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

Optimal. Leaf size=206 \[ \frac {\tanh (e+f x) \text {sech}^2(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}-\frac {b \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(2 a-b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

1/3*(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)/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)-1/3*b
*(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)/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)+1/3*sech(f*x+e)^
2*(a+b*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)/f

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Rubi [A]  time = 0.17, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3192, 412, 525, 418, 411} \[ \frac {\tanh (e+f x) \text {sech}^2(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}-\frac {b \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(2 a-b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[e + f*x]^4*Sqrt[a + b*Sinh[e + f*x]^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*(a - b)*f*S
qrt[(Sech[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*(a - b)*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*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 412

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

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 3.01, size = 204, normalized size = 0.99 \[ \frac {\sqrt {2} \tanh (e+f x) \text {sech}^2(e+f x) \left (\left (8 a^2-4 b^2\right ) \cosh (2 (e+f x))+(2 a-b) (8 a+b \cosh (4 (e+f x))-5 b)\right )-16 i a (a-b) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )+8 i a (2 a-b) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )}{24 f (a-b) \sqrt {2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*I)*a*(2*a - b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] - (16*I)*a*(a - b)*Sqrt
[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] + Sqrt[2]*((8*a^2 - 4*b^2)*Cosh[2*(e + f*x)] +
(2*a - b)*(8*a - 5*b + b*Cosh[4*(e + f*x)]))*Sech[e + f*x]^2*Tanh[e + f*x])/(24*(a - b)*f*Sqrt[2*a - b + b*Co
sh[2*(e + f*x)]])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sinh \left (f x + e\right )^{2} + a} \operatorname {sech}\left (f x + e\right )^{4}, x\right ) \]

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(sqrt(b*sinh(f*x + e)^2 + a)*sech(f*x + e)^4, 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(sech(f*x+e)^4*(a+b*sinh(f*x+e)^2)^(1/2),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 = 0.25, size = 318, normalized size = 1.54 \[ \frac {\left (2 \sqrt {-\frac {b}{a}}\, a b -\sqrt {-\frac {b}{a}}\, b^{2}\right ) \sinh \left (f x +e \right ) \left (\cosh ^{4}\left (f x +e \right )\right )+\left (2 \sqrt {-\frac {b}{a}}\, a^{2}-2 \sqrt {-\frac {b}{a}}\, a b \right ) \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )+\left (\sqrt {-\frac {b}{a}}\, a^{2}-2 \sqrt {-\frac {b}{a}}\, a b +\sqrt {-\frac {b}{a}}\, b^{2}\right ) \sinh \left (f x +e \right )+\sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, b \left (a \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-b \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-2 \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a +b \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )\right ) \left (\cosh ^{2}\left (f x +e \right )\right )}{3 \cosh \left (f x +e \right )^{3} \left (a -b \right ) \sqrt {-\frac {b}{a}}\, \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((2*(-1/a*b)^(1/2)*a*b-(-1/a*b)^(1/2)*b^2)*sinh(f*x+e)*cosh(f*x+e)^4+(2*(-1/a*b)^(1/2)*a^2-2*(-1/a*b)^(1/2
)*a*b)*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)+(b/a
*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*b*(a*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-b*E
llipticF(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+b*Ellip
ticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2)))*cosh(f*x+e)^2)/cosh(f*x+e)^3/(a-b)/(-1/a*b)^(1/2)/(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 \[ \int \sqrt {b \sinh \left (f x + e\right )^{2} + a} \operatorname {sech}\left (f x + e\right )^{4}\,{d x} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sinh ^{2}{\left (e + f x \right )}} \operatorname {sech}^{4}{\left (e + f x \right )}\, dx \]

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

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