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

Optimal. Leaf size=26 \[ -\frac {(a+b x)^2}{2 a x \sqrt {c x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 37} \[ -\frac {(a+b x)^2}{2 a x \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/(x^2*Sqrt[c*x^2]),x]

[Out]

-(a + b*x)^2/(2*a*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 23, normalized size = 0.88 \[ \frac {c x (-a-2 b x)}{2 \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*(-a - 2*b*x))/(2*(c*x^2)^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 21, normalized size = 0.81 \[ -\frac {\sqrt {c x^{2}} {\left (2 \, b x + a\right )}}{2 \, c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(c*x^2)*(2*b*x + a)/(c*x^3)

________________________________________________________________________________________

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)/x^2/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

________________________________________________________________________________________

maple [A]  time = 0.00, size = 19, normalized size = 0.73 \[ -\frac {2 b x +a}{2 \sqrt {c \,x^{2}}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.25, size = 19, normalized size = 0.73 \[ -\frac {b}{\sqrt {c} x} - \frac {a}{2 \, \sqrt {c} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b/(sqrt(c)*x) - 1/2*a/(sqrt(c)*x^2)

________________________________________________________________________________________

mupad [B]  time = 0.16, size = 25, normalized size = 0.96 \[ -\frac {2\,b\,x^3+a\,x^2}{2\,\sqrt {c}\,x\,{\left (x^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)/(x^2*(c*x^2)^(1/2)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.54, size = 31, normalized size = 1.19 \[ - \frac {a}{2 \sqrt {c} x \sqrt {x^{2}}} - \frac {b}{\sqrt {c} \sqrt {x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________