3.2 \(\int (c+d x)^2 \tanh (e+f x) \, dx\)

Optimal. Leaf size=84 \[ \frac {d (c+d x) \text {Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}+\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {(c+d x)^3}{3 d}-\frac {d^2 \text {Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3718, 2190, 2531, 2282, 6589} \[ \frac {d (c+d x) \text {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac {d^2 \text {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {(c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tanh[e + f*x],x]

[Out]

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

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

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Mathematica [A]  time = 1.46, size = 143, normalized size = 1.70 \[ \frac {1}{3} x \tanh (e) \left (3 c^2+3 c d x+d^2 x^2\right )+\frac {e^{2 e} \left (-\frac {3 d \left (e^{-2 e}+1\right ) \left (2 f (c+d x) \text {Li}_2\left (-e^{-2 (e+f x)}\right )+d \text {Li}_3\left (-e^{-2 (e+f x)}\right )\right )}{f^3}+\frac {6 \left (e^{-2 e}+1\right ) (c+d x)^2 \log \left (e^{-2 (e+f x)}+1\right )}{f}+\frac {4 e^{-2 e} (c+d x)^3}{d}\right )}{6 \left (e^{2 e}+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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))/(6*(1 + E^(2
*e))) + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Tanh[e])/3

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fricas [C]  time = 0.50, size = 329, normalized size = 3.92 \[ -\frac {d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2} + 3 \, c^{2} f^{3} x + 6 \, d^{2} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 6 \, d^{2} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 6 \, {\left (d^{2} f x + c d f\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 6 \, {\left (d^{2} f x + c d f\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 3 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) - 3 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) - 3 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) - 3 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right )}{3 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \tanh \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.19, size = 234, normalized size = 2.79 \[ -\frac {d^{2} x^{3}}{3}-c d \,x^{2}+c^{2} x +\frac {c^{2} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right )}{f}-\frac {2 c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}-\frac {2 d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 d^{2} e^{2} x}{f^{2}}+\frac {4 d^{2} e^{3}}{3 f^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x^{2}}{f}+\frac {d^{2} \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {d^{2} \polylog \left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {4 c d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {4 d c e x}{f}-\frac {2 d c \,e^{2}}{f^{2}}+\frac {2 c d \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x}{f}+\frac {c d \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.48, size = 176, normalized size = 2.10 \[ \frac {1}{3} \, d^{2} x^{3} + c d x^{2} + \frac {c^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}{2 \, f} + \frac {c^{2} \log \left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}{2 \, f} + \frac {{\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} c d}{f^{2}} + \frac {{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} d^{2}}{2 \, f^{3}} - \frac {2 \, {\left (d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2}\right )}}{3 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*d^2*x^3 + c*d*x^2 + 1/2*c^2*log(e^(2*f*x + 2*e) + 1)/f + 1/2*c^2*log(e^(-2*f*x - 2*e) + 1)/f + (2*f*x*log(
e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*c*d/f^2 + 1/2*(2*f^2*x^2*log(e^(2*f*x + 2*e) + 1) + 2*f*x*dilo
g(-e^(2*f*x + 2*e)) - polylog(3, -e^(2*f*x + 2*e)))*d^2/f^3 - 2/3*(d^2*f^3*x^3 + 3*c*d*f^3*x^2)/f^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {tanh}\left (e+f\,x\right )\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)*(c + d*x)^2,x)

[Out]

int(tanh(e + f*x)*(c + d*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \tanh {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)**2*tanh(e + f*x), x)

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