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

Optimal. Leaf size=38 \[ \frac {\sqrt {c x^2}}{b}-\frac {a \sqrt {c x^2} \log (a+b x)}{b^2 x} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 43} \[ \frac {\sqrt {c x^2}}{b}-\frac {a \sqrt {c x^2} \log (a+b x)}{b^2 x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x^2]/(a + b*x),x]

[Out]

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

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 {\sqrt {c x^2}}{a+b x} \, dx &=\frac {\sqrt {c x^2} \int \frac {x}{a+b x} \, dx}{x}\\ &=\frac {\sqrt {c x^2} \int \left (\frac {1}{b}-\frac {a}{b (a+b x)}\right ) \, dx}{x}\\ &=\frac {\sqrt {c x^2}}{b}-\frac {a \sqrt {c x^2} \log (a+b x)}{b^2 x}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.74 \[ \frac {c x (b x-a \log (a+b x))}{b^2 \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x^2]/(a + b*x),x]

[Out]

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

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fricas [A]  time = 0.41, size = 27, normalized size = 0.71 \[ \frac {\sqrt {c x^{2}} {\left (b x - a \log \left (b x + a\right )\right )}}{b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.96, size = 37, normalized size = 0.97 \[ \sqrt {c} {\left (\frac {x \mathrm {sgn}\relax (x)}{b} - \frac {a \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\relax (x)}{b^{2}} + \frac {a \log \left ({\left | a \right |}\right ) \mathrm {sgn}\relax (x)}{b^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 29, normalized size = 0.76 \[ -\frac {\sqrt {c \,x^{2}}\, \left (a \ln \left (b x +a \right )-b x \right )}{b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.45, size = 74, normalized size = 1.95 \[ -\frac {\left (-1\right )^{\frac {2 \, c x}{b}} a \sqrt {c} \log \left (\frac {2 \, c x}{b}\right )}{b^{2}} - \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} a \sqrt {c} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{2}} + \frac {\sqrt {c x^{2}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {c\,x^2}}{a+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(1/2)/(a + b*x),x)

[Out]

int((c*x^2)^(1/2)/(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c*x**2)/(a + b*x), x)

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