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

Optimal. Leaf size=57 \[ \frac{1}{3} a^2 x^2 \sqrt{c x^2}+\frac{1}{2} a b x^3 \sqrt{c x^2}+\frac{1}{5} b^2 x^4 \sqrt{c x^2} \]

[Out]

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

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Rubi [A]  time = 0.0134847, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {15, 43} \[ \frac{1}{3} a^2 x^2 \sqrt{c x^2}+\frac{1}{2} a b x^3 \sqrt{c x^2}+\frac{1}{5} b^2 x^4 \sqrt{c x^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[c*x^2]*(a + b*x)^2,x]

[Out]

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

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 x \sqrt{c x^2} (a+b x)^2 \, dx &=\frac{\sqrt{c x^2} \int x^2 (a+b x)^2 \, dx}{x}\\ &=\frac{\sqrt{c x^2} \int \left (a^2 x^2+2 a b x^3+b^2 x^4\right ) \, dx}{x}\\ &=\frac{1}{3} a^2 x^2 \sqrt{c x^2}+\frac{1}{2} a b x^3 \sqrt{c x^2}+\frac{1}{5} b^2 x^4 \sqrt{c x^2}\\ \end{align*}

Mathematica [A]  time = 0.0052016, size = 35, normalized size = 0.61 \[ \frac{1}{30} x^2 \sqrt{c x^2} \left (10 a^2+15 a b x+6 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50877, size = 76, normalized size = 1.33 \begin{align*} \frac{1}{30} \,{\left (6 \, b^{2} x^{4} + 15 \, a b x^{3} + 10 \, a^{2} x^{2}\right )} \sqrt{c x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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)

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Sympy [A]  time = 0.347242, size = 60, normalized size = 1.05 \begin{align*} \frac{a^{2} \sqrt{c} x^{2} \sqrt{x^{2}}}{3} + \frac{a b \sqrt{c} x^{3} \sqrt{x^{2}}}{2} + \frac{b^{2} \sqrt{c} x^{4} \sqrt{x^{2}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.05721, size = 47, normalized size = 0.82 \begin{align*} \frac{1}{30} \,{\left (6 \, b^{2} x^{5} \mathrm{sgn}\left (x\right ) + 15 \, a b x^{4} \mathrm{sgn}\left (x\right ) + 10 \, a^{2} x^{3} \mathrm{sgn}\left (x\right )\right )} \sqrt{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(6*b^2*x^5*sgn(x) + 15*a*b*x^4*sgn(x) + 10*a^2*x^3*sgn(x))*sqrt(c)