3.57 \(\int \frac {x^4}{\sqrt {a x+b x^3}} \, dx\)

Optimal. Leaf size=140 \[ \frac {5 a^{7/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a x+b x^3}}-\frac {10 a \sqrt {a x+b x^3}}{21 b^2}+\frac {2 x^2 \sqrt {a x+b x^3}}{7 b} \]

[Out]

-10/21*a*(b*x^3+a*x)^(1/2)/b^2+2/7*x^2*(b*x^3+a*x)^(1/2)/b+5/21*a^(7/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))
)^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2)
)*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(9/4)/(b*x^3+a*x)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2024, 2011, 329, 220} \[ \frac {5 a^{7/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a x+b x^3}}-\frac {10 a \sqrt {a x+b x^3}}{21 b^2}+\frac {2 x^2 \sqrt {a x+b x^3}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a*x + b*x^3],x]

[Out]

(-10*a*Sqrt[a*x + b*x^3])/(21*b^2) + (2*x^2*Sqrt[a*x + b*x^3])/(7*b) + (5*a^(7/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x
)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(21*b^(9/4)*S
qrt[a*x + b*x^3])

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2024

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

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {a x+b x^3}} \, dx &=\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {(5 a) \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx}{7 b}\\ &=-\frac {10 a \sqrt {a x+b x^3}}{21 b^2}+\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}+\frac {\left (5 a^2\right ) \int \frac {1}{\sqrt {a x+b x^3}} \, dx}{21 b^2}\\ &=-\frac {10 a \sqrt {a x+b x^3}}{21 b^2}+\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}+\frac {\left (5 a^2 \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x^2}} \, dx}{21 b^2 \sqrt {a x+b x^3}}\\ &=-\frac {10 a \sqrt {a x+b x^3}}{21 b^2}+\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}+\frac {\left (10 a^2 \sqrt {x} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{21 b^2 \sqrt {a x+b x^3}}\\ &=-\frac {10 a \sqrt {a x+b x^3}}{21 b^2}+\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}+\frac {5 a^{7/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a x+b x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 80, normalized size = 0.57 \[ \frac {2 x \left (5 a^2 \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )-5 a^2-2 a b x^2+3 b^2 x^4\right )}{21 b^2 \sqrt {x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a*x + b*x^3],x]

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{3} + a x} x^{3}}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^3+a*x)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^3 + a*x)*x^3/(b*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {b x^{3} + a x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^3+a*x)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/sqrt(b*x^3 + a*x), x)

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maple [A]  time = 0.07, size = 149, normalized size = 1.06 \[ \frac {2 \sqrt {b \,x^{3}+a x}\, x^{2}}{7 b}+\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a^{2} \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{21 \sqrt {b \,x^{3}+a x}\, b^{3}}-\frac {10 \sqrt {b \,x^{3}+a x}\, a}{21 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^3+a*x)^(1/2),x)

[Out]

2/7*x^2*(b*x^3+a*x)^(1/2)/b-10/21*a*(b*x^3+a*x)^(1/2)/b^2+5/21*a^2/b^3*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)
^(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)/(b*x^3+a*x)^(1/2)*Ell
ipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {b x^{3} + a x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^3+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/sqrt(b*x^3 + a*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{\sqrt {b\,x^3+a\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a*x + b*x^3)^(1/2),x)

[Out]

int(x^4/(a*x + b*x^3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {x \left (a + b x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/sqrt(x*(a + b*x**2)), x)

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