\(\int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx\) [267]

Optimal result

Integrand size = 18, antiderivative size = 132 \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=-\frac {c^2}{3 a^2 x^3}+\frac {c (b c-a d)}{a^3 x^2}-\frac {(b c-a d) (3 b c-a d)}{a^4 x}-\frac {b (b c-a d)^2}{a^4 (a+b x)}-\frac {2 b (b c-a d) (2 b c-a d) \log (x)}{a^5}+\frac {2 b (b c-a d) (2 b c-a d) \log (a+b x)}{a^5} \]

[Out]

-1/3*c^2/a^2/x^3+c*(-a*d+b*c)/a^3/x^2-(-a*d+b*c)*(-a*d+3*b*c)/a^4/x-b*(-a*d+b*c)^2/a^4/(b*x+a)-2*b*(-a*d+b*c)*
(-a*d+2*b*c)*ln(x)/a^5+2*b*(-a*d+b*c)*(-a*d+2*b*c)*ln(b*x+a)/a^5

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 132, 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} \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=-\frac {2 b \log (x) (b c-a d) (2 b c-a d)}{a^5}+\frac {2 b (b c-a d) (2 b c-a d) \log (a+b x)}{a^5}-\frac {(b c-a d) (3 b c-a d)}{a^4 x}-\frac {b (b c-a d)^2}{a^4 (a+b x)}+\frac {c (b c-a d)}{a^3 x^2}-\frac {c^2}{3 a^2 x^3} \]

[In]

Int[(c + d*x)^2/(x^4*(a + b*x)^2),x]

[Out]

-1/3*c^2/(a^2*x^3) + (c*(b*c - a*d))/(a^3*x^2) - ((b*c - a*d)*(3*b*c - a*d))/(a^4*x) - (b*(b*c - a*d)^2)/(a^4*
(a + b*x)) - (2*b*(b*c - a*d)*(2*b*c - a*d)*Log[x])/a^5 + (2*b*(b*c - a*d)*(2*b*c - a*d)*Log[a + b*x])/a^5

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^2 x^4}+\frac {2 c (-b c+a d)}{a^3 x^3}+\frac {(b c-a d) (3 b c-a d)}{a^4 x^2}+\frac {2 b (b c-a d) (-2 b c+a d)}{a^5 x}+\frac {b^2 (-b c+a d)^2}{a^4 (a+b x)^2}+\frac {2 b^2 (b c-a d) (2 b c-a d)}{a^5 (a+b x)}\right ) \, dx \\ & = -\frac {c^2}{3 a^2 x^3}+\frac {c (b c-a d)}{a^3 x^2}-\frac {(b c-a d) (3 b c-a d)}{a^4 x}-\frac {b (b c-a d)^2}{a^4 (a+b x)}-\frac {2 b (b c-a d) (2 b c-a d) \log (x)}{a^5}+\frac {2 b (b c-a d) (2 b c-a d) \log (a+b x)}{a^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=-\frac {\frac {a^3 c^2}{x^3}+\frac {3 a^2 c (-b c+a d)}{x^2}+\frac {3 a \left (3 b^2 c^2-4 a b c d+a^2 d^2\right )}{x}+\frac {3 a b (b c-a d)^2}{a+b x}+6 b \left (2 b^2 c^2-3 a b c d+a^2 d^2\right ) \log (x)-6 b \left (2 b^2 c^2-3 a b c d+a^2 d^2\right ) \log (a+b x)}{3 a^5} \]

[In]

Integrate[(c + d*x)^2/(x^4*(a + b*x)^2),x]

[Out]

-1/3*((a^3*c^2)/x^3 + (3*a^2*c*(-(b*c) + a*d))/x^2 + (3*a*(3*b^2*c^2 - 4*a*b*c*d + a^2*d^2))/x + (3*a*b*(b*c -
a*d)^2)/(a + b*x) + 6*b*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*Log[x] - 6*b*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*Log[
a + b*x])/a^5

