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

Optimal. Leaf size=211 \[ \frac{2 a b d (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac{a b d^2 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}+\frac{b^2 d^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac{a^2 (c+d x)^3}{3 d}+\frac{2 a b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{2 a b (c+d x)^3}{3 d}+\frac{2 b^2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac{b^2 (c+d x)^2 \tanh (e+f x)}{f}-\frac{b^2 (c+d x)^2}{f}+\frac{b^2 (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.396205, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3722, 3718, 2190, 2531, 2282, 6589, 3720, 2279, 2391, 32} \[ \frac{2 a b d (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac{a b d^2 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}+\frac{b^2 d^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac{a^2 (c+d x)^3}{3 d}+\frac{2 a b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{2 a b (c+d x)^3}{3 d}+\frac{2 b^2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac{b^2 (c+d x)^2 \tanh (e+f x)}{f}-\frac{b^2 (c+d x)^2}{f}+\frac{b^2 (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 (a+b \tanh (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)^2+2 a b (c+d x)^2 \tanh (e+f x)+b^2 (c+d x)^2 \tanh ^2(e+f x)\right ) \, dx\\ &=\frac{a^2 (c+d x)^3}{3 d}+(2 a b) \int (c+d x)^2 \tanh (e+f x) \, dx+b^2 \int (c+d x)^2 \tanh ^2(e+f x) \, dx\\ &=\frac{a^2 (c+d x)^3}{3 d}-\frac{2 a b (c+d x)^3}{3 d}-\frac{b^2 (c+d x)^2 \tanh (e+f x)}{f}+(4 a b) \int \frac{e^{2 (e+f x)} (c+d x)^2}{1+e^{2 (e+f x)}} \, dx+b^2 \int (c+d x)^2 \, dx+\frac{\left (2 b^2 d\right ) \int (c+d x) \tanh (e+f x) \, dx}{f}\\ &=-\frac{b^2 (c+d x)^2}{f}+\frac{a^2 (c+d x)^3}{3 d}-\frac{2 a b (c+d x)^3}{3 d}+\frac{b^2 (c+d x)^3}{3 d}+\frac{2 a b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}-\frac{b^2 (c+d x)^2 \tanh (e+f x)}{f}-\frac{(4 a b d) \int (c+d x) \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f}+\frac{\left (4 b^2 d\right ) \int \frac{e^{2 (e+f x)} (c+d x)}{1+e^{2 (e+f x)}} \, dx}{f}\\ &=-\frac{b^2 (c+d x)^2}{f}+\frac{a^2 (c+d x)^3}{3 d}-\frac{2 a b (c+d x)^3}{3 d}+\frac{b^2 (c+d x)^3}{3 d}+\frac{2 b^2 d (c+d x) \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac{2 a b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{2 a b d (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}-\frac{b^2 (c+d x)^2 \tanh (e+f x)}{f}-\frac{\left (2 a b d^2\right ) \int \text{Li}_2\left (-e^{2 (e+f x)}\right ) \, dx}{f^2}-\frac{\left (2 b^2 d^2\right ) \int \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=-\frac{b^2 (c+d x)^2}{f}+\frac{a^2 (c+d x)^3}{3 d}-\frac{2 a b (c+d x)^3}{3 d}+\frac{b^2 (c+d x)^3}{3 d}+\frac{2 b^2 d (c+d x) \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac{2 a b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{2 a b d (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}-\frac{b^2 (c+d x)^2 \tanh (e+f x)}{f}-\frac{\left (a b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{f^3}-\frac{\left (b^2 d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{f^3}\\ &=-\frac{b^2 (c+d x)^2}{f}+\frac{a^2 (c+d x)^3}{3 d}-\frac{2 a b (c+d x)^3}{3 d}+\frac{b^2 (c+d x)^3}{3 d}+\frac{2 b^2 d (c+d x) \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac{2 a b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{b^2 d^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{f^3}+\frac{2 a b d (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right )}{f^2}-\frac{a b d^2 \text{Li}_3\left (-e^{2 (e+f x)}\right )}{f^3}-\frac{b^2 (c+d x)^2 \tanh (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 5.51275, size = 232, normalized size = 1.1 \[ \frac{1}{3} \left (\frac{b \left (-3 d \text{PolyLog}\left (2,-e^{-2 (e+f x)}\right ) (2 a f (c+d x)+b d)-3 a d^2 \text{PolyLog}\left (3,-e^{-2 (e+f x)}\right )+2 f \left (\frac{f x \left (2 a f \left (-3 c^2 e^{2 e}+3 c d x+d^2 x^2\right )+3 b d \left (d x-2 c e^{2 e}\right )\right )}{e^{2 e}+1}+3 d x \log \left (e^{-2 (e+f x)}+1\right ) (a f (2 c+d x)+b d)+3 c \log \left (e^{2 (e+f x)}+1\right ) (a c f+b d)\right )\right )}{f^3}+x \left (3 c^2+3 c d x+d^2 x^2\right ) \left (a^2+2 a b \tanh (e)+b^2\right )-\frac{3 b^2 \text{sech}(e) (c+d x)^2 \sinh (f x) \text{sech}(e+f x)}{f}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.087, size = 510, normalized size = 2.4 \begin{align*}{\frac{{b}^{2}{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{{f}^{3}}}+{\frac{{a}^{2}{d}^{2}{x}^{3}}{3}}+{\frac{{b}^{2}{d}^{2}{x}^{3}}{3}}+{c}^{2}{a}^{2}x+{c}^{2}{b}^{2}x-2\,{\frac{{b}^{2}{d}^{2}{x}^{2}}{f}}-2\,{\frac{{b}^{2}{d}^{2}{e}^{2}}{{f}^{3}}}+4\,{\frac{b\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) acdx}{f}}+8\,{\frac{cdbae\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-8\,{\frac{cdbaex}{f}}+{\frac{8\,ab{d}^{2}{e}^{3}}{3\,{f}^{3}}}-4\,{\frac{{b}^{2}{d}^{2}ex}{{f}^{2}}}+2\,{\frac{{b}^{2}{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{{f}^{2}}}-4\,{\frac{ab{c}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}+2\,{\frac{ab{c}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{f}}-4\,{\frac{{b}^{2}cd\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}+2\,{\frac{{b}^{2}cd\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{{f}^{2}}}+4\,{\frac{{b}^{2}{d}^{2}e\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+2\,{\frac{{b}^{2} \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{f \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }}-{\frac{2\,ab{d}^{2}{x}^{3}}{3}}+2\,ab{c}^{2}x-2\,abcd{x}^{2}-{\frac{ab{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) }{{f}^{3}}}+{a}^{2}cd{x}^{2}+{b}^{2}cd{x}^{2}-4\,{\frac{cdba{e}^{2}}{{f}^{2}}}+4\,{\frac{ab{d}^{2}{e}^{2}x}{{f}^{2}}}+2\,{\frac{ab{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{2}}{f}}+2\,{\frac{ab{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{{f}^{2}}}+2\,{\frac{cdba{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{{f}^{2}}}-4\,{\frac{ab{d}^{2}{e}^{2}\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*(a+b*tanh(f*x+e))^2,x)

[Out]

b^2*d^2*polylog(2,-exp(2*f*x+2*e))/f^3+1/3*a^2*d^2*x^3+1/3*b^2*d^2*x^3+c^2*a^2*x+c^2*b^2*x-2*b^2/f*d^2*x^2-2*b
^2/f^3*d^2*e^2+4*b/f*ln(exp(2*f*x+2*e)+1)*a*c*d*x+8*b/f^2*a*c*d*e*ln(exp(f*x+e))-8*b/f*a*c*d*e*x+8/3*b/f^3*a*d
^2*e^3-4*b^2/f^2*d^2*e*x+2*b^2/f^2*d^2*ln(exp(2*f*x+2*e)+1)*x-4*b/f*a*c^2*ln(exp(f*x+e))+2*b/f*a*c^2*ln(exp(2*
f*x+2*e)+1)-4*b^2/f^2*c*d*ln(exp(f*x+e))+2*b^2/f^2*c*d*ln(exp(2*f*x+2*e)+1)+4*b^2/f^3*d^2*e*ln(exp(f*x+e))+2/f
*b^2*(d^2*x^2+2*c*d*x+c^2)/(exp(2*f*x+2*e)+1)-2/3*a*b*d^2*x^3+2*a*b*c^2*x-2*a*b*c*d*x^2-a*b*d^2*polylog(3,-exp
(2*f*x+2*e))/f^3+a^2*c*d*x^2+b^2*c*d*x^2-4*b/f^2*a*c*d*e^2+4*b/f^2*a*d^2*e^2*x+2*b/f*a*d^2*ln(exp(2*f*x+2*e)+1
)*x^2+2*b/f^2*a*d^2*polylog(2,-exp(2*f*x+2*e))*x+2*b/f^2*a*c*d*polylog(2,-exp(2*f*x+2*e))-4*b/f^3*a*d^2*e^2*ln
(exp(f*x+e))

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Maxima [B]  time = 1.46856, size = 555, normalized size = 2.63 \begin{align*} \frac{1}{3} \, a^{2} d^{2} x^{3} + a^{2} c d x^{2} + b^{2} c^{2}{\left (x + \frac{e}{f} - \frac{2}{f{\left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}}\right )} + a^{2} c^{2} x + b^{2} c d{\left (\frac{f x^{2} +{\left (f x^{2} e^{\left (2 \, e\right )} - 4 \, 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{2 \, a b c^{2} \log \left (\cosh \left (f x + e\right )\right )}{f} + \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 )} a b d^{2}}{f^{3}} + \frac{{\left (2 \, a b d^{2} f + b^{2} d^{2} f\right )} x^{3} + 6 \,{\left (a b c d f + b^{2} d^{2}\right )} x^{2} +{\left (6 \, a b c d f x^{2} e^{\left (2 \, e\right )} +{\left (2 \, a b d^{2} f e^{\left (2 \, e\right )} + b^{2} d^{2} f e^{\left (2 \, e\right )}\right )} x^{3}\right )} e^{\left (2 \, f x\right )}}{3 \,{\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )}} + \frac{{\left (2 \, a b c d f + b^{2} d^{2}\right )}{\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 )}}{f^{3}} - \frac{2 \,{\left (2 \, a b d^{2} f^{3} x^{3} + 3 \,{\left (2 \, a b c d f + b^{2} d^{2}\right )} f^{2} 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))^2,x, algorithm="maxima")

[Out]

1/3*a^2*d^2*x^3 + a^2*c*d*x^2 + b^2*c^2*(x + e/f - 2/(f*(e^(-2*f*x - 2*e) + 1))) + a^2*c^2*x + b^2*c*d*((f*x^2
+ (f*x^2*e^(2*e) - 4*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) + 2*a*b*c^2*log(cosh(f*x + e))/f + (2*f^2*x^2*log(e^(2*f*x + 2*e) + 1) + 2*f*x*dilog(-e^(2*f*x + 2*e)) - po
lylog(3, -e^(2*f*x + 2*e)))*a*b*d^2/f^3 + 1/3*((2*a*b*d^2*f + b^2*d^2*f)*x^3 + 6*(a*b*c*d*f + b^2*d^2)*x^2 + (
6*a*b*c*d*f*x^2*e^(2*e) + (2*a*b*d^2*f*e^(2*e) + b^2*d^2*f*e^(2*e))*x^3)*e^(2*f*x))/(f*e^(2*f*x + 2*e) + f) +
(2*a*b*c*d*f + b^2*d^2)*(2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))/f^3 - 2/3*(2*a*b*d^2*f^3*x^
3 + 3*(2*a*b*c*d*f + b^2*d^2)*f^2*x^2)/f^3

