### 3.50 $$\int \frac{\sinh ^{\frac{5}{2}}(a+b x)}{\cosh ^{\frac{5}{2}}(a+b x)} \, dx$$

Optimal. Leaf size=81 $-\frac{2 \sinh ^{\frac{3}{2}}(a+b x)}{3 b \cosh ^{\frac{3}{2}}(a+b x)}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}$

[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
*Sinh[a + b*x]^(3/2))/(3*b*Cosh[a + b*x]^(3/2))

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Rubi [A]  time = 0.0811908, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {2566, 2574, 298, 203, 206} $-\frac{2 \sinh ^{\frac{3}{2}}(a+b x)}{3 b \cosh ^{\frac{3}{2}}(a+b x)}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[a + b*x]^(5/2)/Cosh[a + b*x]^(5/2),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
*Sinh[a + b*x]^(3/2))/(3*b*Cosh[a + b*x]^(3/2))

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]

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{\sinh ^{\frac{5}{2}}(a+b x)}{\cosh ^{\frac{5}{2}}(a+b x)} \, dx &=-\frac{2 \sinh ^{\frac{3}{2}}(a+b x)}{3 b \cosh ^{\frac{3}{2}}(a+b x)}+\int \frac{\sqrt{\sinh (a+b x)}}{\sqrt{\cosh (a+b x)}} \, dx\\ &=-\frac{2 \sinh ^{\frac{3}{2}}(a+b x)}{3 b \cosh ^{\frac{3}{2}}(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 \sinh ^{\frac{3}{2}}(a+b x)}{3 b \cosh ^{\frac{3}{2}}(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 \sinh ^{\frac{3}{2}}(a+b x)}{3 b \cosh ^{\frac{3}{2}}(a+b x)}\\ \end{align*}

Mathematica [C]  time = 0.0479155, size = 59, normalized size = 0.73 $\frac{2 \sinh ^{\frac{7}{2}}(a+b x) \cosh ^2(a+b x)^{3/4} \, _2F_1\left (\frac{7}{4},\frac{7}{4};\frac{11}{4};-\sinh ^2(a+b x)\right )}{7 b \cosh ^{\frac{3}{2}}(a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[a + b*x]^(5/2)/Cosh[a + b*x]^(5/2),x]

[Out]

(2*(Cosh[a + b*x]^2)^(3/4)*Hypergeometric2F1[7/4, 7/4, 11/4, -Sinh[a + b*x]^2]*Sinh[a + b*x]^(7/2))/(7*b*Cosh[
a + b*x]^(3/2))

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sinh \left ( bx+a \right ) \right ) ^{{\frac{5}{2}}} \left ( \cosh \left ( bx+a \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^(5/2)/cosh(b*x+a)^(5/2),x)

[Out]

int(sinh(b*x+a)^(5/2)/cosh(b*x+a)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x + a\right )^{\frac{5}{2}}}{\cosh \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(5/2)/cosh(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(sinh(b*x + a)^(5/2)/cosh(b*x + a)^(5/2), x)

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Fricas [B]  time = 2.68691, size = 1705, normalized size = 21.05 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(5/2)/cosh(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(4*cosh(b*x + a)^4 + 16*cosh(b*x + a)*sinh(b*x + a)^3 + 4*sinh(b*x + a)^4 + 8*(3*cosh(b*x + a)^2 + 1)*sin
h(b*x + a)^2 + 6*(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)*arctan(-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) + 8*cosh(b*x + a)^2 + 3*(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))*sqrt(cosh(b*x + a))*sqrt(sinh(b
*x + a)) - 2*cosh(b*x + a)*sinh(b*x + a) - sinh(b*x + a)^2) + 8*(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))*sqrt(cosh(b*x + a))*sqrt(sinh(
b*x + a)) + 16*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 4)/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sin
h(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*co
sh(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(sinh(b*x+a)**(5/2)/cosh(b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x + a\right )^{\frac{5}{2}}}{\cosh \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(5/2)/cosh(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(sinh(b*x + a)^(5/2)/cosh(b*x + a)^(5/2), x)