3.38 \(\int (c+d x)^2 (a+b \coth (e+f x)) \, dx\)

Optimal. Leaf size=101 \[ \frac{b d (c+d x) \text{PolyLog}\left (2,e^{2 (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac{b (c+d x)^3}{3 d} \]

[Out]

(a*(c + d*x)^3)/(3*d) - (b*(c + d*x)^3)/(3*d) + (b*(c + d*x)^2*Log[1 - E^(2*(e + f*x))])/f + (b*d*(c + d*x)*Po
lyLog[2, E^(2*(e + f*x))])/f^2 - (b*d^2*PolyLog[3, E^(2*(e + f*x))])/(2*f^3)

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Rubi [A]  time = 0.216669, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3722, 3716, 2190, 2531, 2282, 6589} \[ \frac{b d (c+d x) \text{PolyLog}\left (2,e^{2 (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac{b (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*(a + b*Coth[e + f*x]),x]

[Out]

(a*(c + d*x)^3)/(3*d) - (b*(c + d*x)^3)/(3*d) + (b*(c + d*x)^2*Log[1 - E^(2*(e + f*x))])/f + (b*d*(c + d*x)*Po
lyLog[2, E^(2*(e + f*x))])/f^2 - (b*d^2*PolyLog[3, E^(2*(e + f*x))])/(2*f^3)

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3716

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

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int (c+d x)^2 (a+b \coth (e+f x)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \coth (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^3}{3 d}+b \int (c+d x)^2 \coth (e+f x) \, dx\\ &=\frac{a (c+d x)^3}{3 d}-\frac{b (c+d x)^3}{3 d}-(2 b) \int \frac{e^{2 (e+f x)} (c+d x)^2}{1-e^{2 (e+f x)}} \, dx\\ &=\frac{a (c+d x)^3}{3 d}-\frac{b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac{(2 b d) \int (c+d x) \log \left (1-e^{2 (e+f x)}\right ) \, dx}{f}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}+\frac{b d (c+d x) \text{Li}_2\left (e^{2 (e+f x)}\right )}{f^2}-\frac{\left (b d^2\right ) \int \text{Li}_2\left (e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}+\frac{b d (c+d x) \text{Li}_2\left (e^{2 (e+f x)}\right )}{f^2}-\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{2 f^3}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}+\frac{b d (c+d x) \text{Li}_2\left (e^{2 (e+f x)}\right )}{f^2}-\frac{b d^2 \text{Li}_3\left (e^{2 (e+f x)}\right )}{2 f^3}\\ \end{align*}

Mathematica [A]  time = 0.216802, size = 149, normalized size = 1.48 \[ \frac{6 b d f (c+d x) \text{PolyLog}\left (2,e^{2 (e+f x)}\right )-3 b d^2 \text{PolyLog}\left (3,e^{2 (e+f x)}\right )+2 f^2 \left (3 a c^2 f x+3 a c d f x^2+a d^2 f x^3+3 b c^2 \log (\tanh (e+f x))+3 b c^2 \log (\cosh (e+f x))+3 b d x (2 c+d x) \log \left (1-e^{2 (e+f x)}\right )-3 b c d f x^2-b d^2 f x^3\right )}{6 f^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*(a + b*Coth[e + f*x]),x]

[Out]

(2*f^2*(3*a*c^2*f*x + 3*a*c*d*f*x^2 - 3*b*c*d*f*x^2 + a*d^2*f*x^3 - b*d^2*f*x^3 + 3*b*d*x*(2*c + d*x)*Log[1 -
E^(2*(e + f*x))] + 3*b*c^2*Log[Cosh[e + f*x]] + 3*b*c^2*Log[Tanh[e + f*x]]) + 6*b*d*f*(c + d*x)*PolyLog[2, E^(
2*(e + f*x))] - 3*b*d^2*PolyLog[3, E^(2*(e + f*x))])/(6*f^3)

