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

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

[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

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Rubi [A]
time = 0.20, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3813, 2221, 2611, 2320, 6724} \begin {gather*} \frac {b d (c+d x) \text {Li}_2\left (-\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 \text {Li}_3\left (-\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)} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 2.61, size = 212, normalized size = 1.35 \begin {gather*} \frac {b \left (\frac {4 e^{2 e} x \left (3 c^2+3 c d x+d^2 x^2\right )}{b \left (-1+e^{2 e}\right )+a \left (1+e^{2 e}\right )}-\frac {6 (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 (e+f x)}}{a-b}\right )}{(a+b) f}-\frac {6 d (c+d x) \text {PolyLog}\left (2,-\frac {(a+b) e^{2 (e+f x)}}{a-b}\right )}{(a+b) f^2}+\frac {3 d^2 \text {PolyLog}\left (3,-\frac {(a+b) e^{2 (e+f x)}}{a-b}\right )}{(a+b) f^3}\right )}{6 (a-b)}+\frac {x \left (3 c^2+3 c d x+d^2 x^2\right ) \cosh (e)}{3 (a \cosh (e)+b \sinh (e))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(732\) vs. \(2(157)=314\).
time = 4.69, size = 733, normalized size = 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 {b \,c^{2} \ln \left (a \,{\mathrm e}^{2 f x +2 e}+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 {b \,d^{2} e^{2} \ln \left (a \,{\mathrm e}^{2 f x +2 e}+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} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3} \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 {2 b \,d^{2} e^{2} x}{f^{2} \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} \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} \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 {b \,d^{2} \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 {2 b d e c \ln \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}+a -b \right )}{f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {4 b d e c \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {2 b c d \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 {2 b c d \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 {2 b d c \,x^{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 \,e^{2}}{f^{2} \left (a +b \right ) \left (-a +b \right )}+\frac {b c d \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 )}\)	\(733\)

Verification of antiderivative is not currently implemented for this CAS.

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (155) = 310\).
time = 0.38, size = 314, normalized size = 2.00 \begin {gather*} -\frac {{\left (2 \, f x \log \left (\frac {{\left (a + b\right )} e^{\left (2 \, f x + 2 \, e\right )}}{a - b} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (a + b\right )} e^{\left (2 \, f x + 2 \, e\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 + b\right )} e^{\left (2 \, f x + 2 \, e\right )}}{a - b} + 1\right ) + 2 \, f x {\rm Li}_2\left (-\frac {{\left (a + b\right )} e^{\left (2 \, f x + 2 \, e\right )}}{a - b}\right ) - {\rm Li}_{3}(-\frac {{\left (a + b\right )} e^{\left (2 \, f x + 2 \, e\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 666 vs. \(2 (155) = 310\).
time = 0.39, size = 666, normalized size = 4.24 \begin {gather*} \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 + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right )}\right ) + 6 \, b d^{2} {\rm polylog}\left (3, -\sqrt {-\frac {a + b}{a - b}} {\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 )\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 + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\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 + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right )}\right ) - 3 \, {\left (b c^{2} f^{2} - 2 \, b c d f \cosh \left (1\right ) + b d^{2} \cosh \left (1\right )^{2} + b d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (b c d f - b d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + 2 \, {\left (a - b\right )} \sqrt {-\frac {a + b}{a - b}}\right ) - 3 \, {\left (b c^{2} f^{2} - 2 \, b c d f \cosh \left (1\right ) + b d^{2} \cosh \left (1\right )^{2} + b d^{2} \sinh \left (1\right )^{2} - 2 \, {\left (b c d f - b d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\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 + 2 \, b c d f \cosh \left (1\right ) - b d^{2} \cosh \left (1\right )^{2} - b d^{2} \sinh \left (1\right )^{2} + 2 \, {\left (b c d f - b d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (\sqrt {-\frac {a + b}{a - b}} {\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 )\right )} + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + 2 \, b c d f \cosh \left (1\right ) - b d^{2} \cosh \left (1\right )^{2} - b d^{2} \sinh \left (1\right )^{2} + 2 \, {\left (b c d f - b d^{2} \cosh \left (1\right )\right )} \sinh \left (1\right )\right )} \log \left (-\sqrt {-\frac {a + b}{a - b}} {\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 )\right )} + 1\right )}{3 \, {\left (a^{2} - b^{2}\right )} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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