\(\int \frac {(a c+(b c+a d) x+b d x^2)^3}{(a+b x)^6} \, dx\) [1792]

Optimal result

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

[Out]

d^3*x/b^3-1/2*(-a*d+b*c)^3/b^4/(b*x+a)^2-3*d*(-a*d+b*c)^2/b^4/(b*x+a)+3*d^2*(-a*d+b*c)*ln(b*x+a)/b^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} \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^6} \, dx=\frac {3 d^2 (b c-a d) \log (a+b x)}{b^4}-\frac {3 d (b c-a d)^2}{b^4 (a+b x)}-\frac {(b c-a d)^3}{2 b^4 (a+b x)^2}+\frac {d^3 x}{b^3} \]

[In]

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

[Out]

(d^3*x)/b^3 - (b*c - a*d)^3/(2*b^4*(a + b*x)^2) - (3*d*(b*c - a*d)^2)/(b^4*(a + b*x)) + (3*d^2*(b*c - a*d)*Log
[a + b*x])/b^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 {(c+d x)^3}{(a+b x)^3} \, dx \\ & = \int \left (\frac {d^3}{b^3}+\frac {(b c-a d)^3}{b^3 (a+b x)^3}+\frac {3 d (b c-a d)^2}{b^3 (a+b x)^2}+\frac {3 d^2 (b c-a d)}{b^3 (a+b x)}\right ) \, dx \\ & = \frac {d^3 x}{b^3}-\frac {(b c-a d)^3}{2 b^4 (a+b x)^2}-\frac {3 d (b c-a d)^2}{b^4 (a+b x)}+\frac {3 d^2 (b c-a d) \log (a+b x)}{b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^6} \, dx=\frac {-5 a^3 d^3+a^2 b d^2 (9 c-4 d x)+a b^2 d \left (-3 c^2+12 c d x+4 d^2 x^2\right )-b^3 \left (c^3+6 c^2 d x-2 d^3 x^3\right )-6 d^2 (-b c+a d) (a+b x)^2 \log (a+b x)}{2 b^4 (a+b x)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.76 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {d^{3} x}{b^{3}}-\frac {3 d^{2} \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{4}}-\frac {-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}}{2 b^{4} \left (b x +a \right )^{2}}-\frac {3 d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{b^{4} \left (b x +a \right )}\)	\(114\)
risch	\(\frac {d^{3} x}{b^{3}}+\frac {\left (-3 a^{2} d^{3}+6 a b c \,d^{2}-3 b^{2} c^{2} d \right ) x -\frac {5 a^{3} d^{3}-9 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d +b^{3} c^{3}}{2 b}}{b^{3} \left (b x +a \right )^{2}}-\frac {3 d^{3} \ln \left (b x +a \right ) a}{b^{4}}+\frac {3 d^{2} \ln \left (b x +a \right ) c}{b^{3}}\)	\(121\)
parallelrisch	\(-\frac {6 \ln \left (b x +a \right ) x^{2} a \,b^{2} d^{3}-6 \ln \left (b x +a \right ) x^{2} b^{3} c \,d^{2}-2 d^{3} x^{3} b^{3}+12 \ln \left (b x +a \right ) x \,a^{2} b \,d^{3}-12 \ln \left (b x +a \right ) x a \,b^{2} c \,d^{2}+6 \ln \left (b x +a \right ) a^{3} d^{3}-6 \ln \left (b x +a \right ) a^{2} b c \,d^{2}+12 x \,a^{2} b \,d^{3}-12 x a \,b^{2} c \,d^{2}+6 x \,b^{3} c^{2} d +9 a^{3} d^{3}-9 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d +b^{3} c^{3}}{2 b^{4} \left (b x +a \right )^{2}}\)	\(190\)
norman	\(\frac {b^{2} d^{3} x^{6}-\frac {a^{3} \left (15 a^{3} b \,d^{3}-9 a^{2} b^{2} c \,d^{2}+3 a \,b^{3} c^{2} d +b^{4} c^{3}\right )}{2 b^{5}}-\frac {\left (18 a^{2} b \,d^{3}-6 a \,b^{2} c \,d^{2}+3 b^{3} c^{2} d \right ) x^{4}}{b}-\frac {\left (103 a^{3} b \,d^{3}-45 a^{2} b^{2} c \,d^{2}+21 a \,b^{3} c^{2} d +b^{4} c^{3}\right ) x^{3}}{2 b^{2}}-\frac {a \left (123 a^{3} b \,d^{3}-63 a^{2} b^{2} c \,d^{2}+27 a \,b^{3} c^{2} d +3 b^{4} c^{3}\right ) x^{2}}{2 b^{3}}-\frac {a^{2} \left (69 a^{3} b \,d^{3}-39 a^{2} b^{2} c \,d^{2}+15 a \,b^{3} c^{2} d +3 b^{4} c^{3}\right ) x}{2 b^{4}}}{\left (b x +a \right )^{5}}-\frac {3 d^{2} \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{4}}\)	\(272\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

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

Time = 0.30 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.41 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^6} \, dx=\frac {2 \, b^{3} d^{3} x^{3} + 4 \, a b^{2} d^{3} x^{2} - b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3} - 2 \, {\left (3 \, b^{3} c^{2} d - 6 \, a b^{2} c d^{2} + 2 \, a^{2} b d^{3}\right )} x + 6 \, {\left (a^{2} b c d^{2} - a^{3} d^{3} + {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{2} + 2 \, {\left (a b^{2} c d^{2} - a^{2} b d^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^6} \, dx=\frac {- 5 a^{3} d^{3} + 9 a^{2} b c d^{2} - 3 a b^{2} c^{2} d - b^{3} c^{3} + x \left (- 6 a^{2} b d^{3} + 12 a b^{2} c d^{2} - 6 b^{3} c^{2} d\right )}{2 a^{2} b^{4} + 4 a b^{5} x + 2 b^{6} x^{2}} + \frac {d^{3} x}{b^{3}} - \frac {3 d^{2} \left (a d - b c\right ) \log {\left (a + b x \right )}}{b^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^6} \, dx=\frac {d^{3} x}{b^{3}} - \frac {b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3} + 6 \, {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {3 \, {\left (b c d^{2} - a d^{3}\right )} \log \left (b x + a\right )}{b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^6} \, dx=\frac {d^{3} x}{b^{3}} + \frac {3 \, {\left (b c d^{2} - a d^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3} + 6 \, {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.67 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^6} \, dx=\frac {d^3\,x}{b^3}-\frac {\ln \left (a+b\,x\right )\,\left (3\,a\,d^3-3\,b\,c\,d^2\right )}{b^4}-\frac {\frac {5\,a^3\,d^3-9\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d+b^3\,c^3}{2\,b}+x\,\left (3\,a^2\,d^3-6\,a\,b\,c\,d^2+3\,b^2\,c^2\,d\right )}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2} \]

[In]

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

[Out]

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