\(\int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx\) [911]

Optimal result

Integrand size = 20, antiderivative size = 64 \[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=\frac {x^2}{b^2 \sqrt {c x^2}}-\frac {a^2 x}{b^3 \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, 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} \[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=-\frac {a^2 x}{b^3 \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 \sqrt {c x^2}}+\frac {x^2}{b^2 \sqrt {c x^2}} \]

[In]

Int[x^3/(Sqrt[c*x^2]*(a + b*x)^2),x]

[Out]

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

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 {x \int \frac {x^2}{(a+b x)^2} \, dx}{\sqrt {c x^2}} \\ & = \frac {x \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}{\sqrt {c x^2}} \\ & = \frac {x^2}{b^2 \sqrt {c x^2}}-\frac {a^2 x}{b^3 \sqrt {c x^2} (a+b x)}-\frac {2 a x \log (a+b x)}{b^3 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=\frac {x \left (-a^2+a b x+b^2 x^2-2 a (a+b x) \log (a+b x)\right )}{b^3 \sqrt {c x^2} (a+b x)} \]

[In]

Integrate[x^3/(Sqrt[c*x^2]*(a + b*x)^2),x]

[Out]

(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 = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {x^{2}}{b^{2} \sqrt {c \,x^{2}}}-\frac {a^{2} x}{b^{3} \left (b x +a \right ) \sqrt {c \,x^{2}}}-\frac {2 a x \ln \left (b x +a \right )}{b^{3} \sqrt {c \,x^{2}}}\)	\(59\)
default	\(-\frac {x \left (2 \ln \left (b x +a \right ) x a b -b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-a b x +a^{2}\right )}{\sqrt {c \,x^{2}}\, b^{3} \left (b x +a \right )}\)	\(60\)

[In]

int(x^3/(b*x+a)^2/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=\frac {{\left (b^{2} x^{2} + a b x - a^{2} - 2 \, {\left (a b x + a^{2}\right )} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{b^{4} c x^{2} + a b^{3} c x} \]

[In]

integrate(x^3/(b*x+a)^2/(c*x^2)^(1/2),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*c*x^2 + a*b^3*c*x)

Sympy [F]

\[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=\int \frac {x^{3}}{\sqrt {c x^{2}} \left (a + b x\right )^{2}}\, dx \]

[In]

integrate(x**3/(b*x+a)**2/(c*x**2)**(1/2),x)

[Out]

Integral(x**3/(sqrt(c*x**2)*(a + b*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.38 \[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=\frac {\sqrt {c x^{2}} a}{b^{3} c x + a b^{2} c} - \frac {2 \, \left (-1\right )^{\frac {2 \, a c x}{b}} a \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{3} \sqrt {c}} - \frac {2 \, a \log \left (b x\right )}{b^{3} \sqrt {c}} + \frac {\sqrt {c x^{2}}}{b^{2} c} \]

[In]

integrate(x^3/(b*x+a)^2/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12 \[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=\frac {{\left (2 \, a \log \left ({\left | a \right |}\right ) + a\right )} \mathrm {sgn}\left (x\right )}{b^{3} \sqrt {c}} + \frac {x}{b^{2} \sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {2 \, a \log \left ({\left | b x + a \right |}\right )}{b^{3} \sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {a^{2}}{{\left (b x + a\right )} b^{3} \sqrt {c} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^3/(b*x+a)^2/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {c x^2} (a+b x)^2} \, dx=\int \frac {x^3}{\sqrt {c\,x^2}\,{\left (a+b\,x\right )}^2} \,d x \]

[In]

int(x^3/((c*x^2)^(1/2)*(a + b*x)^2),x)

[Out]

int(x^3/((c*x^2)^(1/2)*(a + b*x)^2), x)