3.703 \(\int \frac{\sinh ^3(x)}{(a \cosh (x)+b \sinh (x))^3} \, dx\)
Optimal. Leaf size=104 \[ -\frac{b x \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}+\frac{2 a b}{\left (a^2-b^2\right )^2 (a \coth (x)+b)}-\frac{a}{2 \left (a^2-b^2\right ) (a \coth (x)+b)^2}+\frac{a \left (a^2+3 b^2\right ) \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3} \]
[Out]
-((b*(3*a^2 + b^2)*x)/(a^2 - b^2)^3) - a/(2*(a^2 - b^2)*(b + a*Coth[x])^2) + (2*a*b)/((a^2 - b^2)^2*(b + a*Cot
h[x])) + (a*(a^2 + 3*b^2)*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2)^3
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Rubi [A] time = 0.237618, antiderivative size = 104, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used =
{3085, 3483, 3529, 3531, 3530} \[ -\frac{b x \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3}+\frac{2 a b}{\left (a^2-b^2\right )^2 (a \coth (x)+b)}-\frac{a}{2 \left (a^2-b^2\right ) (a \coth (x)+b)^2}+\frac{a \left (a^2+3 b^2\right ) \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[x]^3/(a*Cosh[x] + b*Sinh[x])^3,x]
[Out]
-((b*(3*a^2 + b^2)*x)/(a^2 - b^2)^3) - a/(2*(a^2 - b^2)*(b + a*Coth[x])^2) + (2*a*b)/((a^2 - b^2)^2*(b + a*Cot
h[x])) + (a*(a^2 + 3*b^2)*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2)^3
Rule 3085
Int[sin[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(b + a*Cot[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]
Rule 3483
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
Rule 3529
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]
Rule 3531
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]
Rule 3530
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]
Rubi steps
\begin{align*} \int \frac{\sinh ^3(x)}{(a \cosh (x)+b \sinh (x))^3} \, dx &=i \int \frac{1}{(-i b-i a \coth (x))^3} \, dx\\ &=-\frac{a}{2 \left (a^2-b^2\right ) (b+a \coth (x))^2}+\frac{i \int \frac{-i b+i a \coth (x)}{(-i b-i a \coth (x))^2} \, dx}{a^2-b^2}\\ &=-\frac{a}{2 \left (a^2-b^2\right ) (b+a \coth (x))^2}+\frac{2 a b}{\left (a^2-b^2\right )^2 (b+a \coth (x))}+\frac{i \int \frac{-a^2-b^2+2 a b \coth (x)}{-i b-i a \coth (x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac{b \left (3 a^2+b^2\right ) x}{\left (a^2-b^2\right )^3}-\frac{a}{2 \left (a^2-b^2\right ) (b+a \coth (x))^2}+\frac{2 a b}{\left (a^2-b^2\right )^2 (b+a \coth (x))}+\frac{\left (i a \left (a^2+3 b^2\right )\right ) \int \frac{-a-b \coth (x)}{-i b-i a \coth (x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=-\frac{b \left (3 a^2+b^2\right ) x}{\left (a^2-b^2\right )^3}-\frac{a}{2 \left (a^2-b^2\right ) (b+a \coth (x))^2}+\frac{2 a b}{\left (a^2-b^2\right )^2 (b+a \coth (x))}+\frac{a \left (a^2+3 b^2\right ) \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3}\\ \end{align*}
Mathematica [A] time = 0.6463, size = 117, normalized size = 1.12 \[ -\frac{b x \left (3 a^2+b^2\right )}{(a-b)^3 (a+b)^3}+\frac{\left (a^3+3 a b^2\right ) \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3}+\frac{a^3}{2 (a-b)^2 (a+b)^2 (a \cosh (x)+b \sinh (x))^2}+\frac{3 a b \sinh (x)}{(a-b)^2 (a+b)^2 (a \cosh (x)+b \sinh (x))} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[x]^3/(a*Cosh[x] + b*Sinh[x])^3,x]
