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

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

[Out]

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

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Rubi [A]  time = 0.0162818, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {15, 43} \[ \frac{a \sqrt{c x^2}}{b^2 x (a+b x)}+\frac{\sqrt{c x^2} \log (a+b x)}{b^2 x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x^2]/(a + b*x)^2,x]

[Out]

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

Mathematica [A]  time = 0.0116019, size = 36, normalized size = 0.77 \[ \frac{c x ((a+b x) \log (a+b x)+a)}{b^2 \sqrt{c x^2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x^2]/(a + b*x)^2,x]

[Out]

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

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Maple [A]  time = 0.008, size = 41, normalized size = 0.9 \begin{align*}{\frac{b\ln \left ( bx+a \right ) x+a\ln \left ( bx+a \right ) +a}{{b}^{2}x \left ( bx+a \right ) }\sqrt{c{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51192, size = 84, normalized size = 1.79 \begin{align*} \frac{\sqrt{c x^{2}}{\left ({\left (b x + a\right )} \log \left (b x + a\right ) + a\right )}}{b^{3} x^{2} + a b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2)*((b*x + a)*log(b*x + a) + a)/(b^3*x^2 + a*b^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}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.07896, size = 62, normalized size = 1.32 \begin{align*} -\sqrt{c}{\left (\frac{{\left (\log \left ({\left | a \right |}\right ) + 1\right )} \mathrm{sgn}\left (x\right )}{b^{2}} - \frac{\log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (x\right )}{b^{2}} - \frac{a \mathrm{sgn}\left (x\right )}{{\left (b x + a\right )} b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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