3.357 \(\int \cosh ^4(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx\)

Optimal. Leaf size=301 \[ -\frac {\left (2 a^2-7 a b-3 b^2\right ) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b^2 f}+\frac {\left (2 a^2-7 a b-3 b^2\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\sinh (e+f x) \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {2 (a-3 b) \sinh (e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}-\frac {(a-9 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

1/5*cosh(f*x+e)*sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(3/2)/b/f-2/15*(a-3*b)*cosh(f*x+e)*sinh(f*x+e)*(a+b*sinh(f*x+e
)^2)^(1/2)/b/f+1/15*(2*a^2-7*a*b-3*b^2)*(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^2/f/(sech(f*x+e)^2*(a+b*sin
h(f*x+e)^2)/a)^(1/2)-1/15*(a-9*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)/b/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)
^2)/a)^(1/2)-1/15*(2*a^2-7*a*b-3*b^2)*(a+b*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)/b^2/f

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Rubi [A]  time = 0.30, antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3192, 416, 528, 531, 418, 492, 411} \[ -\frac {\left (2 a^2-7 a b-3 b^2\right ) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b^2 f}+\frac {\left (2 a^2-7 a b-3 b^2\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\sinh (e+f x) \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {2 (a-3 b) \sinh (e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}-\frac {(a-9 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - 3*b)*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(15*b*f) + (Cosh[e + f*x]*Sinh[e + f*x]
*(a + b*Sinh[e + f*x]^2)^(3/2))/(5*b*f) + ((2*a^2 - 7*a*b - 3*b^2)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*S
ech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(15*b^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - ((a -
9*b)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(15*b*f*Sqrt[(Sech[
e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - ((2*a^2 - 7*a*b - 3*b^2)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/
(15*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 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[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 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 528

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

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 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 \cosh ^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 \left (1+x^2\right )^{3/2} \sqrt {a+b x^2} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac {\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 b f}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {\left (-a+5 b-2 (a-3 b) x^2\right ) \sqrt {a+b x^2}}{\sqrt {1+x^2}} \, dx,x,\sinh (e+f x)\right )}{5 b f}\\ &=-\frac {2 (a-3 b) \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}+\frac {\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 b f}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {-a (a-9 b)+\left (-2 a^2+7 a b+3 b^2\right ) x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{15 b f}\\ &=-\frac {2 (a-3 b) \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}+\frac {\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {\left (a (a-9 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 )}{15 b f}+\frac {\left (\left (-2 a^2+7 a b+3 b^2\right ) \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 )}{15 b f}\\ &=-\frac {2 (a-3 b) \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}+\frac {\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {(a-9 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)}}{15 b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {\left (2 a^2-7 a b-3 b^2\right ) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{15 b^2 f}-\frac {\left (\left (-2 a^2+7 a b+3 b^2\right ) \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 )}{15 b^2 f}\\ &=-\frac {2 (a-3 b) \cosh (e+f x) \sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}+\frac {\cosh (e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{5 b f}+\frac {\left (2 a^2-7 a b-3 b^2\right ) 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)}}{15 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {(a-9 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)}}{15 b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {\left (2 a^2-7 a b-3 b^2\right ) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{15 b^2 f}\\ \end {align*}

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Mathematica [C]  time = 1.41, size = 211, normalized size = 0.70 \[ \frac {\sqrt {2} b \sinh (2 (e+f x)) \left (8 a^2+4 b (4 a+3 b) \cosh (2 (e+f x))+32 a b+3 b^2 \cosh (4 (e+f x))-15 b^2\right )-32 i a \left (a^2-4 a b+3 b^2\right ) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )+16 i a \left (2 a^2-7 a b-3 b^2\right ) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )}{240 b^2 f \sqrt {2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*I)*a*(2*a^2 - 7*a*b - 3*b^2)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] - (32*I)
*a*(a^2 - 4*a*b + 3*b^2)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] + Sqrt[2]*b*(8*a^
2 + 32*a*b - 15*b^2 + 4*b*(4*a + 3*b)*Cosh[2*(e + f*x)] + 3*b^2*Cosh[4*(e + f*x)])*Sinh[2*(e + f*x)])/(240*b^2
*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sinh \left (f x + e\right )^{2} + a} \cosh \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(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)*cosh(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(cosh(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 = 0.19, size = 521, normalized size = 1.73 \[ \frac {3 \sqrt {-\frac {b}{a}}\, b^{2} \sinh \left (f x +e \right ) \left (\cosh ^{6}\left (f x +e \right )\right )+4 \sqrt {-\frac {b}{a}}\, a b \sinh \left (f x +e \right ) \left (\cosh ^{4}\left (f x +e \right )\right )+\left (\sqrt {-\frac {b}{a}}\, a^{2}+2 \sqrt {-\frac {b}{a}}\, a b -3 \sqrt {-\frac {b}{a}}\, b^{2}\right ) \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )+a^{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}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )+2 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 ) b -3 \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}-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^{2}+7 \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 b +3 \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^{2}}{15 b \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(cosh(f*x+e)^4*(a+b*sinh(f*x+e)^2)^(1/2),x)

[Out]

1/15*(3*(-1/a*b)^(1/2)*b^2*sinh(f*x+e)*cosh(f*x+e)^6+4*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)*cosh(f*x+e)^4+((-1/a*b)^
(1/2)*a^2+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)+a^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))+2*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-3*(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^2-2*(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^2+7*(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+3*(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^2)/b/(-1/a*b)^(1/2)/cos
h(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 \[ \int \sqrt {b \sinh \left (f x + e\right )^{2} + a} \cosh \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(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)*cosh(f*x + e)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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