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

Optimal. Leaf size=196 \[ \frac{a b d \text{PolyLog}\left (2,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac{b (-2 a c f-2 a d f x+b d) \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f^2 \left (a^2-b^2\right )^2}+\frac{b (c+d x)}{f \left (a^2-b^2\right ) (a+b \tanh (e+f x))}-\frac{(c+d x)^2}{2 d \left (a^2-b^2\right )}+\frac{(-2 a c f-2 a d f x+b d)^2}{4 a d f^2 (a-b) (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.286424, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {3733, 3732, 2190, 2279, 2391} \[ \frac{a b d \text{PolyLog}\left (2,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac{b (-2 a c f-2 a d f x+b d) \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f^2 \left (a^2-b^2\right )^2}+\frac{b (c+d x)}{f \left (a^2-b^2\right ) (a+b \tanh (e+f x))}-\frac{(c+d x)^2}{2 d \left (a^2-b^2\right )}+\frac{(-2 a c f-2 a d f x+b d)^2}{4 a d f^2 (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3733

Int[((c_.) + (d_.)*(x_))/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[(c + d*x)^2/(2*d*(a^2 +
b^2)), x] + (Dist[1/(f*(a^2 + b^2)), Int[(b*d + 2*a*c*f + 2*a*d*f*x)/(a + b*Tan[e + f*x]), x], x] - Simp[(b*(c
+ d*x))/(f*(a^2 + b^2)*(a + b*Tan[e + f*x])), x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0]

Rule 3732

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^S
imp[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 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]

Rubi steps

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

Mathematica [C]  time = 6.65538, size = 476, normalized size = 2.43 \[ \frac{\text{sech}^2(e+f x) (a \cosh (e+f x)+b \sinh (e+f x)) \left (2 a b d (a \cosh (e+f x)+b \sinh (e+f x)) \left (a \text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac{a}{b}\right )+e+f x\right )}\right )+b \sqrt{1-\frac{a^2}{b^2}} e^{-\tanh ^{-1}\left (\frac{a}{b}\right )} (e+f x)^2-i a \left (\pi -2 i \tanh ^{-1}\left (\frac{a}{b}\right )\right ) (e+f x)-2 a \left (\tanh ^{-1}\left (\frac{a}{b}\right )+e+f x\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{a}{b}\right )+e+f x\right )}\right )+2 a \tanh ^{-1}\left (\frac{a}{b}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac{a}{b}\right )+e+f x\right )\right )+i \pi a \log \left (e^{2 (e+f x)}+1\right )-i \pi a \log (\cosh (e+f x))\right )+2 b^2 f \left (b^2-a^2\right ) (c+d x) \sinh (e+f x)-a \left (a^2-b^2\right ) (e+f x) (d (e-f x)-2 c f) (a \cosh (e+f x)+b \sinh (e+f x))-2 b^2 d (a \cosh (e+f x)+b \sinh (e+f x)) (b (e+f x)-a \log (a \cosh (e+f x)+b \sinh (e+f x)))+4 a b c f (a \cosh (e+f x)+b \sinh (e+f x)) (b (e+f x)-a \log (a \cosh (e+f x)+b \sinh (e+f x)))-4 a b d e (a \cosh (e+f x)+b \sinh (e+f x)) (b (e+f x)-a \log (a \cosh (e+f x)+b \sinh (e+f x)))\right )}{2 a f^2 \left (a^2-b^2\right )^2 (a+b \tanh (e+f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sech[e + f*x]^2*(a*Cosh[e + f*x] + b*Sinh[e + f*x])*(2*b^2*(-a^2 + b^2)*f*(c + d*x)*Sinh[e + f*x] - a*(a^2 -
b^2)*(e + f*x)*(-2*c*f + d*(e - f*x))*(a*Cosh[e + f*x] + b*Sinh[e + f*x]) - 2*b^2*d*(b*(e + f*x) - a*Log[a*Cos
h[e + f*x] + b*Sinh[e + f*x]])*(a*Cosh[e + f*x] + b*Sinh[e + f*x]) - 4*a*b*d*e*(b*(e + f*x) - a*Log[a*Cosh[e +
f*x] + b*Sinh[e + f*x]])*(a*Cosh[e + f*x] + b*Sinh[e + f*x]) + 4*a*b*c*f*(b*(e + f*x) - a*Log[a*Cosh[e + f*x]
+ b*Sinh[e + f*x]])*(a*Cosh[e + f*x] + b*Sinh[e + f*x]) + 2*a*b*d*((Sqrt[1 - a^2/b^2]*b*(e + f*x)^2)/E^ArcTan
h[a/b] - I*a*(e + f*x)*(Pi - (2*I)*ArcTanh[a/b]) + I*a*Pi*Log[1 + E^(2*(e + f*x))] - 2*a*(e + f*x + ArcTanh[a/
b])*Log[1 - E^(-2*(e + f*x + ArcTanh[a/b]))] - I*a*Pi*Log[Cosh[e + f*x]] + 2*a*ArcTanh[a/b]*Log[I*Sinh[e + f*x
+ ArcTanh[a/b]]] + a*PolyLog[2, E^(-2*(e + f*x + ArcTanh[a/b]))])*(a*Cosh[e + f*x] + b*Sinh[e + f*x])))/(2*a*
(a^2 - b^2)^2*f^2*(a + b*Tanh[e + f*x])^2)

