\(\int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx\) [916]

Optimal result

Integrand size = 20, antiderivative size = 103 \[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\frac {2 b}{a^3 \sqrt {c x^2}}-\frac {1}{2 a^2 x \sqrt {c x^2}}+\frac {b^2 x}{a^3 \sqrt {c x^2} (a+b x)}+\frac {3 b^2 x \log (x)}{a^4 \sqrt {c x^2}}-\frac {3 b^2 x \log (a+b x)}{a^4 \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 103, 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, 46} \[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\frac {3 b^2 x \log (x)}{a^4 \sqrt {c x^2}}-\frac {3 b^2 x \log (a+b x)}{a^4 \sqrt {c x^2}}+\frac {b^2 x}{a^3 \sqrt {c x^2} (a+b x)}+\frac {2 b}{a^3 \sqrt {c x^2}}-\frac {1}{2 a^2 x \sqrt {c x^2}} \]

[In]

Int[1/(x^2*Sqrt[c*x^2]*(a + b*x)^2),x]

[Out]

(2*b)/(a^3*Sqrt[c*x^2]) - 1/(2*a^2*x*Sqrt[c*x^2]) + (b^2*x)/(a^3*Sqrt[c*x^2]*(a + b*x)) + (3*b^2*x*Log[x])/(a^
4*Sqrt[c*x^2]) - (3*b^2*x*Log[a + b*x])/(a^4*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 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {x \int \frac {1}{x^3 (a+b x)^2} \, dx}{\sqrt {c x^2}} \\ & = \frac {x \int \left (\frac {1}{a^2 x^3}-\frac {2 b}{a^3 x^2}+\frac {3 b^2}{a^4 x}-\frac {b^3}{a^3 (a+b x)^2}-\frac {3 b^3}{a^4 (a+b x)}\right ) \, dx}{\sqrt {c x^2}} \\ & = \frac {2 b}{a^3 \sqrt {c x^2}}-\frac {1}{2 a^2 x \sqrt {c x^2}}+\frac {b^2 x}{a^3 \sqrt {c x^2} (a+b x)}+\frac {3 b^2 x \log (x)}{a^4 \sqrt {c x^2}}-\frac {3 b^2 x \log (a+b x)}{a^4 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\frac {c x \left (\frac {a \left (-a^2+3 a b x+6 b^2 x^2\right )}{a+b x}+6 b^2 x^2 \log (x)-6 b^2 x^2 \log (a+b x)\right )}{2 a^4 \left (c x^2\right )^{3/2}} \]

[In]

Integrate[1/(x^2*Sqrt[c*x^2]*(a + b*x)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {\frac {3 b^{2} x^{2}}{a^{3}}+\frac {3 b x}{2 a^{2}}-\frac {1}{2 a}}{\sqrt {c \,x^{2}}\, x \left (b x +a \right )}+\frac {3 x \,b^{2} \ln \left (-x \right )}{\sqrt {c \,x^{2}}\, a^{4}}-\frac {3 b^{2} x \ln \left (b x +a \right )}{a^{4} \sqrt {c \,x^{2}}}\)	\(86\)
default	\(\frac {6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a \right ) x^{3}+6 a \,b^{2} \ln \left (x \right ) x^{2}-6 \ln \left (b x +a \right ) x^{2} a \,b^{2}+6 a \,b^{2} x^{2}+3 a^{2} b x -a^{3}}{2 x \sqrt {c \,x^{2}}\, a^{4} \left (b x +a \right )}\)	\(95\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\frac {{\left (6 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3} + 6 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \log \left (\frac {x}{b x + a}\right )\right )} \sqrt {c x^{2}}}{2 \, {\left (a^{4} b c x^{4} + a^{5} c x^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\int \frac {1}{x^{2} \sqrt {c x^{2}} \left (a + b x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\frac {6 \, b^{2} x^{2} + 3 \, a b x - a^{2}}{2 \, {\left (a^{3} b \sqrt {c} x^{3} + a^{4} \sqrt {c} x^{2}\right )}} - \frac {3 \, b^{2} \log \left (b x + a\right )}{a^{4} \sqrt {c}} + \frac {3 \, b^{2} \log \left (x\right )}{a^{4} \sqrt {c}} \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx=\int \frac {1}{x^2\,\sqrt {c\,x^2}\,{\left (a+b\,x\right )}^2} \,d x \]

[In]

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

[Out]

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