\(\int \frac {x^6}{(a x^2+b x^3)^{3/2}} \, dx\) [260]

Optimal result

Integrand size = 19, antiderivative size = 98 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 x^4}{b \sqrt {a x^2+b x^3}}-\frac {16 a \sqrt {a x^2+b x^3}}{5 b^3}+\frac {32 a^2 \sqrt {a x^2+b x^3}}{5 b^4 x}+\frac {12 x \sqrt {a x^2+b x^3}}{5 b^2} \]

[Out]

-2*x^4/b/(b*x^3+a*x^2)^(1/2)-16/5*a*(b*x^3+a*x^2)^(1/2)/b^3+32/5*a^2*(b*x^3+a*x^2)^(1/2)/b^4/x+12/5*x*(b*x^3+a
*x^2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2040, 2041, 1602} \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {32 a^2 \sqrt {a x^2+b x^3}}{5 b^4 x}-\frac {16 a \sqrt {a x^2+b x^3}}{5 b^3}+\frac {12 x \sqrt {a x^2+b x^3}}{5 b^2}-\frac {2 x^4}{b \sqrt {a x^2+b x^3}} \]

[In]

Int[x^6/(a*x^2 + b*x^3)^(3/2),x]

[Out]

(-2*x^4)/(b*Sqrt[a*x^2 + b*x^3]) - (16*a*Sqrt[a*x^2 + b*x^3])/(5*b^3) + (32*a^2*Sqrt[a*x^2 + b*x^3])/(5*b^4*x)
+ (12*x*Sqrt[a*x^2 + b*x^3])/(5*b^2)

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 2040

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*(n - j)*(p + 1))), x] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))
, Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n,
j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

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*} \text {integral}& = -\frac {2 x^4}{b \sqrt {a x^2+b x^3}}+\frac {6 \int \frac {x^3}{\sqrt {a x^2+b x^3}} \, dx}{b} \\ & = -\frac {2 x^4}{b \sqrt {a x^2+b x^3}}+\frac {12 x \sqrt {a x^2+b x^3}}{5 b^2}-\frac {(24 a) \int \frac {x^2}{\sqrt {a x^2+b x^3}} \, dx}{5 b^2} \\ & = -\frac {2 x^4}{b \sqrt {a x^2+b x^3}}-\frac {16 a \sqrt {a x^2+b x^3}}{5 b^3}+\frac {12 x \sqrt {a x^2+b x^3}}{5 b^2}+\frac {\left (16 a^2\right ) \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx}{5 b^3} \\ & = -\frac {2 x^4}{b \sqrt {a x^2+b x^3}}-\frac {16 a \sqrt {a x^2+b x^3}}{5 b^3}+\frac {32 a^2 \sqrt {a x^2+b x^3}}{5 b^4 x}+\frac {12 x \sqrt {a x^2+b x^3}}{5 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.51 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (16 a^3+8 a^2 b x-2 a b^2 x^2+b^3 x^3\right )}{5 b^4 \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[x^6/(a*x^2 + b*x^3)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57

method	result	size
gosper	\(\frac {2 \left (b x +a \right ) \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) x^{3}}{5 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(56\)
default	\(\frac {2 \left (b x +a \right ) \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) x^{3}}{5 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(56\)
trager	\(\frac {2 \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{5 \left (b x +a \right ) b^{4} x}\)	\(58\)
risch	\(\frac {2 \left (b^{2} x^{2}-3 a b x +11 a^{2}\right ) \left (b x +a \right ) x}{5 b^{4} \sqrt {x^{2} \left (b x +a \right )}}+\frac {2 a^{3} x}{b^{4} \sqrt {x^{2} \left (b x +a \right )}}\)	\(62\)
pseudoelliptic	\(\frac {\frac {2}{11} b^{6} x^{6}-\frac {8}{33} a \,x^{5} b^{5}+\frac {80}{231} a^{2} x^{4} b^{4}-\frac {128}{231} a^{3} x^{3} b^{3}+\frac {256}{231} a^{4} x^{2} b^{2}-\frac {1024}{231} a^{5} x b -\frac {2048}{231} a^{6}}{b^{7} \sqrt {b x +a}}\)	\(76\)

[In]

int(x^6/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.61 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{5 \, {\left (b^{5} x^{2} + a b^{4} x\right )}} \]

[In]

integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x^{6}}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**6/(b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(x**6/(x**2*(a + b*x))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.42 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )}}{5 \, \sqrt {b x + a} b^{4}} \]

[In]

integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {32 \, a^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{5 \, b^{4}} + \frac {2 \, a^{3}}{\sqrt {b x + a} b^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} b^{16} - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{16} + 15 \, \sqrt {b x + a} a^{2} b^{16}\right )}}{5 \, b^{20} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

-32/5*a^(5/2)*sgn(x)/b^4 + 2*a^3/(sqrt(b*x + a)*b^4*sgn(x)) + 2/5*((b*x + a)^(5/2)*b^16 - 5*(b*x + a)^(3/2)*a*
b^16 + 15*sqrt(b*x + a)*a^2*b^16)/(b^20*sgn(x))

Mupad [B] (verification not implemented)

Time = 9.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (16\,a^3+8\,a^2\,b\,x-2\,a\,b^2\,x^2+b^3\,x^3\right )}{5\,b^4\,x\,\left (a+b\,x\right )} \]

[In]

int(x^6/(a*x^2 + b*x^3)^(3/2),x)

[Out]

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