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

Optimal. Leaf size=57 \[ \frac{a^2 x^4}{3 \sqrt{c x^2}}+\frac{a b x^5}{2 \sqrt{c x^2}}+\frac{b^2 x^6}{5 \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0132603, 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} \[ \frac{a^2 x^4}{3 \sqrt{c x^2}}+\frac{a b x^5}{2 \sqrt{c x^2}}+\frac{b^2 x^6}{5 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*x)^2)/Sqrt[c*x^2],x]

[Out]

(a^2*x^4)/(3*Sqrt[c*x^2]) + (a*b*x^5)/(2*Sqrt[c*x^2]) + (b^2*x^6)/(5*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^3 (a+b x)^2}{\sqrt{c x^2}} \, dx &=\frac{x \int x^2 (a+b x)^2 \, dx}{\sqrt{c x^2}}\\ &=\frac{x \int \left (a^2 x^2+2 a b x^3+b^2 x^4\right ) \, dx}{\sqrt{c x^2}}\\ &=\frac{a^2 x^4}{3 \sqrt{c x^2}}+\frac{a b x^5}{2 \sqrt{c x^2}}+\frac{b^2 x^6}{5 \sqrt{c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0060105, size = 35, normalized size = 0.61 \[ \frac{x^4 \left (10 a^2+15 a b x+6 b^2 x^2\right )}{30 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(10*a^2 + 15*a*b*x + 6*b^2*x^2))/(30*Sqrt[c*x^2])

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Maple [A]  time = 0.003, size = 32, normalized size = 0.6 \begin{align*}{\frac{{x}^{4} \left ( 6\,{b}^{2}{x}^{2}+15\,abx+10\,{a}^{2} \right ) }{30}{\frac{1}{\sqrt{c{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*x^4*(6*b^2*x^2+15*a*b*x+10*a^2)/(c*x^2)^(1/2)

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Maxima [A]  time = 1.03058, size = 73, normalized size = 1.28 \begin{align*} \frac{\sqrt{c x^{2}} b^{2} x^{4}}{5 \, c} + \frac{\sqrt{c x^{2}} a b x^{3}}{2 \, c} + \frac{\sqrt{c x^{2}} a^{2} x^{2}}{3 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.44338, size = 78, normalized size = 1.37 \begin{align*} \frac{{\left (6 \, b^{2} x^{4} + 15 \, a b x^{3} + 10 \, a^{2} x^{2}\right )} \sqrt{c x^{2}}}{30 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(6*b^2*x^4 + 15*a*b*x^3 + 10*a^2*x^2)*sqrt(c*x^2)/c

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Sympy [A]  time = 0.695983, size = 60, normalized size = 1.05 \begin{align*} \frac{a^{2} x^{4}}{3 \sqrt{c} \sqrt{x^{2}}} + \frac{a b x^{5}}{2 \sqrt{c} \sqrt{x^{2}}} + \frac{b^{2} x^{6}}{5 \sqrt{c} \sqrt{x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**4/(3*sqrt(c)*sqrt(x**2)) + a*b*x**5/(2*sqrt(c)*sqrt(x**2)) + b**2*x**6/(5*sqrt(c)*sqrt(x**2))

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Giac [A]  time = 1.06773, size = 55, normalized size = 0.96 \begin{align*} \frac{1}{30} \, \sqrt{c x^{2}}{\left (3 \,{\left (\frac{2 \, b^{2} x}{c} + \frac{5 \, a b}{c}\right )} x + \frac{10 \, a^{2}}{c}\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(c*x^2)*(3*(2*b^2*x/c + 5*a*b/c)*x + 10*a^2/c)*x^2