\(\int \frac {(a+b x)^6}{(a c+(b c+a d) x+b d x^2)^2} \, dx\) [52]

Optimal result

Integrand size = 29, antiderivative size = 104 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {6 b^2 (b c-a d)^2 x}{d^4}-\frac {(b c-a d)^4}{d^5 (c+d x)}-\frac {2 b^3 (b c-a d) (c+d x)^2}{d^5}+\frac {b^4 (c+d x)^3}{3 d^5}-\frac {4 b (b c-a d)^3 \log (c+d x)}{d^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {12 a^3 b c d^3-3 a^4 d^4+18 a^2 b^2 d^2 \left (-c^2+c d x+d^2 x^2\right )+6 a b^3 d \left (2 c^3-4 c^2 d x-3 c d^2 x^2+d^3 x^3\right )+b^4 \left (-3 c^4+9 c^3 d x+6 c^2 d^2 x^2-2 c d^3 x^3+d^4 x^4\right )-12 b (b c-a d)^3 (c+d x) \log (c+d x)}{3 d^5 (c+d x)} \] Input:

Integrate[(a + b*x)^6/(a*c + (b*c + a*d)*x + b*d*x^2)^2,x]
 

Output:

(12*a^3*b*c*d^3 - 3*a^4*d^4 + 18*a^2*b^2*d^2*(-c^2 + c*d*x + d^2*x^2) + 6* 
a*b^3*d*(2*c^3 - 4*c^2*d*x - 3*c*d^2*x^2 + d^3*x^3) + b^4*(-3*c^4 + 9*c^3* 
d*x + 6*c^2*d^2*x^2 - 2*c*d^3*x^3 + d^4*x^4) - 12*b*(b*c - a*d)^3*(c + d*x 
)*Log[c + d*x])/(3*d^5*(c + d*x))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1121, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+b x)^6}{\left (x (a d+b c)+a c+b d x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1121

\(\displaystyle \int \left (-\frac {4 b^3 (c+d x) (b c-a d)}{d^4}+\frac {6 b^2 (b c-a d)^2}{d^4}-\frac {4 b (b c-a d)^3}{d^4 (c+d x)}+\frac {(a d-b c)^4}{d^4 (c+d x)^2}+\frac {b^4 (c+d x)^2}{d^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b^3 (c+d x)^2 (b c-a d)}{d^5}+\frac {6 b^2 x (b c-a d)^2}{d^4}-\frac {(b c-a d)^4}{d^5 (c+d x)}-\frac {4 b (b c-a d)^3 \log (c+d x)}{d^5}+\frac {b^4 (c+d x)^3}{3 d^5}\)

Input:

Int[(a + b*x)^6/(a*c + (b*c + a*d)*x + b*d*x^2)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 1121

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int 
egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.68

method	result	size
default	\(\frac {b^{2} \left (\frac {1}{3} x^{3} b^{2} d^{2}+2 a b \,d^{2} x^{2}-b^{2} c \,x^{2} d +6 a^{2} d^{2} x -8 a b c d x +3 x \,b^{2} c^{2}\right )}{d^{4}}-\frac {a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}}{d^{5} \left (d x +c \right )}+\frac {4 b \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (d x +c \right )}{d^{5}}\)	\(175\)
risch	\(\frac {b^{4} x^{3}}{3 d^{2}}+\frac {2 b^{3} a \,x^{2}}{d^{2}}-\frac {b^{4} c \,x^{2}}{d^{3}}+\frac {6 b^{2} a^{2} x}{d^{2}}-\frac {8 b^{3} a c x}{d^{3}}+\frac {3 b^{4} x \,c^{2}}{d^{4}}-\frac {a^{4}}{d \left (d x +c \right )}+\frac {4 a^{3} b c}{d^{2} \left (d x +c \right )}-\frac {6 a^{2} b^{2} c^{2}}{d^{3} \left (d x +c \right )}+\frac {4 a \,b^{3} c^{3}}{d^{4} \left (d x +c \right )}-\frac {c^{4} b^{4}}{d^{5} \left (d x +c \right )}+\frac {4 b \ln \left (d x +c \right ) a^{3}}{d^{2}}-\frac {12 b^{2} \ln \left (d x +c \right ) a^{2} c}{d^{3}}+\frac {12 b^{3} \ln \left (d x +c \right ) a \,c^{2}}{d^{4}}-\frac {4 b^{4} \ln \left (d x +c \right ) c^{3}}{d^{5}}\)	\(230\)
norman	\(\frac {\frac {b^{5} x^{5}}{3 d}+\frac {2 b^{3} \left (12 a^{2} d^{2}-10 a b c d +3 b^{2} c^{2}\right ) x^{3}}{3 d^{3}}+\frac {b^{4} \left (7 a d -2 b c \right ) x^{4}}{3 d^{2}}-\frac {a \left (a^{4} b \,d^{4}+2 a^{3} b^{2} c \,d^{3}+6 a^{2} b^{3} c^{2} d^{2}-10 a \,b^{4} c^{3} d +4 b^{5} c^{4}\right )}{d^{5} b}-\frac {\left (7 a^{4} b^{2} d^{4}-4 d^{3} c \,a^{3} b^{3}+8 a^{2} c^{2} d^{2} b^{4}-10 a \,c^{3} d \,b^{5}+4 b^{6} c^{4}\right ) x}{d^{5} b}}{\left (b x +a \right ) \left (d x +c \right )}+\frac {4 b \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (d x +c \right )}{d^{5}}\)	\(258\)
parallelrisch	\(\frac {b^{4} d^{4} x^{4}+6 a \,b^{3} d^{4} x^{3}-2 b^{4} c \,d^{3} x^{3}+12 \ln \left (d x +c \right ) x \,a^{3} b \,d^{4}-36 \ln \left (d x +c \right ) x \,a^{2} b^{2} c \,d^{3}+36 \ln \left (d x +c \right ) x a \,b^{3} c^{2} d^{2}-12 \ln \left (d x +c \right ) x \,b^{4} c^{3} d +18 a^{2} b^{2} d^{4} x^{2}-18 a \,b^{3} c \,d^{3} x^{2}+6 b^{4} c^{2} d^{2} x^{2}+12 \ln \left (d x +c \right ) a^{3} b c \,d^{3}-36 \ln \left (d x +c \right ) a^{2} b^{2} c^{2} d^{2}+36 \ln \left (d x +c \right ) a \,b^{3} c^{3} d -12 \ln \left (d x +c \right ) b^{4} c^{4}-3 a^{4} d^{4}+12 a^{3} b c \,d^{3}-36 a^{2} b^{2} c^{2} d^{2}+36 a \,b^{3} c^{3} d -12 c^{4} b^{4}}{3 d^{5} \left (d x +c \right )}\)	\(275\)

