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

Optimal. Leaf size=196 \[ \frac {b (-2 a c f-2 a d f x+b d) \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {b (c+d x)}{f \left (a^2-b^2\right ) (a+b \coth (e+f x))}-\frac {(c+d x)^2}{2 d \left (a^2-b^2\right )}+\frac {a b d \text {Li}_2\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {(-2 a c f-2 a d f x+b d)^2}{4 a d f^2 (a-b) (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 196, normalized size of antiderivative = 1.00, 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, 3731, 2190, 2279, 2391} \[ \frac {a b d \text {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {b (-2 a c f-2 a d f x+b d) \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {b (c+d x)}{f \left (a^2-b^2\right ) (a+b \coth (e+f x))}-\frac {(c+d x)^2}{2 d \left (a^2-b^2\right )}+\frac {(-2 a c f-2 a d f x+b d)^2}{4 a d f^2 (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

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]

Rule 3731

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + Pi*(k_.) + (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^(2*I*k*Pi)*E^Simp[2*I*(e + f*x), x])/((a +
I*b)^2 + (a^2 + b^2)*E^(2*I*k*Pi)*E^Simp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Integer
Q[4*k] && NeQ[a^2 + b^2, 0] && IGtQ[m, 0]

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]

Rubi steps

\begin {align*} \int \frac {c+d x}{(a+b \coth (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 \coth (e+f x))}-\frac {i \int \frac {-i b d+2 i a c f+2 i a d f x}{a+b \coth (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 \coth (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 (c+d x)}{\left (a^2-b^2\right ) f (a+b \coth (e+f x))}+\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 {(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 (c+d x)}{\left (a^2-b^2\right ) f (a+b \coth (e+f x))}+\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) \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 (c+d x)}{\left (a^2-b^2\right ) f (a+b \coth (e+f x))}+\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}\\ \end {align*}

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Mathematica [C]  time = 6.23, size = 477, normalized size = 2.43 \[ \frac {\text {csch}^2(e+f x) (a \sinh (e+f x)+b \cosh (e+f x)) \left (2 b f \left (b^2-a^2\right ) (c+d x) \sinh (e+f x)-\left (a^2-b^2\right ) (e+f x) (d (e-f x)-2 c f) (a \sinh (e+f x)+b \cosh (e+f x))-2 a d (a \sinh (e+f x)+b \cosh (e+f x)) \left (a \sqrt {1-\frac {b^2}{a^2}} e^{-\tanh ^{-1}\left (\frac {b}{a}\right )} (e+f x)^2+b \text {Li}_2\left (e^{-2 \left (e+f x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )}\right )-i b \left (\pi -2 i \tanh ^{-1}\left (\frac {b}{a}\right )\right ) (e+f x)-2 b \left (\tanh ^{-1}\left (\frac {b}{a}\right )+e+f x\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {b}{a}\right )+e+f x\right )}\right )+2 b \tanh ^{-1}\left (\frac {b}{a}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {b}{a}\right )+e+f x\right )\right )+i \pi b \log \left (e^{2 (e+f x)}+1\right )-i \pi b \log (\cosh (e+f x))\right )-4 a c f (a \sinh (e+f x)+b \cosh (e+f x)) (a (e+f x)-b \log (a \sinh (e+f x)+b \cosh (e+f x)))+2 b d (a \sinh (e+f x)+b \cosh (e+f x)) (a (e+f x)-b \log (a \sinh (e+f x)+b \cosh (e+f x)))+4 a d e (a \sinh (e+f x)+b \cosh (e+f x)) (a (e+f x)-b \log (a \sinh (e+f x)+b \cosh (e+f x)))\right )}{2 f^2 (b-a) (a+b) \left (a^2-b^2\right ) (a+b \coth (e+f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 1797, normalized size = 9.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*coth(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)*cosh
(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)*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))) - 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)*si
nh(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x + c}{{\left (b \coth \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.95, size = 524, normalized size = 2.67 \[ \frac {d \,x^{2}}{2 a^{2}+4 a b +2 b^{2}}+\frac {c x}{a^{2}+2 a b +b^{2}}-\frac {2 b^{2} \left (d x +c \right )}{\left (a -b \right ) f \left (a^{2}+2 a b +b^{2}\right ) \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}+\frac {b^{2} d \ln \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2 b^{2} d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2 b a c \ln \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f \left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {4 b a c \ln \left ({\mathrm e}^{f x +e}\right )}{f \left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {2 b d a e \ln \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {4 b d a e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2 b d a \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x}{f \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2 b d a \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) e}{f^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {2 b d a \,x^{2}}{\left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {4 b d a e x}{f \left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {2 b d a \,e^{2}}{f^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {b d a \polylog \left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right )}{f^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (8 \, a b f \int \frac {x}{a^{4} f e^{\left (2 \, f x + 2 \, e\right )} + 2 \, a^{3} b f e^{\left (2 \, f x + 2 \, e\right )} - 2 \, a b^{3} f e^{\left (2 \, f x + 2 \, e\right )} - b^{4} f e^{\left (2 \, f x + 2 \, e\right )} - a^{4} f + 2 \, a^{2} b^{2} f - b^{4} f}\,{d x} + 2 \, b^{2} {\left (\frac {2 \, {\left (f x + e\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} f^{2}} - \frac {\log \left ({\left (a + b\right )} e^{\left (2 \, f x + 2 \, e\right )} - a + b\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} f^{2}}\right )} + \frac {{\left (a^{2} f e^{\left (2 \, e\right )} - b^{2} f e^{\left (2 \, e\right )}\right )} x^{2} e^{\left (2 \, f x\right )} - 4 \, b^{2} x - {\left (a^{2} f - 2 \, a b f + b^{2} f\right )} x^{2}}{a^{4} f - 2 \, a^{2} b^{2} f + b^{4} f - {\left (a^{4} f e^{\left (2 \, e\right )} + 2 \, a^{3} b f e^{\left (2 \, e\right )} - 2 \, a b^{3} f e^{\left (2 \, e\right )} - b^{4} f e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}\right )} d - c {\left (\frac {2 \, a b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} + a + b\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} f} + \frac {2 \, b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, f x - 2 \, e\right )}\right )} f} - \frac {f x + e}{{\left (a^{2} + 2 \, a b + b^{2}\right )} f}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*coth(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {c+d\,x}{{\left (a+b\,\mathrm {coth}\left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c + d x}{\left (a + b \coth {\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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