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

Optimal. Leaf size=49 \[ \frac{a^2 \sqrt{c x^2} \log (x)}{x}+2 a b \sqrt{c x^2}+\frac{1}{2} b^2 x \sqrt{c x^2} \]

[Out]

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

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Rubi [A]  time = 0.009421, antiderivative size = 49, 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} \[ \frac{a^2 \sqrt{c x^2} \log (x)}{x}+2 a b \sqrt{c x^2}+\frac{1}{2} b^2 x \sqrt{c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c*x^2]*(a + b*x)^2)/x^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0097211, size = 33, normalized size = 0.67 \[ \frac{c x \left (2 a^2 \log (x)+b x (4 a+b x)\right )}{2 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 33, normalized size = 0.7 \begin{align*}{\frac{{b}^{2}{x}^{2}+2\,{a}^{2}\ln \left ( x \right ) +4\,abx}{2\,x}\sqrt{c{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58105, size = 73, normalized size = 1.49 \begin{align*} \frac{{\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \left (x\right )\right )} \sqrt{c x^{2}}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c*x**2)*(a + b*x)**2/x**2, x)

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Giac [A]  time = 1.0843, size = 43, normalized size = 0.88 \begin{align*} \frac{1}{2} \,{\left (b^{2} x^{2} \mathrm{sgn}\left (x\right ) + 4 \, a b x \mathrm{sgn}\left (x\right ) + 2 \, a^{2} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (x\right )\right )} \sqrt{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*x^2*sgn(x) + 4*a*b*x*sgn(x) + 2*a^2*log(abs(x))*sgn(x))*sqrt(c)