3.3.52 \(\int \frac {x^4}{\sqrt {a x^2+b x^3}} \, dx\) [252]

Optimal. Leaf size=103 \[ \frac {16 a^2 \sqrt {a x^2+b x^3}}{35 b^3}-\frac {32 a^3 \sqrt {a x^2+b x^3}}{35 b^4 x}-\frac {12 a x \sqrt {a x^2+b x^3}}{35 b^2}+\frac {2 x^2 \sqrt {a x^2+b x^3}}{7 b} \]

[Out]

16/35*a^2*(b*x^3+a*x^2)^(1/2)/b^3-32/35*a^3*(b*x^3+a*x^2)^(1/2)/b^4/x-12/35*a*x*(b*x^3+a*x^2)^(1/2)/b^2+2/7*x^
2*(b*x^3+a*x^2)^(1/2)/b

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Rubi [A]
time = 0.10, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {32 a^3 \sqrt {a x^2+b x^3}}{35 b^4 x}+\frac {16 a^2 \sqrt {a x^2+b x^3}}{35 b^3}-\frac {12 a x \sqrt {a x^2+b x^3}}{35 b^2}+\frac {2 x^2 \sqrt {a x^2+b x^3}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a*x^2 + b*x^3],x]

[Out]

(16*a^2*Sqrt[a*x^2 + b*x^3])/(35*b^3) - (32*a^3*Sqrt[a*x^2 + b*x^3])/(35*b^4*x) - (12*a*x*Sqrt[a*x^2 + b*x^3])
/(35*b^2) + (2*x^2*Sqrt[a*x^2 + b*x^3])/(7*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^4}{\sqrt {a x^2+b x^3}} \, dx &=\frac {2 x^2 \sqrt {a x^2+b x^3}}{7 b}-\frac {(6 a) \int \frac {x^3}{\sqrt {a x^2+b x^3}} \, dx}{7 b}\\ &=-\frac {12 a x \sqrt {a x^2+b x^3}}{35 b^2}+\frac {2 x^2 \sqrt {a x^2+b x^3}}{7 b}+\frac {\left (24 a^2\right ) \int \frac {x^2}{\sqrt {a x^2+b x^3}} \, dx}{35 b^2}\\ &=\frac {16 a^2 \sqrt {a x^2+b x^3}}{35 b^3}-\frac {12 a x \sqrt {a x^2+b x^3}}{35 b^2}+\frac {2 x^2 \sqrt {a x^2+b x^3}}{7 b}-\frac {\left (16 a^3\right ) \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx}{35 b^3}\\ &=\frac {16 a^2 \sqrt {a x^2+b x^3}}{35 b^3}-\frac {32 a^3 \sqrt {a x^2+b x^3}}{35 b^4 x}-\frac {12 a x \sqrt {a x^2+b x^3}}{35 b^2}+\frac {2 x^2 \sqrt {a x^2+b x^3}}{7 b}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 53, normalized size = 0.51 \begin {gather*} \frac {2 \sqrt {x^2 (a+b x)} \left (-16 a^3+8 a^2 b x-6 a b^2 x^2+5 b^3 x^3\right )}{35 b^4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a*x^2 + b*x^3],x]

[Out]

(2*Sqrt[x^2*(a + b*x)]*(-16*a^3 + 8*a^2*b*x - 6*a*b^2*x^2 + 5*b^3*x^3))/(35*b^4*x)

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Maple [A]
time = 0.37, size = 55, normalized size = 0.53

method	result	size
trager	\(-\frac {2 \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{35 b^{4} x}\)	\(52\)
risch	\(-\frac {2 x \left (b x +a \right ) \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right )}{35 \sqrt {x^{2} \left (b x +a \right )}\, b^{4}}\)	\(53\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right ) x}{35 b^{4} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(55\)
default	\(-\frac {2 \left (b x +a \right ) \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right ) x}{35 b^{4} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(b*x+a)*(-5*b^3*x^3+6*a*b^2*x^2-8*a^2*b*x+16*a^3)*x/b^4/(b*x^3+a*x^2)^(1/2)

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Maxima [A]
time = 0.29, size = 53, normalized size = 0.51 \begin {gather*} \frac {2 \, {\left (5 \, b^{4} x^{4} - a b^{3} x^{3} + 2 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x - 16 \, a^{4}\right )}}{35 \, \sqrt {b x + a} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^4*x^4 - a*b^3*x^3 + 2*a^2*b^2*x^2 - 8*a^3*b*x - 16*a^4)/(sqrt(b*x + a)*b^4)

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Fricas [A]
time = 1.23, size = 51, normalized size = 0.50 \begin {gather*} \frac {2 \, {\left (5 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 8 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{35 \, b^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*x^3 - 6*a*b^2*x^2 + 8*a^2*b*x - 16*a^3)*sqrt(b*x^3 + a*x^2)/(b^4*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.75, size = 64, normalized size = 0.62 \begin {gather*} \frac {32 \, a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )}{35 \, b^{4}} + \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}}{35 \, b^{4} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/35*a^(7/2)*sgn(x)/b^4 + 2/35*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b
*x + a)*a^3)/(b^4*sgn(x))

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Mupad [B]
time = 5.19, size = 51, normalized size = 0.50 \begin {gather*} -\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (16\,a^3-8\,a^2\,b\,x+6\,a\,b^2\,x^2-5\,b^3\,x^3\right )}{35\,b^4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a*x^2 + b*x^3)^(1/2),x)

[Out]

-(2*(a*x^2 + b*x^3)^(1/2)*(16*a^3 - 5*b^3*x^3 + 6*a*b^2*x^2 - 8*a^2*b*x))/(35*b^4*x)

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