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Maple [B]  time = 0.055, size = 447, normalized size = 4.4 \begin{align*}{\frac{b{d}^{2}{e}^{2}\ln \left ({{\rm e}^{fx+e}}-1 \right ) }{{f}^{3}}}+2\,{\frac{b{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{fx+e}} \right ) x}{{f}^{2}}}+{\frac{b{d}^{2}\ln \left ({{\rm e}^{fx+e}}+1 \right ){x}^{2}}{f}}-2\,{\frac{b{d}^{2}{e}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+2\,{\frac{cbd{\it polylog} \left ( 2,{{\rm e}^{fx+e}} \right ) }{{f}^{2}}}+2\,{\frac{b{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{fx+e}} \right ) x}{{f}^{2}}}-{\frac{b{d}^{2}{e}^{2}\ln \left ( 1-{{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+{\frac{b{d}^{2}\ln \left ( 1-{{\rm e}^{fx+e}} \right ){x}^{2}}{f}}+2\,{\frac{cbd{\it polylog} \left ( 2,-{{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-2\,{\frac{b{d}^{2}{\it polylog} \left ( 3,{{\rm e}^{fx+e}} \right ) }{{f}^{3}}}-2\,{\frac{b{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+{\frac{b{c}^{2}\ln \left ({{\rm e}^{fx+e}}+1 \right ) }{f}}-2\,{\frac{b{c}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}-2\,{\frac{cbd{e}^{2}}{{f}^{2}}}+2\,{\frac{b{d}^{2}{e}^{2}x}{{f}^{2}}}+{\frac{b{c}^{2}\ln \left ({{\rm e}^{fx+e}}-1 \right ) }{f}}-bcd{x}^{2}+acd{x}^{2}+{\frac{a{d}^{2}{x}^{3}}{3}}-{\frac{b{d}^{2}{x}^{3}}{3}}+2\,{\frac{cbd\ln \left ( 1-{{\rm e}^{fx+e}} \right ) e}{{f}^{2}}}+4\,{\frac{cbde\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-2\,{\frac{cbde\ln \left ({{\rm e}^{fx+e}}-1 \right ) }{{f}^{2}}}+2\,{\frac{cbd\ln \left ( 1-{{\rm e}^{fx+e}} \right ) x}{f}}-4\,{\frac{cbdex}{f}}+2\,{\frac{cbd\ln \left ({{\rm e}^{fx+e}}+1 \right ) x}{f}}+{\frac{4\,b{d}^{2}{e}^{3}}{3\,{f}^{3}}}+{c}^{2}ax+b{c}^{2}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*(a+b*coth(f*x+e)),x)

[Out]

b/f^3*d^2*e^2*ln(exp(f*x+e)-1)+2*b/f^2*d^2*polylog(2,exp(f*x+e))*x+b/f*d^2*ln(exp(f*x+e)+1)*x^2-2*b/f^3*d^2*e^
2*ln(exp(f*x+e))+2*b/f^2*c*d*polylog(2,exp(f*x+e))+2*b/f^2*d^2*polylog(2,-exp(f*x+e))*x-b/f^3*d^2*e^2*ln(1-exp
(f*x+e))+b/f*d^2*ln(1-exp(f*x+e))*x^2+2*b/f^2*c*d*polylog(2,-exp(f*x+e))-2*b/f^3*d^2*polylog(3,exp(f*x+e))-2*b
/f^3*d^2*polylog(3,-exp(f*x+e))+b/f*c^2*ln(exp(f*x+e)+1)-2*b/f*c^2*ln(exp(f*x+e))-2*b/f^2*c*d*e^2+2*b/f^2*d^2*
e^2*x+b/f*c^2*ln(exp(f*x+e)-1)-b*c*d*x^2+a*c*d*x^2+1/3*a*d^2*x^3-1/3*b*d^2*x^3+2*b/f^2*c*d*ln(1-exp(f*x+e))*e+
4*b/f^2*c*d*e*ln(exp(f*x+e))-2*b/f^2*c*d*e*ln(exp(f*x+e)-1)+2*b/f*c*d*ln(1-exp(f*x+e))*x-4*b/f*c*d*e*x+2*b/f*c
*d*ln(exp(f*x+e)+1)*x+4/3*b/f^3*d^2*e^3+c^2*a*x+b*c^2*x

