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

Optimal. Leaf size=88 \[ \frac{d^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac{2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac{(c+d x)^2 \tanh (e+f x)}{f}-\frac{(c+d x)^2}{f}+\frac{(c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.136787, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3720, 3718, 2190, 2279, 2391, 32} \[ \frac{d^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac{2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac{(c+d x)^2 \tanh (e+f x)}{f}-\frac{(c+d x)^2}{f}+\frac{(c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tanh[e + f*x]^2,x]

[Out]

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

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && 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 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 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [C]  time = 4.6061, size = 208, normalized size = 2.36 \[ \frac{d^2 \left (-\text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}(\coth (e))+f x\right )}\right )+2 f x \log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (e))+f x\right )}\right )+2 \tanh ^{-1}(\coth (e)) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (e))+f x\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}(\coth (e))+f x\right )\right )+f x\right )-i \pi \log \left (e^{2 f x}+1\right )+i \pi (f x+\log (\cosh (f x)))\right )}{f^3}+c^2 x+\frac{2 c d (\log (\cosh (e+f x))-f x \tanh (e))}{f^2}-\frac{\text{sech}(e) (c+d x)^2 \sinh (f x) \text{sech}(e+f x)}{f}+c d x^2-\frac{d^2 x^2 \tanh (e) \sqrt{-\text{csch}^2(e)} e^{-\tanh ^{-1}(\coth (e))}}{f}+\frac{d^2 x^3}{3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

c^2*x + c*d*x^2 + (d^2*x^3)/3 + (d^2*((-I)*Pi*Log[1 + E^(2*f*x)] + 2*f*x*Log[1 - E^(-2*(f*x + ArcTanh[Coth[e]]
))] + I*Pi*(f*x + Log[Cosh[f*x]]) + 2*ArcTanh[Coth[e]]*(f*x + Log[1 - E^(-2*(f*x + ArcTanh[Coth[e]]))] - Log[I
*Sinh[f*x + ArcTanh[Coth[e]]]]) - PolyLog[2, E^(-2*(f*x + ArcTanh[Coth[e]]))]))/f^3 - ((c + d*x)^2*Sech[e]*Sec
h[e + f*x]*Sinh[f*x])/f - (d^2*x^2*Sqrt[-Csch[e]^2]*Tanh[e])/(E^ArcTanh[Coth[e]]*f) + (2*c*d*(Log[Cosh[e + f*x
]] - f*x*Tanh[e]))/f^2

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Maple [B]  time = 0.036, size = 177, normalized size = 2. \begin{align*}{\frac{{d}^{2}{x}^{3}}{3}}+cd{x}^{2}+{c}^{2}x+2\,{\frac{{d}^{2}{x}^{2}+2\,cdx+{c}^{2}}{f \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }}+2\,{\frac{cd\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{{f}^{2}}}-4\,{\frac{cd\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-2\,{\frac{{d}^{2}{x}^{2}}{f}}-4\,{\frac{{d}^{2}ex}{{f}^{2}}}-2\,{\frac{{d}^{2}{e}^{2}}{{f}^{3}}}+2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{{f}^{2}}}+{\frac{{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{{f}^{3}}}+4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{2}{\left (x + \frac{e}{f} - \frac{2}{f{\left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}}\right )} - c d{\left (\frac{2 \, x e^{\left (2 \, f x + 2 \, e\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac{f x^{2} +{\left (f x^{2} e^{\left (2 \, e\right )} - 2 \, x e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac{2 \, \log \left ({\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )} e^{\left (-2 \, e\right )}\right )}{f^{2}}\right )} + \frac{1}{3} \, d^{2}{\left (\frac{f x^{3} e^{\left (2 \, f x + 2 \, e\right )} + f x^{3} + 6 \, x^{2}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - 12 \, \int \frac{x}{f e^{\left (2 \, f x + 2 \, e\right )} + f}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*(x + e/f - 2/(f*(e^(-2*f*x - 2*e) + 1))) - c*d*(2*x*e^(2*f*x + 2*e)/(f*e^(2*f*x + 2*e) + f) - (f*x^2 + (f*
x^2*e^(2*e) - 2*x*e^(2*e))*e^(2*f*x))/(f*e^(2*f*x + 2*e) + f) - 2*log((e^(2*f*x + 2*e) + 1)*e^(-2*e))/f^2) + 1
/3*d^2*((f*x^3*e^(2*f*x + 2*e) + f*x^3 + 6*x^2)/(f*e^(2*f*x + 2*e) + f) - 12*integrate(x/(f*e^(2*f*x + 2*e) +
f), x))

