\(\int \frac {x \sqrt {c x^2}}{a+b x} \, dx\) [854]

Optimal result

Integrand size = 18, antiderivative size = 58 \[ \int \frac {x \sqrt {c x^2}}{a+b x} \, dx=-\frac {a \sqrt {c x^2}}{b^2}+\frac {x \sqrt {c x^2}}{2 b}+\frac {a^2 \sqrt {c x^2} \log (a+b x)}{b^3 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, 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} \[ \int \frac {x \sqrt {c x^2}}{a+b x} \, dx=\frac {a^2 \sqrt {c x^2} \log (a+b x)}{b^3 x}-\frac {a \sqrt {c x^2}}{b^2}+\frac {x \sqrt {c x^2}}{2 b} \]

[In]

Int[(x*Sqrt[c*x^2])/(a + b*x),x]

[Out]

-((a*Sqrt[c*x^2])/b^2) + (x*Sqrt[c*x^2])/(2*b) + (a^2*Sqrt[c*x^2]*Log[a + b*x])/(b^3*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 {x^2}{a+b x} \, dx}{x} \\ & = \frac {\sqrt {c x^2} \int \left (-\frac {a}{b^2}+\frac {x}{b}+\frac {a^2}{b^2 (a+b x)}\right ) \, dx}{x} \\ & = -\frac {a \sqrt {c x^2}}{b^2}+\frac {x \sqrt {c x^2}}{2 b}+\frac {a^2 \sqrt {c x^2} \log (a+b x)}{b^3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int \frac {x \sqrt {c x^2}}{a+b x} \, dx=\sqrt {c x^2} \left (\frac {-2 a+b x}{2 b^2}+\frac {a^2 \log (a+b x)}{b^3 x}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {\sqrt {c \,x^{2}}\, \left (b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-2 a b x \right )}{2 b^{3} x}\)	\(40\)
risch	\(\frac {\sqrt {c \,x^{2}}\, \left (\frac {1}{2} b \,x^{2}-a x \right )}{x \,b^{2}}+\frac {a^{2} \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{3} x}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.67 \[ \int \frac {x \sqrt {c x^2}}{a+b x} \, dx=\frac {{\left (b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{2 \, b^{3} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(x*sqrt(c*x**2)/(a + b*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.57 \[ \int \frac {x \sqrt {c x^2}}{a+b x} \, dx=\frac {\left (-1\right )^{\frac {2 \, c x}{b}} a^{2} \sqrt {c} \log \left (\frac {2 \, c x}{b}\right )}{b^{3}} + \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} a^{2} \sqrt {c} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{3}} + \frac {\sqrt {c x^{2}} x}{2 \, b} - \frac {\sqrt {c x^{2}} a}{b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93 \[ \int \frac {x \sqrt {c x^2}}{a+b x} \, dx=\frac {1}{2} \, \sqrt {c} {\left (\frac {2 \, a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{3}} - \frac {2 \, a^{2} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {b x^{2} \mathrm {sgn}\left (x\right ) - 2 \, a x \mathrm {sgn}\left (x\right )}{b^{2}}\right )} \]

[In]

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

[Out]

1/2*sqrt(c)*(2*a^2*log(abs(b*x + a))*sgn(x)/b^3 - 2*a^2*log(abs(a))*sgn(x)/b^3 + (b*x^2*sgn(x) - 2*a*x*sgn(x))
/b^2)

Mupad [F(-1)]

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

[In]

int((x*(c*x^2)^(1/2))/(a + b*x),x)

[Out]

int((x*(c*x^2)^(1/2))/(a + b*x), x)