3.10.5 \(\int \frac {(c x^2)^{3/2}}{x^3 (a+b x)^2} \, dx\) [905]

Optimal. Leaf size=25 \[ -\frac {c \sqrt {c x^2}}{b x (a+b x)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 32} \begin {gather*} -\frac {c \sqrt {c x^2}}{b x (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*Sqrt[c*x^2])/(b*x*(a + b*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 24, normalized size = 0.96 \begin {gather*} -\frac {\left (c x^2\right )^{3/2}}{b x^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*x^2)^(3/2)/(b*x^3*(a + b*x)))

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Maple [A]
time = 0.12, size = 23, normalized size = 0.92

method	result	size
gosper	\(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}}}{\left (b x +a \right ) b \,x^{3}}\)	\(23\)
default	\(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}}}{\left (b x +a \right ) b \,x^{3}}\)	\(23\)
risch	\(-\frac {c \sqrt {c \,x^{2}}}{b x \left (b x +a \right )}\)	\(24\)
trager	\(\frac {c \left (-1+x \right ) \sqrt {c \,x^{2}}}{\left (b x +a \right ) \left (a +b \right ) x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 16, normalized size = 0.64 \begin {gather*} -\frac {c^{\frac {3}{2}}}{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^3/(b*x+a)^2,x, algorithm="maxima")

[Out]

-c^(3/2)/(b^2*x + a*b)

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Fricas [A]
time = 0.44, size = 24, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {c x^{2}} c}{b^{2} x^{2} + a b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(c*x^2)*c/(b^2*x^2 + a*b*x)

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Sympy [A]
time = 0.58, size = 37, normalized size = 1.48 \begin {gather*} \begin {cases} - \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{a b x^{3} + b^{2} x^{4}} & \text {for}\: b \neq 0 \\\frac {\left (c x^{2}\right )^{\frac {3}{2}}}{a^{2} x^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(c*x**2)**(3/2)/(a*b*x**3 + b**2*x**4), Ne(b, 0)), ((c*x**2)**(3/2)/(a**2*x**2), True))

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Giac [A]
time = 0.81, size = 29, normalized size = 1.16 \begin {gather*} -c^{\frac {3}{2}} {\left (\frac {\mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b} - \frac {\mathrm {sgn}\left (x\right )}{a b}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.15, size = 24, normalized size = 0.96 \begin {gather*} -\frac {c^{3/2}\,\sqrt {x^2}}{b^2\,x^2+a\,b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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