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

Optimal. Leaf size=40 \[ \frac{c \sqrt{c x^2}}{b}-\frac{a c \sqrt{c x^2} \log (a+b x)}{b^2 x} \]

[Out]

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

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Rubi [A]  time = 0.0116974, antiderivative size = 40, 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} \[ \frac{c \sqrt{c x^2}}{b}-\frac{a c \sqrt{c x^2} \log (a+b x)}{b^2 x} \]

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(3/2)/(x^2*(a + b*x)),x]

[Out]

(c*Sqrt[c*x^2])/b - (a*c*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{\left (c x^2\right )^{3/2}}{x^2 (a+b x)} \, dx &=\frac{\left (c \sqrt{c x^2}\right ) \int \frac{x}{a+b x} \, dx}{x}\\ &=\frac{\left (c \sqrt{c x^2}\right ) \int \left (\frac{1}{b}-\frac{a}{b (a+b x)}\right ) \, dx}{x}\\ &=\frac{c \sqrt{c x^2}}{b}-\frac{a c \sqrt{c x^2} \log (a+b x)}{b^2 x}\\ \end{align*}

Mathematica [A]  time = 0.0033153, size = 30, normalized size = 0.75 \[ \frac{c^2 x (b x-a \log (a+b x))}{b^2 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 29, normalized size = 0.7 \begin{align*} -{\frac{a\ln \left ( bx+a \right ) -bx}{{x}^{3}{b}^{2}} \left ( c{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(3/2)/x^2/(b*x+a),x)

[Out]

-(c*x^2)^(3/2)*(a*ln(b*x+a)-b*x)/x^3/b^2

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63481, size = 65, normalized size = 1.62 \begin{align*} \frac{{\left (b c x - a c \log \left (b x + a\right )\right )} \sqrt{c x^{2}}}{b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x^{2}\right )^{\frac{3}{2}}}{x^{2} \left (a + b x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.06222, size = 50, normalized size = 1.25 \begin{align*} c^{\frac{3}{2}}{\left (\frac{x \mathrm{sgn}\left (x\right )}{b} - \frac{a \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (x\right )}{b^{2}} + \frac{a \log \left ({\left | a \right |}\right ) \mathrm{sgn}\left (x\right )}{b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^(3/2)*(x*sgn(x)/b - a*log(abs(b*x + a))*sgn(x)/b^2 + a*log(abs(a))*sgn(x)/b^2)