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Fricas [C]  time = 2.90322, size = 4886, normalized size = 23.16 \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*(a+b*tanh(f*x+e))^2,x, algorithm="fricas")

[Out]

1/3*((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 + 3*(a^2 - 2*a*b + b^2)*c*d*f^3*x^2 - 4*a*b*d^2*e^3 + 3*(a^2 - 2*a*b + b^
2)*c^2*f^3*x + 6*b^2*d^2*e^2 - 6*(2*a*b*c^2*e - b^2*c^2)*f^2 + ((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 - 4*a*b*d^2*e^
3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*f^2 - (a^2 - 2*a*b + b^2)*c*d*f^3)*x^2 + 12*(a*b*c*d*e^2 -
b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^2)*c^2*f^3)*x)*cosh(f*x + e)^2 + 2*((a^2 - 2*a*b + b^2)*d^
2*f^3*x^3 - 4*a*b*d^2*e^3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*f^2 - (a^2 - 2*a*b + b^2)*c*d*f^3)
*x^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^2)*c^2*f^3)*x)*cosh(f*x + e)*sinh(
f*x + e) + ((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 - 4*a*b*d^2*e^3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*
f^2 - (a^2 - 2*a*b + b^2)*c*d*f^3)*x^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^
2)*c^2*f^3)*x)*sinh(f*x + e)^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f + 6*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2 + (
2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)^2 + 2*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x +
e)*sinh(f*x + e) + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*sinh(f*x + e)^2)*dilog(I*cosh(f*x + e) + I*sinh(f*x
+ e)) + 6*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2 + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)^2 +
2*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)*sinh(f*x + e) + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2
)*sinh(f*x + e)^2)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) + 6*(a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e + (a*b
*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*cosh(f*x + e)^2 + 2*(a*b*d^2*e^2 + a*b*c^2*f^2
- b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*cosh(f*x + e)*sinh(f*x + e) + (a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e
- (2*a*b*c*d*e - b^2*c*d)*f)*sinh(f*x + e)^2 - (2*a*b*c*d*e - b^2*c*d)*f)*log(cosh(f*x + e) + sinh(f*x + e) +
I) + 6*(a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e + (a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c
*d)*f)*cosh(f*x + e)^2 + 2*(a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*cosh(f*x + e)*s
inh(f*x + e) + (a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*sinh(f*x + e)^2 - (2*a*b*c*
d*e - b^2*c*d)*f)*log(cosh(f*x + e) + sinh(f*x + e) - I) + 6*(a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f +
b^2*d^2*e + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh(f
*x + e)^2 + 2*(a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh
(f*x + e)*sinh(f*x + e) + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^
2*f)*x)*sinh(f*x + e)^2 + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*log(I*cosh(f*x + e) + I*sinh(f*x + e) + 1) + 6*(a*b*d
^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^
2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh(f*x + e)^2 + 2*(a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*
d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh(f*x + e)*sinh(f*x + e) + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*
d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*sinh(f*x + e)^2 + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*log(-I*cos
h(f*x + e) - I*sinh(f*x + e) + 1) - 12*(a*b*d^2*cosh(f*x + e)^2 + 2*a*b*d^2*cosh(f*x + e)*sinh(f*x + e) + a*b*
d^2*sinh(f*x + e)^2 + a*b*d^2)*polylog(3, I*cosh(f*x + e) + I*sinh(f*x + e)) - 12*(a*b*d^2*cosh(f*x + e)^2 + 2
*a*b*d^2*cosh(f*x + e)*sinh(f*x + e) + a*b*d^2*sinh(f*x + e)^2 + a*b*d^2)*polylog(3, -I*cosh(f*x + e) - I*sinh
(f*x + e)))/(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 (a + b \tanh{\left (e + f x \right )}\right )^{2} \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))**2,x)

[Out]

Integral((a + b*tanh(e + f*x))**2*(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 )}^{2}\,{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))^2,x, algorithm="giac")

[Out]

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