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.20

method	result	size
default	\(-\frac {c^{2}}{3 a^{2} x^{3}}-\frac {a^{2} d^{2}-4 a b c d +3 b^{2} c^{2}}{a^{4} x}-\frac {2 b \left (a^{2} d^{2}-3 a b c d +2 b^{2} c^{2}\right ) \ln \left (x \right )}{a^{5}}-\frac {c \left (a d -b c \right )}{a^{3} x^{2}}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b}{a^{4} \left (b x +a \right )}+\frac {2 b \left (a^{2} d^{2}-3 a b c d +2 b^{2} c^{2}\right ) \ln \left (b x +a \right )}{a^{5}}\)	\(158\)
norman	\(\frac {\frac {b \left (2 b \,d^{2} a^{2}-6 b^{2} c d a +4 c^{2} b^{3}\right ) x^{4}}{a^{5}}-\frac {c^{2}}{3 a}-\frac {\left (a^{2} d^{2}-3 a b c d +2 b^{2} c^{2}\right ) x^{2}}{a^{3}}-\frac {c \left (3 a d -2 b c \right ) x}{3 a^{2}}}{x^{3} \left (b x +a \right )}-\frac {2 b \left (a^{2} d^{2}-3 a b c d +2 b^{2} c^{2}\right ) \ln \left (x \right )}{a^{5}}+\frac {2 b \left (a^{2} d^{2}-3 a b c d +2 b^{2} c^{2}\right ) \ln \left (b x +a \right )}{a^{5}}\)	\(166\)
risch	\(\frac {-\frac {2 b \left (a^{2} d^{2}-3 a b c d +2 b^{2} c^{2}\right ) x^{3}}{a^{4}}-\frac {\left (a^{2} d^{2}-3 a b c d +2 b^{2} c^{2}\right ) x^{2}}{a^{3}}-\frac {c \left (3 a d -2 b c \right ) x}{3 a^{2}}-\frac {c^{2}}{3 a}}{x^{3} \left (b x +a \right )}-\frac {2 b \ln \left (x \right ) d^{2}}{a^{3}}+\frac {6 b^{2} \ln \left (x \right ) c d}{a^{4}}-\frac {4 b^{3} \ln \left (x \right ) c^{2}}{a^{5}}+\frac {2 b \ln \left (-b x -a \right ) d^{2}}{a^{3}}-\frac {6 b^{2} \ln \left (-b x -a \right ) c d}{a^{4}}+\frac {4 b^{3} \ln \left (-b x -a \right ) c^{2}}{a^{5}}\)	\(192\)
parallelrisch	\(-\frac {6 \ln \left (x \right ) x^{4} a^{2} b^{2} d^{2}-18 \ln \left (x \right ) x^{4} a \,b^{3} c d +12 \ln \left (x \right ) x^{4} b^{4} c^{2}-6 \ln \left (b x +a \right ) x^{4} a^{2} b^{2} d^{2}+18 \ln \left (b x +a \right ) x^{4} a \,b^{3} c d -12 \ln \left (b x +a \right ) x^{4} b^{4} c^{2}+6 \ln \left (x \right ) x^{3} a^{3} b \,d^{2}-18 \ln \left (x \right ) x^{3} a^{2} b^{2} c d +12 \ln \left (x \right ) x^{3} a \,b^{3} c^{2}-6 \ln \left (b x +a \right ) x^{3} a^{3} b \,d^{2}+18 \ln \left (b x +a \right ) x^{3} a^{2} b^{2} c d -12 \ln \left (b x +a \right ) x^{3} a \,b^{3} c^{2}-6 x^{4} a^{2} b^{2} d^{2}+18 x^{4} a \,b^{3} c d -12 x^{4} b^{4} c^{2}+3 a^{4} d^{2} x^{2}-9 a^{3} b c d \,x^{2}+6 a^{2} b^{2} c^{2} x^{2}+3 a^{4} c d x -2 a^{3} b \,c^{2} x +c^{2} a^{4}}{3 a^{5} x^{3} \left (b x +a \right )}\)	\(308\)

