\(\int \frac {(a+b x)^6}{(a c+(b c+a d) x+b d x^2)^3} \, dx\) [1820]

Optimal result

Integrand size = 29, antiderivative size = 78 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\frac {b^3 x}{d^3}+\frac {(b c-a d)^3}{2 d^4 (c+d x)^2}-\frac {3 b (b c-a d)^2}{d^4 (c+d x)}-\frac {3 b^2 (b c-a d) \log (c+d x)}{d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {640, 45} \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=-\frac {3 b^2 (b c-a d) \log (c+d x)}{d^4}-\frac {3 b (b c-a d)^2}{d^4 (c+d x)}+\frac {(b c-a d)^3}{2 d^4 (c+d x)^2}+\frac {b^3 x}{d^3} \]

[In]

Int[(a + b*x)^6/(a*c + (b*c + a*d)*x + b*d*x^2)^3,x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^3}{(c+d x)^3} \, dx \\ & = \int \left (\frac {b^3}{d^3}+\frac {(-b c+a d)^3}{d^3 (c+d x)^3}+\frac {3 b (b c-a d)^2}{d^3 (c+d x)^2}-\frac {3 b^2 (b c-a d)}{d^3 (c+d x)}\right ) \, dx \\ & = \frac {b^3 x}{d^3}+\frac {(b c-a d)^3}{2 d^4 (c+d x)^2}-\frac {3 b (b c-a d)^2}{d^4 (c+d x)}-\frac {3 b^2 (b c-a d) \log (c+d x)}{d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\frac {-a^3 d^3-3 a^2 b d^2 (c+2 d x)+3 a b^2 c d (3 c+4 d x)+b^3 \left (-5 c^3-4 c^2 d x+4 c d^2 x^2+2 d^3 x^3\right )-6 b^2 (b c-a d) (c+d x)^2 \log (c+d x)}{2 d^4 (c+d x)^2} \]

[In]

Integrate[(a + b*x)^6/(a*c + (b*c + a*d)*x + b*d*x^2)^3,x]

[Out]

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

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {b^{3} x}{d^{3}}-\frac {a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}}{2 d^{4} \left (d x +c \right )^{2}}+\frac {3 b^{2} \left (a d -b c \right ) \ln \left (d x +c \right )}{d^{4}}-\frac {3 b \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{d^{4} \left (d x +c \right )}\)	\(114\)
risch	\(\frac {b^{3} x}{d^{3}}+\frac {\left (-3 a^{2} b \,d^{2}+6 a \,b^{2} c d -3 c^{2} b^{3}\right ) x -\frac {a^{3} d^{3}+3 a^{2} b c \,d^{2}-9 a \,b^{2} c^{2} d +5 b^{3} c^{3}}{2 d}}{d^{3} \left (d x +c \right )^{2}}+\frac {3 b^{2} \ln \left (d x +c \right ) a}{d^{3}}-\frac {3 b^{3} \ln \left (d x +c \right ) c}{d^{4}}\)	\(121\)
parallelrisch	\(\frac {6 \ln \left (d x +c \right ) x^{2} a \,b^{2} d^{3}-6 \ln \left (d x +c \right ) x^{2} b^{3} c \,d^{2}+2 d^{3} x^{3} b^{3}+12 \ln \left (d x +c \right ) x a \,b^{2} c \,d^{2}-12 \ln \left (d x +c \right ) x \,b^{3} c^{2} d +6 \ln \left (d x +c \right ) a \,b^{2} c^{2} d -6 \ln \left (d x +c \right ) b^{3} c^{3}-6 x \,a^{2} b \,d^{3}+12 x a \,b^{2} c \,d^{2}-12 x \,b^{3} c^{2} d -a^{3} d^{3}-3 a^{2} b c \,d^{2}+9 a \,b^{2} c^{2} d -9 b^{3} c^{3}}{2 d^{4} \left (d x +c \right )^{2}}\)	\(191\)
norman	\(\frac {\frac {b^{5} x^{5}}{d}-\frac {\left (17 a^{3} b^{4} d^{3}-5 a^{2} b^{5} c \,d^{2}+19 a \,c^{2} d \,b^{6}+9 c^{3} b^{7}\right ) x^{2}}{2 d^{4} b^{2}}-\frac {2 \left (3 a^{2} b^{4} d^{2}-a c d \,b^{5}+3 c^{2} b^{6}\right ) x^{3}}{d^{3} b}-\frac {\left (a^{3} b^{2} d^{3}+3 a^{2} b^{3} c \,d^{2}-5 a \,b^{4} c^{2} d +9 b^{5} c^{3}\right ) a^{2}}{2 d^{4} b^{2}}-\frac {a \left (4 a^{3} b^{3} d^{3}+a^{2} b^{4} c \,d^{2}+c^{2} d a \,b^{5}+9 c^{3} b^{6}\right ) x}{d^{4} b^{2}}}{\left (b x +a \right )^{2} \left (d x +c \right )^{2}}+\frac {3 b^{2} \left (a d -b c \right ) \ln \left (d x +c \right )}{d^{4}}\)	\(244\)

