2\,{\frac{{a}^{2}B\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{iA{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{2\,iA{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,iB{a}^{2}x-{\frac{2\,iB\cot \left ( dx+c \right ){a}^{2}}{d}}-{\frac{2\,iB{a}^{2}c}{d}}-{\frac{{a}^{2}A \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{a}^{2}B \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.62685, size = 154, normalized size = 1.32 \begin{align*} \frac{6 \,{\left (d x + c\right )}{\left (2 \, A - 2 i \, B\right )} a^{2} + 6 \,{\left (i \, A + B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \,{\left (i \, A + B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac{{\left (12 \, A - 12 i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 3 \,{\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 2 \, A a^{2}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(6*(d*x + c)*(2*A - 2*I*B)*a^2 + 6*(I*A + B)*a^2*log(tan(d*x + c)^2 + 1) - 12*(I*A + B)*a^2*log(tan(d*x +
c)) + ((12*A - 12*I*B)*a^2*tan(d*x + c)^2 + 3*(-2*I*A - B)*a^2*tan(d*x + c) - 2*A*a^2)/tan(d*x + c)^3)/d

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Fricas [A]  time = 1.32167, size = 498, normalized size = 4.26 \begin{align*} \frac{{\left (30 i \, A + 18 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-36 i \, A - 30 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (14 i \, A + 12 \, B\right )} a^{2} +{\left ({\left (-6 i \, A - 6 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (18 i \, A + 18 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-18 i \, A - 18 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (6 i \, A + 6 \, B\right )} a^{2}\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*((30*I*A + 18*B)*a^2*e^(4*I*d*x + 4*I*c) + (-36*I*A - 30*B)*a^2*e^(2*I*d*x + 2*I*c) + (14*I*A + 12*B)*a^2
+ ((-6*I*A - 6*B)*a^2*e^(6*I*d*x + 6*I*c) + (18*I*A + 18*B)*a^2*e^(4*I*d*x + 4*I*c) + (-18*I*A - 18*B)*a^2*e^(
2*I*d*x + 2*I*c) + (6*I*A + 6*B)*a^2)*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)

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Sympy [A]  time = 7.44088, size = 170, normalized size = 1.45 \begin{align*} - \frac{2 a^{2} \left (i A + B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{\frac{\left (10 i A a^{2} + 6 B a^{2}\right ) e^{- 2 i c} e^{4 i d x}}{d} - \frac{\left (12 i A a^{2} + 10 B a^{2}\right ) e^{- 4 i c} e^{2 i d x}}{d} + \frac{\left (14 i A a^{2} + 12 B a^{2}\right ) e^{- 6 i c}}{3 d}}{e^{6 i d x} - 3 e^{- 2 i c} e^{4 i d x} + 3 e^{- 4 i c} e^{2 i d x} - e^{- 6 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

-2*a**2*(I*A + B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + ((10*I*A*a**2 + 6*B*a**2)*exp(-2*I*c)*exp(4*I*d*x)/d - (
12*I*A*a**2 + 10*B*a**2)*exp(-4*I*c)*exp(2*I*d*x)/d + (14*I*A*a**2 + 12*B*a**2)*exp(-6*I*c)/(3*d))/(exp(6*I*d*
x) - 3*exp(-2*I*c)*exp(4*I*d*x) + 3*exp(-4*I*c)*exp(2*I*d*x) - exp(-6*I*c))

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Giac [B]  time = 1.48899, size = 346, normalized size = 2.96 \begin{align*} \frac{A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 27 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 i \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 96 \,{\left (-i \, A a^{2} - B a^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 48 \,{\left (-i \, A a^{2} - B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{-88 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 88 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 27 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 i \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/24*(A*a^2*tan(1/2*d*x + 1/2*c)^3 - 6*I*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 27*A*
a^2*tan(1/2*d*x + 1/2*c) + 24*I*B*a^2*tan(1/2*d*x + 1/2*c) - 96*(-I*A*a^2 - B*a^2)*log(tan(1/2*d*x + 1/2*c) +
I) + 48*(-I*A*a^2 - B*a^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (-88*I*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 88*B*a^2*tan
(1/2*d*x + 1/2*c)^3 - 27*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*I*B*a^2*tan(1/2*d*x + 1/2*c)^2 + 6*I*A*a^2*tan(1/2*
d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) + A*a^2)/tan(1/2*d*x + 1/2*c)^3)/d