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

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

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Rubi [A]  time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 43} \begin {gather*} \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 {gather*}

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*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 0.78 \begin {gather*} \frac {c^2 x ((a+b x) \log (a+b x)+a)}{b^2 \sqrt {c x^2} (a+b x)} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.04, size = 39, normalized size = 0.80 \begin {gather*} \left (c x^2\right )^{3/2} \left (\frac {a}{b^2 x^3 (a+b x)}+\frac {\log (a+b x)}{b^2 x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.41, size = 43, normalized size = 0.88 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.96, size = 46, normalized size = 0.94 \begin {gather*} -c^{\frac {3}{2}} {\left (\frac {{\left (\log \left ({\left | a \right |}\right ) + 1\right )} \mathrm {sgn}\relax (x)}{b^{2}} - \frac {\log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\relax (x)}{b^{2}} - \frac {a \mathrm {sgn}\relax (x)}{{\left (b x + a\right )} b^{2}}\right )} \end {gather*}

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))

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maple [A]  time = 0.00, size = 41, normalized size = 0.84 \begin {gather*} \frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x \ln \left (b x +a \right )+a \ln \left (b x +a \right )+a \right )}{\left (b x +a \right ) b^{2} x^{3}} \end {gather*}

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

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maxima [A]  time = 1.48, size = 80, normalized size = 1.63 \begin {gather*} \frac {\left (-1\right )^{\frac {2 \, c x}{b}} c^{\frac {3}{2}} \log \left (\frac {2 \, c x}{b}\right )}{b^{2}} + \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} c^{\frac {3}{2}} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{2}} - \frac {\sqrt {c x^{2}} c}{b^{2} x + a b} \end {gather*}

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]

(-1)^(2*c*x/b)*c^(3/2)*log(2*c*x/b)/b^2 + (-1)^(2*a*c*x/b)*c^(3/2)*log(-2*a*c*x/(b*abs(b*x + a)))/b^2 - sqrt(c
*x^2)*c/(b^2*x + a*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x^2\right )}^{3/2}}{x^2\,{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{2} \left (a + b x\right )^{2}}\, dx \end {gather*}

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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