[In]

int((d*x+c)^2/x^4/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*c^2/a^2/x^3-(a^2*d^2-4*a*b*c*d+3*b^2*c^2)/a^4/x-2*b*(a^2*d^2-3*a*b*c*d+2*b^2*c^2)/a^5*ln(x)-c*(a*d-b*c)/a
^3/x^2-(a^2*d^2-2*a*b*c*d+b^2*c^2)/a^4*b/(b*x+a)+2*b*(a^2*d^2-3*a*b*c*d+2*b^2*c^2)/a^5*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.92 \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=-\frac {a^{4} c^{2} + 6 \, {\left (2 \, a b^{3} c^{2} - 3 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{3} + 3 \, {\left (2 \, a^{2} b^{2} c^{2} - 3 \, a^{3} b c d + a^{4} d^{2}\right )} x^{2} - {\left (2 \, a^{3} b c^{2} - 3 \, a^{4} c d\right )} x - 6 \, {\left ({\left (2 \, b^{4} c^{2} - 3 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{4} + {\left (2 \, a b^{3} c^{2} - 3 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{3}\right )} \log \left (b x + a\right ) + 6 \, {\left ({\left (2 \, b^{4} c^{2} - 3 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{4} + {\left (2 \, a b^{3} c^{2} - 3 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{3}\right )} \log \left (x\right )}{3 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}} \]

[In]

integrate((d*x+c)^2/x^4/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (119) = 238\).

Time = 0.52 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.47 \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=\frac {- a^{3} c^{2} + x^{3} \left (- 6 a^{2} b d^{2} + 18 a b^{2} c d - 12 b^{3} c^{2}\right ) + x^{2} \left (- 3 a^{3} d^{2} + 9 a^{2} b c d - 6 a b^{2} c^{2}\right ) + x \left (- 3 a^{3} c d + 2 a^{2} b c^{2}\right )}{3 a^{5} x^{3} + 3 a^{4} b x^{4}} - \frac {2 b \left (a d - 2 b c\right ) \left (a d - b c\right ) \log {\left (x + \frac {2 a^{3} b d^{2} - 6 a^{2} b^{2} c d + 4 a b^{3} c^{2} - 2 a b \left (a d - 2 b c\right ) \left (a d - b c\right )}{4 a^{2} b^{2} d^{2} - 12 a b^{3} c d + 8 b^{4} c^{2}} \right )}}{a^{5}} + \frac {2 b \left (a d - 2 b c\right ) \left (a d - b c\right ) \log {\left (x + \frac {2 a^{3} b d^{2} - 6 a^{2} b^{2} c d + 4 a b^{3} c^{2} + 2 a b \left (a d - 2 b c\right ) \left (a d - b c\right )}{4 a^{2} b^{2} d^{2} - 12 a b^{3} c d + 8 b^{4} c^{2}} \right )}}{a^{5}} \]

[In]

integrate((d*x+c)**2/x**4/(b*x+a)**2,x)

[Out]

(-a**3*c**2 + x**3*(-6*a**2*b*d**2 + 18*a*b**2*c*d - 12*b**3*c**2) + x**2*(-3*a**3*d**2 + 9*a**2*b*c*d - 6*a*b
**2*c**2) + x*(-3*a**3*c*d + 2*a**2*b*c**2))/(3*a**5*x**3 + 3*a**4*b*x**4) - 2*b*(a*d - 2*b*c)*(a*d - b*c)*log
(x + (2*a**3*b*d**2 - 6*a**2*b**2*c*d + 4*a*b**3*c**2 - 2*a*b*(a*d - 2*b*c)*(a*d - b*c))/(4*a**2*b**2*d**2 - 1
2*a*b**3*c*d + 8*b**4*c**2))/a**5 + 2*b*(a*d - 2*b*c)*(a*d - b*c)*log(x + (2*a**3*b*d**2 - 6*a**2*b**2*c*d + 4
*a*b**3*c**2 + 2*a*b*(a*d - 2*b*c)*(a*d - b*c))/(4*a**2*b**2*d**2 - 12*a*b**3*c*d + 8*b**4*c**2))/a**5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.34 \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=-\frac {a^{3} c^{2} + 6 \, {\left (2 \, b^{3} c^{2} - 3 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + 3 \, {\left (2 \, a b^{2} c^{2} - 3 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2} - {\left (2 \, a^{2} b c^{2} - 3 \, a^{3} c d\right )} x}{3 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} + \frac {2 \, {\left (2 \, b^{3} c^{2} - 3 \, a b^{2} c d + a^{2} b d^{2}\right )} \log \left (b x + a\right )}{a^{5}} - \frac {2 \, {\left (2 \, b^{3} c^{2} - 3 \, a b^{2} c d + a^{2} b d^{2}\right )} \log \left (x\right )}{a^{5}} \]

