3.72 \(\int \frac{\sinh ^4(x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=108 \[ -\frac{a x \left (2 a^2-b^2\right )}{2 b^4}-\frac{\left (2-\frac{3 a^2}{b^2}\right ) \cosh (x)}{3 b}-\frac{2 a^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^4 \sqrt{a^2+b^2}}-\frac{a \sinh (x) \cosh (x)}{2 b^2}+\frac{\sinh ^2(x) \cosh (x)}{3 b} \]

[Out]

-(a*(2*a^2 - b^2)*x)/(2*b^4) - (2*a^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^4*Sqrt[a^2 + b^2]) - ((2
- (3*a^2)/b^2)*Cosh[x])/(3*b) - (a*Cosh[x]*Sinh[x])/(2*b^2) + (Cosh[x]*Sinh[x]^2)/(3*b)

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Rubi [A]  time = 0.31549, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2793, 3049, 3023, 2735, 2660, 618, 206} \[ -\frac{a x \left (2 a^2-b^2\right )}{2 b^4}-\frac{\left (2-\frac{3 a^2}{b^2}\right ) \cosh (x)}{3 b}-\frac{2 a^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^4 \sqrt{a^2+b^2}}-\frac{a \sinh (x) \cosh (x)}{2 b^2}+\frac{\sinh ^2(x) \cosh (x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(a + b*Sinh[x]),x]

[Out]

-(a*(2*a^2 - b^2)*x)/(2*b^4) - (2*a^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^4*Sqrt[a^2 + b^2]) - ((2
- (3*a^2)/b^2)*Cosh[x])/(3*b) - (a*Cosh[x]*Sinh[x])/(2*b^2) + (Cosh[x]*Sinh[x]^2)/(3*b)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

\begin{align*} \int \frac{\sinh ^4(x)}{a+b \sinh (x)} \, dx &=\frac{\cosh (x) \sinh ^2(x)}{3 b}-\frac{\int \frac{\sinh (x) \left (2 a+2 b \sinh (x)+3 a \sinh ^2(x)\right )}{a+b \sinh (x)} \, dx}{3 b}\\ &=-\frac{a \cosh (x) \sinh (x)}{2 b^2}+\frac{\cosh (x) \sinh ^2(x)}{3 b}-\frac{\int \frac{-3 a^2+a b \sinh (x)-2 \left (3 a^2-2 b^2\right ) \sinh ^2(x)}{a+b \sinh (x)} \, dx}{6 b^2}\\ &=\frac{\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac{a \cosh (x) \sinh (x)}{2 b^2}+\frac{\cosh (x) \sinh ^2(x)}{3 b}-\frac{i \int \frac{3 i a^2 b-3 i a \left (2 a^2-b^2\right ) \sinh (x)}{a+b \sinh (x)} \, dx}{6 b^3}\\ &=-\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac{\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac{a \cosh (x) \sinh (x)}{2 b^2}+\frac{\cosh (x) \sinh ^2(x)}{3 b}+\frac{a^4 \int \frac{1}{a+b \sinh (x)} \, dx}{b^4}\\ &=-\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac{\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac{a \cosh (x) \sinh (x)}{2 b^2}+\frac{\cosh (x) \sinh ^2(x)}{3 b}+\frac{\left (2 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{b^4}\\ &=-\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac{\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac{a \cosh (x) \sinh (x)}{2 b^2}+\frac{\cosh (x) \sinh ^2(x)}{3 b}-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )}{b^4}\\ &=-\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac{2 a^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^4 \sqrt{a^2+b^2}}+\frac{\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac{a \cosh (x) \sinh (x)}{2 b^2}+\frac{\cosh (x) \sinh ^2(x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.431195, size = 105, normalized size = 0.97 \[ \frac{3 b \left (4 a^2-3 b^2\right ) \cosh (x)+3 a \left (\frac{8 a^3 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-4 a^2 x+2 b^2 x-b^2 \sinh (2 x)\right )+b^3 \cosh (3 x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(a + b*Sinh[x]),x]

[Out]

(3*b*(4*a^2 - 3*b^2)*Cosh[x] + b^3*Cosh[3*x] + 3*a*(-4*a^2*x + 2*b^2*x + (8*a^3*ArcTan[(b - a*Tanh[x/2])/Sqrt[
-a^2 - b^2]])/Sqrt[-a^2 - b^2] - b^2*Sinh[2*x]))/(12*b^4)

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Maple [B]  time = 0.027, size = 262, normalized size = 2.4 \begin{align*}{\frac{1}{3\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{\frac{a}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{{a}^{2}}{{b}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{a}^{3}}{{b}^{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{a}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{3\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{\frac{a}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{{a}^{2}}{{b}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{a}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{a}^{3}}{{b}^{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{a}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{{a}^{4}}{{b}^{4}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a+b*sinh(x)),x)

