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

Optimal. Leaf size=103 \[ \frac{b d (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{b (c+d x)^3}{3 d} \]

[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 = 0.213368, antiderivative size = 103, 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, 3718, 2190, 2531, 2282, 6589} \[ \frac{b d (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{b (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*(a + b*Tanh[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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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 \tanh (e+f x)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \tanh (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^3}{3 d}+b \int (c+d x)^2 \tanh (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 = 1.71271, size = 147, normalized size = 1.43 \[ \frac{1}{6} \left (\frac{b e^{2 e} \left (-\frac{3 d \left (e^{-2 e}+1\right ) \left (2 f (c+d x) \text{PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \text{PolyLog}\left (3,-e^{-2 (e+f x)}\right )\right )}{f^3}+\frac{6 \left (e^{-2 e}+1\right ) (c+d x)^2 \log \left (e^{-2 (e+f x)}+1\right )}{f}+\frac{4 e^{-2 e} (c+d x)^3}{d}\right )}{e^{2 e}+1}+2 x \left (3 c^2+3 c d x+d^2 x^2\right ) (a+b \tanh (e))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*E^(2*e)*((4*(c + d*x)^3)/(d*E^(2*e)) + (6*(1 + E^(-2*e))*(c + d*x)^2*Log[1 + E^(-2*(e + f*x))])/f - (3*d*(
1 + E^(-2*e))*(2*f*(c + d*x)*PolyLog[2, -E^(-2*(e + f*x))] + d*PolyLog[3, -E^(-2*(e + f*x))]))/f^3))/(1 + E^(2
*e)) + 2*x*(3*c^2 + 3*c*d*x + d^2*x^2)*(a + b*Tanh[e]))/6

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Maple [B]  time = 0.05, size = 272, normalized size = 2.6 \begin{align*}{\frac{a{d}^{2}{x}^{3}}{3}}-{\frac{b{d}^{2}{x}^{3}}{3}}+acd{x}^{2}-bcd{x}^{2}+{c}^{2}ax+b{c}^{2}x+{\frac{b{c}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{f}}-2\,{\frac{b{c}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}-2\,{\frac{b{d}^{2}{e}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+2\,{\frac{b{d}^{2}{e}^{2}x}{{f}^{2}}}+{\frac{4\,b{d}^{2}{e}^{3}}{3\,{f}^{3}}}+{\frac{b{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{2}}{f}}+{\frac{b{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{{f}^{2}}}-{\frac{b{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{3}}}+4\,{\frac{cbde\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-4\,{\frac{cbdex}{f}}-2\,{\frac{cbd{e}^{2}}{{f}^{2}}}+2\,{\frac{cbd\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{f}}+{\frac{cbd{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{{f}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [A]  time = 1.28981, size = 242, normalized size = 2.35 \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 (\cosh \left (f x + e\right )\right )}{f} + \frac{{\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} b c d}{f^{2}} + \frac{{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x{\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) -{\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} b d^{2}}{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*tanh(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(cosh(f*x + e))/f + (2*f*x*log(e^(2
*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*b*c*d/f^2 + 1/2*(2*f^2*x^2*log(e^(2*f*x + 2*e) + 1) + 2*f*x*dilog(
-e^(2*f*x + 2*e)) - polylog(3, -e^(2*f*x + 2*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 = 2.46372, size = 942, normalized size = 9.15 \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, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 6 \, b d^{2}{\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 6 \,{\left (b d^{2} f x + b c d f\right )}{\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 6 \,{\left (b d^{2} f x + b c d f\right )}{\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\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 ) + i\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 ) - i\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 (i \, \cosh \left (f x + e\right ) + i \, \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 (-i \, \cosh \left (f x + e\right ) - i \, \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*tanh(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, I*cosh(f*x + e) +
I*sinh(f*x + e)) - 6*b*d^2*polylog(3, -I*cosh(f*x + e) - I*sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(I*co
sh(f*x + e) + I*sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) + 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) + I) + 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) - I) + 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(I*c
osh(f*x + e) + I*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(-I*cosh(
f*x + e) - I*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 \tanh{\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*tanh(f*x+e)),x)

[Out]

Integral((a + b*tanh(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 \tanh \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*tanh(f*x+e)),x, algorithm="giac")

[Out]

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