3.851 \(\int \frac{(a+b x)^2}{x^4 (c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=66 \[ -\frac{a^2}{8 c^2 x^7 \sqrt{c x^2}}-\frac{2 a b}{7 c^2 x^6 \sqrt{c x^2}}-\frac{b^2}{6 c^2 x^5 \sqrt{c x^2}} \]

[Out]

-a^2/(8*c^2*x^7*Sqrt[c*x^2]) - (2*a*b)/(7*c^2*x^6*Sqrt[c*x^2]) - b^2/(6*c^2*x^5*Sqrt[c*x^2])

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Rubi [A]  time = 0.0134873, antiderivative size = 66, 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} \[ -\frac{a^2}{8 c^2 x^7 \sqrt{c x^2}}-\frac{2 a b}{7 c^2 x^6 \sqrt{c x^2}}-\frac{b^2}{6 c^2 x^5 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x^4*(c*x^2)^(5/2)),x]

[Out]

-a^2/(8*c^2*x^7*Sqrt[c*x^2]) - (2*a*b)/(7*c^2*x^6*Sqrt[c*x^2]) - b^2/(6*c^2*x^5*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)^2}{x^4 \left (c x^2\right )^{5/2}} \, dx &=\frac{x \int \frac{(a+b x)^2}{x^9} \, dx}{c^2 \sqrt{c x^2}}\\ &=\frac{x \int \left (\frac{a^2}{x^9}+\frac{2 a b}{x^8}+\frac{b^2}{x^7}\right ) \, dx}{c^2 \sqrt{c x^2}}\\ &=-\frac{a^2}{8 c^2 x^7 \sqrt{c x^2}}-\frac{2 a b}{7 c^2 x^6 \sqrt{c x^2}}-\frac{b^2}{6 c^2 x^5 \sqrt{c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0093735, size = 35, normalized size = 0.53 \[ \frac{-21 a^2-48 a b x-28 b^2 x^2}{168 x^3 \left (c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/(x^4*(c*x^2)^(5/2)),x]

[Out]

(-21*a^2 - 48*a*b*x - 28*b^2*x^2)/(168*x^3*(c*x^2)^(5/2))

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Maple [A]  time = 0.005, size = 32, normalized size = 0.5 \begin{align*} -{\frac{28\,{b}^{2}{x}^{2}+48\,abx+21\,{a}^{2}}{168\,{x}^{3}} \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)^2/x^4/(c*x^2)^(5/2),x)

[Out]

-1/168*(28*b^2*x^2+48*a*b*x+21*a^2)/x^3/(c*x^2)^(5/2)

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Maxima [A]  time = 1.01, size = 45, normalized size = 0.68 \begin{align*} -\frac{b^{2}}{6 \, c^{\frac{5}{2}} x^{6}} - \frac{2 \, a b}{7 \, c^{\frac{5}{2}} x^{7}} - \frac{a^{2}}{8 \, c^{\frac{5}{2}} x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b^2/(c^(5/2)*x^6) - 2/7*a*b/(c^(5/2)*x^7) - 1/8*a^2/(c^(5/2)*x^8)

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Fricas [A]  time = 1.5617, size = 85, normalized size = 1.29 \begin{align*} -\frac{{\left (28 \, b^{2} x^{2} + 48 \, a b x + 21 \, a^{2}\right )} \sqrt{c x^{2}}}{168 \, c^{3} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

-1/168*(28*b^2*x^2 + 48*a*b*x + 21*a^2)*sqrt(c*x^2)/(c^3*x^9)

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Sympy [A]  time = 1.83628, size = 61, normalized size = 0.92 \begin{align*} - \frac{a^{2}}{8 c^{\frac{5}{2}} x^{3} \left (x^{2}\right )^{\frac{5}{2}}} - \frac{2 a b}{7 c^{\frac{5}{2}} x^{2} \left (x^{2}\right )^{\frac{5}{2}}} - \frac{b^{2}}{6 c^{\frac{5}{2}} x \left (x^{2}\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/x**4/(c*x**2)**(5/2),x)

[Out]

-a**2/(8*c**(5/2)*x**3*(x**2)**(5/2)) - 2*a*b/(7*c**(5/2)*x**2*(x**2)**(5/2)) - b**2/(6*c**(5/2)*x*(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)^2/x^4/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x