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

Optimal. Leaf size=193 \[ -\frac{b \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(a-b) \tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}+\frac{2 (a+b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

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

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Rubi [A]  time = 0.187938, antiderivative size = 193, 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, 413, 525, 418, 411} \[ \frac{(a-b) \tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}-\frac{b \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{2 (a+b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*f*Sqrt[(Sec
h[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - (b*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a
+ b*Sinh[e + f*x]^2])/(3*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((a - b)*Sech[e + f*x]^2*Sqrt[
a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*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 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[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 \text{sech}^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{\left (a+b x^2\right )^{3/2}}{\left (1+x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{(a-b) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{a (2 a+b)+b (a+2 b) x^2}{\left (1+x^2\right )^{3/2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 f}\\ &=\frac{(a-b) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}-\frac{\left (a 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 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 )}{3 f}\\ &=\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)}}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{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 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(a-b) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}\\ \end{align*}

Mathematica [C]  time = 1.96538, size = 197, normalized size = 1.02 \[ \frac{-2 i a (2 a+b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\frac{\tanh (e+f x) \text{sech}^2(e+f x) \left (\left (4 a^2+6 a b-2 b^2\right ) \cosh (2 (e+f x))+8 a^2+b (a+b) \cosh (4 (e+f x))-3 a b+b^2\right )}{\sqrt{2}}+4 i a (a+b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.116, size = 324, normalized size = 1.7 \begin{align*}{\frac{1}{3\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3}f} \left ( \left ( 2\,\sqrt{-{\frac{b}{a}}}ab+2\,\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \sinh \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( 2\,\sqrt{-{\frac{b}{a}}}{a}^{2}+\sqrt{-{\frac{b}{a}}}ab-3\,\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) + \left ( \sqrt{-{\frac{b}{a}}}{a}^{2}-2\,\sqrt{-{\frac{b}{a}}}ab+\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \sinh \left ( fx+e \right ) +\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}b \left ( a{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) +2\,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-2\,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{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)^(3/2),x)

[Out]

1/3*((2*(-1/a*b)^(1/2)*a*b+2*(-1/a*b)^(1/2)*b^2)*sinh(f*x+e)*cosh(f*x+e)^4+(2*(-1/a*b)^(1/2)*a^2+(-1/a*b)^(1/2
)*a*b-3*(-1/a*b)^(1/2)*b^2)*cosh(f*x+e)^2*sinh(f*x+e)+((-1/a*b)^(1/2)*a^2-2*(-1/a*b)^(1/2)*a*b+(-1/a*b)^(1/2)*
b^2)*sinh(f*x+e)+(cosh(f*x+e)^2)^(1/2)*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*b*(a*EllipticF(sinh(f*x+e)*(-1/a*b)^(
1/2),(a/b)^(1/2))+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-2*b*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2)))*cosh(f*x+e)^2)/cosh(f*x+e)^3/(-1/a*b)^(
1/2)/(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{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \operatorname{sech}\left (f x + e\right )^{4}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \operatorname{sech}\left (f x + e\right )^{4} \sinh \left (f x + e\right )^{2} + a \operatorname{sech}\left (f x + e\right )^{4}\right )} \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)^(3/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \operatorname{sech}\left (f x + e\right )^{4}\,{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)^(3/2),x, algorithm="giac")

[Out]

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