### 3.56 $$\int \frac{\cosh ^{\frac{7}{2}}(a+b x)}{\sinh ^{\frac{7}{2}}(a+b x)} \, dx$$

Optimal. Leaf size=106 $-\frac{2 \cosh ^{\frac{5}{2}}(a+b x)}{5 b \sinh ^{\frac{5}{2}}(a+b x)}-\frac{2 \sqrt{\cosh (a+b x)}}{b \sqrt{\sinh (a+b x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}$

[Out]

-(ArcTan[Sqrt[Sinh[a + b*x]]/Sqrt[Cosh[a + b*x]]]/b) + ArcTanh[Sqrt[Sinh[a + b*x]]/Sqrt[Cosh[a + b*x]]]/b - (2
*Cosh[a + b*x]^(5/2))/(5*b*Sinh[a + b*x]^(5/2)) - (2*Sqrt[Cosh[a + b*x]])/(b*Sqrt[Sinh[a + b*x]])

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Rubi [A]  time = 0.105444, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {2567, 2574, 298, 203, 206} $-\frac{2 \cosh ^{\frac{5}{2}}(a+b x)}{5 b \sinh ^{\frac{5}{2}}(a+b x)}-\frac{2 \sqrt{\cosh (a+b x)}}{b \sqrt{\sinh (a+b x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]^(7/2)/Sinh[a + b*x]^(7/2),x]

[Out]

-(ArcTan[Sqrt[Sinh[a + b*x]]/Sqrt[Cosh[a + b*x]]]/b) + ArcTanh[Sqrt[Sinh[a + b*x]]/Sqrt[Cosh[a + b*x]]]/b - (2
*Cosh[a + b*x]^(5/2))/(5*b*Sinh[a + b*x]^(5/2)) - (2*Sqrt[Cosh[a + b*x]])/(b*Sqrt[Sinh[a + b*x]])

Rule 2567

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

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

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])

Rubi steps

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

Mathematica [C]  time = 0.0383257, size = 59, normalized size = 0.56 $-\frac{2 \cosh ^2(a+b x)^{3/4} \, _2F_1\left (-\frac{5}{4},-\frac{5}{4};-\frac{1}{4};-\sinh ^2(a+b x)\right )}{5 b \sinh ^{\frac{5}{2}}(a+b x) \cosh ^{\frac{3}{2}}(a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]^(7/2)/Sinh[a + b*x]^(7/2),x]

[Out]

(-2*(Cosh[a + b*x]^2)^(3/4)*Hypergeometric2F1[-5/4, -5/4, -1/4, -Sinh[a + b*x]^2])/(5*b*Cosh[a + b*x]^(3/2)*Si
nh[a + b*x]^(5/2))

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Maple [F]  time = 0.061, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cosh \left ( bx+a \right ) \right ) ^{{\frac{7}{2}}} \left ( \sinh \left ( bx+a \right ) \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^(7/2)/sinh(b*x+a)^(7/2),x)

[Out]

int(cosh(b*x+a)^(7/2)/sinh(b*x+a)^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (b x + a\right )^{\frac{7}{2}}}{\sinh \left (b x + a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^(7/2)/sinh(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

integrate(cosh(b*x + a)^(7/2)/sinh(b*x + a)^(7/2), x)

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Fricas [B]  time = 2.38686, size = 2819, normalized size = 26.59 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^(7/2)/sinh(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

-1/10*(24*cosh(b*x + a)^6 + 144*cosh(b*x + a)*sinh(b*x + a)^5 + 24*sinh(b*x + a)^6 + 72*(5*cosh(b*x + a)^2 - 1
)*sinh(b*x + a)^4 - 72*cosh(b*x + a)^4 + 96*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^3 + 72*(5*cosh
(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 10*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 +
sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 3*co
sh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)
^2 + 6*(cosh(b*x + a)^5 - 2*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) - 1)*arctan(-cosh(b*x + a)^2 + 2*(c
osh(b*x + a) + sinh(b*x + a))*sqrt(cosh(b*x + a))*sqrt(sinh(b*x + a)) - 2*cosh(b*x + a)*sinh(b*x + a) - sinh(b
*x + a)^2) + 72*cosh(b*x + a)^2 + 5*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(
5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x
+ a)^3 + 3*(5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^
5 - 2*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))*sqrt(cosh(b*x + a))*sqrt(sinh(b*x + a)) - 2*cosh(b*x + a)*sinh(b*x + a) - sinh(b*x + a)^2) + 16*(3*cosh(b
*x + a)^5 + 15*cosh(b*x + a)*sinh(b*x + a)^4 + 3*sinh(b*x + a)^5 + 2*(15*cosh(b*x + a)^2 - 2)*sinh(b*x + a)^3
- 4*cosh(b*x + a)^3 + 6*(5*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^2 + 3*(5*cosh(b*x + a)^4 - 4*cosh(
b*x + a)^2 + 1)*sinh(b*x + a) + 3*cosh(b*x + a))*sqrt(cosh(b*x + a))*sqrt(sinh(b*x + a)) + 144*(cosh(b*x + a)^
5 - 2*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) - 24)/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a
)^5 + b*sinh(b*x + a)^6 - 3*b*cosh(b*x + a)^4 + 3*(5*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x
+ a)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)^3 + 3*b*cosh(b*x + a)^2 + 3*(5*b*cosh(b*x + a)^4 - 6*b*cosh(b*x + a)
^2 + b)*sinh(b*x + a)^2 + 6*(b*cosh(b*x + a)^5 - 2*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a) - b)

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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(b*x+a)**(7/2)/sinh(b*x+a)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (b x + a\right )^{\frac{7}{2}}}{\sinh \left (b x + a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^(7/2)/sinh(b*x+a)^(7/2),x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)^(7/2)/sinh(b*x + a)^(7/2), x)