3.181 \(\int (c+d x) \cot ^3(a+b x) \, dx\)

Optimal. Leaf size=109 \[ \frac {i d \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {d x}{2 b}+\frac {i (c+d x)^2}{2 d} \]

[Out]

-1/2*d*x/b+1/2*I*(d*x+c)^2/d-1/2*d*cot(b*x+a)/b^2-1/2*(d*x+c)*cot(b*x+a)^2/b-(d*x+c)*ln(1-exp(2*I*(b*x+a)))/b+
1/2*I*d*polylog(2,exp(2*I*(b*x+a)))/b^2

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Rubi [A]  time = 0.13, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3720, 3473, 8, 3717, 2190, 2279, 2391} \[ \frac {i d \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {d x}{2 b}+\frac {i (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Cot[a + b*x]^3,x]

[Out]

-(d*x)/(2*b) + ((I/2)*(c + d*x)^2)/d - (d*Cot[a + b*x])/(2*b^2) - ((c + d*x)*Cot[a + b*x]^2)/(2*b) - ((c + d*x
)*Log[1 - E^((2*I)*(a + b*x))])/b + ((I/2)*d*PolyLog[2, E^((2*I)*(a + b*x))])/b^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 3720

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

Rubi steps

\begin {align*} \int (c+d x) \cot ^3(a+b x) \, dx &=-\frac {(c+d x) \cot ^2(a+b x)}{2 b}+\frac {d \int \cot ^2(a+b x) \, dx}{2 b}-\int (c+d x) \cot (a+b x) \, dx\\ &=\frac {i (c+d x)^2}{2 d}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}+2 i \int \frac {e^{2 i (a+b x)} (c+d x)}{1-e^{2 i (a+b x)}} \, dx-\frac {d \int 1 \, dx}{2 b}\\ &=-\frac {d x}{2 b}+\frac {i (c+d x)^2}{2 d}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {d \int \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {d x}{2 b}+\frac {i (c+d x)^2}{2 d}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^2}\\ &=-\frac {d x}{2 b}+\frac {i (c+d x)^2}{2 d}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {i d \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}\\ \end {align*}

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Mathematica [B]  time = 6.17, size = 240, normalized size = 2.20 \[ \frac {d \csc (a) \sec (a) \left (b^2 x^2 e^{i \tan ^{-1}(\tan (a))}+\frac {\tan (a) \left (i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+i b x \left (2 \tan ^{-1}(\tan (a))-\pi \right )-2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt {\tan ^2(a)+1}}\right )}{2 b^2 \sqrt {\sec ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac {d \csc (a) \sin (b x) \csc (a+b x)}{2 b^2}-\frac {c \left (\cot ^2(a+b x)+2 \log (\tan (a+b x))+2 \log (\cos (a+b x))\right )}{2 b}-\frac {d x \csc ^2(a+b x)}{2 b}-\frac {1}{2} d x^2 \cot (a) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)*Cot[a + b*x]^3,x]

[Out]

-1/2*(d*x^2*Cot[a]) - (d*x*Csc[a + b*x]^2)/(2*b) - (c*(Cot[a + b*x]^2 + 2*Log[Cos[a + b*x]] + 2*Log[Tan[a + b*
x]]))/(2*b) + (d*Csc[a]*Csc[a + b*x]*Sin[b*x])/(2*b^2) + (d*Csc[a]*Sec[a]*(b^2*E^(I*ArcTan[Tan[a]])*x^2 + ((I*
b*x*(-Pi + 2*ArcTan[Tan[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x + ArcTan[Tan[a]])*Log[1 - E^((2*I)*(b*x + A
rcTan[Tan[a]]))] + Pi*Log[Cos[b*x]] + 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] + I*PolyLog[2, E^((2*I)*
(b*x + ArcTan[Tan[a]]))])*Tan[a])/Sqrt[1 + Tan[a]^2]))/(2*b^2*Sqrt[Sec[a]^2*(Cos[a]^2 + Sin[a]^2)])

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fricas [B]  time = 0.46, size = 339, normalized size = 3.11 \[ \frac {4 \, b d x + 4 \, b c + {\left (i \, d \cos \left (2 \, b x + 2 \, a\right ) - i \, d\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + {\left (-i \, d \cos \left (2 \, b x + 2 \, a\right ) + i \, d\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \, {\left (b c - a d - {\left (b c - a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b c - a d - {\left (b c - a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b d x + a d - {\left (b d x + a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left (b d x + a d - {\left (b d x + a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, d \sin \left (2 \, b x + 2 \, a\right )}{4 \, {\left (b^{2} \cos \left (2 \, b x + 2 \, a\right ) - b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*b*d*x + 4*b*c + (I*d*cos(2*b*x + 2*a) - I*d)*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) + (-I*d*cos(2
*b*x + 2*a) + I*d)*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) + 2*(b*c - a*d - (b*c - a*d)*cos(2*b*x + 2*a))
*log(-1/2*cos(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2) + 2*(b*c - a*d - (b*c - a*d)*cos(2*b*x + 2*a))*log(
-1/2*cos(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 2*a) + 1/2) + 2*(b*d*x + a*d - (b*d*x + a*d)*cos(2*b*x + 2*a))*log(-
cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) + 2*(b*d*x + a*d - (b*d*x + a*d)*cos(2*b*x + 2*a))*log(-cos(2*b*x +
2*a) - I*sin(2*b*x + 2*a) + 1) + 2*d*sin(2*b*x + 2*a))/(b^2*cos(2*b*x + 2*a) - b^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \cot \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*cot(b*x + a)^3, x)

