3.57 \(\int \frac {\sinh ^{\frac {7}{3}}(a+b x)}{\cosh ^{\frac {7}{3}}(a+b x)} \, dx\)

Optimal. Leaf size=155 \[ -\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}-\frac {\log \left (1-\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {\log \left (\frac {\sinh ^{\frac {4}{3}}(a+b x)}{\cosh ^{\frac {4}{3}}(a+b x)}+\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}+1\right )}{4 b}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}+1}{\sqrt {3}}\right )}{2 b} \]

[Out]

-1/2*ln(1-sinh(b*x+a)^(2/3)/cosh(b*x+a)^(2/3))/b+1/4*ln(1+sinh(b*x+a)^(2/3)/cosh(b*x+a)^(2/3)+sinh(b*x+a)^(4/3
)/cosh(b*x+a)^(4/3))/b-3/4*sinh(b*x+a)^(4/3)/b/cosh(b*x+a)^(4/3)-1/2*arctan(1/3*(1+2*sinh(b*x+a)^(2/3)/cosh(b*
x+a)^(2/3))*3^(1/2))*3^(1/2)/b

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Rubi [A]  time = 0.18, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2566, 2574, 275, 292, 31, 634, 618, 204, 628} \[ -\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}-\frac {\log \left (1-\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {\log \left (\frac {\sinh ^{\frac {4}{3}}(a+b x)}{\cosh ^{\frac {4}{3}}(a+b x)}+\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}+1\right )}{4 b}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}+1}{\sqrt {3}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^(7/3)/Cosh[a + b*x]^(7/3),x]

[Out]

-(Sqrt[3]*ArcTan[(1 + (2*Sinh[a + b*x]^(2/3))/Cosh[a + b*x]^(2/3))/Sqrt[3]])/(2*b) - Log[1 - Sinh[a + b*x]^(2/
3)/Cosh[a + b*x]^(2/3)]/(2*b) + Log[1 + Sinh[a + b*x]^(2/3)/Cosh[a + b*x]^(2/3) + Sinh[a + b*x]^(4/3)/Cosh[a +
b*x]^(4/3)]/(4*b) - (3*Sinh[a + b*x]^(4/3))/(4*b*Cosh[a + b*x]^(4/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2566

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(a*Sin[e
+ f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Sin[e +
f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Integ
ersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {\sinh ^{\frac {7}{3}}(a+b x)}{\cosh ^{\frac {7}{3}}(a+b x)} \, dx &=-\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}+\int \frac {\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}} \, dx\\ &=-\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}-\frac {3 \operatorname {Subst}\left (\int \frac {x^3}{-1+x^6} \, dx,x,\frac {\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}\right )}{b}\\ &=-\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}-\frac {3 \operatorname {Subst}\left (\int \frac {x}{-1+x^3} \, dx,x,\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x} \, dx,x,\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {-1+x}{1+x+x^2} \, dx,x,\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac {\log \left (1-\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}-\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{4 b}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{4 b}\\ &=-\frac {\log \left (1-\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {\log \left (1+\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}+\frac {\sinh ^{\frac {4}{3}}(a+b x)}{\cosh ^{\frac {4}{3}}(a+b x)}\right )}{4 b}-\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}-\frac {\log \left (1-\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {\log \left (1+\frac {\sinh ^{\frac {2}{3}}(a+b x)}{\cosh ^{\frac {2}{3}}(a+b x)}+\frac {\sinh ^{\frac {4}{3}}(a+b x)}{\cosh ^{\frac {4}{3}}(a+b x)}\right )}{4 b}-\frac {3 \sinh ^{\frac {4}{3}}(a+b x)}{4 b \cosh ^{\frac {4}{3}}(a+b x)}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 59, normalized size = 0.38 \[ \frac {3 \sinh ^{\frac {10}{3}}(a+b x) \cosh ^2(a+b x)^{2/3} \, _2F_1\left (\frac {5}{3},\frac {5}{3};\frac {8}{3};-\sinh ^2(a+b x)\right )}{10 b \cosh ^{\frac {4}{3}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^(7/3)/Cosh[a + b*x]^(7/3),x]

