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

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

[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)

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Rubi [A]  time = 0.209898, 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 = {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 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]

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 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 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{\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*}

Mathematica [A]  time = 0.215353, size = 76, normalized size = 0.86 \[ \frac{4 x \left (8 a^2-4 a b+3 b^2\right )-\frac{32 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a+b}}-8 b (a-b) \sinh (2 x)+b^2 \sinh (4 x)}{32 b^3} \]

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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Maple [B]  time = 0.037, size = 323, normalized size = 3.7 \begin{align*} -{\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 ) ^{-1}}+{\frac{5}{8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\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 ) ^{-2}}+{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{a}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{3}{8\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{2\,{b}^{3}}{a}^{{\frac{5}{2}}}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\sqrt{a}\tanh \left ( x/2 \right ) +\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{1}{2\,{b}^{3}}{a}^{{\frac{5}{2}}}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\sqrt{a}\tanh \left ( x/2 \right ) +\sqrt{a+b} \right ){\frac{1}{\sqrt{a+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 ) ^{-1}}+{\frac{5}{8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\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 ) ^{-2}}-{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{a}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{3}{8\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

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)*a+5/8/b/(tanh(1/2*x)+1)-7/8/b/(tanh(1
/2*x)+1)^2+1/2/b^2/(tanh(1/2*x)+1)^2*a+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))+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)*a+5/8/b/(tanh(1/2*x)-1)+7/8/b/(tanh(1/2*x)-1)^2-1/2/b^2/(tanh(1/2
*x)-1)^2*a-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)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.07442, size = 3320, normalized size = 37.73 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

Timed out

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Giac [B]  time = 1.19982, size = 203, normalized size = 2.31 \begin{align*} -\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}} \end{align*}

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