\(\int \frac {(A+B x) (a+b x+c x^2)^2}{x^7} \, dx\) [864]

Optimal result

Integrand size = 21, antiderivative size = 99 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=-\frac {a^2 A}{6 x^6}-\frac {a (2 A b+a B)}{5 x^5}-\frac {2 a b B+A \left (b^2+2 a c\right )}{4 x^4}-\frac {b^2 B+2 A b c+2 a B c}{3 x^3}-\frac {c (2 b B+A c)}{2 x^2}-\frac {B c^2}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {779} \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=-\frac {a^2 A}{6 x^6}-\frac {A \left (2 a c+b^2\right )+2 a b B}{4 x^4}-\frac {2 a B c+2 A b c+b^2 B}{3 x^3}-\frac {a (a B+2 A b)}{5 x^5}-\frac {c (A c+2 b B)}{2 x^2}-\frac {B c^2}{x} \]

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^7,x]

[Out]

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

Rule 779

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^2 A}{x^7}+\frac {a (2 A b+a B)}{x^6}+\frac {2 a b B+A \left (b^2+2 a c\right )}{x^5}+\frac {b^2 B+2 A b c+2 a B c}{x^4}+\frac {c (2 b B+A c)}{x^3}+\frac {B c^2}{x^2}\right ) \, dx \\ & = -\frac {a^2 A}{6 x^6}-\frac {a (2 A b+a B)}{5 x^5}-\frac {2 a b B+A \left (b^2+2 a c\right )}{4 x^4}-\frac {b^2 B+2 A b c+2 a B c}{3 x^3}-\frac {c (2 b B+A c)}{2 x^2}-\frac {B c^2}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=-\frac {2 a^2 (5 A+6 B x)+2 a x (5 B x (3 b+4 c x)+3 A (4 b+5 c x))+5 x^2 \left (4 B x \left (b^2+3 b c x+3 c^2 x^2\right )+A \left (3 b^2+8 b c x+6 c^2 x^2\right )\right )}{60 x^6} \]

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^7,x]

[Out]

-1/60*(2*a^2*(5*A + 6*B*x) + 2*a*x*(5*B*x*(3*b + 4*c*x) + 3*A*(4*b + 5*c*x)) + 5*x^2*(4*B*x*(b^2 + 3*b*c*x + 3
*c^2*x^2) + A*(3*b^2 + 8*b*c*x + 6*c^2*x^2)))/x^6

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {a^{2} A}{6 x^{6}}-\frac {2 A a c +A \,b^{2}+2 a b B}{4 x^{4}}-\frac {a \left (2 A b +B a \right )}{5 x^{5}}-\frac {c \left (A c +2 B b \right )}{2 x^{2}}-\frac {B \,c^{2}}{x}-\frac {2 A b c +2 B a c +B \,b^{2}}{3 x^{3}}\)	\(90\)
norman	\(\frac {-B \,c^{2} x^{5}+\left (-\frac {1}{2} A \,c^{2}-B b c \right ) x^{4}+\left (-\frac {2}{3} A b c -\frac {2}{3} B a c -\frac {1}{3} B \,b^{2}\right ) x^{3}+\left (-\frac {1}{2} A a c -\frac {1}{4} A \,b^{2}-\frac {1}{2} a b B \right ) x^{2}+\left (-\frac {2}{5} A b a -\frac {1}{5} B \,a^{2}\right ) x -\frac {A \,a^{2}}{6}}{x^{6}}\)	\(93\)
risch	\(\frac {-B \,c^{2} x^{5}+\left (-\frac {1}{2} A \,c^{2}-B b c \right ) x^{4}+\left (-\frac {2}{3} A b c -\frac {2}{3} B a c -\frac {1}{3} B \,b^{2}\right ) x^{3}+\left (-\frac {1}{2} A a c -\frac {1}{4} A \,b^{2}-\frac {1}{2} a b B \right ) x^{2}+\left (-\frac {2}{5} A b a -\frac {1}{5} B \,a^{2}\right ) x -\frac {A \,a^{2}}{6}}{x^{6}}\)	\(93\)
gosper	\(-\frac {60 B \,c^{2} x^{5}+30 A \,c^{2} x^{4}+60 x^{4} B b c +40 x^{3} A b c +40 a B c \,x^{3}+20 B \,b^{2} x^{3}+30 a A c \,x^{2}+15 A \,b^{2} x^{2}+30 B a b \,x^{2}+24 a A b x +12 a^{2} B x +10 A \,a^{2}}{60 x^{6}}\)	\(102\)
parallelrisch	\(-\frac {60 B \,c^{2} x^{5}+30 A \,c^{2} x^{4}+60 x^{4} B b c +40 x^{3} A b c +40 a B c \,x^{3}+20 B \,b^{2} x^{3}+30 a A c \,x^{2}+15 A \,b^{2} x^{2}+30 B a b \,x^{2}+24 a A b x +12 a^{2} B x +10 A \,a^{2}}{60 x^{6}}\)	\(102\)

