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

Optimal. Leaf size=103 \[ \frac {3 b^2 x \log (x)}{a^4 \sqrt {c x^2}}-\frac {3 b^2 x \log (a+b x)}{a^4 \sqrt {c x^2}}+\frac {b^2 x}{a^3 \sqrt {c x^2} (a+b x)}+\frac {2 b}{a^3 \sqrt {c x^2}}-\frac {1}{2 a^2 x \sqrt {c x^2}} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 44} \[ \frac {b^2 x}{a^3 \sqrt {c x^2} (a+b x)}+\frac {3 b^2 x \log (x)}{a^4 \sqrt {c x^2}}-\frac {3 b^2 x \log (a+b x)}{a^4 \sqrt {c x^2}}+\frac {2 b}{a^3 \sqrt {c x^2}}-\frac {1}{2 a^2 x \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[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 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {1}{x^2 \sqrt {c x^2} (a+b x)^2} \, dx &=\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*}

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Mathematica [A]  time = 0.02, size = 81, normalized size = 0.79 \[ \frac {c x \left (a \left (-a^2+3 a b x+6 b^2 x^2\right )+6 b^2 x^2 \log (x) (a+b x)-6 b^2 x^2 (a+b x) \log (a+b x)\right )}{2 a^4 \left (c x^2\right )^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

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

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fricas [A]  time = 0.42, size = 79, normalized size = 0.77 \[ \frac {{\left (6 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3} + 6 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \log \left (\frac {x}{b x + a}\right )\right )} \sqrt {c x^{2}}}{2 \, {\left (a^{4} b c x^{4} + a^{5} c x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.22, size = 152, normalized size = 1.48 \[ -\frac {\frac {6 \, b^{2} \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{4} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )} + \frac {2 \, b^{2}}{{\left (b x + a\right )} a^{3} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )} - \frac {\frac {6 \, a b^{2}}{b x + a} - 5 \, b^{2}}{a^{4} {\left (\frac {a}{b x + a} - 1\right )}^{2} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )}}{2 \, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 95, normalized size = 0.92 \[ \frac {6 b^{3} x^{3} \ln \relax (x )-6 b^{3} x^{3} \ln \left (b x +a \right )+6 a \,b^{2} x^{2} \ln \relax (x )-6 a \,b^{2} x^{2} \ln \left (b x +a \right )+6 a \,b^{2} x^{2}+3 a^{2} b x -a^{3}}{2 \sqrt {c \,x^{2}}\, \left (b x +a \right ) a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.41, size = 76, normalized size = 0.74 \[ \frac {6 \, b^{2} x^{2} + 3 \, a b x - a^{2}}{2 \, {\left (a^{3} b \sqrt {c} x^{3} + a^{4} \sqrt {c} x^{2}\right )}} - \frac {3 \, b^{2} \log \left (b x + a\right )}{a^{4} \sqrt {c}} + \frac {3 \, b^{2} \log \relax (x)}{a^{4} \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,\sqrt {c\,x^2}\,{\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \sqrt {c x^{2}} \left (a + b x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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