3.23 \(\int \frac {\cosh ^4(x)}{a+b \cosh ^2(x)} \, dx\)

Optimal. Leaf size=59 \[ \frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{b^2 \sqrt {a+b}}-\frac {x (2 a-b)}{2 b^2}+\frac {\sinh (x) \cosh (x)}{2 b} \]

[Out]

-1/2*(2*a-b)*x/b^2+1/2*cosh(x)*sinh(x)/b+a^(3/2)*arctanh(a^(1/2)*tanh(x)/(a+b)^(1/2))/b^2/(a+b)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3187, 470, 522, 206, 208} \[ \frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{b^2 \sqrt {a+b}}-\frac {x (2 a-b)}{2 b^2}+\frac {\sinh (x) \cosh (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(a + b*Cosh[x]^2),x]

[Out]

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

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3187

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/2 + p
+ 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 52, normalized size = 0.88 \[ \frac {\frac {4 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+2 x (b-2 a)+b \sinh (2 x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^4/(a + b*Cosh[x]^2),x]

[Out]

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

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fricas [B]  time = 0.58, size = 573, normalized size = 9.71 \[ \left [\frac {b \cosh \relax (x)^{4} + 4 \, b \cosh \relax (x) \sinh \relax (x)^{3} + b \sinh \relax (x)^{4} - 4 \, {\left (2 \, a - b\right )} x \cosh \relax (x)^{2} + 2 \, {\left (3 \, b \cosh \relax (x)^{2} - 2 \, {\left (2 \, a - b\right )} x\right )} \sinh \relax (x)^{2} + 4 \, {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2}\right )} \sqrt {\frac {a}{a + b}} \log \left (\frac {b^{2} \cosh \relax (x)^{4} + 4 \, b^{2} \cosh \relax (x) \sinh \relax (x)^{3} + b^{2} \sinh \relax (x)^{4} + 2 \, {\left (2 \, a b + b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, b^{2} \cosh \relax (x)^{2} + 2 \, a b + b^{2}\right )} \sinh \relax (x)^{2} + 8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (b^{2} \cosh \relax (x)^{3} + {\left (2 \, a b + b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x) - 4 \, {\left ({\left (a b + b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a b + b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a b + b^{2}\right )} \sinh \relax (x)^{2} + 2 \, a^{2} + 3 \, a b + b^{2}\right )} \sqrt {\frac {a}{a + b}}}{b \cosh \relax (x)^{4} + 4 \, b \cosh \relax (x) \sinh \relax (x)^{3} + b \sinh \relax (x)^{4} + 2 \, {\left (2 \, a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, b \cosh \relax (x)^{2} + 2 \, a + b\right )} \sinh \relax (x)^{2} + 4 \, {\left (b \cosh \relax (x)^{3} + {\left (2 \, a + b\right )} \cosh \relax (x)\right )} \sinh \relax (x) + b}\right ) + 4 \, {\left (b \cosh \relax (x)^{3} - 2 \, {\left (2 \, a - b\right )} x \cosh \relax (x)\right )} \sinh \relax (x) - b}{8 \, {\left (b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2}\right )}}, \frac {b \cosh \relax (x)^{4} + 4 \, b \cosh \relax (x) \sinh \relax (x)^{3} + b \sinh \relax (x)^{4} - 4 \, {\left (2 \, a - b\right )} x \cosh \relax (x)^{2} + 2 \, {\left (3 \, b \cosh \relax (x)^{2} - 2 \, {\left (2 \, a - b\right )} x\right )} \sinh \relax (x)^{2} + 8 \, {\left (a \cosh \relax (x)^{2} + 2 \, a \cosh \relax (x) \sinh \relax (x) + a \sinh \relax (x)^{2}\right )} \sqrt {-\frac {a}{a + b}} \arctan \left (\frac {{\left (b \cosh \relax (x)^{2} + 2 \, b \cosh \relax (x) \sinh \relax (x) + b \sinh \relax (x)^{2} + 2 \, a + b\right )} \sqrt {-\frac {a}{a + b}}}{2 \, a}\right ) + 4 \, {\left (b \cosh \relax (x)^{3} - 2 \, {\left (2 \, a - b\right )} x \cosh \relax (x)\right )} \sinh \relax (x) - b}{8 \, {\left (b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+b*cosh(x)^2),x, algorithm="fricas")

[Out]

[1/8*(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 4*(2*a - b)*x*cosh(x)^2 + 2*(3*b*cosh(x)^2 - 2*(2*a
- b)*x)*sinh(x)^2 + 4*(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2)*sqrt(a/(a + b))*log((b^2*cosh(x)^4 + 4
*b^2*cosh(x)*sinh(x)^3 + b^2*sinh(x)^4 + 2*(2*a*b + b^2)*cosh(x)^2 + 2*(3*b^2*cosh(x)^2 + 2*a*b + b^2)*sinh(x)
^2 + 8*a^2 + 8*a*b + b^2 + 4*(b^2*cosh(x)^3 + (2*a*b + b^2)*cosh(x))*sinh(x) - 4*((a*b + b^2)*cosh(x)^2 + 2*(a
*b + b^2)*cosh(x)*sinh(x) + (a*b + b^2)*sinh(x)^2 + 2*a^2 + 3*a*b + b^2)*sqrt(a/(a + b)))/(b*cosh(x)^4 + 4*b*c
osh(x)*sinh(x)^3 + b*sinh(x)^4 + 2*(2*a + b)*cosh(x)^2 + 2*(3*b*cosh(x)^2 + 2*a + b)*sinh(x)^2 + 4*(b*cosh(x)^
3 + (2*a + b)*cosh(x))*sinh(x) + b)) + 4*(b*cosh(x)^3 - 2*(2*a - b)*x*cosh(x))*sinh(x) - b)/(b^2*cosh(x)^2 + 2
*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2), 1/8*(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 4*(2*a - b)*x*
cosh(x)^2 + 2*(3*b*cosh(x)^2 - 2*(2*a - b)*x)*sinh(x)^2 + 8*(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2)*
sqrt(-a/(a + b))*arctan(1/2*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + 2*a + b)*sqrt(-a/(a + b))/a) +
4*(b*cosh(x)^3 - 2*(2*a - b)*x*cosh(x))*sinh(x) - b)/(b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2)]

