3.8.82 \(\int \frac {\sqrt {a+c x^4}}{x^6} \, dx\) [782]

Optimal. Leaf size=258 \[ -\frac {\sqrt {a+c x^4}}{5 x^5}-\frac {2 c \sqrt {a+c x^4}}{5 a x}+\frac {2 c^{3/2} x \sqrt {a+c x^4}}{5 a \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {2 c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+c x^4}}+\frac {c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+c x^4}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {283, 331, 311, 226, 1210} \begin {gather*} \frac {c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+c x^4}}-\frac {2 c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+c x^4}}+\frac {2 c^{3/2} x \sqrt {a+c x^4}}{5 a \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {2 c \sqrt {a+c x^4}}{5 a x}-\frac {\sqrt {a+c x^4}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^4]/x^6,x]

[Out]

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

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+c x^4}}{x^6} \, dx &=-\frac {\sqrt {a+c x^4}}{5 x^5}+\frac {1}{5} (2 c) \int \frac {1}{x^2 \sqrt {a+c x^4}} \, dx\\ &=-\frac {\sqrt {a+c x^4}}{5 x^5}-\frac {2 c \sqrt {a+c x^4}}{5 a x}+\frac {\left (2 c^2\right ) \int \frac {x^2}{\sqrt {a+c x^4}} \, dx}{5 a}\\ &=-\frac {\sqrt {a+c x^4}}{5 x^5}-\frac {2 c \sqrt {a+c x^4}}{5 a x}+\frac {\left (2 c^{3/2}\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{5 \sqrt {a}}-\frac {\left (2 c^{3/2}\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{5 \sqrt {a}}\\ &=-\frac {\sqrt {a+c x^4}}{5 x^5}-\frac {2 c \sqrt {a+c x^4}}{5 a x}+\frac {2 c^{3/2} x \sqrt {a+c x^4}}{5 a \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {2 c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+c x^4}}+\frac {c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+c 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 = 10.01, size = 51, normalized size = 0.20 \begin {gather*} -\frac {\sqrt {a+c x^4} \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {c x^4}{a}\right )}{5 x^5 \sqrt {1+\frac {c x^4}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^4]/x^6,x]

[Out]

-1/5*(Sqrt[a + c*x^4]*Hypergeometric2F1[-5/4, -1/2, -1/4, -((c*x^4)/a)])/(x^5*Sqrt[1 + (c*x^4)/a])

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 130, normalized size = 0.50

method	result	size
risch	\(-\frac {\sqrt {x^{4} c +a}\, \left (2 x^{4} c +a \right )}{5 x^{5} a}+\frac {2 i c^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\)	\(123\)
default	\(-\frac {\sqrt {x^{4} c +a}}{5 x^{5}}-\frac {2 c \sqrt {x^{4} c +a}}{5 a x}+\frac {2 i c^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\)	\(130\)
elliptic	\(-\frac {\sqrt {x^{4} c +a}}{5 x^{5}}-\frac {2 c \sqrt {x^{4} c +a}}{5 a x}+\frac {2 i c^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(1/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

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

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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((c*x^4+a)^(1/2)/x^6,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + a)/x^6, x)

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Fricas [A]
time = 0.09, size = 87, normalized size = 0.34 \begin {gather*} -\frac {2 \, \sqrt {a} c x^{5} \left (-\frac {c}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 2 \, \sqrt {a} c x^{5} \left (-\frac {c}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (2 \, c x^{4} + a\right )} \sqrt {c x^{4} + a}}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x^6,x, algorithm="fricas")

[Out]

-1/5*(2*sqrt(a)*c*x^5*(-c/a)^(3/4)*elliptic_e(arcsin(x*(-c/a)^(1/4)), -1) - 2*sqrt(a)*c*x^5*(-c/a)^(3/4)*ellip
tic_f(arcsin(x*(-c/a)^(1/4)), -1) + (2*c*x^4 + a)*sqrt(c*x^4 + a))/(a*x^5)

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Sympy [C] Result contains complex when optimal does not.
time = 0.47, size = 46, normalized size = 0.18 \begin {gather*} \frac {\sqrt {a} \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 {c x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(1/2)/x**6,x)

[Out]

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

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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((c*x^4+a)^(1/2)/x^6,x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + a)/x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^4+a}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(1/2)/x^6,x)

[Out]

int((a + c*x^4)^(1/2)/x^6, x)

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