3.12 \(\int (c+d x)^2 \tanh ^3(e+f x) \, dx\)

Optimal. Leaf size=157 \[ \frac{d (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac{d^2 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}-\frac{d (c+d x) \tanh (e+f x)}{f^2}+\frac{(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{(c+d x)^2 \tanh ^2(e+f x)}{2 f}+\frac{c d x}{f}-\frac{(c+d x)^3}{3 d}+\frac{d^2 \log (\cosh (e+f x))}{f^3}+\frac{d^2 x^2}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.24834, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3720, 3475, 3718, 2190, 2531, 2282, 6589} \[ \frac{d (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac{d^2 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}-\frac{d (c+d x) \tanh (e+f x)}{f^2}+\frac{(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{(c+d x)^2 \tanh ^2(e+f x)}{2 f}+\frac{c d x}{f}-\frac{(c+d x)^3}{3 d}+\frac{d^2 \log (\cosh (e+f x))}{f^3}+\frac{d^2 x^2}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tanh[e + f*x]^3,x]

[Out]

(c*d*x)/f + (d^2*x^2)/(2*f) - (c + d*x)^3/(3*d) + ((c + d*x)^2*Log[1 + E^(2*(e + f*x))])/f + (d^2*Log[Cosh[e +
f*x]])/f^3 + (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) - (d
*(c + d*x)*Tanh[e + f*x])/f^2 - ((c + d*x)^2*Tanh[e + f*x]^2)/(2*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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [C]  time = 7.8762, size = 460, normalized size = 2.93 \[ -\frac{c d \text{csch}(e) \text{sech}(e) \left (-f^2 x^2 e^{-\tanh ^{-1}(\coth (e))}+\frac{i \coth (e) \left (i \text{PolyLog}\left (2,e^{2 i \left (i \tanh ^{-1}(\coth (e))+i f x\right )}\right )-f x \left (-\pi +2 i \tanh ^{-1}(\coth (e))\right )-2 \left (i \tanh ^{-1}(\coth (e))+i f x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (e))+i f x\right )}\right )+2 i \tanh ^{-1}(\coth (e)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (e))+f x\right )\right )-\pi \log \left (e^{2 f x}+1\right )+\pi \log (\cosh (f x))\right )}{\sqrt{1-\coth ^2(e)}}\right )}{f^2 \sqrt{\text{csch}^2(e) \left (\sinh ^2(e)-\cosh ^2(e)\right )}}+\frac{d^2 e^{-e} \text{sech}(e) \left (-6 \left (e^{2 e}+1\right ) f x \text{PolyLog}\left (2,-e^{-2 (e+f x)}\right )-3 \left (e^{2 e}+1\right ) \text{PolyLog}\left (3,-e^{-2 (e+f x)}\right )+2 f^2 x^2 \left (3 \left (e^{2 e}+1\right ) \log \left (e^{-2 (e+f x)}+1\right )+2 f x\right )\right )}{12 f^3}+\frac{1}{3} x \tanh (e) \left (3 c^2+3 c d x+d^2 x^2\right )+\frac{c^2 \text{sech}(e) (\cosh (e) \log (\sinh (e) \sinh (f x)+\cosh (e) \cosh (f x))-f x \sinh (e))}{f \left (\cosh ^2(e)-\sinh ^2(e)\right )}+\frac{\text{sech}(e) \text{sech}(e+f x) \left (d^2 (-x) \sinh (f x)-c d \sinh (f x)\right )}{f^2}+\frac{(c+d x)^2 \text{sech}^2(e+f x)}{2 f}+\frac{d^2 \text{sech}(e) (\cosh (e) \log (\sinh (e) \sinh (f x)+\cosh (e) \cosh (f x))-f x \sinh (e))}{f^3 \left (\cosh ^2(e)-\sinh ^2(e)\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Tanh[e + f*x]^3,x]

[Out]

