3.1.2 \(\int (c+d x)^2 \tanh (e+f x) \, dx\) [2]

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

[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 = 0.11, 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 = {3799, 2221, 2611, 2320, 6724} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tanh[e + f*x],x]

[Out]

-1/3*(c + d*x)^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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3799

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 6724

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 \text {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 = 1.13, size = 93, normalized size = 1.11 \begin {gather*} -\frac {1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right )+\frac {(c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(x*(3*c^2 + 3*c*d*x + d^2*x^2)) + ((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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(241\) vs. \(2(80)=160\).
time = 1.42, size = 242, normalized size = 2.88

method	result	size
risch	\(-\frac {d^{2} x^{3}}{3}-d c \,x^{2}+c^{2} x +\frac {c^{3}}{3 d}-\frac {2 c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {c^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {2 d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 d^{2} e^{2} x}{f^{2}}+\frac {4 d^{2} e^{3}}{3 f^{3}}+\frac {d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}+\frac {d^{2} \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {d^{2} \polylog \left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {4 d e c \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {4 d c e x}{f}-\frac {2 d c \,e^{2}}{f^{2}}+\frac {2 c d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}+\frac {c d \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (82) = 164\).
time = 0.52, size = 183, normalized size = 2.18 \begin {gather*} \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}} \end {gather*}

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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Fricas [C] Result contains complex when optimal does not.
time = 0.52, size = 483, normalized size = 5.75 \begin {gather*} -\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 + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right ) + 6 \, d^{2} {\rm polylog}\left (3, -i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right ) - 6 \, {\left (d^{2} f x + c d f\right )} {\rm Li}_2\left (i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right ) - 6 \, {\left (d^{2} f x + c d f\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right ) - 3 \, {\left (c^{2} f^{2} - 2 \, c d f \cosh \left (1\right ) + d^{2} \cosh \left (1\right )^{2} + d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (c d f - d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (\cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + i\right ) - 3 \, {\left (c^{2} f^{2} - 2 \, c d f \cosh \left (1\right ) + d^{2} \cosh \left (1\right )^{2} + d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (c d f - d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (\cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - i\right ) - 3 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + 2 \, c d f \cosh \left (1\right ) - d^{2} \cosh \left (1\right )^{2} - d^{2} \sinh \left (1\right )^{2} + 2 \, {\left (c d f - d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + 1\right ) - 3 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + 2 \, c d f \cosh \left (1\right ) - d^{2} \cosh \left (1\right )^{2} - d^{2} \sinh \left (1\right )^{2} + 2 \, {\left (c d f - d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (-i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + 1\right )}{3 \, f^{3}} \end {gather*}

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

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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