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

Optimal. Leaf size=305 \[ \frac{(3 a-8 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\tanh ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a-2 b) \sinh ^2(e+f x) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{f}+\frac{8 (a-2 b) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}-\frac{(3 a-8 b) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}-\frac{8 (a-2 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

-((3*a - 8*b)*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*f) - (8*(a - 2*b)*EllipticE[ArcTan[S
inh[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]) + ((3*a - 8*b)*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]) + (8*(a - 2*b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e +
f*x])/(3*f) + ((a - 2*b)*Sinh[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/f - ((a + b*Sinh[e + f*x]
^2)^(3/2)*Tanh[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.379704, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3196, 467, 577, 582, 531, 418, 492, 411} \[ -\frac{\tanh ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a-2 b) \sinh ^2(e+f x) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{f}+\frac{8 (a-2 b) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}-\frac{(3 a-8 b) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}+\frac{(3 a-8 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{8 (a-2 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a - 8*b)*Cosh[e + f*x]*Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*f) - (8*(a - 2*b)*EllipticE[ArcTan[S
inh[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]) + ((3*a - 8*b)*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]) + (8*(a - 2*b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e +
f*x])/(3*f) + ((a - 2*b)*Sinh[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/f - ((a + b*Sinh[e + f*x]
^2)^(3/2)*Tanh[e + f*x]^3)/(3*f)

Rule 3196

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), 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)/(1 - ff^2*x^2)^((m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rule 467

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*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 577

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

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 \left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^4(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 \left (a+b x^2\right )^{3/2}}{\left (1+x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^3(e+f x)}{3 f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2 \sqrt{a+b x^2} \left (3 a+6 b x^2\right )}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 f}\\ &=\frac{(a-2 b) \sinh ^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}-\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^3(e+f x)}{3 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 (a-3 b)+3 (3 a-8 b) b x^2\right )}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 f}\\ &=-\frac{(3 a-8 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}+\frac{(a-2 b) \sinh ^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}-\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^3(e+f x)}{3 f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{3 a (3 a-8 b) b+24 (a-2 b) b^2 x^2}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{9 b f}\\ &=-\frac{(3 a-8 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}+\frac{(a-2 b) \sinh ^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}-\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^3(e+f x)}{3 f}+\frac{\left (a (3 a-8 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 )}{3 f}+\frac{\left (8 (a-2 b) b \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 )}{3 f}\\ &=-\frac{(3 a-8 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}+\frac{(3 a-8 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)}}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{8 (a-2 b) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}+\frac{(a-2 b) \sinh ^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}-\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^3(e+f x)}{3 f}-\frac{\left (8 (a-2 b) \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 )}{3 f}\\ &=-\frac{(3 a-8 b) \cosh (e+f x) \sinh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f}-\frac{8 (a-2 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 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(3 a-8 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)}}{3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{8 (a-2 b) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 f}+\frac{(a-2 b) \sinh ^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{f}-\frac{\left (a+b \sinh ^2(e+f x)\right )^{3/2} \tanh ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [C]  time = 2.79562, size = 224, normalized size = 0.73 \[ \frac{4 i a (5 a-8 b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-\frac{\tanh (e+f x) \text{sech}^2(e+f x) \left (\left (64 a^2-160 a b+17 b^2\right ) \cosh (2 (e+f x))+32 a^2+2 b (6 a-17 b) \cosh (4 (e+f x))-108 a b-b^2 \cosh (6 (e+f x))+18 b^2\right )}{4 \sqrt{2}}-32 i a (a-2 b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{12 f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-32*I)*a*(a - 2*b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] + (4*I)*a*(5*a - 8*b)
*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] - ((32*a^2 - 108*a*b + 18*b^2 + (64*a^2 -
160*a*b + 17*b^2)*Cosh[2*(e + f*x)] + 2*(6*a - 17*b)*b*Cosh[4*(e + f*x)] - b^2*Cosh[6*(e + f*x)])*Sech[e + f*
x]^2*Tanh[e + f*x])/(4*Sqrt[2]))/(12*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.167, size = 389, normalized size = 1.3 \begin{align*} -{\frac{1}{3\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3}f} \left ( -\sqrt{-{\frac{b}{a}}}{b}^{2}\sinh \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{6}+ \left ( 3\,\sqrt{-{\frac{b}{a}}}ab-7\,\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{4}\sinh \left ( fx+e \right ) + \left ( 4\,\sqrt{-{\frac{b}{a}}}{a}^{2}-13\,\sqrt{-{\frac{b}{a}}}ab+9\,\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) + \left ( -\sqrt{-{\frac{b}{a}}}{a}^{2}+2\,\sqrt{-{\frac{b}{a}}}ab-\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \sinh \left ( fx+e \right ) -\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}} \left ( 3\,{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ){a}^{2}-16\,{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) ab+16\,{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ){b}^{2}+8\,{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) ab-16\,{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ){b}^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-(-1/a*b)^(1/2)*b^2*sinh(f*x+e)*cosh(f*x+e)^6+(3*(-1/a*b)^(1/2)*a*b-7*(-1/a*b)^(1/2)*b^2)*cosh(f*x+e)^4*
sinh(f*x+e)+(4*(-1/a*b)^(1/2)*a^2-13*(-1/a*b)^(1/2)*a*b+9*(-1/a*b)^(1/2)*b^2)*cosh(f*x+e)^2*sinh(f*x+e)+(-(-1/
a*b)^(1/2)*a^2+2*(-1/a*b)^(1/2)*a*b-(-1/a*b)^(1/2)*b^2)*sinh(f*x+e)-(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*
x+e)^2)^(1/2)*(3*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a^2-16*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2)
,(a/b)^(1/2))*a*b+16*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^2+8*EllipticE(sinh(f*x+e)*(-1/a*b)^(1
/2),(a/b)^(1/2))*a*b-16*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^2)*cosh(f*x+e)^2)/(-1/a*b)^(1/2)/c
osh(f*x+e)^3/(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}} \tanh \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(3/2)*tanh(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 )^{2} + a\right )}^{\frac{3}{2}} \tanh \left (f x + e\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sinh(f*x + e)^2 + a)^(3/2)*tanh(f*x + e)^4, 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((a+b*sinh(f*x+e)**2)**(3/2)*tanh(f*x+e)**4,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}} \tanh \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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