(d^2*(2*f^2*x^2*(2*f*x + 3*(1 + E^(2*e))*Log[1 + E^(-2*(e + f*x))]) - 6*(1 + E^(2*e))*f*x*PolyLog[2, -E^(-2*(e
+ f*x))] - 3*(1 + E^(2*e))*PolyLog[3, -E^(-2*(e + f*x))])*Sech[e])/(12*E^e*f^3) + ((c + d*x)^2*Sech[e + f*x]^
2)/(2*f) + (d^2*Sech[e]*(Cosh[e]*Log[Cosh[e]*Cosh[f*x] + Sinh[e]*Sinh[f*x]] - f*x*Sinh[e]))/(f^3*(Cosh[e]^2 -
Sinh[e]^2)) + (c^2*Sech[e]*(Cosh[e]*Log[Cosh[e]*Cosh[f*x] + Sinh[e]*Sinh[f*x]] - f*x*Sinh[e]))/(f*(Cosh[e]^2 -
Sinh[e]^2)) - (c*d*Csch[e]*(-((f^2*x^2)/E^ArcTanh[Coth[e]]) + (I*Coth[e]*(-(f*x*(-Pi + (2*I)*ArcTanh[Coth[e]]
)) - Pi*Log[1 + E^(2*f*x)] - 2*(I*f*x + I*ArcTanh[Coth[e]])*Log[1 - E^((2*I)*(I*f*x + I*ArcTanh[Coth[e]]))] +
Pi*Log[Cosh[f*x]] + (2*I)*ArcTanh[Coth[e]]*Log[I*Sinh[f*x + ArcTanh[Coth[e]]]] + I*PolyLog[2, E^((2*I)*(I*f*x
+ I*ArcTanh[Coth[e]]))]))/Sqrt[1 - Coth[e]^2])*Sech[e])/(f^2*Sqrt[Csch[e]^2*(-Cosh[e]^2 + Sinh[e]^2)]) + (Sech
[e]*Sech[e + f*x]*(-(c*d*Sinh[f*x]) - d^2*x*Sinh[f*x]))/f^2 + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Tanh[e])/3

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Maple [B]  time = 0.059, size = 367, normalized size = 2.3 \begin{align*} -{\frac{{d}^{2}{x}^{3}}{3}}-cd{x}^{2}+{c}^{2}x+2\,{\frac{{d}^{2}f{x}^{2}{{\rm e}^{2\,fx+2\,e}}+2\,cdfx{{\rm e}^{2\,fx+2\,e}}+{c}^{2}f{{\rm e}^{2\,fx+2\,e}}+{d}^{2}x{{\rm e}^{2\,fx+2\,e}}+cd{{\rm e}^{2\,fx+2\,e}}+{d}^{2}x+cd}{{f}^{2} \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) ^{2}}}+{\frac{4\,{d}^{2}{e}^{3}}{3\,{f}^{3}}}-{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{3}}}-2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{{f}^{3}}}+4\,{\frac{cde\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}+{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{2}}{f}}+{\frac{{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{{f}^{2}}}-2\,{\frac{cd{e}^{2}}{{f}^{2}}}+2\,{\frac{{d}^{2}{e}^{2}x}{{f}^{2}}}+{\frac{cd{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{{f}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{f}}-2\,{\frac{{c}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{f}}-2\,{\frac{{d}^{2}{e}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}-4\,{\frac{cdex}{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*tanh(f*x+e)^3,x)

[Out]

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

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Maxima [B]  time = 1.89031, size = 529, normalized size = 3.37 \begin{align*} c^{2}{\left (x + \frac{e}{f} + \frac{\log \left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}{f} + \frac{2 \, e^{\left (-2 \, f x - 2 \, e\right )}}{f{\left (2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1\right )}}\right )} + \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{2 \, d^{2} x}{f^{2}} + \frac{d^{2} f^{2} x^{3} + 3 \, c d f^{2} x^{2} + 6 \, d^{2} x + 6 \, c d +{\left (d^{2} f^{2} x^{3} e^{\left (4 \, e\right )} + 3 \, c d f^{2} x^{2} e^{\left (4 \, e\right )}\right )} e^{\left (4 \, f x\right )} + 2 \,{\left (d^{2} f^{2} x^{3} e^{\left (2 \, e\right )} + 3 \,{\left (c d f^{2} e^{\left (2 \, e\right )} + d^{2} f e^{\left (2 \, e\right )}\right )} x^{2} + 3 \, c d e^{\left (2 \, e\right )} + 3 \,{\left (2 \, c d f e^{\left (2 \, e\right )} + d^{2} e^{\left (2 \, e\right )}\right )} x\right )} e^{\left (2 \, f x\right )}}{3 \,{\left (f^{2} e^{\left (4 \, f x + 4 \, e\right )} + 2 \, f^{2} e^{\left (2 \, f x + 2 \, e\right )} + f^{2}\right )}} + \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{d^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}{f^{3}} - \frac{2 \,{\left (d^{2} f^{3} x^{3} + 3 \, 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*tanh(f*x+e)^3,x, algorithm="maxima")

