3.82 \(\int \sinh ^4(e+f x) (a+b \sinh ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=367 \[ -\frac{\left (a^2-11 a b+8 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{35 b f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 (a-2 b) \left (a^2+4 a b-4 b^2\right ) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b^2 f}+\frac{\left (a^2-11 a b+8 b^2\right ) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b f}+\frac{2 (a-2 b) \left (a^2+4 a b-4 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{35 b^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{b \sinh ^5(e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}+\frac{2 (4 a-3 b) \sinh ^3(e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 f} \]

[Out]

((a^2 - 11*a*b + 8*b^2)*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(35*b*f) + (2*(4*a - 3*b)*Cos
h[e + f*x]*Sinh[e + f*x]^3*Sqrt[a + b*Sinh[e + f*x]^2])/(35*f) + (b*Cosh[e + f*x]*Sinh[e + f*x]^5*Sqrt[a + b*S
inh[e + f*x]^2])/(7*f) + (2*(a - 2*b)*(a^2 + 4*a*b - 4*b^2)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e +
f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(35*b^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - ((a^2 - 11*a
*b + 8*b^2)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(35*b*f*Sqrt[
(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - (2*(a - 2*b)*(a^2 + 4*a*b - 4*b^2)*Sqrt[a + b*Sinh[e + f*x]^2]
*Tanh[e + f*x])/(35*b^2*f)

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Rubi [A]  time = 0.470834, antiderivative size = 367, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3188, 477, 582, 531, 418, 492, 411} \[ -\frac{2 (a-2 b) \left (a^2+4 a b-4 b^2\right ) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b^2 f}+\frac{\left (a^2-11 a b+8 b^2\right ) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b f}-\frac{\left (a^2-11 a b+8 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{35 b f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{2 (a-2 b) \left (a^2+4 a b-4 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{35 b^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{b \sinh ^5(e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}+\frac{2 (4 a-3 b) \sinh ^3(e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 f} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[e + f*x]^4*(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

((a^2 - 11*a*b + 8*b^2)*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(35*b*f) + (2*(4*a - 3*b)*Cos
h[e + f*x]*Sinh[e + f*x]^3*Sqrt[a + b*Sinh[e + f*x]^2])/(35*f) + (b*Cosh[e + f*x]*Sinh[e + f*x]^5*Sqrt[a + b*S
inh[e + f*x]^2])/(7*f) + (2*(a - 2*b)*(a^2 + 4*a*b - 4*b^2)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e +
f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(35*b^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - ((a^2 - 11*a
*b + 8*b^2)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(35*b*f*Sqrt[
(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) - (2*(a - 2*b)*(a^2 + 4*a*b - 4*b^2)*Sqrt[a + b*Sinh[e + f*x]^2]
*Tanh[e + f*x])/(35*b^2*f)

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]

Rule 477

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*(e*x)
^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(b*e*(m + n*(p + q) + 1)), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 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 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 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 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]

Rubi steps

\begin{align*} \int \sinh ^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx &=\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^2\right )^{3/2}}{\sqrt{1+x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{b \cosh (e+f x) \sinh ^5(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (a (7 a-5 b)+2 (4 a-3 b) b x^2\right )}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{7 f}\\ &=\frac{2 (4 a-3 b) \cosh (e+f x) \sinh ^3(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 f}+\frac{b \cosh (e+f x) \sinh ^5(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}-\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (6 a (4 a-3 b) b-3 b \left (a^2-11 a b+8 b^2\right ) x^2\right )}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{35 b f}\\ &=\frac{\left (a^2-11 a b+8 b^2\right ) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b f}+\frac{2 (4 a-3 b) \cosh (e+f x) \sinh ^3(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 f}+\frac{b \cosh (e+f x) \sinh ^5(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{-3 a b \left (a^2-11 a b+8 b^2\right )-6 b \left (a^3+2 a^2 b-12 a b^2+8 b^3\right ) x^2}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{105 b^2 f}\\ &=\frac{\left (a^2-11 a b+8 b^2\right ) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b f}+\frac{2 (4 a-3 b) \cosh (e+f x) \sinh ^3(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 f}+\frac{b \cosh (e+f x) \sinh ^5(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}-\frac{\left (a \left (a^2-11 a b+8 b^2\right ) \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 )}{35 b f}-\frac{\left (2 \left (a^3+2 a^2 b-12 a b^2+8 b^3\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 )}{35 b f}\\ &=\frac{\left (a^2-11 a b+8 b^2\right ) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b f}+\frac{2 (4 a-3 b) \cosh (e+f x) \sinh ^3(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 f}+\frac{b \cosh (e+f x) \sinh ^5(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}-\frac{\left (a^2-11 a b+8 b^2\right ) 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)}}{35 b f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 \left (a^3+2 a^2 b-12 a b^2+8 b^3\right ) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{35 b^2 f}+\frac{\left (2 \left (a^3+2 a^2 b-12 a b^2+8 b^3\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 )}{35 b^2 f}\\ &=\frac{\left (a^2-11 a b+8 b^2\right ) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 b f}+\frac{2 (4 a-3 b) \cosh (e+f x) \sinh ^3(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{35 f}+\frac{b \cosh (e+f x) \sinh ^5(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{7 f}+\frac{2 \left (a^3+2 a^2 b-12 a b^2+8 b^3\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)}}{35 b^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\left (a^2-11 a b+8 b^2\right ) 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)}}{35 b f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 \left (a^3+2 a^2 b-12 a b^2+8 b^3\right ) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{35 b^2 f}\\ \end{align*}