[Out]

(3*(Cosh[a + b*x]^2)^(2/3)*Hypergeometric2F1[5/3, 5/3, 8/3, -Sinh[a + b*x]^2]*Sinh[a + b*x]^(10/3))/(10*b*Cosh
[a + b*x]^(4/3))

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fricas [B]  time = 0.59, size = 1042, normalized size = 6.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(7/3)/cosh(b*x+a)^(7/3),x, algorithm="fricas")

[Out]

-1/4*(2*(sqrt(3)*cosh(b*x + a)^4 + 4*sqrt(3)*cosh(b*x + a)*sinh(b*x + a)^3 + sqrt(3)*sinh(b*x + a)^4 + 2*(3*sq
rt(3)*cosh(b*x + a)^2 + sqrt(3))*sinh(b*x + a)^2 + 2*sqrt(3)*cosh(b*x + a)^2 + 4*(sqrt(3)*cosh(b*x + a)^3 + sq
rt(3)*cosh(b*x + a))*sinh(b*x + a) + sqrt(3))*arctan(1/3*(sqrt(3)*cosh(b*x + a)^2 + 2*sqrt(3)*cosh(b*x + a)*si
nh(b*x + a) + sqrt(3)*sinh(b*x + a)^2 + 4*(sqrt(3)*cosh(b*x + a) + sqrt(3)*sinh(b*x + a))*cosh(b*x + a)^(1/3)*
sinh(b*x + a)^(2/3) + sqrt(3))/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) - (cos
h(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 +
2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*log((cosh(b*x + a)^4 + 4*cosh(b*x
+ a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 2*(co
sh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - co
sh(b*x + a))*cosh(b*x + a)^(2/3)*sinh(b*x + a)^(1/3) + 2*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 +
sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + cosh(b*x + a))*cosh(b*x + a)^(1/3)*sinh(b*x + a)^(2/
3) + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)/(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3
+ sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh
(b*x + a))*sinh(b*x + a) + 1)) + 2*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3
*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)
+ 1)*log(-(cosh(b*x + a)^2 - 2*(cosh(b*x + a) + sinh(b*x + a))*cosh(b*x + a)^(1/3)*sinh(b*x + a)^(2/3) + 2*cos
h(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x +
a)^2 + 1)) + 6*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 - 1)*
sinh(b*x + a) - cosh(b*x + a))*cosh(b*x + a)^(2/3)*sinh(b*x + a)^(1/3))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)
*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 + 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 4*(
b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a) + b)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x + a\right )^{\frac {7}{3}}}{\cosh \left (b x + a\right )^{\frac {7}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(7/3)/cosh(b*x+a)^(7/3),x, algorithm="giac")

[Out]

integrate(sinh(b*x + a)^(7/3)/cosh(b*x + a)^(7/3), x)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{\frac {7}{3}}\left (b x +a \right )}{\cosh \left (b x +a \right )^{\frac {7}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^(7/3)/cosh(b*x+a)^(7/3),x)

[Out]

int(sinh(b*x+a)^(7/3)/cosh(b*x+a)^(7/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x + a\right )^{\frac {7}{3}}}{\cosh \left (b x + a\right )^{\frac {7}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(7/3)/cosh(b*x+a)^(7/3),x, algorithm="maxima")

[Out]

integrate(sinh(b*x + a)^(7/3)/cosh(b*x + a)^(7/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^{7/3}}{{\mathrm {cosh}\left (a+b\,x\right )}^{7/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^(7/3)/cosh(a + b*x)^(7/3),x)

[Out]

int(sinh(a + b*x)^(7/3)/cosh(a + b*x)^(7/3), x)

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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(b*x+a)**(7/3)/cosh(b*x+a)**(7/3),x)

[Out]

Timed out

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