\(\int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx\) [69]

Optimal result

Integrand size = 20, antiderivative size = 157 \[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=\frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1+\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}+\frac {b d^2 \operatorname {PolyLog}\left (3,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3813, 2221, 2611, 2320, 6724} \[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )}-\frac {b (c+d x)^2 \log \left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f \left (a^2-b^2\right )}+\frac {b d^2 \operatorname {PolyLog}\left (3,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^3 \left (a^2-b^2\right )}+\frac {(c+d x)^3}{3 d (a+b)} \]

[In]

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

[Out]

(c + d*x)^3/(3*(a + b)*d) - (b*(c + d*x)^2*Log[1 + (a - b)/((a + b)*E^(2*(e + f*x)))])/((a^2 - b^2)*f) + (b*d*
(c + d*x)*PolyLog[2, -((a - b)/((a + b)*E^(2*(e + f*x))))])/((a^2 - b^2)*f^2) + (b*d^2*PolyLog[3, -((a - b)/((
a + b)*E^(2*(e + f*x))))])/(2*(a^2 - b^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 3813

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Si
mp[2*I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && 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*} \text {integral}& = \frac {(c+d x)^3}{3 (a+b) d}+(2 b) \int \frac {e^{-2 (e+f x)} (c+d x)^2}{(a+b)^2+\left (a^2-b^2\right ) e^{-2 (e+f x)}} \, dx \\ & = \frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1+\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {(2 b d) \int (c+d x) \log \left (1+\frac {\left (a^2-b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{\left (a^2-b^2\right ) f} \\ & = \frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1+\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}-\frac {\left (b d^2\right ) \int \operatorname {PolyLog}\left (2,-\frac {\left (a^2-b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{\left (a^2-b^2\right ) f^2} \\ & = \frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1+\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}+\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\frac {(-a+b) x}{a+b}\right )}{x} \, dx,x,e^{-2 (e+f x)}\right )}{2 \left (a^2-b^2\right ) f^3} \\ & = \frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1+\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}+\frac {b d^2 \operatorname {PolyLog}\left (3,-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.27 \[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=\frac {1}{6} \left (-\frac {4 b (c+d x)^3}{(a+b) d \left (b \left (-1+e^{2 e}\right )+a \left (1+e^{2 e}\right )\right )}-\frac {6 b (c+d x)^2 \log \left (1+\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{(a-b) (a+b) f}+\frac {3 b d \left (2 f (c+d x) \operatorname {PolyLog}\left (2,\frac {(-a+b) e^{-2 (e+f x)}}{a+b}\right )+d \operatorname {PolyLog}\left (3,\frac {(-a+b) e^{-2 (e+f x)}}{a+b}\right )\right )}{(a-b) (a+b) f^3}+\frac {2 x \left (3 c^2+3 c d x+d^2 x^2\right ) \cosh (e)}{a \cosh (e)+b \sinh (e)}\right ) \]

[In]

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

[Out]

((-4*b*(c + d*x)^3)/((a + b)*d*(b*(-1 + E^(2*e)) + a*(1 + E^(2*e)))) - (6*b*(c + d*x)^2*Log[1 + (a - b)/((a +
b)*E^(2*(e + f*x)))])/((a - b)*(a + b)*f) + (3*b*d*(2*f*(c + d*x)*PolyLog[2, (-a + b)/((a + b)*E^(2*(e + f*x))
)] + d*PolyLog[3, (-a + b)/((a + b)*E^(2*(e + f*x)))]))/((a - b)*(a + b)*f^3) + (2*x*(3*c^2 + 3*c*d*x + d^2*x^
2)*Cosh[e])/(a*Cosh[e] + b*Sinh[e]))/6

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(732\) vs. \(2(157)=314\).

