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

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

[Out]

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

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Rubi [A]  time = 0.21, 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 = {3187, 470, 578, 522, 206, 208} \[ \frac {x \left (8 a^2-4 a b+3 b^2\right )}{8 b^3}-\frac {a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{b^3 \sqrt {a+b}}-\frac {(4 a-3 b) \sinh (x) \cosh (x)}{8 b^2}+\frac {\sinh (x) \cosh ^3(x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^6/(a + b*Cosh[x]^2),x]

[Out]

((8*a^2 - 4*a*b + 3*b^2)*x)/(8*b^3) - (a^(5/2)*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(b^3*Sqrt[a + b]) - ((4
*a - 3*b)*Cosh[x]*Sinh[x])/(8*b^2) + (Cosh[x]^3*Sinh[x])/(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 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 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

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

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Mathematica [A]  time = 0.22, size = 76, normalized size = 0.86 \[ \frac {-\frac {32 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+4 x \left (8 a^2-4 a b+3 b^2\right )-8 b (a-b) \sinh (2 x)+b^2 \sinh (4 x)}{32 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 1245, normalized size = 14.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(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 - b^2)*cosh(x)^6 + 4*(7*b^2*cosh(x)^2
- 2*a*b + 2*b^2)*sinh(x)^6 + 8*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x)^4 + 8*(7*b^2*cosh(x)^3 - 6*(a*b - b^2)*cosh(x
))*sinh(x)^5 + 2*(35*b^2*cosh(x)^4 - 60*(a*b - b^2)*cosh(x)^2 + 4*(8*a^2 - 4*a*b + 3*b^2)*x)*sinh(x)^4 + 8*(7*
b^2*cosh(x)^5 - 20*(a*b - b^2)*cosh(x)^3 + 4*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x))*sinh(x)^3 + 8*(a*b - b^2)*cosh
(x)^2 + 4*(7*b^2*cosh(x)^6 - 30*(a*b - b^2)*cosh(x)^4 + 12*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x)^2 + 2*a*b - 2*b^2
)*sinh(x)^2 + 32*(a^2*cosh(x)^4 + 4*a^2*cosh(x)^3*sinh(x) + 6*a^2*cosh(x)^2*sinh(x)^2 + 4*a^2*cosh(x)*sinh(x)^
3 + a^2*sinh(x)^4)*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*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*c
osh(x)^7 - 6*(a*b - b^2)*cosh(x)^5 + 4*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x)^3 + 2*(a*b - 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 - b^2)*cosh(x)^6 + 4*(7*b^2*cosh(x)^2
- 2*a*b + 2*b^2)*sinh(x)^6 + 8*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x)^4 + 8*(7*b^2*cosh(x)^3 - 6*(a*b - b^2)*cosh(
x))*sinh(x)^5 + 2*(35*b^2*cosh(x)^4 - 60*(a*b - b^2)*cosh(x)^2 + 4*(8*a^2 - 4*a*b + 3*b^2)*x)*sinh(x)^4 + 8*(7
*b^2*cosh(x)^5 - 20*(a*b - b^2)*cosh(x)^3 + 4*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x))*sinh(x)^3 + 8*(a*b - b^2)*cos
h(x)^2 + 4*(7*b^2*cosh(x)^6 - 30*(a*b - b^2)*cosh(x)^4 + 12*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x)^2 + 2*a*b - 2*b^
2)*sinh(x)^2 - 64*(a^2*cosh(x)^4 + 4*a^2*cosh(x)^3*sinh(x) + 6*a^2*cosh(x)^2*sinh(x)^2 + 4*a^2*cosh(x)*sinh(x)
^3 + a^2*sinh(x)^4)*sqrt(-a/(a + b))*arctan(1/2*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + 2*a + b)*sq
rt(-a/(a + b))/a) - b^2 + 8*(b^2*cosh(x)^7 - 6*(a*b - b^2)*cosh(x)^5 + 4*(8*a^2 - 4*a*b + 3*b^2)*x*cosh(x)^3 +
2*(a*b - 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 = 150, normalized size = 1.70 \[ -\frac {a^{3} \arctan \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} b^{3}} + \frac {b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} + 8 \, b e^{\left (2 \, x\right )}}{64 \, b^{2}} + \frac {{\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} x}{8 \, b^{3}} - \frac {{\left (48 \, a^{2} e^{\left (4 \, x\right )} - 24 \, a b e^{\left (4 \, x\right )} + 18 \, b^{2} e^{\left (4 \, x\right )} - 8 \, a b e^{\left (2 \, x\right )} + 8 \, b^{2} e^{\left (2 \, x\right )} + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*b^3) + 1/64*(b*e^(4*x) - 8*a*e^(2*x)
+ 8*b*e^(2*x))/b^2 + 1/8*(8*a^2 - 4*a*b + 3*b^2)*x/b^3 - 1/64*(48*a^2*e^(4*x) - 24*a*b*e^(4*x) + 18*b^2*e^(4*
x) - 8*a*b*e^(2*x) + 8*b^2*e^(2*x) + b^2)*e^(-4*x)/b^3

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maple [B]  time = 0.11, size = 326, normalized size = 3.70 \[ \frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {7}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {5}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{b^{3}}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{2}}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 b}-\frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {7}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {5}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{2}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 b}+\frac {a^{\frac {5}{2}} \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \sqrt {a}\, \tanh \left (\frac {x}{2}\right )-\sqrt {a +b}\right )}{2 b^{3} \sqrt {a +b}}-\frac {a^{\frac {5}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \sqrt {a}\, \tanh \left (\frac {x}{2}\right )+\sqrt {a +b}\right )}{2 b^{3} \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.50, 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(cosh(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 = 1.33, size = 178, normalized size = 2.02 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,b}+\frac {x\,\left (8\,a^2-4\,a\,b+3\,b^2\right )}{8\,b^3}+\frac {{\mathrm {e}}^{-2\,x}\,\left (a-b\right )}{8\,b^2}-\frac {{\mathrm {e}}^{2\,x}\,\left (a-b\right )}{8\,b^2}+\frac {a^{5/2}\,\ln \left (\frac {4\,a^3\,{\mathrm {e}}^{2\,x}}{b^4}-\frac {2\,a^{5/2}\,\left (b+2\,a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{b^4\,\sqrt {a+b}}\right )}{2\,b^3\,\sqrt {a+b}}-\frac {a^{5/2}\,\ln \left (\frac {4\,a^3\,{\mathrm {e}}^{2\,x}}{b^4}+\frac {2\,a^{5/2}\,\left (b+2\,a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{b^4\,\sqrt {a+b}}\right )}{2\,b^3\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^6/(a + b*cosh(x)^2),x)

[Out]

exp(4*x)/(64*b) - exp(-4*x)/(64*b) + (x*(8*a^2 - 4*a*b + 3*b^2))/(8*b^3) + (exp(-2*x)*(a - b))/(8*b^2) - (exp(
2*x)*(a - b))/(8*b^2) + (a^(5/2)*log((4*a^3*exp(2*x))/b^4 - (2*a^(5/2)*(b + 2*a*exp(2*x) + b*exp(2*x)))/(b^4*(
a + b)^(1/2))))/(2*b^3*(a + b)^(1/2)) - (a^(5/2)*log((4*a^3*exp(2*x))/b^4 + (2*a^(5/2)*(b + 2*a*exp(2*x) + b*e
xp(2*x)))/(b^4*(a + b)^(1/2))))/(2*b^3*(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)**6/(a+b*cosh(x)**2),x)

[Out]

Timed out

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