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]

1/8*(8*a^2+20*a*b+15*b^2)*x/b^3-1/8*(4*a+7*b)*cosh(x)*sinh(x)/b^2+1/4*cosh(x)*sinh(x)^3/b-(a+b)^(5/2)*arctanh(
a^(1/2)*tanh(x)/(a+b)^(1/2))/b^3/a^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.17, 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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fricas [B]  time = 0.51, size = 1308, normalized size = 14.86 \[ \text {result too large to display} \]

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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giac [B]  time = 0.13, size = 166, normalized size = 1.89 \[ \frac {b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 16 \, b e^{\left (2 \, x\right )}}{64 \, b^{2}} + \frac {{\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x}{8 \, b^{3}} - \frac {{\left (48 \, a^{2} e^{\left (4 \, x\right )} + 120 \, a b e^{\left (4 \, x\right )} + 90 \, b^{2} e^{\left (4 \, x\right )} - 8 \, a b e^{\left (2 \, x\right )} - 16 \, b^{2} e^{\left (2 \, x\right )} + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, b^{3}} - \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} b^{3}} \]

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)

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maple [B]  time = 0.13, size = 575, normalized size = 6.53 \[ \frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {a^{\frac {5}{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^{3} \sqrt {a +b}}+\frac {a^{\frac {5}{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^{3} \sqrt {a +b}}+\frac {3 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 {3 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 {3 \sqrt {a}\, \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 \sqrt {a +b}}+\frac {3 \sqrt {a}\, \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 \sqrt {a +b}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {5}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {7}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {5}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {7}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {15 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 b}+\frac {15 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 b}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{b^{3}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}+\frac {\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 \sqrt {a}\, \sqrt {a +b}}-\frac {\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 \sqrt {a}\, \sqrt {a +b}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {5 a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{2}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {5 a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{2}}+\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}} \]

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+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^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))+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))+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/(tanh(1/2*x)-1)^3-5/8/b/(tanh(1/2*x)-1)^2-7/8/b/(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/(tanh(1/2*x)+1)-15/8/b*ln(tanh(1/2*x)-1)+15/8/b
*ln(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+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))-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-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 [B]  time = 0.45, size = 651, normalized size = 7.40 \[ -\frac {15 \, {\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 )}{64 \, \sqrt {{\left (a + b\right )} a} b} + \frac {5 \, \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}} + \frac {3 \, {\left (2 \, a + b\right )} x}{2 \, b^{2}} + \frac {15 \, x}{16 \, b} - \frac {{\left (4 \, {\left (2 \, a + b\right )} e^{\left (-2 \, x\right )} - b\right )} e^{\left (4 \, x\right )}}{64 \, b^{2}} - \frac {3 \, e^{\left (2 \, x\right )}}{16 \, b} + \frac {3 \, e^{\left (-2 \, x\right )}}{16 \, b} + \frac {{\left (4 \, {\left (2 \, a + b\right )} e^{\left (2 \, x\right )} - b\right )} e^{\left (-4 \, x\right )}}{64 \, b^{2}} - \frac {3 \, {\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 )}{16 \, b^{2}} + \frac {3 \, {\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 )}{16 \, b^{2}} - \frac {3 \, {\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 )}{64 \, \sqrt {{\left (a + b\right )} a} b^{2}} + \frac {3 \, {\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 )}{64 \, \sqrt {{\left (a + b\right )} a} b^{2}} + \frac {{\left (16 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} x}{8 \, b^{3}} - \frac {{\left (16 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left (b e^{\left (4 \, x\right )} + 2 \, {\left (2 \, a + b\right )} e^{\left (2 \, x\right )} + b\right )}{64 \, b^{3}} + \frac {{\left (16 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left (2 \, {\left (2 \, a + b\right )} e^{\left (-2 \, x\right )} + b e^{\left (-4 \, x\right )} + b\right )}{64 \, b^{3}} - \frac {{\left (32 \, a^{3} + 48 \, a^{2} b + 18 \, a b^{2} + b^{3}\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 )}{128 \, \sqrt {{\left (a + b\right )} a} b^{3}} + \frac {{\left (32 \, a^{3} + 48 \, a^{2} b + 18 \, a b^{2} + b^{3}\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 )}{128 \, \sqrt {{\left (a + b\right )} a} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/64*(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)))/(sqr
t((a + b)*a)*b) + 5/32*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) + 3/2*(2*a + b)*x/b^2 + 15/16*x/b - 1/64*(4*(2*a + b)*e^(-2*x) - b)*e^(4*x)/b^2 - 3/16*e^(
2*x)/b + 3/16*e^(-2*x)/b + 1/64*(4*(2*a + b)*e^(2*x) - b)*e^(-4*x)/b^2 - 3/16*(2*a + b)*log(b*e^(4*x) + 2*(2*a
+ b)*e^(2*x) + b)/b^2 + 3/16*(2*a + b)*log(2*(2*a + b)*e^(-2*x) + b*e^(-4*x) + b)/b^2 - 3/64*(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) + 3/64*(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*s
qrt((a + b)*a)))/(sqrt((a + b)*a)*b^2) + 1/8*(16*a^2 + 16*a*b + 3*b^2)*x/b^3 - 1/64*(16*a^2 + 16*a*b + 3*b^2)*
log(b*e^(4*x) + 2*(2*a + b)*e^(2*x) + b)/b^3 + 1/64*(16*a^2 + 16*a*b + 3*b^2)*log(2*(2*a + b)*e^(-2*x) + b*e^(
-4*x) + b)/b^3 - 1/128*(32*a^3 + 48*a^2*b + 18*a*b^2 + b^3)*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) + 1/128*(32*a^3 + 48*a^2*b + 18*a*b^2 + b^3)*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)

