3.4.58 \(\int \cosh ^2(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx\) [358]

Optimal. Leaf size=223 \[ \frac {\cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f}-\frac {(a+b) E\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {2 F\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(a+b) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 b f} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 223, 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 = {3271, 428, 545, 429, 506, 422} \begin {gather*} \frac {2 \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right )}{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}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right )}{3 b 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) \sqrt {a+b \sinh ^2(e+f x)}}{3 b f}+\frac {\sinh (e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

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

Rule 422

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

Rule 428

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

Rule 429

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

Rule 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[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 545

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 3271

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.83, size = 168, normalized size = 0.75 \begin {gather*} \frac {-2 i \sqrt {2} a (a+b) \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )+2 i \sqrt {2} a (a-b) \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )+b (2 a-b+b \cosh (2 (e+f x))) \sinh (2 (e+f x))}{6 b f \sqrt {4 a-2 b+2 b \cosh (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[2]*a*(a + b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] + (2*I)*Sqrt[2]*
a*(a - b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] + b*(2*a - b + b*Cosh[2*(e + f*x
)])*Sinh[2*(e + f*x)])/(6*b*f*Sqrt[4*a - 2*b + 2*b*Cosh[2*(e + f*x)]])

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Maple [A]
time = 1.45, size = 351, normalized size = 1.57

method	result	size
default	\(\frac {\sqrt {-\frac {b}{a}}\, b \left (\cosh ^{4}\left (f x +e \right )\right ) \sinh \left (f x +e \right )+\sqrt {-\frac {b}{a}}\, a \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )-\sqrt {-\frac {b}{a}}\, b \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )+a \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b +\sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a +\sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b}{3 \sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f}\)	\(351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((-1/a*b)^(1/2)*b*cosh(f*x+e)^4*sinh(f*x+e)+(-1/a*b)^(1/2)*a*cosh(f*x+e)^2*sinh(f*x+e)-(-1/a*b)^(1/2)*b*co
sh(f*x+e)^2*sinh(f*x+e)+a*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(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)*EllipticF(sinh(f*x+e)*(-1/a*b)^(
1/2),(a/b)^(1/2))*b+(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))*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)/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sinh(f*x + e)^2 + a)*cosh(f*x + e)^2, x)

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Fricas [F]
time = 0.10, size = 25, normalized size = 0.11 \begin {gather*} {\rm integral}\left (\sqrt {b \sinh \left (f x + e\right )^{2} + a} \cosh \left (f x + e\right )^{2}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \sinh ^{2}{\left (e + f x \right )}} \cosh ^{2}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sinh(e + f*x)**2)*cosh(e + f*x)**2, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^2*(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,sageVARx):;OUTP
UT:Error: Bad Argument Type

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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