3.7 \(\int \frac {\sinh ^5(x)}{a+b \cosh ^2(x)} \, dx\)

Optimal. Leaf size=54 \[ \frac {(a+b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}-\frac {(a+2 b) \cosh (x)}{b^2}+\frac {\cosh ^3(x)}{3 b} \]

[Out]

-(a+2*b)*cosh(x)/b^2+1/3*cosh(x)^3/b+(a+b)^2*arctan(cosh(x)*b^(1/2)/a^(1/2))/b^(5/2)/a^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3190, 390, 205} \[ -\frac {(a+2 b) \cosh (x)}{b^2}+\frac {(a+b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}+\frac {\cosh ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^5/(a + b*Cosh[x]^2),x]

[Out]

((a + b)^2*ArcTan[(Sqrt[b]*Cosh[x])/Sqrt[a]])/(Sqrt[a]*b^(5/2)) - ((a + 2*b)*Cosh[x])/b^2 + Cosh[x]^3/(3*b)

Rule 205

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3190

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

Rubi steps

\begin {align*} \int \frac {\sinh ^5(x)}{a+b \cosh ^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{a+b x^2} \, dx,x,\cosh (x)\right )\\ &=\operatorname {Subst}\left (\int \left (-\frac {a+2 b}{b^2}+\frac {x^2}{b}+\frac {a^2+2 a b+b^2}{b^2 \left (a+b x^2\right )}\right ) \, dx,x,\cosh (x)\right )\\ &=-\frac {(a+2 b) \cosh (x)}{b^2}+\frac {\cosh ^3(x)}{3 b}+\frac {(a+b)^2 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\cosh (x)\right )}{b^2}\\ &=\frac {(a+b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}-\frac {(a+2 b) \cosh (x)}{b^2}+\frac {\cosh ^3(x)}{3 b}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 120, normalized size = 2.22 \[ \frac {-3 \sqrt {b} (4 a+7 b) \cosh (x)+\frac {12 (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {b}-i \sqrt {a+b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {12 (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {b}+i \sqrt {a+b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a}}+b^{3/2} \cosh (3 x)}{12 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^5/(a + b*Cosh[x]^2),x]

[Out]

((12*(a + b)^2*ArcTan[(Sqrt[b] - I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]])/Sqrt[a] + (12*(a + b)^2*ArcTan[(Sqrt[b] +
I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]])/Sqrt[a] - 3*Sqrt[b]*(4*a + 7*b)*Cosh[x] + b^(3/2)*Cosh[3*x])/(12*b^(5/2))

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fricas [B]  time = 0.54, size = 1064, normalized size = 19.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)^2),x, algorithm="fricas")

[Out]

