\(\int \frac {(a c+(b c+a d) x+b d x^2)^3}{(a+b x)^9} \, dx\) [1795]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 29, antiderivative size = 58 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=-\frac {(c+d x)^4}{5 (b c-a d) (a+b x)^5}+\frac {d (c+d x)^4}{20 (b c-a d)^2 (a+b x)^4} \]

[Out]

-1/5*(d*x+c)^4/(-a*d+b*c)/(b*x+a)^5+1/20*d*(d*x+c)^4/(-a*d+b*c)^2/(b*x+a)^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {640, 47, 37} \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=\frac {d (c+d x)^4}{20 (a+b x)^4 (b c-a d)^2}-\frac {(c+d x)^4}{5 (a+b x)^5 (b c-a d)} \]

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^9,x]

[Out]

-1/5*(c + d*x)^4/((b*c - a*d)*(a + b*x)^5) + (d*(c + d*x)^4)/(20*(b*c - a*d)^2*(a + b*x)^4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
 NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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)^6} \, dx \\ & = -\frac {(c+d x)^4}{5 (b c-a d) (a+b x)^5}-\frac {d \int \frac {(c+d x)^3}{(a+b x)^5} \, dx}{5 (b c-a d)} \\ & = -\frac {(c+d x)^4}{5 (b c-a d) (a+b x)^5}+\frac {d (c+d x)^4}{20 (b c-a d)^2 (a+b x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.67 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=-\frac {a^3 d^3+a^2 b d^2 (2 c+5 d x)+a b^2 d \left (3 c^2+10 c d x+10 d^2 x^2\right )+b^3 \left (4 c^3+15 c^2 d x+20 c d^2 x^2+10 d^3 x^3\right )}{20 b^4 (a+b x)^5} \]

[In]

Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^9,x]

[Out]

-1/20*(a^3*d^3 + a^2*b*d^2*(2*c + 5*d*x) + a*b^2*d*(3*c^2 + 10*c*d*x + 10*d^2*x^2) + b^3*(4*c^3 + 15*c^2*d*x +
 20*c*d^2*x^2 + 10*d^3*x^3))/(b^4*(a + b*x)^5)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(109\) vs. \(2(54)=108\).

Time = 2.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.90

method result size
risch \(\frac {-\frac {d^{3} x^{3}}{2 b}-\frac {d^{2} \left (a d +2 b c \right ) x^{2}}{2 b^{2}}-\frac {d \left (a^{2} d^{2}+2 a b c d +3 b^{2} c^{2}\right ) x}{4 b^{3}}-\frac {a^{3} d^{3}+2 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d +4 b^{3} c^{3}}{20 b^{4}}}{\left (b x +a \right )^{5}}\) \(110\)
gosper \(-\frac {10 d^{3} x^{3} b^{3}+10 x^{2} a \,b^{2} d^{3}+20 x^{2} b^{3} c \,d^{2}+5 x \,a^{2} b \,d^{3}+10 x a \,b^{2} c \,d^{2}+15 x \,b^{3} c^{2} d +a^{3} d^{3}+2 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d +4 b^{3} c^{3}}{20 b^{4} \left (b x +a \right )^{5}}\) \(115\)
default \(\frac {d^{2} \left (a d -b c \right )}{b^{4} \left (b x +a \right )^{3}}-\frac {-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}}{5 b^{4} \left (b x +a \right )^{5}}-\frac {3 d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{4 b^{4} \left (b x +a \right )^{4}}-\frac {d^{3}}{2 b^{4} \left (b x +a \right )^{2}}\) \(121\)
parallelrisch \(\frac {-10 d^{3} x^{3} b^{4}-10 a \,b^{3} d^{3} x^{2}-20 b^{4} c \,d^{2} x^{2}-5 x \,a^{2} d^{3} b^{2}-10 x a \,b^{3} c \,d^{2}-15 x \,b^{4} c^{2} d -a^{3} b \,d^{3}-2 a^{2} b^{2} c \,d^{2}-3 a \,b^{3} c^{2} d -4 b^{4} c^{3}}{20 b^{5} \left (b x +a \right )^{5}}\) \(121\)
norman \(\frac {\frac {\left (-2 a \,b^{4} d^{3}-b^{5} d^{2} c \right ) x^{5}}{b^{3}}+\frac {a^{3} \left (-a^{3} b^{4} d^{3}-2 a^{2} b^{5} c \,d^{2}-3 a \,c^{2} d \,b^{6}-4 c^{3} b^{7}\right )}{20 b^{8}}-\frac {b^{2} d^{3} x^{6}}{2}+\frac {\left (-13 d^{3} a^{2} b^{4}-14 c \,d^{2} a \,b^{5}-3 c^{2} d \,b^{6}\right ) x^{4}}{4 b^{4}}+\frac {\left (-14 a^{3} b^{4} d^{3}-23 a^{2} b^{5} c \,d^{2}-12 a \,c^{2} d \,b^{6}-c^{3} b^{7}\right ) x^{3}}{5 b^{5}}+\frac {a \left (-14 a^{3} b^{4} d^{3}-28 a^{2} b^{5} c \,d^{2}-27 a \,c^{2} d \,b^{6}-6 c^{3} b^{7}\right ) x^{2}}{10 b^{6}}+\frac {a^{2} \left (-2 a^{3} b^{4} d^{3}-4 a^{2} b^{5} c \,d^{2}-6 a \,c^{2} d \,b^{6}-3 c^{3} b^{7}\right ) x}{5 b^{7}}}{\left (b x +a \right )^{8}}\) \(288\)

[In]

int((b*d*x^2+(a*d+b*c)*x+a*c)^3/(b*x+a)^9,x,method=_RETURNVERBOSE)

[Out]

(-1/2/b*d^3*x^3-1/2/b^2*d^2*(a*d+2*b*c)*x^2-1/4/b^3*d*(a^2*d^2+2*a*b*c*d+3*b^2*c^2)*x-1/20/b^4*(a^3*d^3+2*a^2*
b*c*d^2+3*a*b^2*c^2*d+4*b^3*c^3))/(b*x+a)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (54) = 108\).