Mathematica [C]  time = 2.87267, size = 262, normalized size = 0.71 \[ \frac{-64 i a \left (3 a^2 b+2 a^3-13 a b^2+8 b^3\right ) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\sqrt{2} b \sinh (2 (e+f x)) \left (b \left (144 a^2-480 a b+299 b^2\right ) \cosh (2 (e+f x))-496 a^2 b+32 a^3+2 b^2 (26 a-27 b) \cosh (4 (e+f x))+684 a b^2+5 b^3 \cosh (6 (e+f x))-250 b^3\right )+128 i a \left (2 a^2 b+a^3-12 a b^2+8 b^3\right ) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{2240 b^2 f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[e + f*x]^4*(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

((128*I)*a*(a^3 + 2*a^2*b - 12*a*b^2 + 8*b^3)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b
/a] - (64*I)*a*(2*a^3 + 3*a^2*b - 13*a*b^2 + 8*b^3)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f
*x), b/a] + Sqrt[2]*b*(32*a^3 - 496*a^2*b + 684*a*b^2 - 250*b^3 + b*(144*a^2 - 480*a*b + 299*b^2)*Cosh[2*(e +
f*x)] + 2*(26*a - 27*b)*b^2*Cosh[4*(e + f*x)] + 5*b^3*Cosh[6*(e + f*x)])*Sinh[2*(e + f*x)])/(2240*b^2*f*Sqrt[2
*a - b + b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.074, size = 743, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x+e)^4*(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

1/35*(5*(-1/a*b)^(1/2)*b^3*sinh(f*x+e)*cosh(f*x+e)^8+(13*(-1/a*b)^(1/2)*a*b^2-21*(-1/a*b)^(1/2)*b^3)*cosh(f*x+
e)^6*sinh(f*x+e)+(9*(-1/a*b)^(1/2)*a^2*b-43*(-1/a*b)^(1/2)*a*b^2+35*(-1/a*b)^(1/2)*b^3)*cosh(f*x+e)^4*sinh(f*x
+e)+((-1/a*b)^(1/2)*a^3-20*(-1/a*b)^(1/2)*a^2*b+38*(-1/a*b)^(1/2)*a*b^2-19*(-1/a*b)^(1/2)*b^3)*cosh(f*x+e)^2*s
inh(f*x+e)+(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))*a^3+15*(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))*a^2*b-32*(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))*a*b^2+16*(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^3-2*(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^3-4*(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*b+24*(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^2-16*(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^3)/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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sinh \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^4*(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(3/2)*sinh(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sinh \left (f x + e\right )^{6} + a \sinh \left (f x + e\right )^{4}\right )} \sqrt{b \sinh \left (f x + e\right )^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^4*(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral((b*sinh(f*x + e)^6 + a*sinh(f*x + e)^4)*sqrt(b*sinh(f*x + e)^2 + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)**4*(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sinh \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^4*(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(3/2)*sinh(f*x + e)^4, x)