[1/24*(a*b^2*cosh(x)^6 + 6*a*b^2*cosh(x)*sinh(x)^5 + a*b^2*sinh(x)^6 - 3*(4*a^2*b + 7*a*b^2)*cosh(x)^4 + 3*(5*
a*b^2*cosh(x)^2 - 4*a^2*b - 7*a*b^2)*sinh(x)^4 + 4*(5*a*b^2*cosh(x)^3 - 3*(4*a^2*b + 7*a*b^2)*cosh(x))*sinh(x)
^3 + a*b^2 - 3*(4*a^2*b + 7*a*b^2)*cosh(x)^2 + 3*(5*a*b^2*cosh(x)^4 - 4*a^2*b - 7*a*b^2 - 6*(4*a^2*b + 7*a*b^2
)*cosh(x)^2)*sinh(x)^2 - 12*((a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(x)^2*sinh(x) + 3*(a^2
+ 2*a*b + b^2)*cosh(x)*sinh(x)^2 + (a^2 + 2*a*b + b^2)*sinh(x)^3)*sqrt(-a*b)*log((b*cosh(x)^4 + 4*b*cosh(x)*si
nh(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) - 4*(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (3*cosh(x)^2 + 1)*sinh(x) + cosh(x))*
sqrt(-a*b) + b)/(b*cosh(x)^4 + 4*b*cosh(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)) + 6*(a*b^2*cosh(x)^5 - 2*(4*a^2*b + 7
*a*b^2)*cosh(x)^3 - (4*a^2*b + 7*a*b^2)*cosh(x))*sinh(x))/(a*b^3*cosh(x)^3 + 3*a*b^3*cosh(x)^2*sinh(x) + 3*a*b
^3*cosh(x)*sinh(x)^2 + a*b^3*sinh(x)^3), 1/24*(a*b^2*cosh(x)^6 + 6*a*b^2*cosh(x)*sinh(x)^5 + a*b^2*sinh(x)^6 -
3*(4*a^2*b + 7*a*b^2)*cosh(x)^4 + 3*(5*a*b^2*cosh(x)^2 - 4*a^2*b - 7*a*b^2)*sinh(x)^4 + 4*(5*a*b^2*cosh(x)^3
- 3*(4*a^2*b + 7*a*b^2)*cosh(x))*sinh(x)^3 + a*b^2 - 3*(4*a^2*b + 7*a*b^2)*cosh(x)^2 + 3*(5*a*b^2*cosh(x)^4 -
4*a^2*b - 7*a*b^2 - 6*(4*a^2*b + 7*a*b^2)*cosh(x)^2)*sinh(x)^2 + 24*((a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 +
2*a*b + b^2)*cosh(x)^2*sinh(x) + 3*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^2 + (a^2 + 2*a*b + b^2)*sinh(x)^3)*sqrt
(a*b)*arctan(1/2*sqrt(a*b)*(cosh(x) + sinh(x))/a) - 24*((a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 + 2*a*b + b^2)*
cosh(x)^2*sinh(x) + 3*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^2 + (a^2 + 2*a*b + b^2)*sinh(x)^3)*sqrt(a*b)*arctan(
1/2*(b*cosh(x)^3 + 3*b*cosh(x)*sinh(x)^2 + b*sinh(x)^3 + (4*a + b)*cosh(x) + (3*b*cosh(x)^2 + 4*a + b)*sinh(x)
)*sqrt(a*b)/(a*b)) + 6*(a*b^2*cosh(x)^5 - 2*(4*a^2*b + 7*a*b^2)*cosh(x)^3 - (4*a^2*b + 7*a*b^2)*cosh(x))*sinh(
x))/(a*b^3*cosh(x)^3 + 3*a*b^3*cosh(x)^2*sinh(x) + 3*a*b^3*cosh(x)*sinh(x)^2 + a*b^3*sinh(x)^3)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [a,b]=[-97,37]Warning, need to choose a branch for the root of a polynomial with parameters. Thi
s might be wrong.The choice was done assuming [a,b]=[-81,22]Undef/Unsigned Inf encountered in limitLimit: Max
order reached or unable to make series expansion Error: Bad Argument Value

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maple [B]  time = 0.08, size = 214, normalized size = 3.96 \[ -\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {3}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {3}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\arctan \left (\frac {2 \left (a +b \right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 a +2 b}{4 \sqrt {a b}}\right ) a^{2}}{b^{2} \sqrt {a b}}+\frac {2 \arctan \left (\frac {2 \left (a +b \right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 a +2 b}{4 \sqrt {a b}}\right ) a}{b \sqrt {a b}}+\frac {\arctan \left (\frac {2 \left (a +b \right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 a +2 b}{4 \sqrt {a b}}\right )}{\sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^5/(a+b*cosh(x)^2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b e^{\left (6 \, x\right )} - 3 \, {\left (4 \, a + 7 \, b\right )} e^{\left (4 \, x\right )} - 3 \, {\left (4 \, a + 7 \, b\right )} e^{\left (2 \, x\right )} + b\right )} e^{\left (-3 \, x\right )}}{24 \, b^{2}} + \frac {1}{32} \, \int \frac {64 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (3 \, x\right )} - {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{x}\right )}}{b^{3} e^{\left (4 \, x\right )} + b^{3} + 2 \, {\left (2 \, a b^{2} + b^{3}\right )} e^{\left (2 \, x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

1/24*(b*e^(6*x) - 3*(4*a + 7*b)*e^(4*x) - 3*(4*a + 7*b)*e^(2*x) + b)*e^(-3*x)/b^2 + 1/32*integrate(64*((a^2 +
2*a*b + b^2)*e^(3*x) - (a^2 + 2*a*b + b^2)*e^x)/(b^3*e^(4*x) + b^3 + 2*(2*a*b^2 + b^3)*e^(2*x)), x)

