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

Optimal. Leaf size=97 \[ -\frac{a x \left (2 a^2+3 b^2\right )}{2 b^4}-\frac{2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^4}+\frac{\cosh (x) \left (2 \left (a^2+b^2\right )-a b \sinh (x)\right )}{2 b^3}+\frac{\cosh ^3(x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.240779, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2695, 2865, 2735, 2660, 618, 206} \[ -\frac{a x \left (2 a^2+3 b^2\right )}{2 b^4}-\frac{2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^4}+\frac{\cosh (x) \left (2 \left (a^2+b^2\right )-a b \sinh (x)\right )}{2 b^3}+\frac{\cosh ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(a + b*Sinh[x]),x]

[Out]

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

Rule 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

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

Mathematica [C]  time = 3.49875, size = 651, normalized size = 6.71 \[ \frac{b \cosh ^3(x) \left (\sqrt{a+i b} \left (2 \sqrt{b^2} \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}} \left (\sqrt{a-i b} \left (3 a^2+4 b^2\right ) \sqrt{1+i \sinh (x)} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}-3 i \sqrt{b} \left ((1+i) \sqrt{2} a^2-(-1)^{3/4} a b+(1+i) \sqrt{2} b^2\right ) \sin ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{a-i b} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}}{\sqrt{b}}\right )\right )-12 i b \sqrt{a-i b} \left (a^2+b^2\right ) \sqrt{1+i \sinh (x)} \tan ^{-1}\left (\frac{\sqrt{-i b} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}}{\sqrt{i b} \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}}}\right )+2 \left (b^2\right )^{3/2} \sqrt{a-i b} \sqrt{1+i \sinh (x)} \sinh ^2(x) \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}-3 a b \sqrt{b^2} \sqrt{a-i b} \sqrt{1+i \sinh (x)} \sinh (x) \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}\right )+12 \sqrt{b^2} (a+i b) (a-i b)^2 \sqrt{1+i \sinh (x)} \tanh ^{-1}\left (\frac{\sqrt{a-i b} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}}{\sqrt{a+i b} \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}}}\right )\right )}{6 \left (b^2\right )^{3/2} (a-i b)^{3/2} (a+i b)^{3/2} \sqrt{1+i \sinh (x)} \left (-\frac{b (\sinh (x)-i)}{a+i b}\right )^{3/2} \left (-\frac{b (\sinh (x)+i)}{a-i b}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Cosh[x]^3*(12*(a - I*b)^2*(a + I*b)*Sqrt[b^2]*ArcTanh[(Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/
(Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))])]*Sqrt[1 + I*Sinh[x]] + Sqrt[a + I*b]*((-12*I)*Sqrt[a - I
*b]*b*(a^2 + b^2)*ArcTan[(Sqrt[(-I)*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/(Sqrt[I*b]*Sqrt[-((b*(-I + Sinh[x
]))/(a + I*b))])]*Sqrt[1 + I*Sinh[x]] - 3*a*Sqrt[a - I*b]*b*Sqrt[b^2]*Sqrt[1 + I*Sinh[x]]*Sinh[x]*Sqrt[-((b*(-
I + Sinh[x]))/(a + I*b))]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))] + 2*Sqrt[a - I*b]*(b^2)^(3/2)*Sqrt[1 + I*Sinh[x
]]*Sinh[x]^2*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))] + 2*Sqrt[b^2]*Sqrt[-((
b*(-I + Sinh[x]))/(a + I*b))]*((-3*I)*Sqrt[b]*((1 + I)*Sqrt[2]*a^2 - (-1)^(3/4)*a*b + (1 + I)*Sqrt[2]*b^2)*Arc
Sin[((1/2 + I/2)*Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]] + Sqrt[a - I*b]*(3*a^2 + 4*b^2)*
Sqrt[1 + I*Sinh[x]]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))]))))/(6*(a - I*b)^(3/2)*(a + I*b)^(3/2)*(b^2)^(3/2)*Sq
rt[1 + I*Sinh[x]]*(-((b*(-I + Sinh[x]))/(a + I*b)))^(3/2)*(-((b*(I + Sinh[x]))/(a - I*b)))^(3/2))

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Maple [B]  time = 0.032, size = 336, normalized size = 3.5 \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{3}{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{3\,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{3}{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{3\,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 ) }+4\,{\frac{{a}^{2}}{{b}^{2}\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 ) }+2\,{\frac{1}{\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(cosh(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+3/2/b/(tanh(1/2*x)+1)-a^3/b^4*ln(tanh(1/2*x)+1)-3/2*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-3/2/b/(tanh(1/2*x)-1)+a^3/b^4*ln(tanh(1/2*x)-1)+3/2*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))+4*a^2/b^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+
b^2)^(1/2))+2/(a^2+b^2)^(1/2)*arctanh(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(cosh(x)^4/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.93333, size = 1523, normalized size = 15.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b^3*cosh(x)^6 + b^3*sinh(x)^6 - 3*a*b^2*cosh(x)^5 + 3*(2*b^3*cosh(x) - a*b^2)*sinh(x)^5 - 12*(2*a^3 + 3*
a*b^2)*x*cosh(x)^3 + 3*(4*a^2*b + 5*b^3)*cosh(x)^4 + 3*(5*b^3*cosh(x)^2 - 5*a*b^2*cosh(x) + 4*a^2*b + 5*b^3)*s
inh(x)^4 + 3*a*b^2*cosh(x) + 2*(10*b^3*cosh(x)^3 - 15*a*b^2*cosh(x)^2 - 6*(2*a^3 + 3*a*b^2)*x + 6*(4*a^2*b + 5
*b^3)*cosh(x))*sinh(x)^3 + b^3 + 3*(4*a^2*b + 5*b^3)*cosh(x)^2 + 3*(5*b^3*cosh(x)^4 - 10*a*b^2*cosh(x)^3 + 4*a
^2*b + 5*b^3 - 12*(2*a^3 + 3*a*b^2)*x*cosh(x) + 6*(4*a^2*b + 5*b^3)*cosh(x)^2)*sinh(x)^2 + 24*((a^2 + b^2)*cos
h(x)^3 + 3*(a^2 + b^2)*cosh(x)^2*sinh(x) + 3*(a^2 + b^2)*cosh(x)*sinh(x)^2 + (a^2 + b^2)*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*sqr
t(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x)
- b)) + 3*(2*b^3*cosh(x)^5 - 5*a*b^2*cosh(x)^4 - 12*(2*a^3 + 3*a*b^2)*x*cosh(x)^2 + 4*(4*a^2*b + 5*b^3)*cosh(
x)^3 + a*b^2 + 2*(4*a^2*b + 5*b^3)*cosh(x))*sinh(x))/(b^4*cosh(x)^3 + 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*cosh(x)*
sinh(x)^2 + b^4*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(cosh(x)**4/(a+b*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.24197, size = 227, normalized size = 2.34 \begin{align*} \frac{b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} + 15 \, b^{2} e^{x}}{24 \, b^{3}} - \frac{{\left (2 \, a^{3} + 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 + 5 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} + \frac{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b^2*e^(3*x) - 3*a*b*e^(2*x) + 12*a^2*e^x + 15*b^2*e^x)/b^3 - 1/2*(2*a^3 + 3*a*b^2)*x/b^4 + 1/24*(3*a*b^2
*e^x + b^3 + 3*(4*a^2*b + 5*b^3)*e^(2*x))*e^(-3*x)/b^4 + (a^4 + 2*a^2*b^2 + b^4)*log(abs(2*b*e^x + 2*a - 2*sqr
t(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4)