3.835 \(\int \frac {(a+b x)^2}{x^4 \sqrt {c x^2}} \, dx\)

Optimal. Leaf size=57 \[ -\frac {a^2}{4 x^3 \sqrt {c x^2}}-\frac {2 a b}{3 x^2 \sqrt {c x^2}}-\frac {b^2}{2 x \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 57, 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} \[ -\frac {a^2}{4 x^3 \sqrt {c x^2}}-\frac {2 a b}{3 x^2 \sqrt {c x^2}}-\frac {b^2}{2 x \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x^4*Sqrt[c*x^2]),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.61 \[ \frac {-3 a^2-8 a b x-6 b^2 x^2}{12 x^3 \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2 - 8*a*b*x - 6*b^2*x^2)/(12*x^3*Sqrt[c*x^2])

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fricas [A]  time = 0.42, size = 34, normalized size = 0.60 \[ -\frac {{\left (6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}\right )} \sqrt {c x^{2}}}{12 \, c x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)*sqrt(c*x^2)/(c*x^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.01, size = 32, normalized size = 0.56 \[ -\frac {6 b^{2} x^{2}+8 a b x +3 a^{2}}{12 \sqrt {c \,x^{2}}\, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/x^4/(c*x^2)^(1/2),x)

[Out]

-1/12*(6*b^2*x^2+8*a*b*x+3*a^2)/x^3/(c*x^2)^(1/2)

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maxima [A]  time = 1.33, size = 33, normalized size = 0.58 \[ -\frac {b^{2}}{2 \, \sqrt {c} x^{2}} - \frac {2 \, a b}{3 \, \sqrt {c} x^{3}} - \frac {a^{2}}{4 \, \sqrt {c} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.19, size = 42, normalized size = 0.74 \[ -\frac {3\,a^2\,\sqrt {x^2}+6\,b^2\,x^2\,\sqrt {x^2}+8\,a\,b\,x\,\sqrt {x^2}}{12\,\sqrt {c}\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(x^4*(c*x^2)^(1/2)),x)

[Out]

-(3*a^2*(x^2)^(1/2) + 6*b^2*x^2*(x^2)^(1/2) + 8*a*b*x*(x^2)^(1/2))/(12*c^(1/2)*x^5)

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sympy [A]  time = 0.82, size = 61, normalized size = 1.07 \[ - \frac {a^{2}}{4 \sqrt {c} x^{3} \sqrt {x^{2}}} - \frac {2 a b}{3 \sqrt {c} x^{2} \sqrt {x^{2}}} - \frac {b^{2}}{2 \sqrt {c} x \sqrt {x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/x**4/(c*x**2)**(1/2),x)

[Out]

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

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