3.13 \(\int \frac{\sinh ^6(x)}{a+b \cosh ^2(x)} \, dx\)

Optimal. Leaf size=88 \[ \frac{x \left (8 a^2+20 a b+15 b^2\right )}{8 b^3}-\frac{(a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} b^3}-\frac{(4 a+7 b) \sinh (x) \cosh (x)}{8 b^2}+\frac{\sinh ^3(x) \cosh (x)}{4 b} \]

[Out]

((8*a^2 + 20*a*b + 15*b^2)*x)/(8*b^3) - ((a + b)^(5/2)*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(Sqrt[a]*b^3) -
((4*a + 7*b)*Cosh[x]*Sinh[x])/(8*b^2) + (Cosh[x]*Sinh[x]^3)/(4*b)

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Rubi [A]  time = 0.168095, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3191, 414, 527, 522, 206, 208} \[ \frac{x \left (8 a^2+20 a b+15 b^2\right )}{8 b^3}-\frac{(a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} b^3}-\frac{(4 a+7 b) \sinh (x) \cosh (x)}{8 b^2}+\frac{\sinh ^3(x) \cosh (x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^6/(a + b*Cosh[x]^2),x]

[Out]

((8*a^2 + 20*a*b + 15*b^2)*x)/(8*b^3) - ((a + b)^(5/2)*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(Sqrt[a]*b^3) -
((4*a + 7*b)*Cosh[x]*Sinh[x])/(8*b^2) + (Cosh[x]*Sinh[x]^3)/(4*b)

Rule 3191

Int[cos[(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/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

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

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 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]

Rubi steps

\begin{align*} \int \frac{\sinh ^6(x)}{a+b \cosh ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a-(a+b) x^2\right )} \, dx,x,\coth (x)\right )\\ &=\frac{\cosh (x) \sinh ^3(x)}{4 b}+\frac{\operatorname{Subst}\left (\int \frac{-a-4 b-3 (a+b) x^2}{\left (1-x^2\right )^2 \left (a+(-a-b) x^2\right )} \, dx,x,\coth (x)\right )}{4 b}\\ &=-\frac{(4 a+7 b) \cosh (x) \sinh (x)}{8 b^2}+\frac{\cosh (x) \sinh ^3(x)}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{4 a^2+9 a b+8 b^2+(a+b) (4 a+7 b) x^2}{\left (1-x^2\right ) \left (a+(-a-b) x^2\right )} \, dx,x,\coth (x)\right )}{8 b^2}\\ &=-\frac{(4 a+7 b) \cosh (x) \sinh (x)}{8 b^2}+\frac{\cosh (x) \sinh ^3(x)}{4 b}-\frac{(a+b)^3 \operatorname{Subst}\left (\int \frac{1}{a+(-a-b) x^2} \, dx,x,\coth (x)\right )}{b^3}+\frac{\left (8 a^2+20 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\coth (x)\right )}{8 b^3}\\ &=\frac{\left (8 a^2+20 a b+15 b^2\right ) x}{8 b^3}-\frac{(a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} b^3}-\frac{(4 a+7 b) \cosh (x) \sinh (x)}{8 b^2}+\frac{\cosh (x) \sinh ^3(x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.162487, size = 76, normalized size = 0.86 \[ \frac{4 x \left (8 a^2+20 a b+15 b^2\right )-8 b (a+2 b) \sinh (2 x)-\frac{32 (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a}}+b^2 \sinh (4 x)}{32 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^6/(a + b*Cosh[x]^2),x]

[Out]

(4*(8*a^2 + 20*a*b + 15*b^2)*x - (32*(a + b)^(5/2)*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/Sqrt[a] - 8*b*(a +
2*b)*Sinh[2*x] + b^2*Sinh[4*x])/(32*b^3)

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Maple [B]  time = 0.059, size = 563, normalized size = 6.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^6/(a+b*cosh(x)^2),x)

[Out]

1/4/b/(tanh(1/2*x)-1)^4-1/4/b/(tanh(1/2*x)+1)^4-1/2/a^(1/2)/(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/a^(1/2)/(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))+15/8/b*ln(tanh(1/2*x)+1)-15/8/b*ln(tanh(1/2*x)-1)+1/2/b/(tanh(1/2*x)+1)^3+5/8/b/(tanh(1/2*x)+1)^2-7/8/b/(t
anh(1/2*x)+1)+1/2/b/(tanh(1/2*x)-1)^3-5/8/b/(tanh(1/2*x)-1)^2-7/8/b/(tanh(1/2*x)-1)+1/b^3*ln(tanh(1/2*x)+1)*a^
2-1/b^3*ln(tanh(1/2*x)-1)*a^2-3/2/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))-3/2/b*a^(1/2)/(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^3*
a^(5/2)/(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))+3/2/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))+3/2/b*a^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tan
h(1/2*x)^2-2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))-1/2/b^2/(tanh(1/2*x)-1)*a-5/2*a/b^2*ln(tanh(1/2*x)-1)-1/2/b^3*a^
(5/2)/(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/(tanh(1/2*x)+1)^2*a-
1/2/b^2/(tanh(1/2*x)+1)*a+5/2*a/b^2*ln(tanh(1/2*x)+1)-1/2/b^2/(tanh(1/2*x)-1)^2*a

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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)^6/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.0685, size = 3571, normalized size = 40.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^6/(a+b*cosh(x)^2),x, algorithm="fricas")

[Out]

