3.4.57 \(\int \cosh ^4(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \, dx\) [357]

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

[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 = 0.20, 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 = {3271, 427, 542, 545, 429, 506, 422} \begin {gather*} \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 (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right )}{15 b^2 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 ) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b^2 f}-\frac {(a-9 b) \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 )}{15 b 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} \end {gather*}

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 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 427

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 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 542

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 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 ^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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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] Result contains complex when optimal does not.
time = 1.01, size = 211, normalized size = 0.70 \begin {gather*} \frac {16 i a \left (2 a^2-7 a b-3 b^2\right ) \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )-32 i a \left (a^2-4 a b+3 b^2\right ) \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )+\sqrt {2} b \left (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))\right ) \sinh (2 (e+f x))}{240 b^2 f \sqrt {2 a-b+b \cosh (2 (e+f x))}} \end {gather*}

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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Maple [A]
time = 1.75, size = 521, normalized size = 1.73

method	result	size
default	\(\frac {3 \sqrt {-\frac {b}{a}}\, b^{2} \left (\cosh ^{6}\left (f x +e \right )\right ) \sinh \left (f x +e \right )+4 \sqrt {-\frac {b}{a}}\, a b \left (\cosh ^{4}\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 -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}\)	\(521\)

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,method=_RETURNVERBOSE)

[Out]

1/15*(3*(-1/a*b)^(1/2)*b^2*cosh(f*x+e)^6*sinh(f*x+e)+4*(-1/a*b)^(1/2)*a*b*cosh(f*x+e)^4*sinh(f*x+e)+((-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.10, size = 25, normalized size = 0.08 \begin {gather*} {\rm integral}\left (\sqrt {b \sinh \left (f x + e\right )^{2} + a} \cosh \left (f x + e\right )^{4}, x\right ) \end {gather*}

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

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]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

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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)^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,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 )}^4\,\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)^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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