3.2.60 \(\int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=144 \[ -\frac {a x^2 (a+b x)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^3 (a+b x)}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {646, 43} \begin {gather*} \frac {a^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^2 (a+b x)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^3 (a+b x)}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

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*} \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {x^3}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {a^2}{b^4}-\frac {a x}{b^3}+\frac {x^2}{b^2}-\frac {a^3}{b^4 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {a^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^2 (a+b x)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^3 (a+b x)}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 57, normalized size = 0.40 \begin {gather*} \frac {(a+b x) \left (b x \left (6 a^2-3 a b x+2 b^2 x^2\right )-6 a^3 \log (a+b x)\right )}{6 b^4 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.30, size = 188, normalized size = 1.31 \begin {gather*} \frac {-6 a^2 x+3 a b x^2-2 b^2 x^3}{12 \left (b^2\right )^{3/2}}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (11 a^2-5 a b x+2 b^2 x^2\right )}{12 b^4}+\frac {a^3 \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )}{2 b^5}+\frac {a^3 \left (\sqrt {b^2}-b\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(11*a^2 - 5*a*b*x + 2*b^2*x^2))/(12*b^4) + (-6*a^2*x + 3*a*b*x^2 - 2*b^2*x^3)/(
12*(b^2)^(3/2)) + (a^3*(b + Sqrt[b^2])*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^5) + (a^3*(
-b + Sqrt[b^2])*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^5)

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 41, normalized size = 0.28 \begin {gather*} \frac {2 \, b^{3} x^{3} - 3 \, a b^{2} x^{2} + 6 \, a^{2} b x - 6 \, a^{3} \log \left (b x + a\right )}{6 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 67, normalized size = 0.47 \begin {gather*} -\frac {a^{3} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{b^{4}} + \frac {2 \, b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} x \mathrm {sgn}\left (b x + a\right )}{6 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 56, normalized size = 0.39 \begin {gather*} -\frac {\left (b x +a \right ) \left (-2 b^{3} x^{3}+3 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )-6 a^{2} b x \right )}{6 \sqrt {\left (b x +a \right )^{2}}\, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.41, size = 87, normalized size = 0.60 \begin {gather*} -\frac {5 \, a x^{2}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} x^{2}}{3 \, b^{2}} + \frac {5 \, a^{2} x}{3 \, b^{3}} - \frac {a^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{3 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a + b*x)^2)^(1/2),x)

[Out]

int(x^3/((a + b*x)^2)^(1/2), x)

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 37, normalized size = 0.26 \begin {gather*} - \frac {a^{3} \log {\left (a + b x \right )}}{b^{4}} + \frac {a^{2} x}{b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {x^{3}}{3 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**3*log(a + b*x)/b**4 + a**2*x/b**3 - a*x**2/(2*b**2) + x**3/(3*b)

________________________________________________________________________________________