Time = 0.34 (sec) , antiderivative size = 733, normalized size of antiderivative = 4.67

method	result	size
risch	\(\frac {d^{2} x^{3}}{3 a +3 b}+\frac {d c \,x^{2}}{a +b}+\frac {c^{2} x}{a +b}+\frac {c^{3}}{3 \left (a +b \right ) d}+\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {b \,d^{2} e^{2} \ln \left ({\mathrm e}^{2 f x +2 e} a +b \,{\mathrm e}^{2 f x +2 e}+a -b \right )}{f^{3} \left (a +b \right ) \left (a -b \right )}+\frac {2 b \,d^{2} e^{2} x}{f^{2} \left (a +b \right ) \left (-a +b \right )}-\frac {b \,d^{2} \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{-a +b}\right ) e^{2}}{f^{3} \left (a +b \right ) \left (-a +b \right )}+\frac {b \,d^{2} \operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{-a +b}\right ) x}{f^{2} \left (a +b \right ) \left (-a +b \right )}-\frac {2 b d c \,x^{2}}{\left (a +b \right ) \left (-a +b \right )}-\frac {2 b d c \,e^{2}}{f^{2} \left (a +b \right ) \left (-a +b \right )}+\frac {2 b d c \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{-a +b}\right ) x}{f \left (a +b \right ) \left (-a +b \right )}+\frac {b d c \operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{-a +b}\right )}{f^{2} \left (a +b \right ) \left (-a +b \right )}-\frac {4 b d c e x}{f \left (a +b \right ) \left (-a +b \right )}+\frac {2 b d c \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{-a +b}\right ) e}{f^{2} \left (a +b \right ) \left (-a +b \right )}-\frac {b \,c^{2} \ln \left ({\mathrm e}^{2 f x +2 e} a +b \,{\mathrm e}^{2 f x +2 e}+a -b \right )}{f \left (a +b \right ) \left (a -b \right )}+\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f \left (a +b \right ) \left (a -b \right )}-\frac {2 b \,d^{2} x^{3}}{3 \left (a +b \right ) \left (-a +b \right )}+\frac {4 b \,d^{2} e^{3}}{3 f^{3} \left (a +b \right ) \left (-a +b \right )}+\frac {b \,d^{2} \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{-a +b}\right ) x^{2}}{f \left (a +b \right ) \left (-a +b \right )}-\frac {b \,d^{2} \operatorname {polylog}\left (3, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{-a +b}\right )}{2 f^{3} \left (a +b \right ) \left (-a +b \right )}-\frac {4 b d c e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {2 b d c e \ln \left ({\mathrm e}^{2 f x +2 e} a +b \,{\mathrm e}^{2 f x +2 e}+a -b \right )}{f^{2} \left (a +b \right ) \left (a -b \right )}\)	\(733\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 500 vs. \(2 (152) = 304\).

Time = 0.27 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.18 \[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=\frac {{\left (a + b\right )} d^{2} f^{3} x^{3} + 3 \, {\left (a + b\right )} c d f^{3} x^{2} + 3 \, {\left (a + b\right )} c^{2} f^{3} x + 6 \, b d^{2} {\rm polylog}\left (3, \sqrt {-\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) + 6 \, b d^{2} {\rm polylog}\left (3, -\sqrt {-\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (\sqrt {-\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (-\sqrt {-\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) + 2 \, {\left (a - b\right )} \sqrt {-\frac {a + b}{a - b}}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) - 2 \, {\left (a - b\right )} \sqrt {-\frac {a + b}{a - b}}\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (\sqrt {-\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\sqrt {-\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right )}{3 \, {\left (a^{2} - b^{2}\right )} f^{3}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=\int \frac {\left (c + d x\right )^{2}}{a + b \tanh {\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (152) = 304\).

Time = 0.31 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.15 \[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=-\frac {{\left (2 \, f x \log \left (\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right )\right )} b c d}{a^{2} f^{2} - b^{2} f^{2}} - \frac {{\left (2 \, f^{2} x^{2} \log \left (\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + 2 \, f x {\rm Li}_2\left (-\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right ) - {\rm Li}_{3}(-\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b})\right )} b d^{2}}{2 \, {\left (a^{2} f^{3} - b^{2} f^{3}\right )}} - c^{2} {\left (\frac {b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} - a - b\right )}{{\left (a^{2} - b^{2}\right )} f} - \frac {f x + e}{{\left (a + b\right )} f}\right )} + \frac {2 \, {\left (b d^{2} f^{3} x^{3} + 3 \, b c d f^{3} x^{2}\right )}}{3 \, {\left (a^{2} f^{3} - b^{2} f^{3}\right )}} + \frac {d^{2} x^{3} + 3 \, c d x^{2}}{3 \, {\left (a + b\right )}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{b \tanh \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{a+b \tanh (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{a+b\,\mathrm {tanh}\left (e+f\,x\right )} \,d x \]

[In]

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

[Out]

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