3.3.84 \(\int \frac {x^9}{\sqrt {a x^2+b x^5}} \, dx\) [284]

Optimal. Leaf size=80 \[ \frac {16 a^2 \sqrt {a x^2+b x^5}}{45 b^3 x}-\frac {8 a x^2 \sqrt {a x^2+b x^5}}{45 b^2}+\frac {2 x^5 \sqrt {a x^2+b x^5}}{15 b} \]

[Out]

16/45*a^2*(b*x^5+a*x^2)^(1/2)/b^3/x-8/45*a*x^2*(b*x^5+a*x^2)^(1/2)/b^2+2/15*x^5*(b*x^5+a*x^2)^(1/2)/b

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Rubi [A]
time = 0.08, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 1602} \begin {gather*} \frac {16 a^2 \sqrt {a x^2+b x^5}}{45 b^3 x}-\frac {8 a x^2 \sqrt {a x^2+b x^5}}{45 b^2}+\frac {2 x^5 \sqrt {a x^2+b x^5}}{15 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/Sqrt[a*x^2 + b*x^5],x]

[Out]

(16*a^2*Sqrt[a*x^2 + b*x^5])/(45*b^3*x) - (8*a*x^2*Sqrt[a*x^2 + b*x^5])/(45*b^2) + (2*x^5*Sqrt[a*x^2 + b*x^5])
/(15*b)

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
 1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 2041

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

Rubi steps

\begin {align*} \int \frac {x^9}{\sqrt {a x^2+b x^5}} \, dx &=\frac {2 x^5 \sqrt {a x^2+b x^5}}{15 b}-\frac {(4 a) \int \frac {x^6}{\sqrt {a x^2+b x^5}} \, dx}{5 b}\\ &=-\frac {8 a x^2 \sqrt {a x^2+b x^5}}{45 b^2}+\frac {2 x^5 \sqrt {a x^2+b x^5}}{15 b}+\frac {\left (8 a^2\right ) \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx}{15 b^2}\\ &=\frac {16 a^2 \sqrt {a x^2+b x^5}}{45 b^3 x}-\frac {8 a x^2 \sqrt {a x^2+b x^5}}{45 b^2}+\frac {2 x^5 \sqrt {a x^2+b x^5}}{15 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 46, normalized size = 0.58 \begin {gather*} \frac {2 \sqrt {x^2 \left (a+b x^3\right )} \left (8 a^2-4 a b x^3+3 b^2 x^6\right )}{45 b^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*Sqrt[x^2*(a + b*x^3)]*(8*a^2 - 4*a*b*x^3 + 3*b^2*x^6))/(45*b^3*x)

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Maple [A]
time = 0.37, size = 48, normalized size = 0.60

method result size
trager \(\frac {2 \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right ) \sqrt {b \,x^{5}+a \,x^{2}}}{45 b^{3} x}\) \(43\)
gosper \(\frac {2 \left (b \,x^{3}+a \right ) \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right ) x}{45 b^{3} \sqrt {b \,x^{5}+a \,x^{2}}}\) \(48\)
default \(\frac {2 \left (b \,x^{3}+a \right ) \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right ) x}{45 b^{3} \sqrt {b \,x^{5}+a \,x^{2}}}\) \(48\)
risch \(\frac {2 x \left (b \,x^{3}+a \right ) \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right )}{45 \sqrt {x^{2} \left (b \,x^{3}+a \right )}\, b^{3}}\) \(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^5+a*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/45*(b*x^3+a)*(3*b^2*x^6-4*a*b*x^3+8*a^2)*x/b^3/(b*x^5+a*x^2)^(1/2)

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Maxima [A]
time = 0.30, size = 46, normalized size = 0.58 \begin {gather*} \frac {2 \, {\left (3 \, b^{3} x^{9} - a b^{2} x^{6} + 4 \, a^{2} b x^{3} + 8 \, a^{3}\right )}}{45 \, \sqrt {b x^{3} + a} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

2/45*(3*b^3*x^9 - a*b^2*x^6 + 4*a^2*b*x^3 + 8*a^3)/(sqrt(b*x^3 + a)*b^3)

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Fricas [A]
time = 1.23, size = 42, normalized size = 0.52 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{6} - 4 \, a b x^{3} + 8 \, a^{2}\right )} \sqrt {b x^{5} + a x^{2}}}{45 \, b^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

2/45*(3*b^2*x^6 - 4*a*b*x^3 + 8*a^2)*sqrt(b*x^5 + a*x^2)/(b^3*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{9}}{\sqrt {x^{2} \left (a + b x^{3}\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**5+a*x**2)**(1/2),x)

[Out]

Integral(x**9/sqrt(x**2*(a + b*x**3)), x)

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Giac [A]
time = 1.86, size = 65, normalized size = 0.81 \begin {gather*} -\frac {16 \, a^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{45 \, b^{3}} + \frac {2 \, \sqrt {b x^{3} + a} a^{2}}{3 \, b^{3} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a\right )}}{45 \, b^{3} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")

[Out]

-16/45*a^(5/2)*sgn(x)/b^3 + 2/3*sqrt(b*x^3 + a)*a^2/(b^3*sgn(x)) + 2/45*(3*(b*x^3 + a)^(5/2) - 10*(b*x^3 + a)^
(3/2)*a)/(b^3*sgn(x))

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Mupad [B]
time = 5.18, size = 42, normalized size = 0.52 \begin {gather*} \frac {2\,\sqrt {b\,x^5+a\,x^2}\,\left (8\,a^2-4\,a\,b\,x^3+3\,b^2\,x^6\right )}{45\,b^3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(a*x^2 + b*x^5)^(1/2),x)

[Out]

(2*(a*x^2 + b*x^5)^(1/2)*(8*a^2 + 3*b^2*x^6 - 4*a*b*x^3))/(45*b^3*x)

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