Input:

int((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

b^2/d^4*(1/3*x^3*b^2*d^2+2*a*b*d^2*x^2-b^2*c*x^2*d+6*a^2*d^2*x-8*a*b*c*d*x 
+3*x*b^2*c^2)-(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d^2-4*a*b^3*c^3*d+b^4*c 
^4)/d^5/(d*x+c)+4*b/d^5*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)*ln(d 
*x+c)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (102) = 204\).

Time = 0.08 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.57 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b^{4} d^{4} x^{4} - 3 \, b^{4} c^{4} + 12 \, a b^{3} c^{3} d - 18 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} - 3 \, a^{4} d^{4} - 2 \, {\left (b^{4} c d^{3} - 3 \, a b^{3} d^{4}\right )} x^{3} + 6 \, {\left (b^{4} c^{2} d^{2} - 3 \, a b^{3} c d^{3} + 3 \, a^{2} b^{2} d^{4}\right )} x^{2} + 3 \, {\left (3 \, b^{4} c^{3} d - 8 \, a b^{3} c^{2} d^{2} + 6 \, a^{2} b^{2} c d^{3}\right )} x - 12 \, {\left (b^{4} c^{4} - 3 \, a b^{3} c^{3} d + 3 \, a^{2} b^{2} c^{2} d^{2} - a^{3} b c d^{3} + {\left (b^{4} c^{3} d - 3 \, a b^{3} c^{2} d^{2} + 3 \, a^{2} b^{2} c d^{3} - a^{3} b d^{4}\right )} x\right )} \log \left (d x + c\right )}{3 \, {\left (d^{6} x + c d^{5}\right )}} \] Input:

integrate((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.49 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b^{4} x^{3}}{3 d^{2}} + \frac {4 b \left (a d - b c\right )^{3} \log {\left (c + d x \right )}}{d^{5}} + x^{2} \cdot \left (\frac {2 a b^{3}}{d^{2}} - \frac {b^{4} c}{d^{3}}\right ) + x \left (\frac {6 a^{2} b^{2}}{d^{2}} - \frac {8 a b^{3} c}{d^{3}} + \frac {3 b^{4} c^{2}}{d^{4}}\right ) + \frac {- a^{4} d^{4} + 4 a^{3} b c d^{3} - 6 a^{2} b^{2} c^{2} d^{2} + 4 a b^{3} c^{3} d - b^{4} c^{4}}{c d^{5} + d^{6} x} \] Input:

integrate((b*x+a)**6/(a*c+(a*d+b*c)*x+b*d*x**2)**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.76 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{d^{6} x + c d^{5}} + \frac {b^{4} d^{2} x^{3} - 3 \, {\left (b^{4} c d - 2 \, a b^{3} d^{2}\right )} x^{2} + 3 \, {\left (3 \, b^{4} c^{2} - 8 \, a b^{3} c d + 6 \, a^{2} b^{2} d^{2}\right )} x}{3 \, d^{4}} - \frac {4 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \log \left (d x + c\right )}{d^{5}} \] Input:

integrate((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.81 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {4 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \log \left ({\left | d x + c \right |}\right )}{d^{5}} + \frac {b^{4} d^{4} x^{3} - 3 \, b^{4} c d^{3} x^{2} + 6 \, a b^{3} d^{4} x^{2} + 9 \, b^{4} c^{2} d^{2} x - 24 \, a b^{3} c d^{3} x + 18 \, a^{2} b^{2} d^{4} x}{3 \, d^{6}} - \frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{{\left (d x + c\right )} d^{5}} \] Input:

integrate((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="giac")
 