[In]

int((B*x+A)*(c*x^2+b*x+a)^2/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=-\frac {60 \, B c^{2} x^{5} + 30 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 20 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + 10 \, A a^{2} + 15 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 12 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{6}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^7,x, algorithm="fricas")

[Out]

-1/60*(60*B*c^2*x^5 + 30*(2*B*b*c + A*c^2)*x^4 + 20*(B*b^2 + 2*(B*a + A*b)*c)*x^3 + 10*A*a^2 + 15*(2*B*a*b + A
*b^2 + 2*A*a*c)*x^2 + 12*(B*a^2 + 2*A*a*b)*x)/x^6

Sympy [A] (verification not implemented)

Time = 7.62 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=\frac {- 10 A a^{2} - 60 B c^{2} x^{5} + x^{4} \left (- 30 A c^{2} - 60 B b c\right ) + x^{3} \left (- 40 A b c - 40 B a c - 20 B b^{2}\right ) + x^{2} \left (- 30 A a c - 15 A b^{2} - 30 B a b\right ) + x \left (- 24 A a b - 12 B a^{2}\right )}{60 x^{6}} \]

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**2/x**7,x)

[Out]

(-10*A*a**2 - 60*B*c**2*x**5 + x**4*(-30*A*c**2 - 60*B*b*c) + x**3*(-40*A*b*c - 40*B*a*c - 20*B*b**2) + x**2*(
-30*A*a*c - 15*A*b**2 - 30*B*a*b) + x*(-24*A*a*b - 12*B*a**2))/(60*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=-\frac {60 \, B c^{2} x^{5} + 30 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 20 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + 10 \, A a^{2} + 15 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 12 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{6}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^7,x, algorithm="maxima")

[Out]

-1/60*(60*B*c^2*x^5 + 30*(2*B*b*c + A*c^2)*x^4 + 20*(B*b^2 + 2*(B*a + A*b)*c)*x^3 + 10*A*a^2 + 15*(2*B*a*b + A
*b^2 + 2*A*a*c)*x^2 + 12*(B*a^2 + 2*A*a*b)*x)/x^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=-\frac {60 \, B c^{2} x^{5} + 60 \, B b c x^{4} + 30 \, A c^{2} x^{4} + 20 \, B b^{2} x^{3} + 40 \, B a c x^{3} + 40 \, A b c x^{3} + 30 \, B a b x^{2} + 15 \, A b^{2} x^{2} + 30 \, A a c x^{2} + 12 \, B a^{2} x + 24 \, A a b x + 10 \, A a^{2}}{60 \, x^{6}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^7,x, algorithm="giac")

[Out]

-1/60*(60*B*c^2*x^5 + 60*B*b*c*x^4 + 30*A*c^2*x^4 + 20*B*b^2*x^3 + 40*B*a*c*x^3 + 40*A*b*c*x^3 + 30*B*a*b*x^2
+ 15*A*b^2*x^2 + 30*A*a*c*x^2 + 12*B*a^2*x + 24*A*a*b*x + 10*A*a^2)/x^6

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^7} \, dx=-\frac {x^4\,\left (\frac {A\,c^2}{2}+B\,b\,c\right )+\frac {A\,a^2}{6}+x^2\,\left (\frac {A\,b^2}{4}+\frac {B\,a\,b}{2}+\frac {A\,a\,c}{2}\right )+x^3\,\left (\frac {B\,b^2}{3}+\frac {2\,A\,c\,b}{3}+\frac {2\,B\,a\,c}{3}\right )+x\,\left (\frac {B\,a^2}{5}+\frac {2\,A\,b\,a}{5}\right )+B\,c^2\,x^5}{x^6} \]

[In]

int(((A + B*x)*(a + b*x + c*x^2)^2)/x^7,x)

[Out]

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