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

Optimal. Leaf size=300 \[ -\frac {\left (2 a^2+3 a b-8 b^2\right ) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b^2 f}+\frac {\left (2 a^2+3 a b-8 b^2\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(a-4 b) \sinh (e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}+\frac {\sinh ^3(e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{5 f}-\frac {(a-4 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

1/15*(a-4*b)*cosh(f*x+e)*sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/b/f+1/5*cosh(f*x+e)*sinh(f*x+e)^3*(a+b*sinh(f*x
+e)^2)^(1/2)/f+1/15*(2*a^2+3*a*b-8*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-4*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+3*a*b-8*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 = 0.33, antiderivative size = 300, 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 = {3188, 478, 582, 531, 418, 492, 411} \[ -\frac {\left (2 a^2+3 a b-8 b^2\right ) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b^2 f}+\frac {\left (2 a^2+3 a b-8 b^2\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\sinh ^3(e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{5 f}+\frac {(a-4 b) \sinh (e+f x) \cosh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{15 b f}-\frac {(a-4 b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{15 b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - 4*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]^3*
Sqrt[a + b*Sinh[e + f*x]^2])/(5*f) + ((2*a^2 + 3*a*b - 8*b^2)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[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 - 4*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 + 3*a*b - 8*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 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 478

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

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 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 582

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

Rule 3188

Int[sin[(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^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/Sqrt[1 - ff^2*x^2], x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rubi steps

\begin {align*} \int \sinh ^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 {x^4 \sqrt {a+b x^2}}{\sqrt {1+x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac {\cosh (e+f x) \sinh ^3(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{5 f}-\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (3 a+(-a+4 b) x^2\right )}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{5 f}\\ &=\frac {(a-4 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 ^3(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{5 f}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {-a (a-4 b)+\left (-2 a^2-3 a b+8 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 {(a-4 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 ^3(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{5 f}-\frac {\left (a (a-4 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^2-3 a b+8 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 {(a-4 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 ^3(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{5 f}-\frac {(a-4 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+3 a b-8 b^2\right ) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{15 b^2 f}-\frac {\left (\left (-2 a^2-3 a b+8 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 {(a-4 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 ^3(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{5 f}+\frac {\left (2 a^2+3 a b-8 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-4 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+3 a b-8 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 = 1.38, size = 210, normalized size = 0.70 \[ \frac {\sqrt {2} b \sinh (2 (e+f x)) \left (8 a^2+4 b (4 a-7 b) \cosh (2 (e+f x))-48 a b+3 b^2 \cosh (4 (e+f x))+25 b^2\right )-32 i a \left (a^2+a b-2 b^2\right ) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )+16 i a \left (2 a^2+3 a b-8 b^2\right ) \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )}{240 b^2 f \sqrt {2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*I)*a*(2*a^2 + 3*a*b - 8*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 + a*b - 2*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
- 48*a*b + 25*b^2 + 4*(4*a - 7*b)*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 = 3.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sinh \left (f x + e\right )^{2} + a} \sinh \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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)*sinh(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(sinh(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 = 0.14, size = 512, normalized size = 1.71 \[ \frac {3 \sqrt {-\frac {b}{a}}\, b^{2} \left (\sinh ^{7}\left (f x +e \right )\right )+4 \sqrt {-\frac {b}{a}}\, a b \left (\sinh ^{5}\left (f x +e \right )\right )-\sqrt {-\frac {b}{a}}\, b^{2} \left (\sinh ^{5}\left (f x +e \right )\right )+\sqrt {-\frac {b}{a}}\, a^{2} \left (\sinh ^{3}\left (f x +e \right )\right )-4 \sqrt {-\frac {b}{a}}\, b^{2} \left (\sinh ^{3}\left (f x +e \right )\right )+a^{2} \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{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 )+7 a \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{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 -8 \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{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 {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{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}-3 \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{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 +8 \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{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}+\sqrt {-\frac {b}{a}}\, a^{2} \sinh \left (f x +e \right )-4 \sqrt {-\frac {b}{a}}\, a b \sinh \left (f x +e \right )}{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(sinh(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)^7+4*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^5-(-1/a*b)^(1/2)*b^2*sinh(f*x+e)^5+(
-1/a*b)^(1/2)*a^2*sinh(f*x+e)^3-4*(-1/a*b)^(1/2)*b^2*sinh(f*x+e)^3+a^2*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x
+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))+7*a*((a+b*sinh(f*x+e)^2)/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-8*((a+b*sinh(f*x+e)^2)/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*((a+b*sinh(f*x+e)^2)/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^2-3*((a+b*sinh(f*x+e)^2)/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+8*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*E
llipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^2+(-1/a*b)^(1/2)*a^2*sinh(f*x+e)-4*(-1/a*b)^(1/2)*a*b*sinh(
f*x+e))/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 \[ \int \sqrt {b \sinh \left (f x + e\right )^{2} + a} \sinh \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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)*sinh(f*x + e)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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