3.381 \(\int \frac{\text{sech}^4(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=219 \[ -\frac{b (a-3 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 a f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f (a-b)}+\frac{2 (a-2 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

(2*(a - 2*b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*(a - b)^2
*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - ((a - 3*b)*b*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]
*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a*(a - b)^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]
) + (Sech[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*(a - b)*f)

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Rubi [A]  time = 0.194929, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3192, 414, 525, 418, 411} \[ \frac{\tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f (a-b)}-\frac{b (a-3 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{2 (a-2 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[e + f*x]^4/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

(2*(a - 2*b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*(a - b)^2
*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - ((a - 3*b)*b*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]
*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a*(a - b)^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]
) + (Sech[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*(a - b)*f)

Rule 3192

Int[cos[(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*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2
)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !IntegerQ
[p]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, 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 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]

Rubi steps

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

Mathematica [C]  time = 2.1519, size = 219, normalized size = 1. \[ \frac{-2 i \left (2 a^2-5 a b+3 b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\frac{\tanh (e+f x) \text{sech}^2(e+f x) \left (\left (4 a^2-6 a b-2 b^2\right ) \cosh (2 (e+f x))+8 a^2+b (a-2 b) \cosh (4 (e+f x))-15 a b+4 b^2\right )}{\sqrt{2}}+4 i a (a-2 b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 f (a-b)^2 \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[e + f*x]^4/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

((4*I)*a*(a - 2*b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] - (2*I)*(2*a^2 - 5*a*b
+ 3*b^2)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] + ((8*a^2 - 15*a*b + 4*b^2 + (4*a
^2 - 6*a*b - 2*b^2)*Cosh[2*(e + f*x)] + (a - 2*b)*b*Cosh[4*(e + f*x)])*Sech[e + f*x]^2*Tanh[e + f*x])/Sqrt[2])
/(6*(a - b)^2*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.22, size = 343, normalized size = 1.6 \begin{align*}{\frac{1}{3\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3} \left ({a}^{2}-2\,ab+{b}^{2} \right ) f}\sqrt{ \left ( a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}} \left ( 2\, \left ( \cosh \left ( fx+e \right ) \right ) ^{4}\sqrt{-{\frac{b}{a}}}b \left ( a-2\,b \right ) \sinh \left ( fx+e \right ) + \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sqrt{-{\frac{b}{a}}} \left ( 2\,{a}^{2}-5\,ab+3\,{b}^{2} \right ) \sinh \left ( fx+e \right ) +\sqrt{-{\frac{b}{a}}} \left ({a}^{2}-2\,ab+{b}^{2} \right ) \sinh \left ( fx+e \right ) +\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}b \left ( a{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -b{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -2\,{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) a+4\,b{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{b \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( a-b \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),x)

[Out]

1/3*((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2)/cosh(f*x+e)^3/(-1/a*b)^(1/2)/(b*cosh(f*x+e)^4+(a-b)*cosh(f*x+e)^
2)^(1/2)/(a^2-2*a*b+b^2)*(2*cosh(f*x+e)^4*(-1/a*b)^(1/2)*b*(a-2*b)*sinh(f*x+e)+cosh(f*x+e)^2*(-1/a*b)^(1/2)*(2
*a^2-5*a*b+3*b^2)*sinh(f*x+e)+(-1/a*b)^(1/2)*(a^2-2*a*b+b^2)*sinh(f*x+e)+(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(co
sh(f*x+e)^2)^(1/2)*b*(a*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-b*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/
2),(a/b)^(1/2))-2*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a+4*b*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2)
,(a/b)^(1/2)))*cosh(f*x+e)^2)/(a+b*sinh(f*x+e)^2)^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}\left (f x + e\right )^{4}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{sech}\left (f x + e\right )^{4}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\left (e + f x \right )}}{\sqrt{a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)**4/(a+b*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(sech(e + f*x)**4/sqrt(a + b*sinh(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}\left (f x + e\right )^{4}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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