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Maxima [B]  time = 1.38171, size = 324, normalized size = 3.21 \begin{align*} \frac{1}{3} \, a d^{2} x^{3} + \frac{1}{3} \, b d^{2} x^{3} + a c d x^{2} + b c d x^{2} + a c^{2} x + \frac{b c^{2} \log \left (\sinh \left (f x + e\right )\right )}{f} + \frac{2 \,{\left (f x \log \left (e^{\left (f x + e\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (f x + e\right )}\right )\right )} b c d}{f^{2}} + \frac{2 \,{\left (f x \log \left (-e^{\left (f x + e\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (f x + e\right )}\right )\right )} b c d}{f^{2}} + \frac{{\left (f^{2} x^{2} \log \left (e^{\left (f x + e\right )} + 1\right ) + 2 \, f x{\rm Li}_2\left (-e^{\left (f x + e\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (f x + e\right )})\right )} b d^{2}}{f^{3}} + \frac{{\left (f^{2} x^{2} \log \left (-e^{\left (f x + e\right )} + 1\right ) + 2 \, f x{\rm Li}_2\left (e^{\left (f x + e\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (f x + e\right )})\right )} b d^{2}}{f^{3}} - \frac{2 \,{\left (b d^{2} f^{3} x^{3} + 3 \, b c d f^{3} x^{2}\right )}}{3 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*coth(f*x+e)),x, algorithm="maxima")

[Out]

1/3*a*d^2*x^3 + 1/3*b*d^2*x^3 + a*c*d*x^2 + b*c*d*x^2 + a*c^2*x + b*c^2*log(sinh(f*x + e))/f + 2*(f*x*log(e^(f
*x + e) + 1) + dilog(-e^(f*x + e)))*b*c*d/f^2 + 2*(f*x*log(-e^(f*x + e) + 1) + dilog(e^(f*x + e)))*b*c*d/f^2 +
(f^2*x^2*log(e^(f*x + e) + 1) + 2*f*x*dilog(-e^(f*x + e)) - 2*polylog(3, -e^(f*x + e)))*b*d^2/f^3 + (f^2*x^2*
log(-e^(f*x + e) + 1) + 2*f*x*dilog(e^(f*x + e)) - 2*polylog(3, e^(f*x + e)))*b*d^2/f^3 - 2/3*(b*d^2*f^3*x^3 +
3*b*c*d*f^3*x^2)/f^3

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Fricas [C]  time = 2.24311, size = 782, normalized size = 7.74 \begin{align*} \frac{{\left (a - b\right )} d^{2} f^{3} x^{3} + 3 \,{\left (a - b\right )} c d f^{3} x^{2} + 3 \,{\left (a - b\right )} c^{2} f^{3} x - 6 \, b d^{2}{\rm polylog}\left (3, \cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) - 6 \, b d^{2}{\rm polylog}\left (3, -\cosh \left (f x + e\right ) - \sinh \left (f x + e\right )\right ) + 6 \,{\left (b d^{2} f x + b c d f\right )}{\rm Li}_2\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) + 6 \,{\left (b d^{2} f x + b c d f\right )}{\rm Li}_2\left (-\cosh \left (f x + e\right ) - \sinh \left (f x + e\right )\right ) + 3 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + 1\right ) + 3 \,{\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - 1\right ) + 3 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\cosh \left (f x + e\right ) - \sinh \left (f x + e\right ) + 1\right )}{3 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*coth(f*x+e)),x, algorithm="fricas")

[Out]

1/3*((a - b)*d^2*f^3*x^3 + 3*(a - b)*c*d*f^3*x^2 + 3*(a - b)*c^2*f^3*x - 6*b*d^2*polylog(3, cosh(f*x + e) + si
nh(f*x + e)) - 6*b*d^2*polylog(3, -cosh(f*x + e) - sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(cosh(f*x + e
) + sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(-cosh(f*x + e) - sinh(f*x + e)) + 3*(b*d^2*f^2*x^2 + 2*b*c*
d*f^2*x + b*c^2*f^2)*log(cosh(f*x + e) + sinh(f*x + e) + 1) + 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(cosh
(f*x + e) + sinh(f*x + e) - 1) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(-cosh(f*x + e
) - sinh(f*x + e) + 1))/f^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \coth{\left (e + f x \right )}\right ) \left (c + d x\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*(a+b*coth(f*x+e)),x)

[Out]

Integral((a + b*coth(e + f*x))*(c + d*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2}{\left (b \coth \left (f x + e\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*coth(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*x + c)^2*(b*coth(f*x + e) + a), x)