3.799 \(\int \frac{a+b x}{(c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{a}{4 c^2 x^3 \sqrt{c x^2}}-\frac{b}{3 c^2 x^2 \sqrt{c x^2}} \]

[Out]

-a/(4*c^2*x^3*Sqrt[c*x^2]) - b/(3*c^2*x^2*Sqrt[c*x^2])

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Rubi [A]  time = 0.0072754, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {15, 43} \[ -\frac{a}{4 c^2 x^3 \sqrt{c x^2}}-\frac{b}{3 c^2 x^2 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/(c*x^2)^(5/2),x]

[Out]

-a/(4*c^2*x^3*Sqrt[c*x^2]) - b/(3*c^2*x^2*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 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{a+b x}{\left (c x^2\right )^{5/2}} \, dx &=\frac{x \int \frac{a+b x}{x^5} \, dx}{c^2 \sqrt{c x^2}}\\ &=\frac{x \int \left (\frac{a}{x^5}+\frac{b}{x^4}\right ) \, dx}{c^2 \sqrt{c x^2}}\\ &=-\frac{a}{4 c^2 x^3 \sqrt{c x^2}}-\frac{b}{3 c^2 x^2 \sqrt{c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0027211, size = 27, normalized size = 0.66 \[ -\frac{\sqrt{c x^2} (3 a+4 b x)}{12 c^3 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/(c*x^2)^(5/2),x]

[Out]

-(Sqrt[c*x^2]*(3*a + 4*b*x))/(12*c^3*x^5)

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Maple [A]  time = 0.002, size = 19, normalized size = 0.5 \begin{align*} -{\frac{x \left ( 4\,bx+3\,a \right ) }{12} \left ( c{x}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*x*(4*b*x+3*a)/(c*x^2)^(5/2)

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Maxima [A]  time = 1.06456, size = 31, normalized size = 0.76 \begin{align*} -\frac{b}{3 \, \left (c x^{2}\right )^{\frac{3}{2}} c} - \frac{a}{4 \, c^{\frac{5}{2}} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b/((c*x^2)^(3/2)*c) - 1/4*a/(c^(5/2)*x^4)

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Fricas [A]  time = 1.54431, size = 58, normalized size = 1.41 \begin{align*} -\frac{\sqrt{c x^{2}}{\left (4 \, b x + 3 \, a\right )}}{12 \, c^{3} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(c*x^2)*(4*b*x + 3*a)/(c^3*x^5)

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Sympy [A]  time = 0.879129, size = 36, normalized size = 0.88 \begin{align*} - \frac{a x}{4 c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}} - \frac{b x^{2}}{3 c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x