\(\int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx\) [761]

Optimal result

Integrand size = 18, antiderivative size = 28 \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx=b \sqrt {c x^2}+\frac {a \sqrt {c x^2} \log (x)}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 45} \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx=\frac {a \sqrt {c x^2} \log (x)}{x}+b \sqrt {c x^2} \]

[In]

Int[(Sqrt[c*x^2]*(a + b*x))/x^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx=\frac {c x (b x+a \log (x))}{\sqrt {c x^2}} \]

[In]

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

[Out]

(c*x*(b*x + a*Log[x]))/Sqrt[c*x^2]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {\sqrt {c \,x^{2}}\, \left (b x +a \ln \left (x \right )\right )}{x}\)	\(20\)
risch	\(b \sqrt {c \,x^{2}}+\frac {a \ln \left (x \right ) \sqrt {c \,x^{2}}}{x}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx=\frac {\sqrt {c x^{2}} {\left (b x + a \log \left (x\right )\right )}}{x} \]

[In]

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

[Out]

sqrt(c*x^2)*(b*x + a*log(x))/x

Sympy [A] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx=\frac {a \sqrt {c x^{2}} \log {\left (x \right )}}{x} + b \sqrt {c x^{2}} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx={\left (b x \mathrm {sgn}\left (x\right ) + a \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c x^2} (a+b x)}{x^2} \, dx=\int \frac {\sqrt {c\,x^2}\,\left (a+b\,x\right )}{x^2} \,d x \]

[In]

int(((c*x^2)^(1/2)*(a + b*x))/x^2,x)

[Out]

int(((c*x^2)^(1/2)*(a + b*x))/x^2, x)