3.112 \(\int \frac {\sinh ^4(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=256 \[ \frac {(2 a-b) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{b^2 f (a-b)}-\frac {(2 a-b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{b^2 f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {a \sinh (e+f x) \cosh (e+f x)}{b f (a-b) \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{b f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

-a*cosh(f*x+e)*sinh(f*x+e)/(a-b)/b/f/(a+b*sinh(f*x+e)^2)^(1/2)-(2*a-b)*(1/(1+sinh(f*x+e)^2))^(1/2)*(1+sinh(f*x
+e)^2)^(1/2)*EllipticE(sinh(f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/2))*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2
)/(a-b)/b^2/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)+(1/(1+sinh(f*x+e)^2))^(1/2)*(1+sinh(f*x+e)^2)^(1/2)*
EllipticF(sinh(f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/2))*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/(a-b)/b/f/(
sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)+(2*a-b)*(a+b*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)/(a-b)/b^2/f

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Rubi [A]  time = 0.24, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3188, 470, 531, 418, 492, 411} \[ \frac {(2 a-b) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{b^2 f (a-b)}-\frac {(2 a-b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{b^2 f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {a \sinh (e+f x) \cosh (e+f x)}{b f (a-b) \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{b f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[e + f*x]^4/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-((a*Cosh[e + f*x]*Sinh[e + f*x])/((a - b)*b*f*Sqrt[a + b*Sinh[e + f*x]^2])) - ((2*a - b)*EllipticE[ArcTan[Sin
h[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/((a - b)*b^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*S
inh[e + f*x]^2))/a]) + (EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(
(a - b)*b*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((2*a - b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e
+ f*x])/((a - b)*b^2*f)

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

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 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 3188

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/Sqrt[1 - ff^2*x^2], x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sinh ^4(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac {a \cosh (e+f x) \sinh (e+f x)}{(a-b) b f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {a+(2 a-b) x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{(a-b) b f}\\ &=-\frac {a \cosh (e+f x) \sinh (e+f x)}{(a-b) b f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\left (a \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{(a-b) b f}+\frac {\left ((2 a-b) \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{(a-b) b f}\\ &=-\frac {a \cosh (e+f x) \sinh (e+f x)}{(a-b) b f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{(a-b) b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(2 a-b) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{(a-b) b^2 f}-\frac {\left ((2 a-b) \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{(a-b) b^2 f}\\ &=-\frac {a \cosh (e+f x) \sinh (e+f x)}{(a-b) b f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {(2 a-b) E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{(a-b) b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{(a-b) b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(2 a-b) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{(a-b) b^2 f}\\ \end {align*}

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Mathematica [C]  time = 1.05, size = 156, normalized size = 0.61 \[ \frac {a \left (4 i (a-b) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )-2 i (2 a-b) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )-\sqrt {2} b \sinh (2 (e+f x))\right )}{2 b^2 f (a-b) \sqrt {2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[e + f*x]^4/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

(a*((-2*I)*(2*a - b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] + (4*I)*(a - b)*Sqrt[
(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] - Sqrt[2]*b*Sinh[2*(e + f*x)]))/(2*(a - b)*b^2*
f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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fricas [F]  time = 1.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sinh \left (f x + e\right )^{2} + a} \sinh \left (f x + e\right )^{4}}{b^{2} \sinh \left (f x + e\right )^{4} + 2 \, a b \sinh \left (f x + e\right )^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)*sinh(f*x + e)^4/(b^2*sinh(f*x + e)^4 + 2*a*b*sinh(f*x + e)^2 + a^2), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [A]  time = 0.13, size = 313, normalized size = 1.22 \[ -\frac {\sqrt {-\frac {b}{a}}\, a \sinh \left (f x +e \right ) \left (\cosh ^{2}\left (f x +e \right )\right )+a \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b -2 \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a +\sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b}{b \left (a -b \right ) \sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

-1/b*((-1/a*b)^(1/2)*a*sinh(f*x+e)*cosh(f*x+e)^2+a*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*Ell
ipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*Ellipti
cF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b-2*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*Ellipti
cE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a+(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE
(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b)/(a-b)/(-1/a*b)^(1/2)/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (f x + e\right )^{4}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(sinh(f*x + e)^4/(b*sinh(f*x + e)^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {sinh}\left (e+f\,x\right )}^4}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(e + f*x)^4/(a + b*sinh(e + f*x)^2)^(3/2),x)

[Out]

int(sinh(e + f*x)^4/(a + b*sinh(e + f*x)^2)^(3/2), 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(f*x+e)**4/(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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