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

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

[Out]

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

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Rubi [A]  time = 0.0145969, 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} \[ \frac{a c \sqrt{c x^2}}{b^2 x (a+b x)}+\frac{c \sqrt{c x^2} \log (a+b x)}{b^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0088697, size = 38, normalized size = 0.78 \[ \frac{c^2 x ((a+b x) \log (a+b x)+a)}{b^2 \sqrt{c x^2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*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 = 0.003, size = 41, normalized size = 0.8 \begin{align*}{\frac{b\ln \left ( bx+a \right ) x+a\ln \left ( bx+a \right ) +a}{{x}^{3}{b}^{2} \left ( bx+a \right ) } \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)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.21269, size = 92, normalized size = 1.88 \begin{align*} \frac{\sqrt{c x^{2}}{\left (a c +{\left (b c x + a c\right )} \log \left (b x + a\right )\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)^(3/2)/x^2/(b*x+a)^2,x, algorithm="fricas")

[Out]

sqrt(c*x^2)*(a*c + (b*c*x + a*c)*log(b*x + 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{\left (c x^{2}\right )^{\frac{3}{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)**(3/2)/x**2/(b*x+a)**2,x)

[Out]

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

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Giac [A]  time = 1.06821, size = 62, normalized size = 1.27 \begin{align*} -c^{\frac{3}{2}}{\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)^(3/2)/x^2/(b*x+a)^2,x, algorithm="giac")

[Out]

-c^(3/2)*((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))