Output:

-4*(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*log(abs(d*x + c 
))/d^5 + 1/3*(b^4*d^4*x^3 - 3*b^4*c*d^3*x^2 + 6*a*b^3*d^4*x^2 + 9*b^4*c^2* 
d^2*x - 24*a*b^3*c*d^3*x + 18*a^2*b^2*d^4*x)/d^6 - (b^4*c^4 - 4*a*b^3*c^3* 
d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/((d*x + c)*d^5)
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.95 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=x^2\,\left (\frac {2\,a\,b^3}{d^2}-\frac {b^4\,c}{d^3}\right )-x\,\left (\frac {2\,c\,\left (\frac {4\,a\,b^3}{d^2}-\frac {2\,b^4\,c}{d^3}\right )}{d}-\frac {6\,a^2\,b^2}{d^2}+\frac {b^4\,c^2}{d^4}\right )+\frac {b^4\,x^3}{3\,d^2}-\frac {\ln \left (c+d\,x\right )\,\left (-4\,a^3\,b\,d^3+12\,a^2\,b^2\,c\,d^2-12\,a\,b^3\,c^2\,d+4\,b^4\,c^3\right )}{d^5}-\frac {a^4\,d^4-4\,a^3\,b\,c\,d^3+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d+b^4\,c^4}{d\,\left (x\,d^5+c\,d^4\right )} \] Input:

int((a + b*x)^6/(a*c + x*(a*d + b*c) + b*d*x^2)^2,x)
 

Output:

x^2*((2*a*b^3)/d^2 - (b^4*c)/d^3) - x*((2*c*((4*a*b^3)/d^2 - (2*b^4*c)/d^3 
))/d - (6*a^2*b^2)/d^2 + (b^4*c^2)/d^4) + (b^4*x^3)/(3*d^2) - (log(c + d*x 
)*(4*b^4*c^3 - 4*a^3*b*d^3 + 12*a^2*b^2*c*d^2 - 12*a*b^3*c^2*d))/d^5 - (a^ 
4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3)/(d*(c 
*d^4 + d^5*x))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.86 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {12 \,\mathrm {log}\left (d x +c \right ) a^{3} b \,c^{2} d^{3}+12 \,\mathrm {log}\left (d x +c \right ) a^{3} b c \,d^{4} x -36 \,\mathrm {log}\left (d x +c \right ) a^{2} b^{2} c^{3} d^{2}-36 \,\mathrm {log}\left (d x +c \right ) a^{2} b^{2} c^{2} d^{3} x +36 \,\mathrm {log}\left (d x +c \right ) a \,b^{3} c^{4} d +36 \,\mathrm {log}\left (d x +c \right ) a \,b^{3} c^{3} d^{2} x -12 \,\mathrm {log}\left (d x +c \right ) b^{4} c^{5}-12 \,\mathrm {log}\left (d x +c \right ) b^{4} c^{4} d x +3 a^{4} d^{5} x -12 a^{3} b c \,d^{4} x +36 a^{2} b^{2} c^{2} d^{3} x +18 a^{2} b^{2} c \,d^{4} x^{2}-36 a \,b^{3} c^{3} d^{2} x -18 a \,b^{3} c^{2} d^{3} x^{2}+6 a \,b^{3} c \,d^{4} x^{3}+12 b^{4} c^{4} d x +6 b^{4} c^{3} d^{2} x^{2}-2 b^{4} c^{2} d^{3} x^{3}+b^{4} c \,d^{4} x^{4}}{3 c \,d^{5} \left (d x +c \right )} \] Input:

int((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x)
 

Output:

(12*log(c + d*x)*a**3*b*c**2*d**3 + 12*log(c + d*x)*a**3*b*c*d**4*x - 36*l 
og(c + d*x)*a**2*b**2*c**3*d**2 - 36*log(c + d*x)*a**2*b**2*c**2*d**3*x + 
36*log(c + d*x)*a*b**3*c**4*d + 36*log(c + d*x)*a*b**3*c**3*d**2*x - 12*lo 
g(c + d*x)*b**4*c**5 - 12*log(c + d*x)*b**4*c**4*d*x + 3*a**4*d**5*x - 12* 
a**3*b*c*d**4*x + 36*a**2*b**2*c**2*d**3*x + 18*a**2*b**2*c*d**4*x**2 - 36 
*a*b**3*c**3*d**2*x - 18*a*b**3*c**2*d**3*x**2 + 6*a*b**3*c*d**4*x**3 + 12 
*b**4*c**4*d*x + 6*b**4*c**3*d**2*x**2 - 2*b**4*c**2*d**3*x**3 + b**4*c*d* 
*4*x**4)/(3*c*d**5*(c + d*x))