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

Optimal. Leaf size=65 \[ -\frac{\sqrt{c x^2} \log (a+b x)}{a^2 x}+\frac{\sqrt{c x^2} \log (x)}{a^2 x}+\frac{\sqrt{c x^2}}{a x (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0179407, antiderivative size = 65, 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, 44} \[ -\frac{\sqrt{c x^2} \log (a+b x)}{a^2 x}+\frac{\sqrt{c x^2} \log (x)}{a^2 x}+\frac{\sqrt{c x^2}}{a x (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x^2]/(x^2*(a + b*x)^2),x]

[Out]

Sqrt[c*x^2]/(a*x*(a + b*x)) + (Sqrt[c*x^2]*Log[x])/(a^2*x) - (Sqrt[c*x^2]*Log[a + b*x])/(a^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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.014482, size = 45, normalized size = 0.69 \[ \frac{c x (\log (x) (a+b x)-(a+b x) \log (a+b x)+a)}{a^2 \sqrt{c x^2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 52, normalized size = 0.8 \begin{align*}{\frac{b\ln \left ( x \right ) x-b\ln \left ( bx+a \right ) x+a\ln \left ( x \right ) -a\ln \left ( bx+a \right ) +a}{{a}^{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)/x^2/(b*x+a)^2,x)

[Out]

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

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Maxima [A]  time = 1.03193, size = 51, normalized size = 0.78 \begin{align*} \frac{\sqrt{c}}{a b x + a^{2}} - \frac{\sqrt{c} \log \left (b x + a\right )}{a^{2}} + \frac{\sqrt{c} \log \left (x\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.43205, size = 89, normalized size = 1.37 \begin{align*} \frac{\sqrt{c x^{2}}{\left ({\left (b x + a\right )} \log \left (\frac{x}{b x + a}\right ) + a\right )}}{a^{2} b x^{2} + a^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2)*((b*x + a)*log(x/(b*x + a)) + a)/(a^2*b*x^2 + a^3*x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError