3.81 \(\int \frac {\sinh ^3(x)}{a+b \tanh (x)} \, dx\)
Optimal. Leaf size=137 \[ -\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a \cosh (x)}{a^2-b^2}-\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a^3 b \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]
[Out]
-a^3*b*arctan((b*cosh(x)+a*sinh(x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)-a*b^2*cosh(x)/(a^2-b^2)^2-a*cosh(x)/(a^2-
b^2)+1/3*a*cosh(x)^3/(a^2-b^2)+a^2*b*sinh(x)/(a^2-b^2)^2-1/3*b*sinh(x)^3/(a^2-b^2)
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Rubi [A] time = 0.28, antiderivative size = 137, normalized size of antiderivative = 1.00,
number of steps used = 10, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used
= {3518, 3109, 2564, 30, 2633, 3099, 3074, 206, 2638} \[ -\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a \cosh (x)}{a^2-b^2}-\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a^3 b \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[x]^3/(a + b*Tanh[x]),x]
[Out]
-((a^3*b*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2)) - (a*b^2*Cosh[x])/(a^2 - b^2)^2 -
(a*Cosh[x])/(a^2 - b^2) + (a*Cosh[x]^3)/(3*(a^2 - b^2)) + (a^2*b*Sinh[x])/(a^2 - b^2)^2 - (b*Sinh[x]^3)/(3*(a
^2 - b^2))
Rule 30
Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2564
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])
Rule 2633
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Rule 2638
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Rule 3074
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]
Rule 3099
Int[sin[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(a*Sin[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a^2/(a^2 + b^2), Int[Sin[c + d*x]^(m - 2)/
(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x] + Dist[b/(a^2 + b^2), Int[Sin[c + d*x]^(m - 1), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]
Rule 3109
Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]
Rule 3518
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))
Rubi steps
\begin {align*} \int \frac {\sinh ^3(x)}{a+b \tanh (x)} \, dx &=\int \frac {\cosh (x) \sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx\\ &=\frac {a \int \sinh ^3(x) \, dx}{a^2-b^2}-\frac {b \int \cosh (x) \sinh ^2(x) \, dx}{a^2-b^2}+\frac {(a b) \int \frac {\sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=\frac {a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {\left (a^3 b\right ) \int \frac {1}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {\left (a b^2\right ) \int \sinh (x) \, dx}{\left (a^2-b^2\right )^2}-\frac {a \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (x)\right )}{a^2-b^2}-\frac {(i b) \operatorname {Subst}\left (\int x^2 \, dx,x,i \sinh (x)\right )}{a^2-b^2}\\ &=-\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {\left (i a^3 b\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )}{\left (a^2-b^2\right )^2}\\ &=-\frac {a^3 b \tan ^{-1}\left (\frac {b \cosh (x)+a \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {a \cosh (x)}{a^2-b^2}+\frac {a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \sinh ^3(x)}{3 \left (a^2-b^2\right )}\\ \end {align*}
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Mathematica [A] time = 1.23, size = 180, normalized size = 1.31 \[ \frac {-3 a \sqrt {a-b} \sqrt {a+b} \left (3 a^2+b^2\right ) \cosh (x)+a \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right ) \cosh (3 x)+b \left (-24 a^3 \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )+3 \sqrt {a-b} \sqrt {a+b} \left (5 a^2-b^2\right ) \sinh (x)-\sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right ) \sinh (3 x)\right )}{12 (a-b)^{5/2} (a+b)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[x]^3/(a + b*Tanh[x]),x]
[Out]
(-3*a*Sqrt[a - b]*Sqrt[a + b]*(3*a^2 + b^2)*Cosh[x] + a*Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*Cosh[3*x] + b*(-24
*a^3*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])] + 3*Sqrt[a - b]*Sqrt[a + b]*(5*a^2 - b^2)*Sinh[x] - S
qrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*Sinh[3*x]))/(12*(a - b)^(5/2)*(a + b)^(5/2))
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fricas [B] time = 0.78, size = 1861, normalized size = 13.58 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a+b*tanh(x)),x, algorithm="fricas")
[Out]
[1/24*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3
+ a*b^4 - b^5)*cosh(x)*sinh(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 + a^5 + a^4*b
- 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 - 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^4 -
3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b
^5)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(3*a^5 - 5*a
^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 -
a*b^4 + b^5)*cosh(x)^2 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2
+ 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 + 6*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^2)*s
inh(x)^2 - 24*(a^3*b*cosh(x)^3 + 3*a^3*b*cosh(x)^2*sinh(x) + 3*a^3*b*cosh(x)*sinh(x)^2 + a^3*b*sinh(x)^3)*sqrt
(-a^2 + b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + 2*sqrt(-a^2 + b^2)*(cosh
(x) + sinh(x)) - a + b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)) + 6*((a^5
- a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 - 2*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4
- b^5)*cosh(x)^3 - (3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^
2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2
+ 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^3), 1/24*((a^5 - a^4*b - 2*
a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)*s
inh(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 + a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3
+ a*b^4 + b^5 - 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^4 - 3*(3*a^5 - 5*a^4*b - 2*
a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^4
+ 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2
*b^3 - a*b^4 - b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5)*cosh(x)^2 -
3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b
^5)*cosh(x)^4 + 6*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^2)*sinh(x)^2 + 48*(a^3*b*cos
h(x)^3 + 3*a^3*b*cosh(x)^2*sinh(x) + 3*a^3*b*cosh(x)*sinh(x)^2 + a^3*b*sinh(x)^3)*sqrt(a^2 - b^2)*arctan(sqrt(
a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + 6*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(
x)^5 - 2*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^3 - (3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*
a^2*b^3 - a*b^4 + b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 +
3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b
^2 + 3*a^2*b^4 - b^6)*sinh(x)^3)]
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giac [A] time = 0.14, size = 163, normalized size = 1.19 \[ -\frac {2 \, a^{3} b \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (9 \, a e^{\left (2 \, x\right )} - 3 \, b e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-3 \, x\right )}}{24 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {a^{2} e^{\left (3 \, x\right )} + 2 \, a b e^{\left (3 \, x\right )} + b^{2} e^{\left (3 \, x\right )} - 9 \, a^{2} e^{x} - 12 \, a b e^{x} - 3 \, b^{2} e^{x}}{24 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a+b*tanh(x)),x, algorithm="giac")
[Out]
-2*a^3*b*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) - 1/24*(9*a*e^(2*x)
- 3*b*e^(2*x) - a + b)*e^(-3*x)/(a^2 - 2*a*b + b^2) + 1/24*(a^2*e^(3*x) + 2*a*b*e^(3*x) + b^2*e^(3*x) - 9*a^2
*e^x - 12*a*b*e^x - 3*b^2*e^x)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)
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maple [A] time = 0.12, size = 166, normalized size = 1.21 \[ -\frac {2 a^{3} b \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {16}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}-\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {8}{\left (16 a -16 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {16}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3} \left (16 a -16 b \right )}-\frac {a}{2 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(x)^3/(a+b*tanh(x)),x)
[Out]
-2*a^3*b/(a-b)^2/(a+b)^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/2*x)+2*b)/(a^2-b^2)^(1/2))-16/3/(tanh(1/2*x)-1
)^3/(16*a+16*b)-8/(16*a+16*b)/(tanh(1/2*x)-1)^2+1/2/(a+b)^2/(tanh(1/2*x)-1)*a-8/(16*a-16*b)/(tanh(1/2*x)+1)^2+
16/3/(tanh(1/2*x)+1)^3/(16*a-16*b)-1/2/(a-b)^2/(tanh(1/2*x)+1)*a
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)^3/(a+b*tanh(x)),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?
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mupad [B] time = 1.46, size = 261, normalized size = 1.91 \[ \frac {{\mathrm {e}}^{-3\,x}}{24\,a-24\,b}+\frac {{\mathrm {e}}^{3\,x}}{24\,a+24\,b}-\frac {{\mathrm {e}}^x\,\left (3\,a+b\right )}{8\,{\left (a+b\right )}^2}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a-b\right )}{8\,{\left (a-b\right )}^2}-\frac {2\,\mathrm {atan}\left (\frac {a^3\,b\,{\mathrm {e}}^x\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}{a^5\,\sqrt {a^6\,b^2}-b^5\,\sqrt {a^6\,b^2}+2\,a^2\,b^3\,\sqrt {a^6\,b^2}-2\,a^3\,b^2\,\sqrt {a^6\,b^2}+a\,b^4\,\sqrt {a^6\,b^2}-a^4\,b\,\sqrt {a^6\,b^2}}\right )\,\sqrt {a^6\,b^2}}{\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(x)^3/(a + b*tanh(x)),x)
[Out]
exp(-3*x)/(24*a - 24*b) + exp(3*x)/(24*a + 24*b) - (exp(x)*(3*a + b))/(8*(a + b)^2) - (exp(-x)*(3*a - b))/(8*(
a - b)^2) - (2*atan((a^3*b*exp(x)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2))/(a^5*
(a^6*b^2)^(1/2) - b^5*(a^6*b^2)^(1/2) + 2*a^2*b^3*(a^6*b^2)^(1/2) - 2*a^3*b^2*(a^6*b^2)^(1/2) + a*b^4*(a^6*b^2
)^(1/2) - a^4*b*(a^6*b^2)^(1/2)))*(a^6*b^2)^(1/2))/(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*
b^2)^(1/2)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}{\relax (x )}}{a + b \tanh {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(x)**3/(a+b*tanh(x)),x)
[Out]
Integral(sinh(x)**3/(a + b*tanh(x)), x)
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