\(\int \frac {x^2 \sqrt {c x^2}}{(a+b x)^2} \, dx\) [894]

Optimal result

Integrand size = 20, antiderivative size = 85 \[ \int \frac {x^2 \sqrt {c x^2}}{(a+b x)^2} \, dx=-\frac {2 a \sqrt {c x^2}}{b^3}+\frac {x \sqrt {c x^2}}{2 b^2}+\frac {a^3 \sqrt {c x^2}}{b^4 x (a+b x)}+\frac {3 a^2 \sqrt {c x^2} \log (a+b x)}{b^4 x} \]

[Out]

-2*a*(c*x^2)^(1/2)/b^3+1/2*x*(c*x^2)^(1/2)/b^2+a^3*(c*x^2)^(1/2)/b^4/x/(b*x+a)+3*a^2*ln(b*x+a)*(c*x^2)^(1/2)/b
^4/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {x^2 \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {a^3 \sqrt {c x^2}}{b^4 x (a+b x)}+\frac {3 a^2 \sqrt {c x^2} \log (a+b x)}{b^4 x}-\frac {2 a \sqrt {c x^2}}{b^3}+\frac {x \sqrt {c x^2}}{2 b^2} \]

[In]

Int[(x^2*Sqrt[c*x^2])/(a + b*x)^2,x]

[Out]

(-2*a*Sqrt[c*x^2])/b^3 + (x*Sqrt[c*x^2])/(2*b^2) + (a^3*Sqrt[c*x^2])/(b^4*x*(a + b*x)) + (3*a^2*Sqrt[c*x^2]*Lo
g[a + b*x])/(b^4*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^3}{(a+b x)^2} \, dx}{x} \\ & = \frac {\sqrt {c x^2} \int \left (-\frac {2 a}{b^3}+\frac {x}{b^2}-\frac {a^3}{b^3 (a+b x)^2}+\frac {3 a^2}{b^3 (a+b x)}\right ) \, dx}{x} \\ & = -\frac {2 a \sqrt {c x^2}}{b^3}+\frac {x \sqrt {c x^2}}{2 b^2}+\frac {a^3 \sqrt {c x^2}}{b^4 x (a+b x)}+\frac {3 a^2 \sqrt {c x^2} \log (a+b x)}{b^4 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int \frac {x^2 \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {c x \left (2 a^3-4 a^2 b x-3 a b^2 x^2+b^3 x^3+6 a^2 (a+b x) \log (a+b x)\right )}{2 b^4 \sqrt {c x^2} (a+b x)} \]

[In]

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

[Out]

(c*x*(2*a^3 - 4*a^2*b*x - 3*a*b^2*x^2 + b^3*x^3 + 6*a^2*(a + b*x)*Log[a + b*x]))/(2*b^4*Sqrt[c*x^2]*(a + b*x))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {\sqrt {c \,x^{2}}\, \left (\frac {1}{2} b \,x^{2}-2 a x \right )}{x \,b^{3}}+\frac {a^{3} \sqrt {c \,x^{2}}}{b^{4} x \left (b x +a \right )}+\frac {3 a^{2} \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{4} x}\)	\(75\)
default	\(\frac {\sqrt {c \,x^{2}}\, \left (b^{3} x^{3}+6 \ln \left (b x +a \right ) a^{2} b x -3 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )-4 a^{2} b x +2 a^{3}\right )}{2 x \,b^{4} \left (b x +a \right )}\)	\(76\)

[In]

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

[Out]

(c*x^2)^(1/2)/x*(1/2*b*x^2-2*a*x)/b^3+a^3*(c*x^2)^(1/2)/b^4/x/(b*x+a)+3*a^2*ln(b*x+a)*(c*x^2)^(1/2)/b^4/x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \frac {x^2 \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {{\left (b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3} + 6 \, {\left (a^{2} b x + a^{3}\right )} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{2 \, {\left (b^{5} x^{2} + a b^{4} x\right )}} \]

[In]

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

[Out]

1/2*(b^3*x^3 - 3*a*b^2*x^2 - 4*a^2*b*x + 2*a^3 + 6*(a^2*b*x + a^3)*log(b*x + a))*sqrt(c*x^2)/(b^5*x^2 + a*b^4*
x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 \sqrt {c x^2}}{(a+b x)^2} \, dx=-\frac {\sqrt {c x^{2}} a^{2}}{b^{4} x + a b^{3}} + \frac {3 \, \left (-1\right )^{\frac {2 \, c x}{b}} a^{2} \sqrt {c} \log \left (\frac {2 \, c x}{b}\right )}{b^{4}} + \frac {3 \, \left (-1\right )^{\frac {2 \, a c x}{b}} a^{2} \sqrt {c} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{4}} + \frac {\sqrt {c x^{2}} x}{2 \, b^{2}} - \frac {2 \, \sqrt {c x^{2}} a}{b^{3}} \]

[In]

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

[Out]

-sqrt(c*x^2)*a^2/(b^4*x + a*b^3) + 3*(-1)^(2*c*x/b)*a^2*sqrt(c)*log(2*c*x/b)/b^4 + 3*(-1)^(2*a*c*x/b)*a^2*sqrt
(c)*log(-2*a*c*x/(b*abs(b*x + a)))/b^4 + 1/2*sqrt(c*x^2)*x/b^2 - 2*sqrt(c*x^2)*a/b^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \sqrt {c x^2}}{(a+b x)^2} \, dx=\frac {1}{2} \, \sqrt {c} {\left (\frac {6 \, a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {2 \, a^{3} \mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b^{4}} - \frac {2 \, {\left (3 \, a^{2} \log \left ({\left | a \right |}\right ) + a^{2}\right )} \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {b^{2} x^{2} \mathrm {sgn}\left (x\right ) - 4 \, a b x \mathrm {sgn}\left (x\right )}{b^{4}}\right )} \]

[In]

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

[Out]

1/2*sqrt(c)*(6*a^2*log(abs(b*x + a))*sgn(x)/b^4 + 2*a^3*sgn(x)/((b*x + a)*b^4) - 2*(3*a^2*log(abs(a)) + a^2)*s
gn(x)/b^4 + (b^2*x^2*sgn(x) - 4*a*b*x*sgn(x))/b^4)

Mupad [F(-1)]

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

[In]

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

[Out]

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