\(\int \frac {x^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [201]

Optimal result

Integrand size = 24, antiderivative size = 37 \[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {x^4}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 37} \[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {x^4}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[x^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

x^4/(4*a*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {x^3}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {x^4}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(147\) vs. \(2(37)=74\).

Time = 0.78 (sec) , antiderivative size = 147, normalized size of antiderivative = 3.97 \[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {x^4 \left (a^5+a b^4 x^4-a^3 \sqrt {a^2} \sqrt {(a+b x)^2}-a \sqrt {a^2} b^2 x^2 \sqrt {(a+b x)^2}+\sqrt {a^2} b x \sqrt {(a+b x)^2} \left (a^2+b^2 x^2\right )\right )}{4 a^5 (a+b x)^3 \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )} \]

[In]

Integrate[x^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (4 b^{3} x^{3}+6 a \,b^{2} x^{2}+4 a^{2} b x +a^{3}\right )}{4 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(48\)
default	\(-\frac {\left (b x +a \right ) \left (4 b^{3} x^{3}+6 a \,b^{2} x^{2}+4 a^{2} b x +a^{3}\right )}{4 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(48\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {x^{3}}{b}-\frac {3 a \,x^{2}}{2 b^{2}}-\frac {a^{2} x}{b^{3}}-\frac {a^{3}}{4 b^{4}}\right )}{\left (b x +a \right )^{5}}\)	\(53\)

[In]

int(x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (24) = 48\).

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.05 \[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 4 \, a^{2} b x + a^{3}}{4 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \]

[In]

integrate(x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/4*(4*b^3*x^3 + 6*a*b^2*x^2 + 4*a^2*b*x + a^3)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^
4)

Sympy [F]

\[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**3/((a + b*x)**2)**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (24) = 48\).

Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.76 \[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, a^{2}}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} - \frac {a}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a^{2}}{3 \, b^{7} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {a^{3}}{4 \, b^{8} {\left (x + \frac {a}{b}\right )}^{4}} \]

[In]

integrate(x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

-x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 2/3*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) - 1/2*a/(b^6*(x + a
/b)^2) + 2/3*a^2/(b^7*(x + a/b)^3) + 1/4*a^3/(b^8*(x + a/b)^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 4 \, a^{2} b x + a^{3}}{4 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate(x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

-1/4*(4*b^3*x^3 + 6*a*b^2*x^2 + 4*a^2*b*x + a^3)/((b*x + a)^4*b^4*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 9.42 (sec) , antiderivative size = 128, normalized size of antiderivative = 3.46 \[ \int \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^4\,{\left (a+b\,x\right )}^5}-\frac {a^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^4}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^2}+\frac {3\,a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,b^4\,{\left (a+b\,x\right )}^3} \]

[In]

int(x^3/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

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