3.776 \(\int x^4 \sqrt {a+c x^4} \, dx\)

Optimal. Leaf size=127 \[ -\frac {a^{7/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{21 c^{5/4} \sqrt {a+c x^4}}+\frac {2 a x \sqrt {a+c x^4}}{21 c}+\frac {1}{7} x^5 \sqrt {a+c x^4} \]

[Out]

2/21*a*x*(c*x^4+a)^(1/2)/c+1/7*x^5*(c*x^4+a)^(1/2)-1/21*a^(7/4)*(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)/c^(5/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {279, 321, 220} \[ -\frac {a^{7/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{21 c^{5/4} \sqrt {a+c x^4}}+\frac {1}{7} x^5 \sqrt {a+c x^4}+\frac {2 a x \sqrt {a+c x^4}}{21 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[a + c*x^4],x]

[Out]

(2*a*x*Sqrt[a + c*x^4])/(21*c) + (x^5*Sqrt[a + c*x^4])/7 - (a^(7/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(
Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(21*c^(5/4)*Sqrt[a + c*x^4])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 279

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

Rule 321

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

Rubi steps

\begin {align*} \int x^4 \sqrt {a+c x^4} \, dx &=\frac {1}{7} x^5 \sqrt {a+c x^4}+\frac {1}{7} (2 a) \int \frac {x^4}{\sqrt {a+c x^4}} \, dx\\ &=\frac {2 a x \sqrt {a+c x^4}}{21 c}+\frac {1}{7} x^5 \sqrt {a+c x^4}-\frac {\left (2 a^2\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{21 c}\\ &=\frac {2 a x \sqrt {a+c x^4}}{21 c}+\frac {1}{7} x^5 \sqrt {a+c x^4}-\frac {a^{7/4} \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 )}{21 c^{5/4} \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 62, normalized size = 0.49 \[ \frac {x \sqrt {a+c x^4} \left (-\frac {a \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^4}{a}\right )}{\sqrt {\frac {c x^4}{a}+1}}+a+c x^4\right )}{7 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[a + c*x^4],x]

[Out]

(x*Sqrt[a + c*x^4]*(a + c*x^4 - (a*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^4)/a)])/Sqrt[1 + (c*x^4)/a]))/(7*c
)

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{4} + a} x^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + a)*x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{4} + a} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^4 + a)*x^4, x)

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maple [C]  time = 0.05, size = 108, normalized size = 0.85 \[ \frac {\sqrt {c \,x^{4}+a}\, x^{5}}{7}-\frac {2 \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a^{2} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{21 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, c}+\frac {2 \sqrt {c \,x^{4}+a}\, a x}{21 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*x^5*(c*x^4+a)^(1/2)+2/21*a*x*(c*x^4+a)^(1/2)/c-2/21*a^2/c/(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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{4} + a} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^4 + a)*x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\sqrt {c\,x^4+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + c*x^4)^(1/2),x)

[Out]

int(x^4*(a + c*x^4)^(1/2), x)

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sympy [C]  time = 2.03, size = 39, normalized size = 0.31 \[ \frac {\sqrt {a} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), c*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4))

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