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Maple [B]  time = 0.224, size = 661, normalized size = 3.4 \begin{align*}{\frac{d{x}^{2}}{2\,{a}^{2}+4\,ab+2\,{b}^{2}}}+{\frac{cx}{{a}^{2}+2\,ab+{b}^{2}}}+2\,{\frac{{b}^{2} \left ( dx+c \right ) }{ \left ( a-b \right ) f \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( a{{\rm e}^{2\,fx+2\,e}}+b{{\rm e}^{2\,fx+2\,e}}+a-b \right ) }}-2\,{\frac{{b}^{2}d\ln \left ({{\rm e}^{fx+e}} \right ) }{ \left ( a-b \right ) ^{2}{f}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{{b}^{2}d\ln \left ( a{{\rm e}^{2\,fx+2\,e}}+b{{\rm e}^{2\,fx+2\,e}}+a-b \right ) }{ \left ( a-b \right ) ^{2}{f}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+4\,{\frac{abc\ln \left ({{\rm e}^{fx+e}} \right ) }{ \left ( a-b \right ) ^{2}f \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-2\,{\frac{abc\ln \left ( a{{\rm e}^{2\,fx+2\,e}}+b{{\rm e}^{2\,fx+2\,e}}+a-b \right ) }{ \left ( a-b \right ) ^{2}f \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-4\,{\frac{abde\ln \left ({{\rm e}^{fx+e}} \right ) }{ \left ( a-b \right ) ^{2}{f}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+2\,{\frac{abde\ln \left ( a{{\rm e}^{2\,fx+2\,e}}+b{{\rm e}^{2\,fx+2\,e}}+a-b \right ) }{ \left ( a-b \right ) ^{2}{f}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+2\,{\frac{abdx}{ \left ( a-b \right ) f \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( -a+b \right ) }\ln \left ( 1-{\frac{ \left ( a+b \right ){{\rm e}^{2\,fx+2\,e}}}{-a+b}} \right ) }+2\,{\frac{abde}{ \left ( a-b \right ){f}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( -a+b \right ) }\ln \left ( 1-{\frac{ \left ( a+b \right ){{\rm e}^{2\,fx+2\,e}}}{-a+b}} \right ) }-2\,{\frac{abd{x}^{2}}{ \left ( a-b \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( -a+b \right ) }}-4\,{\frac{abdex}{ \left ( a-b \right ) f \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( -a+b \right ) }}-2\,{\frac{abd{e}^{2}}{ \left ( a-b \right ){f}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( -a+b \right ) }}+{\frac{bda}{ \left ( a-b \right ){f}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( -a+b \right ) }{\it polylog} \left ( 2,{\frac{ \left ( a+b \right ){{\rm e}^{2\,fx+2\,e}}}{-a+b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (8 \, a b f \int \frac{x}{a^{4} f e^{\left (2 \, f x + 2 \, e\right )} + 2 \, a^{3} b f e^{\left (2 \, f x + 2 \, e\right )} - 2 \, a b^{3} f e^{\left (2 \, f x + 2 \, e\right )} - b^{4} f e^{\left (2 \, f x + 2 \, e\right )} + a^{4} f - 2 \, a^{2} b^{2} f + b^{4} f}\,{d x} - 2 \, b^{2}{\left (\frac{2 \,{\left (f x + e\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} f^{2}} - \frac{\log \left ({\left (a + b\right )} e^{\left (2 \, f x + 2 \, e\right )} + a - b\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} f^{2}}\right )} + \frac{{\left (a^{2} f e^{\left (2 \, e\right )} - b^{2} f e^{\left (2 \, e\right )}\right )} x^{2} e^{\left (2 \, f x\right )} + 4 \, b^{2} x +{\left (a^{2} f - 2 \, a b f + b^{2} f\right )} x^{2}}{a^{4} f - 2 \, a^{2} b^{2} f + b^{4} f +{\left (a^{4} f e^{\left (2 \, e\right )} + 2 \, a^{3} b f e^{\left (2 \, e\right )} - 2 \, a b^{3} f e^{\left (2 \, e\right )} - b^{4} f e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}\right )} d - c{\left (\frac{2 \, a b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} - a - b\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} f} + \frac{2 \, b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} +{\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, f x - 2 \, e\right )}\right )} f} - \frac{f x + e}{{\left (a^{2} + 2 \, a b + b^{2}\right )} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.84712, size = 4065, normalized size = 20.74 \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)/(a+b*tanh(f*x+e))^2,x, algorithm="fricas")

[Out]

