3.8.84 \(\int \frac {a+b x}{x \sqrt {c x^2}} \, dx\) [784]

Optimal. Leaf size=27 \[ -\frac {a}{\sqrt {c x^2}}+\frac {b x \log (x)}{\sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 45} \begin {gather*} \frac {b x \log (x)}{\sqrt {c x^2}}-\frac {a}{\sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/(x*Sqrt[c*x^2]),x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 23, normalized size = 0.85 \begin {gather*} \frac {c x^2 (-a+b x \log (x))}{\left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 18, normalized size = 0.67

method	result	size
default	\(\frac {b x \ln \left (x \right )-a}{\sqrt {c \,x^{2}}}\)	\(18\)
risch	\(-\frac {a}{\sqrt {c \,x^{2}}}+\frac {b x \ln \left (x \right )}{\sqrt {c \,x^{2}}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 17, normalized size = 0.63 \begin {gather*} \frac {b \log \left (x\right )}{\sqrt {c}} - \frac {a}{\sqrt {c} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(x)/sqrt(c) - a/(sqrt(c)*x)

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Fricas [A]
time = 0.86, size = 23, normalized size = 0.85 \begin {gather*} \frac {\sqrt {c x^{2}} {\left (b x \log \left (x\right ) - a\right )}}{c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2)*(b*x*log(x) - a)/(c*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.69, size = 26, normalized size = 0.96 \begin {gather*} \frac {b \log \left ({\left | x \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {a}{\sqrt {c} x \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(abs(x))/(sqrt(c)*sgn(x)) - a/(sqrt(c)*x*sgn(x))

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Mupad [B]
time = 1.22, size = 22, normalized size = 0.81 \begin {gather*} -\frac {\frac {a}{\sqrt {x^2}}-b\,\ln \left (c\,x\right )\,\mathrm {sign}\left (x\right )}{\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)/(x*(c*x^2)^(1/2)),x)

[Out]

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

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