[1/64*(b^2*cosh(x)^8 + 8*b^2*cosh(x)*sinh(x)^7 + b^2*sinh(x)^8 - 8*(a*b + 2*b^2)*cosh(x)^6 + 4*(7*b^2*cosh(x)^
2 - 2*a*b - 4*b^2)*sinh(x)^6 + 8*(8*a^2 + 20*a*b + 15*b^2)*x*cosh(x)^4 + 8*(7*b^2*cosh(x)^3 - 6*(a*b + 2*b^2)*
cosh(x))*sinh(x)^5 + 2*(35*b^2*cosh(x)^4 - 60*(a*b + 2*b^2)*cosh(x)^2 + 4*(8*a^2 + 20*a*b + 15*b^2)*x)*sinh(x)
^4 + 8*(7*b^2*cosh(x)^5 - 20*(a*b + 2*b^2)*cosh(x)^3 + 4*(8*a^2 + 20*a*b + 15*b^2)*x*cosh(x))*sinh(x)^3 + 8*(a
*b + 2*b^2)*cosh(x)^2 + 4*(7*b^2*cosh(x)^6 - 30*(a*b + 2*b^2)*cosh(x)^4 + 12*(8*a^2 + 20*a*b + 15*b^2)*x*cosh(
x)^2 + 2*a*b + 4*b^2)*sinh(x)^2 + 32*((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)^3*sinh(x)
+ 6*(a^2 + 2*a*b + b^2)*cosh(x)^2*sinh(x)^2 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2 + 2*a*b + b^2)*si
nh(x)^4)*sqrt((a + b)/a)*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)*cos
h(x))*sinh(x) + 4*(a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 + 2*a^2 + a*b)*sqrt((a + b)/a))/(b*co
sh(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)) - b^2 + 8*(b^2*cosh(x)^7 - 6*(a*b + 2*b^2)*cosh(x)^5 + 4*(
8*a^2 + 20*a*b + 15*b^2)*x*cosh(x)^3 + 2*(a*b + 2*b^2)*cosh(x))*sinh(x))/(b^3*cosh(x)^4 + 4*b^3*cosh(x)^3*sinh
(x) + 6*b^3*cosh(x)^2*sinh(x)^2 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x)^4), 1/64*(b^2*cosh(x)^8 + 8*b^2*cosh(x
)*sinh(x)^7 + b^2*sinh(x)^8 - 8*(a*b + 2*b^2)*cosh(x)^6 + 4*(7*b^2*cosh(x)^2 - 2*a*b - 4*b^2)*sinh(x)^6 + 8*(8
*a^2 + 20*a*b + 15*b^2)*x*cosh(x)^4 + 8*(7*b^2*cosh(x)^3 - 6*(a*b + 2*b^2)*cosh(x))*sinh(x)^5 + 2*(35*b^2*cosh
(x)^4 - 60*(a*b + 2*b^2)*cosh(x)^2 + 4*(8*a^2 + 20*a*b + 15*b^2)*x)*sinh(x)^4 + 8*(7*b^2*cosh(x)^5 - 20*(a*b +
2*b^2)*cosh(x)^3 + 4*(8*a^2 + 20*a*b + 15*b^2)*x*cosh(x))*sinh(x)^3 + 8*(a*b + 2*b^2)*cosh(x)^2 + 4*(7*b^2*co
sh(x)^6 - 30*(a*b + 2*b^2)*cosh(x)^4 + 12*(8*a^2 + 20*a*b + 15*b^2)*x*cosh(x)^2 + 2*a*b + 4*b^2)*sinh(x)^2 - 6
4*((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)^3*sinh(x) + 6*(a^2 + 2*a*b + b^2)*cosh(x)^2*s
inh(x)^2 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2 + 2*a*b + b^2)*sinh(x)^4)*sqrt(-(a + b)/a)*arctan(1/
2*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + 2*a + b)*sqrt(-(a + b)/a)/(a + b)) - b^2 + 8*(b^2*cosh(x)
^7 - 6*(a*b + 2*b^2)*cosh(x)^5 + 4*(8*a^2 + 20*a*b + 15*b^2)*x*cosh(x)^3 + 2*(a*b + 2*b^2)*cosh(x))*sinh(x))/(
b^3*cosh(x)^4 + 4*b^3*cosh(x)^3*sinh(x) + 6*b^3*cosh(x)^2*sinh(x)^2 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x)^4)
]

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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)**6/(a+b*cosh(x)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.3103, size = 224, normalized size = 2.55 \begin{align*} \frac{b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 16 \, b e^{\left (2 \, x\right )}}{64 \, b^{2}} + \frac{{\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x}{8 \, b^{3}} - \frac{{\left (48 \, a^{2} e^{\left (4 \, x\right )} + 120 \, a b e^{\left (4 \, x\right )} + 90 \, b^{2} e^{\left (4 \, x\right )} - 8 \, a b e^{\left (2 \, x\right )} - 16 \, b^{2} e^{\left (2 \, x\right )} + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, b^{3}} - \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \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^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^6/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

1/64*(b*e^(4*x) - 8*a*e^(2*x) - 16*b*e^(2*x))/b^2 + 1/8*(8*a^2 + 20*a*b + 15*b^2)*x/b^3 - 1/64*(48*a^2*e^(4*x)
+ 120*a*b*e^(4*x) + 90*b^2*e^(4*x) - 8*a*b*e^(2*x) - 16*b^2*e^(2*x) + b^2)*e^(-4*x)/b^3 - (a^3 + 3*a^2*b + 3*
a*b^2 + b^3)*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*b^3)