\(\int \frac {x \sqrt {c x^2}}{(a+b x)^2} \, dx\) [895]

Optimal result

Integrand size = 18, antiderivative size = 65 \[ \int \frac {x \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {\sqrt {c x^2}}{b^2}-\frac {a^2 \sqrt {c x^2}}{b^3 x (a+b x)}-\frac {2 a \sqrt {c x^2} \log (a+b x)}{b^3 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, 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} \[ \int \frac {x \sqrt {c x^2}}{(a+b x)^2} \, dx=-\frac {a^2 \sqrt {c x^2}}{b^3 x (a+b x)}-\frac {2 a \sqrt {c x^2} \log (a+b x)}{b^3 x}+\frac {\sqrt {c x^2}}{b^2} \]

[In]

Int[(x*Sqrt[c*x^2])/(a + b*x)^2,x]

[Out]

Sqrt[c*x^2]/b^2 - (a^2*Sqrt[c*x^2])/(b^3*x*(a + b*x)) - (2*a*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)^2} \, dx}{x} \\ & = \frac {\sqrt {c x^2} \int \left (\frac {1}{b^2}+\frac {a^2}{b^2 (a+b x)^2}-\frac {2 a}{b^2 (a+b x)}\right ) \, dx}{x} \\ & = \frac {\sqrt {c x^2}}{b^2}-\frac {a^2 \sqrt {c x^2}}{b^3 x (a+b x)}-\frac {2 a \sqrt {c x^2} \log (a+b x)}{b^3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int \frac {x \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {c x \left (-a^2+a b x+b^2 x^2-2 a (a+b x) \log (a+b x)\right )}{b^3 \sqrt {c x^2} (a+b x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {\sqrt {c \,x^{2}}}{b^{2}}-\frac {a^{2} \sqrt {c \,x^{2}}}{b^{3} x \left (b x +a \right )}-\frac {2 a \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{3} x}\)	\(60\)
default	\(-\frac {\sqrt {c \,x^{2}}\, \left (2 \ln \left (b x +a \right ) x a b -b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-a b x +a^{2}\right )}{x \,b^{3} \left (b x +a \right )}\)	\(62\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \frac {x \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {{\left (b^{2} x^{2} + a b x - a^{2} - 2 \, {\left (a b x + a^{2}\right )} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{b^{4} x^{2} + a b^{3} x} \]

[In]

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

[Out]

(b^2*x^2 + a*b*x - a^2 - 2*(a*b*x + a^2)*log(b*x + a))*sqrt(c*x^2)/(b^4*x^2 + a*b^3*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.48 \[ \int \frac {x \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {\sqrt {c x^{2}} a}{b^{3} x + a b^{2}} - \frac {2 \, \left (-1\right )^{\frac {2 \, c x}{b}} a \sqrt {c} \log \left (\frac {2 \, c x}{b}\right )}{b^{3}} - \frac {2 \, \left (-1\right )^{\frac {2 \, a c x}{b}} a \sqrt {c} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{3}} + \frac {\sqrt {c x^{2}}}{b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {x \sqrt {c x^2}}{(a+b x)^2} \, dx=\sqrt {c} {\left (\frac {x \mathrm {sgn}\left (x\right )}{b^{2}} - \frac {2 \, a \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {{\left (2 \, a \log \left ({\left | a \right |}\right ) + a\right )} \mathrm {sgn}\left (x\right )}{b^{3}} - \frac {a^{2} \mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b^{3}}\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((x*(c*x^2)^(1/2))/(a + b*x)^2,x)

[Out]

int((x*(c*x^2)^(1/2))/(a + b*x)^2, x)