\(\int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx\) [219]

Optimal result

Integrand size = 18, antiderivative size = 67 \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=-\frac {c^2}{2 a x^2}+\frac {c (b c-2 a d)}{a^2 x}+\frac {(b c-a d)^2 \log (x)}{a^3}-\frac {(b c-a d)^2 \log (a+b x)}{a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {90} \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=\frac {\log (x) (b c-a d)^2}{a^3}-\frac {(b c-a d)^2 \log (a+b x)}{a^3}+\frac {c (b c-2 a d)}{a^2 x}-\frac {c^2}{2 a x^2} \]

[In]

Int[(c + d*x)^2/(x^3*(a + b*x)),x]

[Out]

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

Rule 90

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=-\frac {\frac {a c (a c-2 b c x+4 a d x)}{x^2}-2 (b c-a d)^2 \log (x)+2 (b c-a d)^2 \log (a+b x)}{2 a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.34

method	result	size
default	\(-\frac {c^{2}}{2 a \,x^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (x \right )}{a^{3}}-\frac {c \left (2 a d -b c \right )}{a^{2} x}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{a^{3}}\)	\(90\)
norman	\(\frac {-\frac {c^{2}}{2 a}-\frac {c \left (2 a d -b c \right ) x}{a^{2}}}{x^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (x \right )}{a^{3}}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{a^{3}}\)	\(90\)
risch	\(\frac {-\frac {c^{2}}{2 a}-\frac {c \left (2 a d -b c \right ) x}{a^{2}}}{x^{2}}+\frac {\ln \left (-x \right ) d^{2}}{a}-\frac {2 \ln \left (-x \right ) b c d}{a^{2}}+\frac {\ln \left (-x \right ) b^{2} c^{2}}{a^{3}}-\frac {\ln \left (b x +a \right ) d^{2}}{a}+\frac {2 \ln \left (b x +a \right ) b c d}{a^{2}}-\frac {\ln \left (b x +a \right ) b^{2} c^{2}}{a^{3}}\)	\(113\)
parallelrisch	\(\frac {2 \ln \left (x \right ) x^{2} a^{2} d^{2}-4 \ln \left (x \right ) x^{2} a b c d +2 \ln \left (x \right ) x^{2} b^{2} c^{2}-2 \ln \left (b x +a \right ) x^{2} a^{2} d^{2}+4 \ln \left (b x +a \right ) x^{2} a b c d -2 \ln \left (b x +a \right ) x^{2} b^{2} c^{2}-4 a^{2} c d x +2 a b \,c^{2} x -c^{2} a^{2}}{2 a^{3} x^{2}}\)	\(120\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.39 \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=-\frac {a^{2} c^{2} + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \log \left (b x + a\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \log \left (x\right ) - 2 \, {\left (a b c^{2} - 2 \, a^{2} c d\right )} x}{2 \, a^{3} x^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (58) = 116\).

Time = 0.35 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.79 \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=\frac {- a c^{2} + x \left (- 4 a c d + 2 b c^{2}\right )}{2 a^{2} x^{2}} + \frac {\left (a d - b c\right )^{2} \log {\left (x + \frac {a^{3} d^{2} - 2 a^{2} b c d + a b^{2} c^{2} - a \left (a d - b c\right )^{2}}{2 a^{2} b d^{2} - 4 a b^{2} c d + 2 b^{3} c^{2}} \right )}}{a^{3}} - \frac {\left (a d - b c\right )^{2} \log {\left (x + \frac {a^{3} d^{2} - 2 a^{2} b c d + a b^{2} c^{2} + a \left (a d - b c\right )^{2}}{2 a^{2} b d^{2} - 4 a b^{2} c d + 2 b^{3} c^{2}} \right )}}{a^{3}} \]

[In]

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

[Out]

(-a*c**2 + x*(-4*a*c*d + 2*b*c**2))/(2*a**2*x**2) + (a*d - b*c)**2*log(x + (a**3*d**2 - 2*a**2*b*c*d + a*b**2*
c**2 - a*(a*d - b*c)**2)/(2*a**2*b*d**2 - 4*a*b**2*c*d + 2*b**3*c**2))/a**3 - (a*d - b*c)**2*log(x + (a**3*d**
2 - 2*a**2*b*c*d + a*b**2*c**2 + a*(a*d - b*c)**2)/(2*a**2*b*d**2 - 4*a*b**2*c*d + 2*b**3*c**2))/a**3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31 \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=-\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x + a\right )}{a^{3}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (x\right )}{a^{3}} - \frac {a c^{2} - 2 \, {\left (b c^{2} - 2 \, a c d\right )} x}{2 \, a^{2} x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.51 \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{3} b} - \frac {a^{2} c^{2} - 2 \, {\left (a b c^{2} - 2 \, a^{2} c d\right )} x}{2 \, a^{3} x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.33 \[ \int \frac {(c+d x)^2}{x^3 (a+b x)} \, dx=-\frac {\frac {c^2}{2\,a}+\frac {c\,x\,\left (2\,a\,d-b\,c\right )}{a^2}}{x^2}-\frac {2\,\mathrm {atanh}\left (\frac {{\left (a\,d-b\,c\right )}^2\,\left (a+2\,b\,x\right )}{a\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{a^3} \]

[In]

int((c + d*x)^2/(x^3*(a + b*x)),x)

[Out]

- (c^2/(2*a) + (c*x*(2*a*d - b*c))/a^2)/x^2 - (2*atanh(((a*d - b*c)^2*(a + 2*b*x))/(a*(a^2*d^2 + b^2*c^2 - 2*a
*b*c*d)))*(a*d - b*c)^2)/a^3