[In]

integrate((d*x+c)^2/x^4/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.79 \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=-\frac {2 \, {\left (2 \, b^{4} c^{2} - 3 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{5} b} - \frac {\frac {b^{7} c^{2}}{b x + a} - \frac {2 \, a b^{6} c d}{b x + a} + \frac {a^{2} b^{5} d^{2}}{b x + a}}{a^{4} b^{4}} + \frac {13 \, b^{3} c^{2} - 15 \, a b^{2} c d + 3 \, a^{2} b d^{2} - \frac {3 \, {\left (10 \, a b^{4} c^{2} - 11 \, a^{2} b^{3} c d + 2 \, a^{3} b^{2} d^{2}\right )}}{{\left (b x + a\right )} b} + \frac {3 \, {\left (6 \, a^{2} b^{5} c^{2} - 6 \, a^{3} b^{4} c d + a^{4} b^{3} d^{2}\right )}}{{\left (b x + a\right )}^{2} b^{2}}}{3 \, a^{5} {\left (\frac {a}{b x + a} - 1\right )}^{3}} \]

[In]

integrate((d*x+c)^2/x^4/(b*x+a)^2,x, algorithm="giac")

[Out]

-2*(2*b^4*c^2 - 3*a*b^3*c*d + a^2*b^2*d^2)*log(abs(-a/(b*x + a) + 1))/(a^5*b) - (b^7*c^2/(b*x + a) - 2*a*b^6*c
*d/(b*x + a) + a^2*b^5*d^2/(b*x + a))/(a^4*b^4) + 1/3*(13*b^3*c^2 - 15*a*b^2*c*d + 3*a^2*b*d^2 - 3*(10*a*b^4*c
^2 - 11*a^2*b^3*c*d + 2*a^3*b^2*d^2)/((b*x + a)*b) + 3*(6*a^2*b^5*c^2 - 6*a^3*b^4*c*d + a^4*b^3*d^2)/((b*x + a
)^2*b^2))/(a^5*(a/(b*x + a) - 1)^3)

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.36 \[ \int \frac {(c+d x)^2}{x^4 (a+b x)^2} \, dx=\frac {4\,b\,\mathrm {atanh}\left (\frac {2\,b\,\left (a\,d-b\,c\right )\,\left (a\,d-2\,b\,c\right )\,\left (a+2\,b\,x\right )}{a\,\left (2\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+4\,b^3\,c^2\right )}\right )\,\left (a\,d-b\,c\right )\,\left (a\,d-2\,b\,c\right )}{a^5}-\frac {\frac {c^2}{3\,a}+\frac {x^2\,\left (a^2\,d^2-3\,a\,b\,c\,d+2\,b^2\,c^2\right )}{a^3}+\frac {2\,b\,x^3\,\left (a^2\,d^2-3\,a\,b\,c\,d+2\,b^2\,c^2\right )}{a^4}+\frac {c\,x\,\left (3\,a\,d-2\,b\,c\right )}{3\,a^2}}{b\,x^4+a\,x^3} \]

[In]

int((c + d*x)^2/(x^4*(a + b*x)^2),x)

[Out]

(4*b*atanh((2*b*(a*d - b*c)*(a*d - 2*b*c)*(a + 2*b*x))/(a*(4*b^3*c^2 + 2*a^2*b*d^2 - 6*a*b^2*c*d)))*(a*d - b*c
)*(a*d - 2*b*c))/a^5 - (c^2/(3*a) + (x^2*(a^2*d^2 + 2*b^2*c^2 - 3*a*b*c*d))/a^3 + (2*b*x^3*(a^2*d^2 + 2*b^2*c^
2 - 3*a*b*c*d))/a^4 + (c*x*(3*a*d - 2*b*c))/(3*a^2))/(a*x^3 + b*x^4)