3.1.38 \(\int e^{\text {csch}^{-1}(a x^2)} x^4 \, dx\) [38]

Optimal. Leaf size=202 \[ -\frac {2 \sqrt {1+\frac {1}{a^2 x^4}}}{5 a^2 \left (a+\frac {1}{x^2}\right ) x}+\frac {2 \sqrt {1+\frac {1}{a^2 x^4}} x}{5 a^2}+\frac {x^3}{3 a}+\frac {1}{5} \sqrt {1+\frac {1}{a^2 x^4}} x^5+\frac {2 \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 a^{7/2} \sqrt {1+\frac {1}{a^2 x^4}}}-\frac {\sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 a^{7/2} \sqrt {1+\frac {1}{a^2 x^4}}} \]

[Out]

1/3*x^3/a-2/5*(1+1/a^2/x^4)^(1/2)/a^2/(a+1/x^2)/x+2/5*x*(1+1/a^2/x^4)^(1/2)/a^2+1/5*x^5*(1+1/a^2/x^4)^(1/2)+2/
5*(a+1/x^2)*(cos(2*arccot(x*a^(1/2)))^2)^(1/2)/cos(2*arccot(x*a^(1/2)))*EllipticE(sin(2*arccot(x*a^(1/2))),1/2
*2^(1/2))*((a^2+1/x^4)/(a+1/x^2)^2)^(1/2)/a^(7/2)/(1+1/a^2/x^4)^(1/2)-1/5*(a+1/x^2)*(cos(2*arccot(x*a^(1/2)))^
2)^(1/2)/cos(2*arccot(x*a^(1/2)))*EllipticF(sin(2*arccot(x*a^(1/2))),1/2*2^(1/2))*((a^2+1/x^4)/(a+1/x^2)^2)^(1
/2)/a^(7/2)/(1+1/a^2/x^4)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6471, 30, 342, 283, 331, 311, 226, 1210} \begin {gather*} \frac {2 x \sqrt {\frac {1}{a^2 x^4}+1}}{5 a^2}+\frac {1}{5} x^5 \sqrt {\frac {1}{a^2 x^4}+1}-\frac {2 \sqrt {\frac {1}{a^2 x^4}+1}}{5 a^2 x \left (a+\frac {1}{x^2}\right )}-\frac {\sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 a^{7/2} \sqrt {\frac {1}{a^2 x^4}+1}}+\frac {2 \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 a^{7/2} \sqrt {\frac {1}{a^2 x^4}+1}}+\frac {x^3}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcCsch[a*x^2]*x^4,x]

[Out]

(-2*Sqrt[1 + 1/(a^2*x^4)])/(5*a^2*(a + x^(-2))*x) + (2*Sqrt[1 + 1/(a^2*x^4)]*x)/(5*a^2) + x^3/(3*a) + (Sqrt[1
+ 1/(a^2*x^4)]*x^5)/5 + (2*Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticE[2*ArcCot[Sqrt[a]*x], 1/2
])/(5*a^(7/2)*Sqrt[1 + 1/(a^2*x^4)]) - (Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticF[2*ArcCot[Sq
rt[a]*x], 1/2])/(5*a^(7/2)*Sqrt[1 + 1/(a^2*x^4)])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 283

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

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 6471

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Dist[1/a, Int[x^(m - p), x], x] + Int[x^m*Sqrt[1 + 1/
(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]

