3.41 \(\int (c+d x)^2 \cot (a+b x) \csc (a+b x) \, dx\)

Optimal. Leaf size=90 \[ \frac {2 i d^2 \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^3}-\frac {2 i d^2 \text {Li}_2\left (e^{i (a+b x)}\right )}{b^3}-\frac {4 d (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \csc (a+b x)}{b} \]

[Out]

-4*d*(d*x+c)*arctanh(exp(I*(b*x+a)))/b^2-(d*x+c)^2*csc(b*x+a)/b+2*I*d^2*polylog(2,-exp(I*(b*x+a)))/b^3-2*I*d^2
*polylog(2,exp(I*(b*x+a)))/b^3

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Rubi [A]  time = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4410, 4183, 2279, 2391} \[ \frac {2 i d^2 \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}-\frac {2 i d^2 \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}-\frac {4 d (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cot[a + b*x]*Csc[a + b*x],x]

[Out]

(-4*d*(c + d*x)*ArcTanh[E^(I*(a + b*x))])/b^2 - ((c + d*x)^2*Csc[a + b*x])/b + ((2*I)*d^2*PolyLog[2, -E^(I*(a
+ b*x))])/b^3 - ((2*I)*d^2*PolyLog[2, E^(I*(a + b*x))])/b^3

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 4183

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

Rule 4410

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp
[((c + d*x)^m*Csc[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Fr
eeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int (c+d x)^2 \cot (a+b x) \csc (a+b x) \, dx &=-\frac {(c+d x)^2 \csc (a+b x)}{b}+\frac {(2 d) \int (c+d x) \csc (a+b x) \, dx}{b}\\ &=-\frac {4 d (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \csc (a+b x)}{b}-\frac {\left (2 d^2\right ) \int \log \left (1-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (2 d^2\right ) \int \log \left (1+e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {4 d (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \csc (a+b x)}{b}+\frac {\left (2 i d^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}-\frac {\left (2 i d^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}\\ &=-\frac {4 d (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \csc (a+b x)}{b}+\frac {2 i d^2 \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^3}-\frac {2 i d^2 \text {Li}_2\left (e^{i (a+b x)}\right )}{b^3}\\ \end {align*}

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Mathematica [B]  time = 2.04, size = 234, normalized size = 2.60 \[ \frac {-2 b^2 \csc (a) (c+d x)^2+b^2 \csc \left (\frac {a}{2}\right ) \sin \left (\frac {b x}{2}\right ) (c+d x)^2 \csc \left (\frac {1}{2} (a+b x)\right )-b^2 \sec \left (\frac {a}{2}\right ) \sin \left (\frac {b x}{2}\right ) (c+d x)^2 \sec \left (\frac {1}{2} (a+b x)\right )-8 b c d \tanh ^{-1}\left (\cos (a)-\sin (a) \tan \left (\frac {b x}{2}\right )\right )+4 d^2 \left (2 \tan ^{-1}(\tan (a)) \tanh ^{-1}\left (\cos (a)-\sin (a) \tan \left (\frac {b x}{2}\right )\right )+\frac {\sec (a) \left (i \text {Li}_2\left (-e^{i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )-i \text {Li}_2\left (e^{i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+\left (\tan ^{-1}(\tan (a))+b x\right ) \left (\log \left (1-e^{i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-\log \left (1+e^{i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )\right )\right )}{\sqrt {\sec ^2(a)}}\right )}{2 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Cot[a + b*x]*Csc[a + b*x],x]

[Out]

(-8*b*c*d*ArcTanh[Cos[a] - Sin[a]*Tan[(b*x)/2]] - 2*b^2*(c + d*x)^2*Csc[a] + 4*d^2*(2*ArcTan[Tan[a]]*ArcTanh[C
os[a] - Sin[a]*Tan[(b*x)/2]] + (((b*x + ArcTan[Tan[a]])*(Log[1 - E^(I*(b*x + ArcTan[Tan[a]]))] - Log[1 + E^(I*
(b*x + ArcTan[Tan[a]]))]) + I*PolyLog[2, -E^(I*(b*x + ArcTan[Tan[a]]))] - I*PolyLog[2, E^(I*(b*x + ArcTan[Tan[
a]]))])*Sec[a])/Sqrt[Sec[a]^2]) + b^2*(c + d*x)^2*Csc[a/2]*Csc[(a + b*x)/2]*Sin[(b*x)/2] - b^2*(c + d*x)^2*Sec
[a/2]*Sec[(a + b*x)/2]*Sin[(b*x)/2])/(2*b^3)

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fricas [B]  time = 0.53, size = 375, normalized size = 4.17 \[ -\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + i \, d^{2} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d^{2} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + i \, d^{2} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d^{2} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) - {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) - {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right )}{b^{3} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)*csc(b*x+a)^2,x, algorithm="fricas")

[Out]

