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

Optimal. Leaf size=47 \[ -\frac{a^2}{\sqrt{c x^2}}+\frac{2 a b x \log (x)}{\sqrt{c x^2}}+\frac{b^2 x^2}{\sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0105107, antiderivative size = 47, 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} \[ -\frac{a^2}{\sqrt{c x^2}}+\frac{2 a b x \log (x)}{\sqrt{c x^2}}+\frac{b^2 x^2}{\sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x*Sqrt[c*x^2]),x]

[Out]

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

Mathematica [A]  time = 0.0093845, size = 34, normalized size = 0.72 \[ \frac{c x^2 \left (-a^2+2 a b x \log (x)+b^2 x^2\right )}{\left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2*(-a^2 + b^2*x^2 + 2*a*b*x*Log[x]))/(c*x^2)^(3/2)

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Maple [A]  time = 0.003, size = 29, normalized size = 0.6 \begin{align*}{(2\,ab\ln \left ( x \right ) x+{b}^{2}{x}^{2}-{a}^{2}){\frac{1}{\sqrt{c{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01363, size = 47, normalized size = 1. \begin{align*} \frac{2 \, a b \log \left (x\right )}{\sqrt{c}} + \frac{\sqrt{c x^{2}} b^{2}}{c} - \frac{a^{2}}{\sqrt{c} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5568, size = 73, normalized size = 1.55 \begin{align*} \frac{{\left (b^{2} x^{2} + 2 \, a b x \log \left (x\right ) - a^{2}\right )} \sqrt{c x^{2}}}{c x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.0755, size = 88, normalized size = 1.87 \begin{align*} \frac{\sqrt{c x^{2}} b^{2}}{c} - \frac{2 \,{\left (a b \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2}} \right |}\right ) - \frac{a^{2} \sqrt{c}}{\sqrt{c} x - \sqrt{c x^{2}}}\right )}}{\sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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