[Out]

1/3/b/(tanh(1/2*x)+1)^3+1/2/b^2/(tanh(1/2*x)+1)^2*a-1/2/b/(tanh(1/2*x)+1)^2+1/b^3/(tanh(1/2*x)+1)*a^2-1/2/b^2/
(tanh(1/2*x)+1)*a-1/2/b/(tanh(1/2*x)+1)-a^3/b^4*ln(tanh(1/2*x)+1)+1/2*a/b^2*ln(tanh(1/2*x)+1)-1/3/b/(tanh(1/2*
x)-1)^3-1/2/b^2/(tanh(1/2*x)-1)^2*a-1/2/b/(tanh(1/2*x)-1)^2-1/b^3/(tanh(1/2*x)-1)*a^2-1/2/b^2/(tanh(1/2*x)-1)*
a+1/2/b/(tanh(1/2*x)-1)+a^3/b^4*ln(tanh(1/2*x)-1)-1/2*a/b^2*ln(tanh(1/2*x)-1)+2*a^4/b^4/(a^2+b^2)^(1/2)*arctan
h(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.22059, size = 1947, normalized size = 18.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

1/24*((a^2*b^3 + b^5)*cosh(x)^6 + (a^2*b^3 + b^5)*sinh(x)^6 - 3*(a^3*b^2 + a*b^4)*cosh(x)^5 - 3*(a^3*b^2 + a*b
^4 - 2*(a^2*b^3 + b^5)*cosh(x))*sinh(x)^5 + a^2*b^3 + b^5 - 12*(2*a^5 + a^3*b^2 - a*b^4)*x*cosh(x)^3 + 3*(4*a^
4*b + a^2*b^3 - 3*b^5)*cosh(x)^4 + 3*(4*a^4*b + a^2*b^3 - 3*b^5 + 5*(a^2*b^3 + b^5)*cosh(x)^2 - 5*(a^3*b^2 + a
*b^4)*cosh(x))*sinh(x)^4 + 2*(10*(a^2*b^3 + b^5)*cosh(x)^3 - 15*(a^3*b^2 + a*b^4)*cosh(x)^2 - 6*(2*a^5 + a^3*b
^2 - a*b^4)*x + 6*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x))*sinh(x)^3 + 3*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x)^2 + 3
*(4*a^4*b + a^2*b^3 - 3*b^5 + 5*(a^2*b^3 + b^5)*cosh(x)^4 - 10*(a^3*b^2 + a*b^4)*cosh(x)^3 - 12*(2*a^5 + a^3*b
^2 - a*b^4)*x*cosh(x) + 6*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x)^2)*sinh(x)^2 + 24*(a^4*cosh(x)^3 + 3*a^4*cosh(x)
^2*sinh(x) + 3*a^4*cosh(x)*sinh(x)^2 + a^4*sinh(x)^3)*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a
*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*c
osh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) + 3*(a^3*b^2 + a*b^4)*cosh(x) + 3*(2*(a
^2*b^3 + b^5)*cosh(x)^5 + a^3*b^2 + a*b^4 - 5*(a^3*b^2 + a*b^4)*cosh(x)^4 - 12*(2*a^5 + a^3*b^2 - a*b^4)*x*cos
h(x)^2 + 4*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x)^3 + 2*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x))*sinh(x))/((a^2*b^4 +
b^6)*cosh(x)^3 + 3*(a^2*b^4 + b^6)*cosh(x)^2*sinh(x) + 3*(a^2*b^4 + b^6)*cosh(x)*sinh(x)^2 + (a^2*b^4 + b^6)*
sinh(x)^3)

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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)**4/(a+b*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.34664, size = 211, normalized size = 1.95 \begin{align*} \frac{a^{4} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} b^{4}} + \frac{b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} - 9 \, b^{2} e^{x}}{24 \, b^{3}} - \frac{{\left (2 \, a^{3} - a b^{2}\right )} x}{2 \, b^{4}} + \frac{{\left (3 \, a b^{2} e^{x} + b^{3} + 3 \,{\left (4 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*sinh(x)),x, algorithm="giac")

[Out]

a^4*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4) +
1/24*(b^2*e^(3*x) - 3*a*b*e^(2*x) + 12*a^2*e^x - 9*b^2*e^x)/b^3 - 1/2*(2*a^3 - a*b^2)*x/b^4 + 1/24*(3*a*b^2*e
^x + b^3 + 3*(4*a^2*b - 3*b^3)*e^(2*x))*e^(-3*x)/b^4