-(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + I*d^2*dilog(cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - I*d^2*dilog(
cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + I*d^2*dilog(-cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - I*d^2
*dilog(-cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + (b*d^2*x + b*c*d)*log(cos(b*x + a) + I*sin(b*x + a) + 1)
*sin(b*x + a) + (b*d^2*x + b*c*d)*log(cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) - (b*c*d - a*d^2)*log(-1
/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) - (b*c*d - a*d^2)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b
*x + a) + 1/2)*sin(b*x + a) - (b*d^2*x + a*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + 1)*sin(b*x + a) - (b*d^2*
x + a*d^2)*log(-cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a))/(b^3*sin(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \cos \left (b x + a\right ) \csc \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)*csc(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*cos(b*x + a)*csc(b*x + a)^2, x)

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maple [B]  time = 0.03, size = 233, normalized size = 2.59 \[ -\frac {d^{2} x^{2}}{b \sin \left (b x +a \right )}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}-\frac {2 i d^{2} \dilog \left (1-{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i d^{2} \dilog \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {2 a \,d^{2} \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b^{3}}-\frac {2 c d x}{b \sin \left (b x +a \right )}+\frac {2 c d \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b^{2}}-\frac {c^{2}}{b \sin \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*cos(b*x+a)*csc(b*x+a)^2,x)

[Out]

-1/b*d^2/sin(b*x+a)*x^2+2/b^2*d^2*ln(1-exp(I*(b*x+a)))*x+2/b^3*d^2*ln(1-exp(I*(b*x+a)))*a-2/b^2*d^2*ln(exp(I*(
b*x+a))+1)*x-2/b^3*d^2*ln(exp(I*(b*x+a))+1)*a-2*I/b^3*d^2*dilog(1-exp(I*(b*x+a)))+2*I/b^3*d^2*dilog(exp(I*(b*x
+a))+1)-2/b^3*a*d^2*ln(csc(b*x+a)-cot(b*x+a))-2/b*c*d/sin(b*x+a)*x+2/b^2*c*d*ln(csc(b*x+a)-cot(b*x+a))-1/b*c^2
/sin(b*x+a)

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maxima [B]  time = 0.52, size = 556, normalized size = 6.18 \[ \frac {{\left (2 \, b d^{2} x + 2 \, b c d - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + {\left (2 \, b c d \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b c d \sin \left (2 \, b x + 2 \, a\right ) - 2 \, b c d\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) - {\left (2 \, b d^{2} x \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b d^{2} x \sin \left (2 \, b x + 2 \, a\right ) - 2 \, b d^{2} x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (b x + a\right ) + 2 \, {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 2 \, {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (-i \, b d^{2} x - i \, b c d + {\left (i \, b d^{2} x + i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (2 i \, b^{2} d^{2} x^{2} + 4 i \, b^{2} c d x + 2 i \, b^{2} c^{2}\right )} \sin \left (b x + a\right )}{-i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + b^{3} \sin \left (2 \, b x + 2 \, a\right ) + i \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)*csc(b*x+a)^2,x, algorithm="maxima")

[Out]

((2*b*d^2*x + 2*b*c*d - 2*(b*d^2*x + b*c*d)*cos(2*b*x + 2*a) - (2*I*b*d^2*x + 2*I*b*c*d)*sin(2*b*x + 2*a))*arc
tan2(sin(b*x + a), cos(b*x + a) + 1) + (2*b*c*d*cos(2*b*x + 2*a) + 2*I*b*c*d*sin(2*b*x + 2*a) - 2*b*c*d)*arcta
n2(sin(b*x + a), cos(b*x + a) - 1) - (2*b*d^2*x*cos(2*b*x + 2*a) + 2*I*b*d^2*x*sin(2*b*x + 2*a) - 2*b*d^2*x)*a
rctan2(sin(b*x + a), -cos(b*x + a) + 1) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(b*x + a) + 2*(d^2*cos(2*
b*x + 2*a) + I*d^2*sin(2*b*x + 2*a) - d^2)*dilog(-e^(I*b*x + I*a)) - 2*(d^2*cos(2*b*x + 2*a) + I*d^2*sin(2*b*x
+ 2*a) - d^2)*dilog(e^(I*b*x + I*a)) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(2*b*x + 2*a) + (b*d^
2*x + b*c*d)*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*d^2*x - I*b*c
*d + (I*b*d^2*x + I*b*c*d)*cos(2*b*x + 2*a) - (b*d^2*x + b*c*d)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x
+ a)^2 - 2*cos(b*x + a) + 1) - (2*I*b^2*d^2*x^2 + 4*I*b^2*c*d*x + 2*I*b^2*c^2)*sin(b*x + a))/(-I*b^3*cos(2*b*
x + 2*a) + b^3*sin(2*b*x + 2*a) + I*b^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^2}{{\sin \left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(a + b*x)*(c + d*x)^2)/sin(a + b*x)^2,x)

[Out]

int((cos(a + b*x)*(c + d*x)^2)/sin(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \cos {\left (a + b x \right )} \csc ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*cos(b*x+a)*csc(b*x+a)**2,x)

[Out]

Integral((c + d*x)**2*cos(a + b*x)*csc(a + b*x)**2, x)

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