[Out]
-((b*(3*a^2 + b^2)*x)/((a - b)^3*(a + b)^3)) + ((a^3 + 3*a*b^2)*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2)^3 + a^
3/(2*(a - b)^2*(a + b)^2*(a*Cosh[x] + b*Sinh[x])^2) + (3*a*b*Sinh[x])/((a - b)^2*(a + b)^2*(a*Cosh[x] + b*Sinh
[x]))
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Maple [B] time = 0.092, size = 404, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(x)^3/(a*cosh(x)+b*sinh(x))^3,x)
[Out]
-1/(a-b)^3*ln(tanh(1/2*x)+1)+4*a^4/(a-b)^3/(a+b)^3/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2*tanh(1/2*x)^3*b-4*a^2
/(a-b)^3/(a+b)^3/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2*tanh(1/2*x)^3*b^3-2*a^5/(a-b)^3/(a+b)^3/(a+2*tanh(1/2*x
)*b+a*tanh(1/2*x)^2)^2*tanh(1/2*x)^2+12*a^3/(a-b)^3/(a+b)^3/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2*tanh(1/2*x)^
2*b^2-10*a/(a-b)^3/(a+b)^3/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2*tanh(1/2*x)^2*b^4+4*a^4/(a-b)^3/(a+b)^3/(a+2*
tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2*tanh(1/2*x)*b-4*a^2/(a-b)^3/(a+b)^3/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2*tan
h(1/2*x)*b^3+a^3/(a-b)^3/(a+b)^3*ln(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)+3*a/(a-b)^3/(a+b)^3*ln(a+2*tanh(1/2*x)*
b+a*tanh(1/2*x)^2)*b^2-1/(a+b)^3*ln(tanh(1/2*x)-1)
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Maxima [B] time = 1.33858, size = 390, normalized size = 3.75 \begin{align*} \frac{{\left (a^{3} + 3 \, a b^{2}\right )} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{2 \,{\left (3 \, a^{3} b + 3 \, a^{2} b^{2} +{\left (a^{4} + 2 \, a^{3} b - 3 \, a^{2} b^{2}\right )} e^{\left (-2 \, x\right )}\right )}}{a^{7} + a^{6} b - 3 \, a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6} - b^{7} + 2 \,{\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - 3 \, a^{2} b^{5} - a b^{6} + b^{7}\right )} e^{\left (-2 \, x\right )} +{\left (a^{7} - 3 \, a^{6} b + a^{5} b^{2} + 5 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} e^{\left (-4 \, x\right )}} + \frac{x}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a*cosh(x)+b*sinh(x))^3,x, algorithm="maxima")
[Out]
(a^3 + 3*a*b^2)*log(-(a - b)*e^(-2*x) - a - b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 2*(3*a^3*b + 3*a^2*b^2 +
(a^4 + 2*a^3*b - 3*a^2*b^2)*e^(-2*x))/(a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b
^7 + 2*(a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7)*e^(-2*x) + (a^7 - 3*a^6*b +
a^5*b^2 + 5*a^4*b^3 - 5*a^3*b^4 - a^2*b^5 + 3*a*b^6 - b^7)*e^(-4*x)) + x/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)
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Fricas [B] time = 2.03017, size = 2854, normalized size = 27.44 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a*cosh(x)+b*sinh(x))^3,x, algorithm="fricas")
[Out]
-((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*x*cosh(x)^4 + 4*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a
^2*b^3 + 5*a*b^4 + b^5)*x*cosh(x)*sinh(x)^3 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*x*sinh
(x)^4 + 6*a^4*b - 12*a^3*b^2 + 6*a^2*b^3 - 2*(a^5 - 3*a^4*b - a^3*b^2 + 3*a^2*b^3 - (a^5 + 3*a^4*b + 2*a^3*b^2
