3.218 \(\int \frac{(c+d x)^2}{x^2 (a+b x)} \, dx\)

Optimal. Leaf size=51 \[ -\frac{c \log (x) (b c-2 a d)}{a^2}+\frac{(b c-a d)^2 \log (a+b x)}{a^2 b}-\frac{c^2}{a x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0389579, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {88} \[ -\frac{c \log (x) (b c-2 a d)}{a^2}+\frac{(b c-a d)^2 \log (a+b x)}{a^2 b}-\frac{c^2}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2/(x^2*(a + b*x)),x]

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{(c+d x)^2}{x^2 (a+b x)} \, dx &=\int \left (\frac{c^2}{a x^2}+\frac{c (-b c+2 a d)}{a^2 x}+\frac{(-b c+a d)^2}{a^2 (a+b x)}\right ) \, dx\\ &=-\frac{c^2}{a x}-\frac{c (b c-2 a d) \log (x)}{a^2}+\frac{(b c-a d)^2 \log (a+b x)}{a^2 b}\\ \end{align*}

Mathematica [A]  time = 0.0228893, size = 51, normalized size = 1. \[ \frac{-a b c^2+b c x \log (x) (2 a d-b c)+x (b c-a d)^2 \log (a+b x)}{a^2 b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 73, normalized size = 1.4 \begin{align*} -{\frac{{c}^{2}}{ax}}+2\,{\frac{c\ln \left ( x \right ) d}{a}}-{\frac{{c}^{2}\ln \left ( x \right ) b}{{a}^{2}}}+{\frac{\ln \left ( bx+a \right ){d}^{2}}{b}}-2\,{\frac{\ln \left ( bx+a \right ) cd}{a}}+{\frac{b\ln \left ( bx+a \right ){c}^{2}}{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.45973, size = 86, normalized size = 1.69 \begin{align*} -\frac{c^{2}}{a x} - \frac{{\left (b c^{2} - 2 \, a c d\right )} \log \left (x\right )}{a^{2}} + \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x + a\right )}{a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2/(a*x) - (b*c^2 - 2*a*c*d)*log(x)/a^2 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(b*x + a)/(a^2*b)

________________________________________________________________________________________

Fricas [A]  time = 1.9995, size = 140, normalized size = 2.75 \begin{align*} -\frac{a b c^{2} -{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \log \left (b x + a\right ) +{\left (b^{2} c^{2} - 2 \, a b c d\right )} x \log \left (x\right )}{a^{2} b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*b*c^2 - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*log(b*x + a) + (b^2*c^2 - 2*a*b*c*d)*x*log(x))/(a^2*b*x)

________________________________________________________________________________________

Sympy [B]  time = 1.31208, size = 141, normalized size = 2.76 \begin{align*} - \frac{c^{2}}{a x} + \frac{c \left (2 a d - b c\right ) \log{\left (x + \frac{- 2 a^{2} c d + a b c^{2} + a c \left (2 a d - b c\right )}{a^{2} d^{2} - 4 a b c d + 2 b^{2} c^{2}} \right )}}{a^{2}} + \frac{\left (a d - b c\right )^{2} \log{\left (x + \frac{- 2 a^{2} c d + a b c^{2} + \frac{a \left (a d - b c\right )^{2}}{b}}{a^{2} d^{2} - 4 a b c d + 2 b^{2} c^{2}} \right )}}{a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**2/(a*x) + c*(2*a*d - b*c)*log(x + (-2*a**2*c*d + a*b*c**2 + a*c*(2*a*d - b*c))/(a**2*d**2 - 4*a*b*c*d + 2*
b**2*c**2))/a**2 + (a*d - b*c)**2*log(x + (-2*a**2*c*d + a*b*c**2 + a*(a*d - b*c)**2/b)/(a**2*d**2 - 4*a*b*c*d
+ 2*b**2*c**2))/(a**2*b)

________________________________________________________________________________________

Giac [A]  time = 1.25627, size = 89, normalized size = 1.75 \begin{align*} -\frac{c^{2}}{a x} - \frac{{\left (b c^{2} - 2 \, a c d\right )} \log \left ({\left | x \right |}\right )}{a^{2}} + \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2/(a*x) - (b*c^2 - 2*a*c*d)*log(abs(x))/a^2 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(abs(b*x + a))/(a^2*b)