Rubi steps

\begin {align*} \int e^{\text {csch}^{-1}\left (a x^2\right )} x^4 \, dx &=\frac {\int x^2 \, dx}{a}+\int \sqrt {1+\frac {1}{a^2 x^4}} x^4 \, dx\\ &=\frac {x^3}{3 a}-\text {Subst}\left (\int \frac {\sqrt {1+\frac {x^4}{a^2}}}{x^6} \, dx,x,\frac {1}{x}\right )\\ &=\frac {x^3}{3 a}+\frac {1}{5} \sqrt {1+\frac {1}{a^2 x^4}} x^5-\frac {2 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+\frac {x^4}{a^2}}} \, dx,x,\frac {1}{x}\right )}{5 a^2}\\ &=\frac {2 \sqrt {1+\frac {1}{a^2 x^4}} x}{5 a^2}+\frac {x^3}{3 a}+\frac {1}{5} \sqrt {1+\frac {1}{a^2 x^4}} x^5-\frac {2 \text {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{a^2}}} \, dx,x,\frac {1}{x}\right )}{5 a^4}\\ &=\frac {2 \sqrt {1+\frac {1}{a^2 x^4}} x}{5 a^2}+\frac {x^3}{3 a}+\frac {1}{5} \sqrt {1+\frac {1}{a^2 x^4}} x^5-\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{a^2}}} \, dx,x,\frac {1}{x}\right )}{5 a^3}+\frac {2 \text {Subst}\left (\int \frac {1-\frac {x^2}{a}}{\sqrt {1+\frac {x^4}{a^2}}} \, dx,x,\frac {1}{x}\right )}{5 a^3}\\ &=-\frac {2 \sqrt {1+\frac {1}{a^2 x^4}}}{5 a^2 \left (a+\frac {1}{x^2}\right ) x}+\frac {2 \sqrt {1+\frac {1}{a^2 x^4}} x}{5 a^2}+\frac {x^3}{3 a}+\frac {1}{5} \sqrt {1+\frac {1}{a^2 x^4}} x^5+\frac {2 \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 a^{7/2} \sqrt {1+\frac {1}{a^2 x^4}}}-\frac {\sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 a^{7/2} \sqrt {1+\frac {1}{a^2 x^4}}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.18, size = 112, normalized size = 0.55 \begin {gather*} \frac {4 \sqrt {2} e^{-\text {csch}^{-1}\left (a x^2\right )} \left (\frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{-1+e^{2 \text {csch}^{-1}\left (a x^2\right )}}\right )^{5/2} x^5 \left (-4+7 e^{2 \text {csch}^{-1}\left (a x^2\right )}+4 \left (1-e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )^{5/2} \, _2F_1\left (\frac {3}{4},\frac {7}{2};\frac {7}{4};e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )\right )}{21 \left (a x^2\right )^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCsch[a*x^2]*x^4,x]

[Out]

(4*Sqrt[2]*(E^ArcCsch[a*x^2]/(-1 + E^(2*ArcCsch[a*x^2])))^(5/2)*x^5*(-4 + 7*E^(2*ArcCsch[a*x^2]) + 4*(1 - E^(2
*ArcCsch[a*x^2]))^(5/2)*Hypergeometric2F1[3/4, 7/2, 7/4, E^(2*ArcCsch[a*x^2])]))/(21*E^ArcCsch[a*x^2]*(a*x^2)^
(5/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.08, size = 152, normalized size = 0.75

method	result	size
default	\(-\frac {\sqrt {\frac {a^{2} x^{4}+1}{a^{2} x^{4}}}\, x^{2} \left (-\sqrt {i a}\, a^{3} x^{7}-x^{3} a \sqrt {i a}+2 i \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \EllipticE \left (x \sqrt {i a}, i\right )-2 i \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \EllipticF \left (x \sqrt {i a}, i\right )\right )}{5 \left (a^{2} x^{4}+1\right ) a \sqrt {i a}}+\frac {x^{3}}{3 a}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^4,x,method=_RETURNVERBOSE)

[Out]

-1/5*((a^2*x^4+1)/a^2/x^4)^(1/2)*x^2*(-(I*a)^(1/2)*a^3*x^7-x^3*a*(I*a)^(1/2)+2*I*(1-I*a*x^2)^(1/2)*(1+I*a*x^2)
^(1/2)*EllipticE(x*(I*a)^(1/2),I)-2*I*(1-I*a*x^2)^(1/2)*(1+I*a*x^2)^(1/2)*EllipticF(x*(I*a)^(1/2),I))/(a^2*x^4
+1)/a/(I*a)^(1/2)+1/3*x^3/a

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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((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^4,x, algorithm="maxima")

[Out]

1/3*x^3/a + integrate(sqrt(a^2*x^4 + 1)*x^2, x)/a

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Fricas [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((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^4,x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [C] Result contains complex when optimal does not.
time = 1.44, size = 48, normalized size = 0.24 \begin {gather*} - \frac {x^{5} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 \Gamma \left (- \frac {1}{4}\right )} + \frac {x^{3}}{3 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x**2+(1+1/a**2/x**4)**(1/2))*x**4,x)

[Out]

-x**5*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), exp_polar(I*pi)/(a**2*x**4))/(4*gamma(-1/4)) + x**3/(3*a)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^4,x, algorithm="giac")

[Out]

integrate(x^4*(sqrt(1/(a^2*x^4) + 1) + 1/(a*x^2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^4\,\left (\sqrt {\frac {1}{a^2\,x^4}+1}+\frac {1}{a\,x^2}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*((1/(a^2*x^4) + 1)^(1/2) + 1/(a*x^2)),x)

[Out]

int(x^4*((1/(a^2*x^4) + 1)^(1/2) + 1/(a*x^2)), x)

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