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giac [B]  time = 0.13, size = 95, normalized size = 1.61 \[ \frac {a^{2} \arctan \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} b^{2}} - \frac {{\left (2 \, a - b\right )} x}{2 \, b^{2}} + \frac {e^{\left (2 \, x\right )}}{8 \, b} + \frac {{\left (4 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} - b\right )} e^{\left (-2 \, x\right )}}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

a^2*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*b^2) - 1/2*(2*a - b)*x/b^2 + 1/8*e^(2
*x)/b + 1/8*(4*a*e^(2*x) - 2*b*e^(2*x) - b)*e^(-2*x)/b^2

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maple [B]  time = 0.10, size = 188, normalized size = 3.19 \[ \frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b}-\frac {a^{\frac {3}{2}} \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \sqrt {a}\, \tanh \left (\frac {x}{2}\right )-\sqrt {a +b}\right )}{2 b^{2} \sqrt {a +b}}+\frac {a^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \sqrt {a}\, \tanh \left (\frac {x}{2}\right )+\sqrt {a +b}\right )}{2 b^{2} \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(a+b*cosh(x)^2),x)

[Out]

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

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maxima [B]  time = 0.56, size = 347, normalized size = 5.88 \[ -\frac {{\left (2 \, a + b\right )} \log \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{4 \, \sqrt {{\left (a + b\right )} a} b} - \frac {3 \, \log \left (\frac {b e^{\left (-2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (-2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{16 \, \sqrt {{\left (a + b\right )} a}} - \frac {{\left (2 \, a + b\right )} x}{b^{2}} + \frac {x}{b} + \frac {e^{\left (2 \, x\right )}}{8 \, b} - \frac {e^{\left (-2 \, x\right )}}{8 \, b} + \frac {{\left (2 \, a + b\right )} \log \left (b e^{\left (4 \, x\right )} + 2 \, {\left (2 \, a + b\right )} e^{\left (2 \, x\right )} + b\right )}{8 \, b^{2}} - \frac {{\left (2 \, a + b\right )} \log \left (2 \, {\left (2 \, a + b\right )} e^{\left (-2 \, x\right )} + b e^{\left (-4 \, x\right )} + b\right )}{8 \, b^{2}} + \frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \log \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{32 \, \sqrt {{\left (a + b\right )} a} b^{2}} - \frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \log \left (\frac {b e^{\left (-2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (-2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{32 \, \sqrt {{\left (a + b\right )} a} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

-1/4*(2*a + b)*log((b*e^(2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(2*x) + 2*a + b + 2*sqrt((a + b)*a)))/(sqrt(
(a + b)*a)*b) - 3/16*log((b*e^(-2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(-2*x) + 2*a + b + 2*sqrt((a + b)*a))
)/sqrt((a + b)*a) - (2*a + b)*x/b^2 + x/b + 1/8*e^(2*x)/b - 1/8*e^(-2*x)/b + 1/8*(2*a + b)*log(b*e^(4*x) + 2*(
2*a + b)*e^(2*x) + b)/b^2 - 1/8*(2*a + b)*log(2*(2*a + b)*e^(-2*x) + b*e^(-4*x) + b)/b^2 + 1/32*(8*a^2 + 8*a*b
+ b^2)*log((b*e^(2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(2*x) + 2*a + b + 2*sqrt((a + b)*a)))/(sqrt((a + b)
*a)*b^2) - 1/32*(8*a^2 + 8*a*b + b^2)*log((b*e^(-2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(-2*x) + 2*a + b + 2
*sqrt((a + b)*a)))/(sqrt((a + b)*a)*b^2)

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mupad [B]  time = 1.17, size = 142, normalized size = 2.41 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b}-\frac {x\,\left (2\,a-b\right )}{2\,b^2}+\frac {a^{3/2}\,\ln \left (-\frac {4\,a^2\,{\mathrm {e}}^{2\,x}}{b^3}-\frac {2\,a^{3/2}\,\left (b+2\,a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{b^3\,\sqrt {a+b}}\right )}{2\,b^2\,\sqrt {a+b}}-\frac {a^{3/2}\,\ln \left (\frac {2\,a^{3/2}\,\left (b+2\,a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{b^3\,\sqrt {a+b}}-\frac {4\,a^2\,{\mathrm {e}}^{2\,x}}{b^3}\right )}{2\,b^2\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(a + b*cosh(x)^2),x)

[Out]

exp(2*x)/(8*b) - exp(-2*x)/(8*b) - (x*(2*a - b))/(2*b^2) + (a^(3/2)*log(- (4*a^2*exp(2*x))/b^3 - (2*a^(3/2)*(b
+ 2*a*exp(2*x) + b*exp(2*x)))/(b^3*(a + b)^(1/2))))/(2*b^2*(a + b)^(1/2)) - (a^(3/2)*log((2*a^(3/2)*(b + 2*a*
exp(2*x) + b*exp(2*x)))/(b^3*(a + b)^(1/2)) - (4*a^2*exp(2*x))/b^3))/(2*b^2*(a + b)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**4/(a+b*cosh(x)**2),x)

[Out]

Timed out

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