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Fricas [C]  time = 1.94973, size = 2051, normalized size = 23.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*tanh(f*x+e)^2,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*e^2 - 12*c*d*e*f + 6*c^2*f^2 + (d^2*f^3*x^3 + 6*d^2*e^2
- 12*c*d*e*f + 3*(c*d*f^3 - 2*d^2*f^2)*x^2 + 3*(c^2*f^3 - 4*c*d*f^2)*x)*cosh(f*x + e)^2 + 2*(d^2*f^3*x^3 + 6*
d^2*e^2 - 12*c*d*e*f + 3*(c*d*f^3 - 2*d^2*f^2)*x^2 + 3*(c^2*f^3 - 4*c*d*f^2)*x)*cosh(f*x + e)*sinh(f*x + e) +
(d^2*f^3*x^3 + 6*d^2*e^2 - 12*c*d*e*f + 3*(c*d*f^3 - 2*d^2*f^2)*x^2 + 3*(c^2*f^3 - 4*c*d*f^2)*x)*sinh(f*x + e)
^2 + 6*(d^2*cosh(f*x + e)^2 + 2*d^2*cosh(f*x + e)*sinh(f*x + e) + d^2*sinh(f*x + e)^2 + d^2)*dilog(I*cosh(f*x
+ e) + I*sinh(f*x + e)) + 6*(d^2*cosh(f*x + e)^2 + 2*d^2*cosh(f*x + e)*sinh(f*x + e) + d^2*sinh(f*x + e)^2 + d
^2)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) - 6*(d^2*e - c*d*f + (d^2*e - c*d*f)*cosh(f*x + e)^2 + 2*(d^2*e
- c*d*f)*cosh(f*x + e)*sinh(f*x + e) + (d^2*e - c*d*f)*sinh(f*x + e)^2)*log(cosh(f*x + e) + sinh(f*x + e) + I)
- 6*(d^2*e - c*d*f + (d^2*e - c*d*f)*cosh(f*x + e)^2 + 2*(d^2*e - c*d*f)*cosh(f*x + e)*sinh(f*x + e) + (d^2*e
- c*d*f)*sinh(f*x + e)^2)*log(cosh(f*x + e) + sinh(f*x + e) - I) + 6*(d^2*f*x + d^2*e + (d^2*f*x + d^2*e)*cos
h(f*x + e)^2 + 2*(d^2*f*x + d^2*e)*cosh(f*x + e)*sinh(f*x + e) + (d^2*f*x + d^2*e)*sinh(f*x + e)^2)*log(I*cosh
(f*x + e) + I*sinh(f*x + e) + 1) + 6*(d^2*f*x + d^2*e + (d^2*f*x + d^2*e)*cosh(f*x + e)^2 + 2*(d^2*f*x + d^2*e
)*cosh(f*x + e)*sinh(f*x + e) + (d^2*f*x + d^2*e)*sinh(f*x + e)^2)*log(-I*cosh(f*x + e) - I*sinh(f*x + e) + 1)
)/(f^3*cosh(f*x + e)^2 + 2*f^3*cosh(f*x + e)*sinh(f*x + e) + f^3*sinh(f*x + e)^2 + f^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)**2*tanh(e + f*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} \tanh \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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