[Out]

c^2*(x + e/f + log(e^(-2*f*x - 2*e) + 1)/f + 2*e^(-2*f*x - 2*e)/(f*(2*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1)
)) + (2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*c*d/f^2 - 2*d^2*x/f^2 + 1/3*(d^2*f^2*x^3 + 3*c
*d*f^2*x^2 + 6*d^2*x + 6*c*d + (d^2*f^2*x^3*e^(4*e) + 3*c*d*f^2*x^2*e^(4*e))*e^(4*f*x) + 2*(d^2*f^2*x^3*e^(2*e
) + 3*(c*d*f^2*e^(2*e) + d^2*f*e^(2*e))*x^2 + 3*c*d*e^(2*e) + 3*(2*c*d*f*e^(2*e) + d^2*e^(2*e))*x)*e^(2*f*x))/
(f^2*e^(4*f*x + 4*e) + 2*f^2*e^(2*f*x + 2*e) + 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)))*d^2/f^3 + d^2*log(e^(2*f*x + 2*e) + 1)/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]  time = 2.63755, size = 7159, normalized size = 45.6 \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)^3,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 + 2*d^2*e^3 + 6*c^2*e*f^2 + (d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 2*d
^2*e^3 - 6*c*d*e^2*f + 6*c^2*e*f^2 + 6*d^2*e + 3*(c^2*f^3 + 2*d^2*f)*x)*cosh(f*x + e)^4 + 4*(d^2*f^3*x^3 + 3*c
*d*f^3*x^2 + 2*d^2*e^3 - 6*c*d*e^2*f + 6*c^2*e*f^2 + 6*d^2*e + 3*(c^2*f^3 + 2*d^2*f)*x)*cosh(f*x + e)*sinh(f*x
+ e)^3 + (d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 2*d^2*e^3 - 6*c*d*e^2*f + 6*c^2*e*f^2 + 6*d^2*e + 3*(c^2*f^3 + 2*d^2*
f)*x)*sinh(f*x + e)^4 + 6*d^2*e + 2*(d^2*f^3*x^3 + 2*d^2*e^3 + 6*d^2*e + 3*(2*c^2*e - c^2)*f^2 + 3*(c*d*f^3 -
d^2*f^2)*x^2 - 3*(2*c*d*e^2 + c*d)*f + 3*(c^2*f^3 - 2*c*d*f^2 + d^2*f)*x)*cosh(f*x + e)^2 + 2*(d^2*f^3*x^3 + 2
*d^2*e^3 + 6*d^2*e + 3*(2*c^2*e - c^2)*f^2 + 3*(c*d*f^3 - d^2*f^2)*x^2 + 3*(d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 2*d^
2*e^3 - 6*c*d*e^2*f + 6*c^2*e*f^2 + 6*d^2*e + 3*(c^2*f^3 + 2*d^2*f)*x)*cosh(f*x + e)^2 - 3*(2*c*d*e^2 + c*d)*f
+ 3*(c^2*f^3 - 2*c*d*f^2 + d^2*f)*x)*sinh(f*x + e)^2 - 6*(c*d*e^2 + c*d)*f - 6*((d^2*f*x + c*d*f)*cosh(f*x +
e)^4 + 4*(d^2*f*x + c*d*f)*cosh(f*x + e)*sinh(f*x + e)^3 + (d^2*f*x + c*d*f)*sinh(f*x + e)^4 + d^2*f*x + c*d*f
+ 2*(d^2*f*x + c*d*f)*cosh(f*x + e)^2 + 2*(d^2*f*x + c*d*f + 3*(d^2*f*x + c*d*f)*cosh(f*x + e)^2)*sinh(f*x +
e)^2 + 4*((d^2*f*x + c*d*f)*cosh(f*x + e)^3 + (d^2*f*x + c*d*f)*cosh(f*x + e))*sinh(f*x + e))*dilog(I*cosh(f*x
+ e) + I*sinh(f*x + e)) - 6*((d^2*f*x + c*d*f)*cosh(f*x + e)^4 + 4*(d^2*f*x + c*d*f)*cosh(f*x + e)*sinh(f*x +
e)^3 + (d^2*f*x + c*d*f)*sinh(f*x + e)^4 + d^2*f*x + c*d*f + 2*(d^2*f*x + c*d*f)*cosh(f*x + e)^2 + 2*(d^2*f*x
+ c*d*f + 3*(d^2*f*x + c*d*f)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 4*((d^2*f*x + c*d*f)*cosh(f*x + e)^3 + (d^2*
f*x + c*d*f)*cosh(f*x + e))*sinh(f*x + e))*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 + d^2)*cosh(f*x + e)^4 + 4*(d^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f*x + e)*sinh(f*x + e)^3 + (d
^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*sinh(f*x + e)^4 + d^2*e^2 - 2*c*d*e*f + c^2*f^2 + 2*(d^2*e^2 - 2*c*d*e*f +
c^2*f^2 + d^2)*cosh(f*x + e)^2 + 2*(d^2*e^2 - 2*c*d*e*f + c^2*f^2 + 3*(d^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*c
osh(f*x + e)^2 + d^2)*sinh(f*x + e)^2 + d^2 + 4*((d^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f*x + e)^3 + (d^2*
e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f*x + e))*sinh(f*x + e))*log(cosh(f*x + e) + sinh(f*x + e) + I) - 3*((d^
2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f*x + e)^4 + 4*(d^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f*x + e)*sin
