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

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

[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])

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Rubi [A]  time = 0.0466961, antiderivative size = 144, normalized size of antiderivative = 1., 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} \[ \frac{a^2 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a x^2 (a+b x)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^3 (a+b x)}{3 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a^3 (a+b x) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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]

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])

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 = 0.0183377, size = 57, normalized size = 0.4 \[ \frac{(a+b x) \left (b x \left (6 a^2-3 a b x+2 b^2 x^2\right )-6 a^3 \log (a+b x)\right )}{6 b^4 \sqrt{(a+b x)^2}} \]

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])

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Maple [A]  time = 0.225, size = 56, normalized size = 0.4 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( -2\,{b}^{3}{x}^{3}+3\,a{b}^{2}{x}^{2}+6\,{a}^{3}\ln \left ( bx+a \right ) -6\,b{a}^{2}x \right ) }{6\,{b}^{4}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

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

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Maxima [A]  time = 1.22456, size = 159, normalized size = 1.1 \begin{align*} -\frac{5 \, a^{3} b \log \left (x + \frac{a}{b}\right )}{3 \,{\left (b^{2}\right )}^{\frac{5}{2}}} + \frac{5 \, a^{2} x}{3 \,{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{5 \, a x^{2}}{6 \, \sqrt{b^{2}} b} + \frac{2 \, a^{3} \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{3 \, b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} x^{2}}{3 \, b^{2}} - \frac{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.6327, size = 92, normalized size = 0.64 \begin{align*} \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{align*}

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

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Sympy [A]  time = 0.626093, size = 37, normalized size = 0.26 \begin{align*} - \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{align*}

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)

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Giac [A]  time = 1.29153, size = 90, normalized size = 0.62 \begin{align*} -\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{align*}

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