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mupad [B]  time = 1.72, size = 248, normalized size = 2.82 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,b}+\frac {x\,\left (8\,a^2+20\,a\,b+15\,b^2\right )}{8\,b^3}+\frac {{\mathrm {e}}^{-2\,x}\,\left (a+2\,b\right )}{8\,b^2}-\frac {{\mathrm {e}}^{2\,x}\,\left (a+2\,b\right )}{8\,b^2}+\frac {\ln \left (\frac {4\,{\left (a+b\right )}^5\,\left (2\,a\,b+8\,a^2\,{\mathrm {e}}^{2\,x}+b^2\,{\mathrm {e}}^{2\,x}+b^2+8\,a\,b\,{\mathrm {e}}^{2\,x}\right )}{a\,b^8}-\frac {8\,{\left (a+b\right )}^{11/2}\,\left (b+4\,a\,{\mathrm {e}}^{2\,x}+2\,b\,{\mathrm {e}}^{2\,x}\right )}{\sqrt {a}\,b^8}\right )\,{\left (a+b\right )}^{5/2}}{2\,\sqrt {a}\,b^3}-\frac {\ln \left (\frac {8\,{\left (a+b\right )}^{11/2}\,\left (b+4\,a\,{\mathrm {e}}^{2\,x}+2\,b\,{\mathrm {e}}^{2\,x}\right )}{\sqrt {a}\,b^8}+\frac {4\,{\left (a+b\right )}^5\,\left (2\,a\,b+8\,a^2\,{\mathrm {e}}^{2\,x}+b^2\,{\mathrm {e}}^{2\,x}+b^2+8\,a\,b\,{\mathrm {e}}^{2\,x}\right )}{a\,b^8}\right )\,{\left (a+b\right )}^{5/2}}{2\,\sqrt {a}\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^6/(a + b*cosh(x)^2),x)

[Out]

exp(4*x)/(64*b) - exp(-4*x)/(64*b) + (x*(20*a*b + 8*a^2 + 15*b^2))/(8*b^3) + (exp(-2*x)*(a + 2*b))/(8*b^2) - (
exp(2*x)*(a + 2*b))/(8*b^2) + (log((4*(a + b)^5*(2*a*b + 8*a^2*exp(2*x) + b^2*exp(2*x) + b^2 + 8*a*b*exp(2*x))
)/(a*b^8) - (8*(a + b)^(11/2)*(b + 4*a*exp(2*x) + 2*b*exp(2*x)))/(a^(1/2)*b^8))*(a + b)^(5/2))/(2*a^(1/2)*b^3)
- (log((8*(a + b)^(11/2)*(b + 4*a*exp(2*x) + 2*b*exp(2*x)))/(a^(1/2)*b^8) + (4*(a + b)^5*(2*a*b + 8*a^2*exp(2
*x) + b^2*exp(2*x) + b^2 + 8*a*b*exp(2*x)))/(a*b^8))*(a + b)^(5/2))/(2*a^(1/2)*b^3)

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

[Out]

Timed out

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