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mupad [B]  time = 1.30, size = 548, normalized size = 10.15 \[ \frac {{\mathrm {e}}^{-3\,x}}{24\,b}+\frac {{\mathrm {e}}^{3\,x}}{24\,b}-\frac {{\mathrm {e}}^{-x}\,\left (4\,a+7\,b\right )}{8\,b^2}+\frac {\left (2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,{\left (a+b\right )}^2\,\sqrt {a\,b^5}}{2\,a\,b^2\,\sqrt {{\left (a+b\right )}^4}}\right )-2\,\mathrm {atan}\left (\frac {a\,b^6\,{\mathrm {e}}^x\,\left (\frac {4\,\left (6\,a^2\,b^4\,\sqrt {a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4}+6\,a^3\,b^3\,\sqrt {a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4}+2\,a^4\,b^2\,\sqrt {a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4}+2\,a\,b^5\,\sqrt {a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4}\right )}{a^2\,b^{11}\,{\left (a+b\right )}^2}+\frac {2\,\left (a^5\,\sqrt {a\,b^5}+b^5\,\sqrt {a\,b^5}+5\,a\,b^4\,\sqrt {a\,b^5}+5\,a^4\,b\,\sqrt {a\,b^5}+10\,a^2\,b^3\,\sqrt {a\,b^5}+10\,a^3\,b^2\,\sqrt {a\,b^5}\right )}{a^2\,b^8\,\sqrt {a\,b^5}\,\sqrt {{\left (a+b\right )}^4}}\right )\,\sqrt {a\,b^5}}{4\,a^3+12\,a^2\,b+12\,a\,b^2+4\,b^3}+\frac {2\,{\mathrm {e}}^{3\,x}\,\left (a^5\,\sqrt {a\,b^5}+b^5\,\sqrt {a\,b^5}+5\,a\,b^4\,\sqrt {a\,b^5}+5\,a^4\,b\,\sqrt {a\,b^5}+10\,a^2\,b^3\,\sqrt {a\,b^5}+10\,a^3\,b^2\,\sqrt {a\,b^5}\right )}{a\,b^2\,\sqrt {{\left (a+b\right )}^4}\,\left (4\,a^3+12\,a^2\,b+12\,a\,b^2+4\,b^3\right )}\right )\right )\,\sqrt {a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4}}{2\,\sqrt {a\,b^5}}-\frac {{\mathrm {e}}^x\,\left (4\,a+7\,b\right )}{8\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^5/(a + b*cosh(x)^2),x)

[Out]

exp(-3*x)/(24*b) + exp(3*x)/(24*b) - (exp(-x)*(4*a + 7*b))/(8*b^2) + ((2*atan((exp(x)*(a + b)^2*(a*b^5)^(1/2))
/(2*a*b^2*((a + b)^4)^(1/2))) - 2*atan((a*b^6*exp(x)*((4*(6*a^2*b^4*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2
)^(1/2) + 6*a^3*b^3*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)^(1/2) + 2*a^4*b^2*(4*a*b^3 + 4*a^3*b + a^4 + b
^4 + 6*a^2*b^2)^(1/2) + 2*a*b^5*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)^(1/2)))/(a^2*b^11*(a + b)^2) + (2*
(a^5*(a*b^5)^(1/2) + b^5*(a*b^5)^(1/2) + 5*a*b^4*(a*b^5)^(1/2) + 5*a^4*b*(a*b^5)^(1/2) + 10*a^2*b^3*(a*b^5)^(1
/2) + 10*a^3*b^2*(a*b^5)^(1/2)))/(a^2*b^8*(a*b^5)^(1/2)*((a + b)^4)^(1/2)))*(a*b^5)^(1/2))/(12*a*b^2 + 12*a^2*
b + 4*a^3 + 4*b^3) + (2*exp(3*x)*(a^5*(a*b^5)^(1/2) + b^5*(a*b^5)^(1/2) + 5*a*b^4*(a*b^5)^(1/2) + 5*a^4*b*(a*b
^5)^(1/2) + 10*a^2*b^3*(a*b^5)^(1/2) + 10*a^3*b^2*(a*b^5)^(1/2)))/(a*b^2*((a + b)^4)^(1/2)*(12*a*b^2 + 12*a^2*
b + 4*a^3 + 4*b^3))))*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)^(1/2))/(2*(a*b^5)^(1/2)) - (exp(x)*(4*a + 7*
b))/(8*b^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(sinh(x)**5/(a+b*cosh(x)**2),x)

[Out]

Timed out

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