Time = 0.27 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.76 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=-\frac {10 \, b^{3} d^{3} x^{3} + 4 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3} + 10 \, {\left (2 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} + 5 \, {\left (3 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{20 \, {\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^9,x, algorithm="fricas")

[Out]

-1/20*(10*b^3*d^3*x^3 + 4*b^3*c^3 + 3*a*b^2*c^2*d + 2*a^2*b*c*d^2 + a^3*d^3 + 10*(2*b^3*c*d^2 + a*b^2*d^3)*x^2
 + 5*(3*b^3*c^2*d + 2*a*b^2*c*d^2 + a^2*b*d^3)*x)/(b^9*x^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10*a^3*b^6*x^2 + 5
*a^4*b^5*x + a^5*b^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (46) = 92\).

Time = 10.87 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.97 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=\frac {- a^{3} d^{3} - 2 a^{2} b c d^{2} - 3 a b^{2} c^{2} d - 4 b^{3} c^{3} - 10 b^{3} d^{3} x^{3} + x^{2} \left (- 10 a b^{2} d^{3} - 20 b^{3} c d^{2}\right ) + x \left (- 5 a^{2} b d^{3} - 10 a b^{2} c d^{2} - 15 b^{3} c^{2} d\right )}{20 a^{5} b^{4} + 100 a^{4} b^{5} x + 200 a^{3} b^{6} x^{2} + 200 a^{2} b^{7} x^{3} + 100 a b^{8} x^{4} + 20 b^{9} x^{5}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x**2)**3/(b*x+a)**9,x)

[Out]

(-a**3*d**3 - 2*a**2*b*c*d**2 - 3*a*b**2*c**2*d - 4*b**3*c**3 - 10*b**3*d**3*x**3 + x**2*(-10*a*b**2*d**3 - 20
*b**3*c*d**2) + x*(-5*a**2*b*d**3 - 10*a*b**2*c*d**2 - 15*b**3*c**2*d))/(20*a**5*b**4 + 100*a**4*b**5*x + 200*
a**3*b**6*x**2 + 200*a**2*b**7*x**3 + 100*a*b**8*x**4 + 20*b**9*x**5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (54) = 108\).

Time = 0.20 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.76 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=-\frac {10 \, b^{3} d^{3} x^{3} + 4 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3} + 10 \, {\left (2 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} + 5 \, {\left (3 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{20 \, {\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^9,x, algorithm="maxima")

[Out]

-1/20*(10*b^3*d^3*x^3 + 4*b^3*c^3 + 3*a*b^2*c^2*d + 2*a^2*b*c*d^2 + a^3*d^3 + 10*(2*b^3*c*d^2 + a*b^2*d^3)*x^2
 + 5*(3*b^3*c^2*d + 2*a*b^2*c*d^2 + a^2*b*d^3)*x)/(b^9*x^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10*a^3*b^6*x^2 + 5
*a^4*b^5*x + a^5*b^4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (54) = 108\).

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=-\frac {10 \, b^{3} d^{3} x^{3} + 20 \, b^{3} c d^{2} x^{2} + 10 \, a b^{2} d^{3} x^{2} + 15 \, b^{3} c^{2} d x + 10 \, a b^{2} c d^{2} x + 5 \, a^{2} b d^{3} x + 4 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3}}{20 \, {\left (b x + a\right )}^{5} b^{4}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^9,x, algorithm="giac")

[Out]

-1/20*(10*b^3*d^3*x^3 + 20*b^3*c*d^2*x^2 + 10*a*b^2*d^3*x^2 + 15*b^3*c^2*d*x + 10*a*b^2*c*d^2*x + 5*a^2*b*d^3*
x + 4*b^3*c^3 + 3*a*b^2*c^2*d + 2*a^2*b*c*d^2 + a^3*d^3)/((b*x + a)^5*b^4)

Mupad [B] (verification not implemented)

Time = 9.82 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.66 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^9} \, dx=-\frac {\frac {a^3\,d^3+2\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d+4\,b^3\,c^3}{20\,b^4}+\frac {d^3\,x^3}{2\,b}+\frac {d\,x\,\left (a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2\right )}{4\,b^3}+\frac {d^2\,x^2\,\left (a\,d+2\,b\,c\right )}{2\,b^2}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5} \]

[In]

int((a*c + x*(a*d + b*c) + b*d*x^2)^3/(a + b*x)^9,x)

[Out]

-((a^3*d^3 + 4*b^3*c^3 + 3*a*b^2*c^2*d + 2*a^2*b*c*d^2)/(20*b^4) + (d^3*x^3)/(2*b) + (d*x*(a^2*d^2 + 3*b^2*c^2
 + 2*a*b*c*d))/(4*b^3) + (d^2*x^2*(a*d + 2*b*c))/(2*b^2))/(a^5 + b^5*x^5 + 5*a*b^4*x^4 + 10*a^3*b^2*x^2 + 10*a
^2*b^3*x^3 + 5*a^4*b*x)