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

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

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Rubi [A]  time = 0.01, antiderivative size = 66, 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^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 {gather*}

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*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.53 \begin {gather*} \frac {-21 a^2-48 a b x-28 b^2 x^2}{168 x^3 \left (c x^2\right )^{5/2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.03, size = 35, normalized size = 0.53 \begin {gather*} \frac {-21 a^2-48 a b x-28 b^2 x^2}{168 x^3 \left (c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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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fricas [A]  time = 1.20, size = 34, normalized size = 0.52 \begin {gather*} -\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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

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

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maple [A]  time = 0.01, size = 32, normalized size = 0.48 \begin {gather*} -\frac {28 b^{2} x^{2}+48 a b x +21 a^{2}}{168 \left (c \,x^{2}\right )^{\frac {5}{2}} x^{3}} \end {gather*}

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 = 1.35, size = 33, normalized size = 0.50 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.18, size = 42, normalized size = 0.64 \begin {gather*} -\frac {21\,a^2\,\sqrt {x^2}+28\,b^2\,x^2\,\sqrt {x^2}+48\,a\,b\,x\,\sqrt {x^2}}{168\,c^{5/2}\,x^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(21*a^2*(x^2)^(1/2) + 28*b^2*x^2*(x^2)^(1/2) + 48*a*b*x*(x^2)^(1/2))/(168*c^(5/2)*x^9)

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sympy [A]  time = 2.03, size = 61, normalized size = 0.92 \begin {gather*} - \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 {gather*}

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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