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maple [B]  time = 0.09, size = 281, normalized size = 2.58 \[ \frac {i d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-i c x +\frac {2 b d x \,{\mathrm e}^{2 i \left (b x +a \right )}+2 b c \,{\mathrm e}^{2 i \left (b x +a \right )}-i d \,{\mathrm e}^{2 i \left (b x +a \right )}+i d}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}-\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}+\frac {2 c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {i d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d \,x^{2}}{2}+\frac {i d \,a^{2}}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {2 i d a x}{b}-\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {2 d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*cot(b*x+a)^3,x)

[Out]

I*d*polylog(2,-exp(I*(b*x+a)))/b^2-I*c*x+(2*b*d*x*exp(2*I*(b*x+a))+2*b*c*exp(2*I*(b*x+a))-I*d*exp(2*I*(b*x+a))
+I*d)/b^2/(exp(2*I*(b*x+a))-1)^2-1/b*c*ln(exp(I*(b*x+a))-1)-1/b*c*ln(exp(I*(b*x+a))+1)+2/b*c*ln(exp(I*(b*x+a))
)+I/b^2*d*polylog(2,exp(I*(b*x+a)))+1/2*I*d*x^2+I/b^2*d*a^2-1/b*d*ln(exp(I*(b*x+a))+1)*x+2*I/b*d*a*x-1/b*d*ln(
1-exp(I*(b*x+a)))*x-1/b^2*d*ln(1-exp(I*(b*x+a)))*a+1/b^2*d*a*ln(exp(I*(b*x+a))-1)-2/b^2*d*a*ln(exp(I*(b*x+a)))

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maxima [B]  time = 0.53, size = 839, normalized size = 7.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*d*x^2 + 2*b^2*c*x - (2*b*d*x + 2*b*c + 2*(b*d*x + b*c)*cos(4*b*x + 4*a) - 4*(b*d*x + b*c)*cos(2*b*x + 2*a
) + (2*I*b*d*x + 2*I*b*c)*sin(4*b*x + 4*a) + (-4*I*b*d*x - 4*I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), co
s(b*x + a) + 1) - (2*b*c*cos(4*b*x + 4*a) - 4*b*c*cos(2*b*x + 2*a) + 2*I*b*c*sin(4*b*x + 4*a) - 4*I*b*c*sin(2*
b*x + 2*a) + 2*b*c)*arctan2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d*x*cos(4*b*x + 4*a) - 4*b*d*x*cos(2*b*x +
2*a) + 2*I*b*d*x*sin(4*b*x + 4*a) - 4*I*b*d*x*sin(2*b*x + 2*a) + 2*b*d*x)*arctan2(sin(b*x + a), -cos(b*x + a)
+ 1) + (b^2*d*x^2 + 2*b^2*c*x)*cos(4*b*x + 4*a) - (2*b^2*d*x^2 + 4*I*b*c + (4*b^2*c + 4*I*b*d)*x + 2*d)*cos(2*
b*x + 2*a) + (2*d*cos(4*b*x + 4*a) - 4*d*cos(2*b*x + 2*a) + 2*I*d*sin(4*b*x + 4*a) - 4*I*d*sin(2*b*x + 2*a) +
2*d)*dilog(-e^(I*b*x + I*a)) + (2*d*cos(4*b*x + 4*a) - 4*d*cos(2*b*x + 2*a) + 2*I*d*sin(4*b*x + 4*a) - 4*I*d*s
in(2*b*x + 2*a) + 2*d)*dilog(e^(I*b*x + I*a)) - (-I*b*d*x - I*b*c + (-I*b*d*x - I*b*c)*cos(4*b*x + 4*a) + (2*I
*b*d*x + 2*I*b*c)*cos(2*b*x + 2*a) + (b*d*x + b*c)*sin(4*b*x + 4*a) - 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*log(co
s(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (-I*b*d*x - I*b*c + (-I*b*d*x - I*b*c)*cos(4*b*x + 4*a)
+ (2*I*b*d*x + 2*I*b*c)*cos(2*b*x + 2*a) + (b*d*x + b*c)*sin(4*b*x + 4*a) - 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*
log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - (-I*b^2*d*x^2 - 2*I*b^2*c*x)*sin(4*b*x + 4*a) - (2
*I*b^2*d*x^2 - 4*b*c - 4*(-I*b^2*c + b*d)*x + 2*I*d)*sin(2*b*x + 2*a) + 2*d)/(-2*I*b^2*cos(4*b*x + 4*a) + 4*I*
b^2*cos(2*b*x + 2*a) + 2*b^2*sin(4*b*x + 4*a) - 4*b^2*sin(2*b*x + 2*a) - 2*I*b^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (a+b\,x\right )}^3\,\left (c+d\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(a + b*x)^3*(c + d*x),x)

[Out]

int(cot(a + b*x)^3*(c + d*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \cot ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cot(b*x+a)**3,x)

[Out]

Integral((c + d*x)*cot(a + b*x)**3, x)

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