[In]

int((b*x+a)^6/(b*d*x^2+(a*d+b*c)*x+a*c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (76) = 152\).

Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.41 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\frac {2 \, b^{3} d^{3} x^{3} + 4 \, b^{3} c d^{2} x^{2} - 5 \, b^{3} c^{3} + 9 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} - a^{3} d^{3} - 2 \, {\left (2 \, b^{3} c^{2} d - 6 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} x - 6 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d + {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{2} + 2 \, {\left (b^{3} c^{2} d - a b^{2} c d^{2}\right )} x\right )} \log \left (d x + c\right )}{2 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} \]

[In]

integrate((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\frac {b^{3} x}{d^{3}} + \frac {3 b^{2} \left (a d - b c\right ) \log {\left (c + d x \right )}}{d^{4}} + \frac {- a^{3} d^{3} - 3 a^{2} b c d^{2} + 9 a b^{2} c^{2} d - 5 b^{3} c^{3} + x \left (- 6 a^{2} b d^{3} + 12 a b^{2} c d^{2} - 6 b^{3} c^{2} d\right )}{2 c^{2} d^{4} + 4 c d^{5} x + 2 d^{6} x^{2}} \]

[In]

integrate((b*x+a)**6/(a*c+(a*d+b*c)*x+b*d*x**2)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.60 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\frac {b^{3} x}{d^{3}} - \frac {5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3} + 6 \, {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{2 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} - \frac {3 \, {\left (b^{3} c - a b^{2} d\right )} \log \left (d x + c\right )}{d^{4}} \]

[In]

integrate((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\frac {b^{3} x}{d^{3}} - \frac {3 \, {\left (b^{3} c - a b^{2} d\right )} \log \left ({\left | d x + c \right |}\right )}{d^{4}} - \frac {5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3} + 6 \, {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{2 \, {\left (d x + c\right )}^{2} d^{4}} \]

[In]

integrate((b*x+a)^6/(a*c+(a*d+b*c)*x+b*d*x^2)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.73 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.67 \[ \int \frac {(a+b x)^6}{\left (a c+(b c+a d) x+b d x^2\right )^3} \, dx=\frac {b^3\,x}{d^3}-\frac {\ln \left (c+d\,x\right )\,\left (3\,b^3\,c-3\,a\,b^2\,d\right )}{d^4}-\frac {\frac {a^3\,d^3+3\,a^2\,b\,c\,d^2-9\,a\,b^2\,c^2\,d+5\,b^3\,c^3}{2\,d}+x\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )}{c^2\,d^3+2\,c\,d^4\,x+d^5\,x^2} \]

[In]

int((a + b*x)^6/(a*c + x*(a*d + b*c) + b*d*x^2)^3,x)

[Out]

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