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

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

[Out]

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

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Rubi [A]  time = 0.0113116, 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} \[ -\frac{a^2 \sqrt{c x^2}}{x^2}+\frac{2 a b \sqrt{c x^2} \log (x)}{x}+b^2 \sqrt{c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c*x^2]*(a + b*x)^2)/x^3,x]

[Out]

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

Mathematica [A]  time = 0.0119556, size = 31, normalized size = 0.63 \[ \frac{c \left (-a^2+2 a b x \log (x)+b^2 x^2\right )}{\sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 32, normalized size = 0.7 \begin{align*}{\frac{2\,ab\ln \left ( x \right ) x+{b}^{2}{x}^{2}-{a}^{2}}{{x}^{2}}\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^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.73218, size = 68, normalized size = 1.39 \begin{align*} \frac{{\left (b^{2} x^{2} + 2 \, a b x \log \left (x\right ) - a^{2}\right )} \sqrt{c x^{2}}}{x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

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

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Giac [A]  time = 1.06002, size = 42, normalized size = 0.86 \begin{align*}{\left (b^{2} x \mathrm{sgn}\left (x\right ) + 2 \, a b \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (x\right ) - \frac{a^{2} \mathrm{sgn}\left (x\right )}{x}\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^3,x, algorithm="giac")

[Out]

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