3.9.33 \(\int \frac {(a+b x)^2}{x^2 \sqrt {c x^2}} \, dx\) [833]

Optimal. Leaf size=49 \[ -\frac {2 a b}{\sqrt {c x^2}}-\frac {a^2}{2 x \sqrt {c x^2}}+\frac {b^2 x \log (x)}{\sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 49, 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, 45} \begin {gather*} -\frac {a^2}{2 x \sqrt {c x^2}}-\frac {2 a b}{\sqrt {c x^2}}+\frac {b^2 x \log (x)}{\sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x^2*Sqrt[c*x^2]),x]

[Out]

(-2*a*b)/Sqrt[c*x^2] - a^2/(2*x*Sqrt[c*x^2]) + (b^2*x*Log[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 45

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

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Mathematica [A]
time = 0.01, size = 35, normalized size = 0.71 \begin {gather*} \frac {c x \left (-a (a+4 b x)+2 b^2 x^2 \log (x)\right )}{2 \left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 34, normalized size = 0.69

method	result	size
default	\(\frac {2 b^{2} \ln \left (x \right ) x^{2}-4 a b x -a^{2}}{2 x \sqrt {c \,x^{2}}}\)	\(34\)
risch	\(\frac {-\frac {1}{2} a^{2}-2 a b x}{\sqrt {c \,x^{2}}\, x}+\frac {b^{2} x \ln \left (x \right )}{\sqrt {c \,x^{2}}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/x^2/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 31, normalized size = 0.63 \begin {gather*} \frac {b^{2} \log \left (x\right )}{\sqrt {c}} - \frac {2 \, a b}{\sqrt {c} x} - \frac {a^{2}}{2 \, \sqrt {c} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 36, normalized size = 0.73 \begin {gather*} \frac {{\left (2 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a b x - a^{2}\right )} \sqrt {c x^{2}}}{2 \, c x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{2}}{x^{2} \sqrt {c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**2/(x**2*sqrt(c*x**2)), x)

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Giac [A]
time = 1.75, size = 43, normalized size = 0.88 \begin {gather*} \frac {b^{2} \log \left ({\left | x \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {4 \, a b \sqrt {c} x + a^{2} \sqrt {c}}{2 \, c x^{2} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*log(abs(x))/(sqrt(c)*sgn(x)) - 1/2*(4*a*b*sqrt(c)*x + a^2*sqrt(c))/(c*x^2*sgn(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^2}{x^2\,\sqrt {c\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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