3.489 \(\int \frac{\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=80 \[ \frac{\left (a^2+b^2\right ) \log (\sinh (c+d x))}{a^3 d}-\frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^3 d}+\frac{b \text{csch}(c+d x)}{a^2 d}-\frac{\text{csch}^2(c+d x)}{2 a d} \]

[Out]

(b*Csch[c + d*x])/(a^2*d) - Csch[c + d*x]^2/(2*a*d) + ((a^2 + b^2)*Log[Sinh[c + d*x]])/(a^3*d) - ((a^2 + b^2)*
Log[a + b*Sinh[c + d*x]])/(a^3*d)

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Rubi [A]  time = 0.105165, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2721, 894} \[ \frac{\left (a^2+b^2\right ) \log (\sinh (c+d x))}{a^3 d}-\frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^3 d}+\frac{b \text{csch}(c+d x)}{a^2 d}-\frac{\text{csch}^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^3/(a + b*Sinh[c + d*x]),x]

[Out]

(b*Csch[c + d*x])/(a^2*d) - Csch[c + d*x]^2/(2*a*d) + ((a^2 + b^2)*Log[Sinh[c + d*x]])/(a^3*d) - ((a^2 + b^2)*
Log[a + b*Sinh[c + d*x]])/(a^3*d)

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\coth ^3(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{-b^2-x^2}{x^3 (a+x)} \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{b^2}{a x^3}+\frac{b^2}{a^2 x^2}+\frac{-a^2-b^2}{a^3 x}+\frac{a^2+b^2}{a^3 (a+x)}\right ) \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=\frac{b \text{csch}(c+d x)}{a^2 d}-\frac{\text{csch}^2(c+d x)}{2 a d}+\frac{\left (a^2+b^2\right ) \log (\sinh (c+d x))}{a^3 d}-\frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.11834, size = 64, normalized size = 0.8 \[ \frac{2 \left (a^2+b^2\right ) (\log (\sinh (c+d x))-\log (a+b \sinh (c+d x)))-a^2 \text{csch}^2(c+d x)+2 a b \text{csch}(c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^3/(a + b*Sinh[c + d*x]),x]

[Out]

(2*a*b*Csch[c + d*x] - a^2*Csch[c + d*x]^2 + 2*(a^2 + b^2)*(Log[Sinh[c + d*x]] - Log[a + b*Sinh[c + d*x]]))/(2
*a^3*d)

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Maple [B]  time = 0.001, size = 194, normalized size = 2.4 \begin{align*} -{\frac{1}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{2\,d{a}^{2}}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,d{a}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{da}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }-{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x)

[Out]

-1/8/d/a*tanh(1/2*d*x+1/2*c)^2-1/2/d/a^2*tanh(1/2*d*x+1/2*c)*b-1/8/d/a/tanh(1/2*d*x+1/2*c)^2+1/d/a*ln(tanh(1/2
*d*x+1/2*c))+1/d/a^3*ln(tanh(1/2*d*x+1/2*c))*b^2+1/2/d*b/a^2/tanh(1/2*d*x+1/2*c)-1/d/a*ln(tanh(1/2*d*x+1/2*c)^
2*a-2*tanh(1/2*d*x+1/2*c)*b-a)-1/d/a^3*ln(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)*b^2

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Maxima [B]  time = 1.29418, size = 234, normalized size = 2.92 \begin{align*} -\frac{2 \,{\left (b e^{\left (-d x - c\right )} - a e^{\left (-2 \, d x - 2 \, c\right )} - b e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{{\left (2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a^{2} e^{\left (-4 \, d x - 4 \, c\right )} - a^{2}\right )} d} - \frac{{\left (a^{2} + b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a^{3} d} + \frac{{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{3} d} + \frac{{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.84773, size = 1554, normalized size = 19.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(2*a*b*cosh(d*x + c)^3 + 2*a*b*sinh(d*x + c)^3 - 2*a^2*cosh(d*x + c)^2 - 2*a*b*cosh(d*x + c) + 2*(3*a*b*cosh(d
*x + c) - a^2)*sinh(d*x + c)^2 - ((a^2 + b^2)*cosh(d*x + c)^4 + 4*(a^2 + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 +
(a^2 + b^2)*sinh(d*x + c)^4 - 2*(a^2 + b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + b^2)*cosh(d*x + c)^2 - a^2 - b^2)*si
nh(d*x + c)^2 + a^2 + b^2 + 4*((a^2 + b^2)*cosh(d*x + c)^3 - (a^2 + b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*(
b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + ((a^2 + b^2)*cosh(d*x + c)^4 + 4*(a^2 + b^2)*cosh(d*x
+ c)*sinh(d*x + c)^3 + (a^2 + b^2)*sinh(d*x + c)^4 - 2*(a^2 + b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + b^2)*cosh(d*x
+ c)^2 - a^2 - b^2)*sinh(d*x + c)^2 + a^2 + b^2 + 4*((a^2 + b^2)*cosh(d*x + c)^3 - (a^2 + b^2)*cosh(d*x + c))
*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 2*(3*a*b*cosh(d*x + c)^2 - 2*a^2*cosh(d
*x + c) - a*b)*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*d*sinh(d*x
+ c)^4 - 2*a^3*d*cosh(d*x + c)^2 + a^3*d + 2*(3*a^3*d*cosh(d*x + c)^2 - a^3*d)*sinh(d*x + c)^2 + 4*(a^3*d*cosh
(d*x + c)^3 - a^3*d*cosh(d*x + c))*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**3/(a+b*sinh(d*x+c)),x)

[Out]

Integral(coth(c + d*x)**3/(a + b*sinh(c + d*x)), x)

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Giac [B]  time = 1.41714, size = 224, normalized size = 2.8 \begin{align*} \frac{\frac{{\left (a^{2} e^{c} + b^{2} e^{c}\right )} e^{\left (-c\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{3}} + \frac{{\left (a^{2} e^{c} + b^{2} e^{c}\right )} e^{\left (-c\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{3}} - \frac{{\left (a^{2} + b^{2}\right )} \log \left ({\left | b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b \right |}\right )}{a^{3}} + \frac{2 \,{\left (a b e^{\left (3 \, d x + 3 \, c\right )} - a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b e^{\left (d x + c\right )}\right )}}{a^{3}{\left (e^{\left (d x + c\right )} + 1\right )}^{2}{\left (e^{\left (d x + c\right )} - 1\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

((a^2*e^c + b^2*e^c)*e^(-c)*log(e^(d*x + c) + 1)/a^3 + (a^2*e^c + b^2*e^c)*e^(-c)*log(abs(e^(d*x + c) - 1))/a^
3 - (a^2 + b^2)*log(abs(b*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - b))/a^3 + 2*(a*b*e^(3*d*x + 3*c) - a^2*e^(2*d*x
+ 2*c) - a*b*e^(d*x + c))/(a^3*(e^(d*x + c) + 1)^2*(e^(d*x + c) - 1)^2))/d