### 3.191 $$\int \frac{x^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=133 $\frac{a^3}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a (a+b x) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

(-3*a^2)/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + a^3/(2*b^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x*(a + b
*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*(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 = 0.0580191, antiderivative size = 133, 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} $\frac{a^3}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a (a+b x) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*a^2)/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + a^3/(2*b^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x*(a + b
*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*(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}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{x^3}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac{1}{b^6}-\frac{a^3}{b^6 (a+b x)^3}+\frac{3 a^2}{b^6 (a+b x)^2}-\frac{3 a}{b^6 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{3 a^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^3}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a (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 = 0.02345, size = 71, normalized size = 0.53 $\frac{-4 a^2 b x-5 a^3+4 a b^2 x^2-6 a (a+b x)^2 \log (a+b x)+2 b^3 x^3}{2 b^4 (a+b x) \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.224, size = 89, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 6\,\ln \left ( bx+a \right ){x}^{2}a{b}^{2}-2\,{b}^{3}{x}^{3}+12\,\ln \left ( bx+a \right ) x{a}^{2}b-4\,a{b}^{2}{x}^{2}+6\,{a}^{3}\ln \left ( bx+a \right ) +4\,b{a}^{2}x+5\,{a}^{3} \right ) \left ( bx+a \right ) }{2\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*ln(b*x+a)*x^2*a*b^2-2*b^3*x^3+12*ln(b*x+a)*x*a^2*b-4*a*b^2*x^2+6*a^3*ln(b*x+a)+4*b*a^2*x+5*a^3)*(b*x+a
)/b^4/((b*x+a)^2)^(3/2)

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Maxima [A]  time = 1.18906, size = 180, normalized size = 1.35 \begin{align*} \frac{x^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac{3 \, a \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b} - \frac{9 \, a^{3} b}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{6 \, a^{2} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \, a^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac{a^{3}}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 3*a*log(x + a/b)/((b^2)^(3/2)*b) - 9/2*a^3*b/((b^2)^(7/2)*(x + a/b)^
2) - 6*a^2*x/((b^2)^(5/2)*(x + a/b)^2) + 2*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - a^3/((b^2)^(3/2)*b^3*(x +
a/b)^2)

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Fricas [A]  time = 1.67851, size = 176, normalized size = 1.32 \begin{align*} \frac{2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x