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

Optimal. Leaf size=57 \[ \frac{a^2 x^3}{2 \sqrt{c x^2}}+\frac{2 a b x^4}{3 \sqrt{c x^2}}+\frac{b^2 x^5}{4 \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0121125, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 43} \[ \frac{a^2 x^3}{2 \sqrt{c x^2}}+\frac{2 a b x^4}{3 \sqrt{c x^2}}+\frac{b^2 x^5}{4 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^2)/Sqrt[c*x^2],x]

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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

Mathematica [A]  time = 0.0048516, size = 35, normalized size = 0.61 \[ \frac{x^3 \left (6 a^2+8 a b x+3 b^2 x^2\right )}{12 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(6*a^2 + 8*a*b*x + 3*b^2*x^2))/(12*Sqrt[c*x^2])

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Maple [A]  time = 0.003, size = 32, normalized size = 0.6 \begin{align*}{\frac{{x}^{3} \left ( 3\,{b}^{2}{x}^{2}+8\,abx+6\,{a}^{2} \right ) }{12}{\frac{1}{\sqrt{c{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^3*(3*b^2*x^2+8*a*b*x+6*a^2)/(c*x^2)^(1/2)

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Maxima [A]  time = 1.04988, size = 63, normalized size = 1.11 \begin{align*} \frac{\sqrt{c x^{2}} b^{2} x^{3}}{4 \, c} + \frac{2 \, \sqrt{c x^{2}} a b x^{2}}{3 \, c} + \frac{a^{2} x^{2}}{2 \, \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.52525, size = 73, normalized size = 1.28 \begin{align*} \frac{{\left (3 \, b^{2} x^{3} + 8 \, a b x^{2} + 6 \, a^{2} x\right )} \sqrt{c x^{2}}}{12 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^2*x^3 + 8*a*b*x^2 + 6*a^2*x)*sqrt(c*x^2)/c

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Sympy [A]  time = 0.59411, size = 61, normalized size = 1.07 \begin{align*} \frac{a^{2} x^{3}}{2 \sqrt{c} \sqrt{x^{2}}} + \frac{2 a b x^{4}}{3 \sqrt{c} \sqrt{x^{2}}} + \frac{b^{2} x^{5}}{4 \sqrt{c} \sqrt{x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**3/(2*sqrt(c)*sqrt(x**2)) + 2*a*b*x**4/(3*sqrt(c)*sqrt(x**2)) + b**2*x**5/(4*sqrt(c)*sqrt(x**2))

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Giac [A]  time = 1.08845, size = 51, normalized size = 0.89 \begin{align*} \frac{1}{12} \, \sqrt{c x^{2}}{\left ({\left (\frac{3 \, b^{2} x}{c} + \frac{8 \, a b}{c}\right )} x + \frac{6 \, a^{2}}{c}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(c*x^2)*((3*b^2*x/c + 8*a*b/c)*x + 6*a^2/c)*x