3.8.97 \(\int \frac {x^2 (a+b x)}{(c x^2)^{5/2}} \, dx\) [797]

Optimal. Leaf size=29 \[ -\frac {(a+b x)^2}{2 a c^2 x \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 37} \begin {gather*} -\frac {(a+b x)^2}{2 a c^2 x \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x))/(c*x^2)^(5/2),x]

[Out]

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

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin {align*} \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx &=\frac {x \int \frac {a+b x}{x^3} \, dx}{c^2 \sqrt {c x^2}}\\ &=-\frac {(a+b x)^2}{2 a c^2 x \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 24, normalized size = 0.83 \begin {gather*} \frac {x^3 (-a-2 b x)}{2 \left (c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x))/(c*x^2)^(5/2),x]

[Out]

(x^3*(-a - 2*b*x))/(2*(c*x^2)^(5/2))

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Maple [A]
time = 0.02, size = 19, normalized size = 0.66

method	result	size
gosper	\(-\frac {x^{3} \left (2 b x +a \right )}{2 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(19\)
default	\(-\frac {x^{3} \left (2 b x +a \right )}{2 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(19\)
risch	\(\frac {-b x -\frac {a}{2}}{c^{2} x \sqrt {c \,x^{2}}}\)	\(23\)
trager	\(\frac {\left (-1+x \right ) \left (a x +2 b x +a \right ) \sqrt {c \,x^{2}}}{2 c^{3} x^{3}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)/(c*x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*x^3*(2*b*x+a)/(c*x^2)^(5/2)

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Maxima [A]
time = 0.28, size = 26, normalized size = 0.90 \begin {gather*} -\frac {b x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}} c} - \frac {a}{2 \, c^{\frac {5}{2}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

-b*x^2/((c*x^2)^(3/2)*c) - 1/2*a/(c^(5/2)*x^2)

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Fricas [A]
time = 1.21, size = 21, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {c x^{2}} {\left (2 \, b x + a\right )}}{2 \, c^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.28, size = 29, normalized size = 1.00 \begin {gather*} - \frac {a x^{3}}{2 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x+a)/(c*x**2)**(5/2),x)

[Out]

-a*x**3/(2*(c*x**2)**(5/2)) - b*x**4/(c*x**2)**(5/2)

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Giac [A]
time = 1.52, size = 18, normalized size = 0.62 \begin {gather*} -\frac {2 \, b x + a}{2 \, c^{\frac {5}{2}} x^{2} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x+a)/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

-1/2*(2*b*x + a)/(c^(5/2)*x^2*sgn(x))

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Mupad [B]
time = 0.15, size = 25, normalized size = 0.86 \begin {gather*} -\frac {2\,b\,x^3+a\,x^2}{2\,c^{5/2}\,x\,{\left (x^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*x))/(c*x^2)^(5/2),x)

[Out]

-(a*x^2 + 2*b*x^3)/(2*c^(5/2)*x*(x^2)^(3/2))

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