- 2*a^2*b^3 - 3*a*b^4 - b^5)*x)*cosh(x)^2 - 2*(a^5 - 3*a^4*b - a^3*b^2 + 3*a^2*b^3 - 3*(a^5 + 5*a^4*b + 10*a^
3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*x*cosh(x)^2 - (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*x)*s
inh(x)^2 + (a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*x - (a^5 - 2*a^4*b + 4*a^3*b^2 - 6*a^2*b^3 + 3*
a*b^4 + (a^5 + 2*a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 3*a*b^4)*cosh(x)^4 + 4*(a^5 + 2*a^4*b + 4*a^3*b^2 + 6*a^2*b^3
+ 3*a*b^4)*cosh(x)*sinh(x)^3 + (a^5 + 2*a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 3*a*b^4)*sinh(x)^4 + 2*(a^5 + 2*a^3*b
^2 - 3*a*b^4)*cosh(x)^2 + 2*(a^5 + 2*a^3*b^2 - 3*a*b^4 + 3*(a^5 + 2*a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 3*a*b^4)*c
osh(x)^2)*sinh(x)^2 + 4*((a^5 + 2*a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 3*a*b^4)*cosh(x)^3 + (a^5 + 2*a^3*b^2 - 3*a*
b^4)*cosh(x))*sinh(x))*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))) + 4*((a^5 + 5*a^4*b + 10*a^3*b^2 + 1
0*a^2*b^3 + 5*a*b^4 + b^5)*x*cosh(x)^3 - (a^5 - 3*a^4*b - a^3*b^2 + 3*a^2*b^3 - (a^5 + 3*a^4*b + 2*a^3*b^2 - 2
*a^2*b^3 - 3*a*b^4 - b^5)*x)*cosh(x))*sinh(x))/(a^8 - 2*a^7*b - 2*a^6*b^2 + 6*a^5*b^3 - 6*a^3*b^5 + 2*a^2*b^6
+ 2*a*b^7 - b^8 + (a^8 + 2*a^7*b - 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*cosh(x)^4 +
4*(a^8 + 2*a^7*b - 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*cosh(x)*sinh(x)^3 + (a^8 + 2
*a^7*b - 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*sinh(x)^4 + 2*(a^8 - 4*a^6*b^2 + 6*a^4
*b^4 - 4*a^2*b^6 + b^8)*cosh(x)^2 + 2*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8 + 3*(a^8 + 2*a^7*b - 2*a^
6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*cosh(x)^2)*sinh(x)^2 + 4*((a^8 + 2*a^7*b - 2*a^6*b^
2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*cosh(x)^3 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 +
b^8)*cosh(x))*sinh(x))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)**3/(a*cosh(x)+b*sinh(x))**3,x)
[Out]
Timed out
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Giac [B] time = 1.14466, size = 339, normalized size = 3.26 \begin{align*} \frac{{\left (a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac{x}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{3 \, a^{4} e^{\left (4 \, x\right )} + 3 \, a^{3} b e^{\left (4 \, x\right )} + 9 \, a^{2} b^{2} e^{\left (4 \, x\right )} + 9 \, a b^{3} e^{\left (4 \, x\right )} + 2 \, a^{4} e^{\left (2 \, x\right )} + 10 \, a^{3} b e^{\left (2 \, x\right )} + 6 \, a^{2} b^{2} e^{\left (2 \, x\right )} - 18 \, a b^{3} e^{\left (2 \, x\right )} + 3 \, a^{4} + 3 \, a^{3} b - 15 \, a^{2} b^{2} + 9 \, a b^{3}}{2 \,{\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )}{\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a*cosh(x)+b*sinh(x))^3,x, algorithm="giac")
[Out]
(a^3 + 3*a*b^2)*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - x/(a^3 - 3*a^2*b
+ 3*a*b^2 - b^3) - 1/2*(3*a^4*e^(4*x) + 3*a^3*b*e^(4*x) + 9*a^2*b^2*e^(4*x) + 9*a*b^3*e^(4*x) + 2*a^4*e^(2*x)
+ 10*a^3*b*e^(2*x) + 6*a^2*b^2*e^(2*x) - 18*a*b^3*e^(2*x) + 3*a^4 + 3*a^3*b - 15*a^2*b^2 + 9*a*b^3)/((a^5 - a
^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*(a*e^(2*x) + b*e^(2*x) + a - b)^2)