3.8.83 \(\int \frac {a+b x}{\sqrt {c x^2}} \, dx\) [783]

Optimal. Leaf size=29 \[ \frac {b x^2}{\sqrt {c x^2}}+\frac {a x \log (x)}{\sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {15, 45} \begin {gather*} \frac {a x \log (x)}{\sqrt {c x^2}}+\frac {b x^2}{\sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/Sqrt[c*x^2],x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 19, normalized size = 0.66 \begin {gather*} \frac {x (b x+a \log (x))}{\sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/Sqrt[c*x^2],x]

[Out]

(x*(b*x + a*Log[x]))/Sqrt[c*x^2]

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Maple [A]
time = 0.03, size = 18, normalized size = 0.62

method	result	size
default	\(\frac {x \left (b x +a \ln \left (x \right )\right )}{\sqrt {c \,x^{2}}}\)	\(18\)
risch	\(\frac {b \,x^{2}}{\sqrt {c \,x^{2}}}+\frac {a x \ln \left (x \right )}{\sqrt {c \,x^{2}}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 20, normalized size = 0.69 \begin {gather*} \frac {a \log \left (x\right )}{\sqrt {c}} + \frac {\sqrt {c x^{2}} b}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x)/sqrt(c) + sqrt(c*x^2)*b/c

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Fricas [A]
time = 1.08, size = 22, normalized size = 0.76 \begin {gather*} \frac {\sqrt {c x^{2}} {\left (b x + a \log \left (x\right )\right )}}{c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2)*(b*x + a*log(x))/(c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)/sqrt(c*x**2), x)

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Giac [A]
time = 2.02, size = 23, normalized size = 0.79 \begin {gather*} \frac {b x}{\sqrt {c} \mathrm {sgn}\left (x\right )} + \frac {a \log \left ({\left | x \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.51, size = 17, normalized size = 0.59 \begin {gather*} \frac {b\,\left |x\right |+a\,\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)/(c*x^2)^(1/2),x)

[Out]

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

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