1/2*((a^3 + a^2*b - a*b^2 - b^3)*d*f^2*x^2 + 2*(a^3 + a^2*b - a*b^2 - b^3)*c*f^2*x - 4*(a^2*b - a*b^2)*d*e^2 -
4*(a*b^2 - b^3)*d*e + ((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*f^2*x^2 - 4*(a^2*b + a*b^2)*d*e^2 + 8*(a^2*b + a*b^2
)*c*e*f - 4*(a*b^2 + b^3)*d*e + 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*c*f^2 - 2*(a*b^2 + b^3)*d*f)*x)*cosh(f*x +
e)^2 + 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*f^2*x^2 - 4*(a^2*b + a*b^2)*d*e^2 + 8*(a^2*b + a*b^2)*c*e*f - 4*(a
*b^2 + b^3)*d*e + 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*c*f^2 - 2*(a*b^2 + b^3)*d*f)*x)*cosh(f*x + e)*sinh(f*x +
e) + ((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*f^2*x^2 - 4*(a^2*b + a*b^2)*d*e^2 + 8*(a^2*b + a*b^2)*c*e*f - 4*(a*b^2
+ b^3)*d*e + 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*c*f^2 - 2*(a*b^2 + b^3)*d*f)*x)*sinh(f*x + e)^2 + 4*(2*(a^2*b
- a*b^2)*c*e + (a*b^2 - b^3)*c)*f - 4*((a^2*b + a*b^2)*d*cosh(f*x + e)^2 + 2*(a^2*b + a*b^2)*d*cosh(f*x + e)*
sinh(f*x + e) + (a^2*b + a*b^2)*d*sinh(f*x + e)^2 + (a^2*b - a*b^2)*d)*dilog(sqrt(-(a + b)/(a - b))*(cosh(f*x
+ e) + sinh(f*x + e))) - 4*((a^2*b + a*b^2)*d*cosh(f*x + e)^2 + 2*(a^2*b + a*b^2)*d*cosh(f*x + e)*sinh(f*x + e
) + (a^2*b + a*b^2)*d*sinh(f*x + e)^2 + (a^2*b - a*b^2)*d)*dilog(-sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + sinh
(f*x + e))) + 2*(2*(a^2*b - a*b^2)*d*e - 2*(a^2*b - a*b^2)*c*f + (2*(a^2*b + a*b^2)*d*e - 2*(a^2*b + a*b^2)*c*
f + (a*b^2 + b^3)*d)*cosh(f*x + e)^2 + 2*(2*(a^2*b + a*b^2)*d*e - 2*(a^2*b + a*b^2)*c*f + (a*b^2 + b^3)*d)*cos
h(f*x + e)*sinh(f*x + e) + (2*(a^2*b + a*b^2)*d*e - 2*(a^2*b + a*b^2)*c*f + (a*b^2 + b^3)*d)*sinh(f*x + e)^2 +
(a*b^2 - b^3)*d)*log(2*(a + b)*cosh(f*x + e) + 2*(a + b)*sinh(f*x + e) + 2*(a - b)*sqrt(-(a + b)/(a - b))) +
2*(2*(a^2*b - a*b^2)*d*e - 2*(a^2*b - a*b^2)*c*f + (2*(a^2*b + a*b^2)*d*e - 2*(a^2*b + a*b^2)*c*f + (a*b^2 + b
^3)*d)*cosh(f*x + e)^2 + 2*(2*(a^2*b + a*b^2)*d*e - 2*(a^2*b + a*b^2)*c*f + (a*b^2 + b^3)*d)*cosh(f*x + e)*sin
h(f*x + e) + (2*(a^2*b + a*b^2)*d*e - 2*(a^2*b + a*b^2)*c*f + (a*b^2 + b^3)*d)*sinh(f*x + e)^2 + (a*b^2 - b^3)
*d)*log(2*(a + b)*cosh(f*x + e) + 2*(a + b)*sinh(f*x + e) - 2*(a - b)*sqrt(-(a + b)/(a - b))) - 4*((a^2*b - a*
b^2)*d*f*x + (a^2*b - a*b^2)*d*e + ((a^2*b + a*b^2)*d*f*x + (a^2*b + a*b^2)*d*e)*cosh(f*x + e)^2 + 2*((a^2*b +
a*b^2)*d*f*x + (a^2*b + a*b^2)*d*e)*cosh(f*x + e)*sinh(f*x + e) + ((a^2*b + a*b^2)*d*f*x + (a^2*b + a*b^2)*d*
e)*sinh(f*x + e)^2)*log(sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e)) + 1) - 4*((a^2*b - a*b^2)*d*f*x
+ (a^2*b - a*b^2)*d*e + ((a^2*b + a*b^2)*d*f*x + (a^2*b + a*b^2)*d*e)*cosh(f*x + e)^2 + 2*((a^2*b + a*b^2)*d*
f*x + (a^2*b + a*b^2)*d*e)*cosh(f*x + e)*sinh(f*x + e) + ((a^2*b + a*b^2)*d*f*x + (a^2*b + a*b^2)*d*e)*sinh(f*
x + e)^2)*log(-sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e)) + 1))/((a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*
b^3 + a*b^4 + b^5)*f^2*cosh(f*x + e)^2 + 2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*f^2*cosh(f*x +
e)*sinh(f*x + e) + (a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*f^2*sinh(f*x + e)^2 + (a^5 - a^4*b - 2*
a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{{\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)/(a+b*tanh(f*x+e))^2,x, algorithm="giac")

[Out]

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