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

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

[Out]

-arctan(cosh(b*x+a)^(1/2)/sinh(b*x+a)^(1/2))/b+arctanh(cosh(b*x+a)^(1/2)/sinh(b*x+a)^(1/2))/b-2/5*sinh(b*x+a)^
(5/2)/b/cosh(b*x+a)^(5/2)-2*sinh(b*x+a)^(1/2)/b/cosh(b*x+a)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 = {2566, 2575, 298, 203, 206} \[ -\frac {2 \sinh ^{\frac {5}{2}}(a+b x)}{5 b \cosh ^{\frac {5}{2}}(a+b x)}-\frac {2 \sqrt {\sinh (a+b x)}}{b \sqrt {\cosh (a+b x)}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}}\right )}{b}+\frac {\tanh ^{-1}\left (\frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 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*Cos[e + f*x])^(1/k)/(b*Si
n[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}{2}}(a+b x)}{\cosh ^{\frac {7}{2}}(a+b x)} \, dx &=-\frac {2 \sinh ^{\frac {5}{2}}(a+b x)}{5 b \cosh ^{\frac {5}{2}}(a+b x)}+\int \frac {\sinh ^{\frac {3}{2}}(a+b x)}{\cosh ^{\frac {3}{2}}(a+b x)} \, dx\\ &=-\frac {2 \sqrt {\sinh (a+b x)}}{b \sqrt {\cosh (a+b x)}}-\frac {2 \sinh ^{\frac {5}{2}}(a+b x)}{5 b \cosh ^{\frac {5}{2}}(a+b x)}+\int \frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}} \, dx\\ &=-\frac {2 \sqrt {\sinh (a+b x)}}{b \sqrt {\cosh (a+b x)}}-\frac {2 \sinh ^{\frac {5}{2}}(a+b x)}{5 b \cosh ^{\frac {5}{2}}(a+b x)}+\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}}\right )}{b}\\ &=-\frac {2 \sqrt {\sinh (a+b x)}}{b \sqrt {\cosh (a+b x)}}-\frac {2 \sinh ^{\frac {5}{2}}(a+b x)}{5 b \cosh ^{\frac {5}{2}}(a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}}\right )}{b}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}}\right )}{b}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}}\right )}{b}+\frac {\tanh ^{-1}\left (\frac {\sqrt {\cosh (a+b x)}}{\sqrt {\sinh (a+b x)}}\right )}{b}-\frac {2 \sqrt {\sinh (a+b x)}}{b \sqrt {\cosh (a+b x)}}-\frac {2 \sinh ^{\frac {5}{2}}(a+b x)}{5 b \cosh ^{\frac {5}{2}}(a+b x)}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 59, normalized size = 0.56 \[ \frac {2 \sinh ^{\frac {9}{2}}(a+b x) \sqrt [4]{\cosh ^2(a+b x)} \, _2F_1\left (\frac {9}{4},\frac {9}{4};\frac {13}{4};-\sinh ^2(a+b x)\right )}{9 b \sqrt {\cosh (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cosh[a + b*x]^2)^(1/4)*Hypergeometric2F1[9/4, 9/4, 13/4, -Sinh[a + b*x]^2]*Sinh[a + b*x]^(9/2))/(9*b*Sqrt[
Cosh[a + b*x]])

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fricas [B]  time = 0.47, size = 997, normalized size = 9.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(7/2)/cosh(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x + a\right )^{\frac {7}{2}}}{\cosh \left (b x + a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{\frac {7}{2}}\left (b x +a \right )}{\cosh \left (b x +a \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x + a\right )^{\frac {7}{2}}}{\cosh \left (b x + a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^{7/2}}{{\mathrm {cosh}\left (a+b\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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