h(f*x + e)^3 + (d^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*sinh(f*x + e)^4 + d^2*e^2 - 2*c*d*e*f + c^2*f^2 + 2*(d^2*
e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f*x + e)^2 + 2*(d^2*e^2 - 2*c*d*e*f + c^2*f^2 + 3*(d^2*e^2 - 2*c*d*e*f +
c^2*f^2 + d^2)*cosh(f*x + e)^2 + d^2)*sinh(f*x + e)^2 + d^2 + 4*((d^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f
*x + e)^3 + (d^2*e^2 - 2*c*d*e*f + c^2*f^2 + d^2)*cosh(f*x + e))*sinh(f*x + e))*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*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e)^4 +
4*(d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e)*sinh(f*x + e)^3 + (d^2*f^2*x^2 + 2*c*d*f^2*x
- d^2*e^2 + 2*c*d*e*f)*sinh(f*x + e)^4 - d^2*e^2 + 2*c*d*e*f + 2*(d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d
*e*f)*cosh(f*x + e)^2 + 2*(d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f + 3*(d^2*f^2*x^2 + 2*c*d*f^2*x - d^
2*e^2 + 2*c*d*e*f)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 4*((d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cos
h(f*x + e)^3 + (d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e))*sinh(f*x + e))*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*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)
*cosh(f*x + e)^4 + 4*(d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e)*sinh(f*x + e)^3 + (d^2*f^
2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*sinh(f*x + e)^4 - d^2*e^2 + 2*c*d*e*f + 2*(d^2*f^2*x^2 + 2*c*d*f^2*
x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e)^2 + 2*(d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f + 3*(d^2*f^2*x^2
+ 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 4*((d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e
^2 + 2*c*d*e*f)*cosh(f*x + e)^3 + (d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e))*sinh(f*x +
e))*log(-I*cosh(f*x + e) - I*sinh(f*x + e) + 1) + 6*(d^2*cosh(f*x + e)^4 + 4*d^2*cosh(f*x + e)*sinh(f*x + e)^3
+ d^2*sinh(f*x + e)^4 + 2*d^2*cosh(f*x + e)^2 + 2*(3*d^2*cosh(f*x + e)^2 + d^2)*sinh(f*x + e)^2 + d^2 + 4*(d^
2*cosh(f*x + e)^3 + d^2*cosh(f*x + e))*sinh(f*x + e))*polylog(3, I*cosh(f*x + e) + I*sinh(f*x + e)) + 6*(d^2*c
osh(f*x + e)^4 + 4*d^2*cosh(f*x + e)*sinh(f*x + e)^3 + d^2*sinh(f*x + e)^4 + 2*d^2*cosh(f*x + e)^2 + 2*(3*d^2*
cosh(f*x + e)^2 + d^2)*sinh(f*x + e)^2 + d^2 + 4*(d^2*cosh(f*x + e)^3 + d^2*cosh(f*x + e))*sinh(f*x + e))*poly
log(3, -I*cosh(f*x + e) - I*sinh(f*x + e)) + 4*((d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 2*d^2*e^3 - 6*c*d*e^2*f + 6*c^2
*e*f^2 + 6*d^2*e + 3*(c^2*f^3 + 2*d^2*f)*x)*cosh(f*x + e)^3 + (d^2*f^3*x^3 + 2*d^2*e^3 + 6*d^2*e + 3*(2*c^2*e
- c^2)*f^2 + 3*(c*d*f^3 - d^2*f^2)*x^2 - 3*(2*c*d*e^2 + c*d)*f + 3*(c^2*f^3 - 2*c*d*f^2 + d^2*f)*x)*cosh(f*x +
e))*sinh(f*x + e))/(f^3*cosh(f*x + e)^4 + 4*f^3*cosh(f*x + e)*sinh(f*x + e)^3 + f^3*sinh(f*x + e)^4 + 2*f^3*c
osh(f*x + e)^2 + f^3 + 2*(3*f^3*cosh(f*x + e)^2 + f^3)*sinh(f*x + e)^2 + 4*(f^3*cosh(f*x + e)^3 + f^3*cosh(f*x
+ e))*sinh(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{2} \tanh ^{3}{\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)**3,x)

[Out]

Integral((c + d*x)**2*tanh(e + f*x)**3, 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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*tanh(f*x+e)^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*tanh(f*x + e)^3, x)