\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^3}{x^8} \, dx\) [552]

Optimal result

Integrand size = 27, antiderivative size = 101 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=-\frac {a^6 B}{6 x^6}-\frac {6 a^5 b B}{5 x^5}-\frac {15 a^4 b^2 B}{4 x^4}-\frac {20 a^3 b^3 B}{3 x^3}-\frac {15 a^2 b^4 B}{2 x^2}-\frac {6 a b^5 B}{x}-\frac {A (a+b x)^7}{7 a x^7}+b^6 B \log (x) \]

[Out]

-1/6*a^6*B/x^6-6/5*a^5*b*B/x^5-15/4*a^4*b^2*B/x^4-20/3*a^3*b^3*B/x^3-15/2*a^2*b^4*B/x^2-6*a*b^5*B/x-1/7*A*(b*x
+a)^7/a/x^7+b^6*B*ln(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {27, 79, 45} \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=-\frac {a^6 B}{6 x^6}-\frac {6 a^5 b B}{5 x^5}-\frac {15 a^4 b^2 B}{4 x^4}-\frac {20 a^3 b^3 B}{3 x^3}-\frac {15 a^2 b^4 B}{2 x^2}-\frac {A (a+b x)^7}{7 a x^7}-\frac {6 a b^5 B}{x}+b^6 B \log (x) \]

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^8,x]

[Out]

-1/6*(a^6*B)/x^6 - (6*a^5*b*B)/(5*x^5) - (15*a^4*b^2*B)/(4*x^4) - (20*a^3*b^3*B)/(3*x^3) - (15*a^2*b^4*B)/(2*x
^2) - (6*a*b^5*B)/x - (A*(a + b*x)^7)/(7*a*x^7) + b^6*B*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^6 (A+B x)}{x^8} \, dx \\ & = -\frac {A (a+b x)^7}{7 a x^7}+B \int \frac {(a+b x)^6}{x^7} \, dx \\ & = -\frac {A (a+b x)^7}{7 a x^7}+B \int \left (\frac {a^6}{x^7}+\frac {6 a^5 b}{x^6}+\frac {15 a^4 b^2}{x^5}+\frac {20 a^3 b^3}{x^4}+\frac {15 a^2 b^4}{x^3}+\frac {6 a b^5}{x^2}+\frac {b^6}{x}\right ) \, dx \\ & = -\frac {a^6 B}{6 x^6}-\frac {6 a^5 b B}{5 x^5}-\frac {15 a^4 b^2 B}{4 x^4}-\frac {20 a^3 b^3 B}{3 x^3}-\frac {15 a^2 b^4 B}{2 x^2}-\frac {6 a b^5 B}{x}-\frac {A (a+b x)^7}{7 a x^7}+b^6 B \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=-\frac {A b^6}{x}-\frac {3 a b^5 (A+2 B x)}{x^2}-\frac {5 a^2 b^4 (2 A+3 B x)}{2 x^3}-\frac {5 a^3 b^3 (3 A+4 B x)}{3 x^4}-\frac {3 a^4 b^2 (4 A+5 B x)}{4 x^5}-\frac {a^5 b (5 A+6 B x)}{5 x^6}-\frac {a^6 (6 A+7 B x)}{42 x^7}+b^6 B \log (x) \]

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^8,x]

[Out]

-((A*b^6)/x) - (3*a*b^5*(A + 2*B*x))/x^2 - (5*a^2*b^4*(2*A + 3*B*x))/(2*x^3) - (5*a^3*b^3*(3*A + 4*B*x))/(3*x^
4) - (3*a^4*b^2*(4*A + 5*B*x))/(4*x^5) - (a^5*b*(5*A + 6*B*x))/(5*x^6) - (a^6*(6*A + 7*B*x))/(42*x^7) + b^6*B*
Log[x]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.25

method	result	size
default	\(-\frac {a^{5} \left (6 A b +B a \right )}{6 x^{6}}-\frac {5 a^{3} b^{2} \left (4 A b +3 B a \right )}{4 x^{4}}-\frac {3 a^{4} b \left (5 A b +2 B a \right )}{5 x^{5}}+b^{6} B \ln \left (x \right )-\frac {3 a \,b^{4} \left (2 A b +5 B a \right )}{2 x^{2}}-\frac {A \,a^{6}}{7 x^{7}}-\frac {b^{5} \left (A b +6 B a \right )}{x}-\frac {5 a^{2} b^{3} \left (3 A b +4 B a \right )}{3 x^{3}}\)	\(126\)
norman	\(\frac {\left (-3 A a \,b^{5}-\frac {15}{2} B \,b^{4} a^{2}\right ) x^{5}+\left (-5 A \,a^{3} b^{3}-\frac {15}{4} B \,a^{4} b^{2}\right ) x^{3}+\left (-3 A \,a^{4} b^{2}-\frac {6}{5} B \,a^{5} b \right ) x^{2}+\left (-A \,a^{5} b -\frac {1}{6} B \,a^{6}\right ) x +\left (-5 A \,b^{4} a^{2}-\frac {20}{3} B \,a^{3} b^{3}\right ) x^{4}+\left (-A \,b^{6}-6 B a \,b^{5}\right ) x^{6}-\frac {A \,a^{6}}{7}}{x^{7}}+b^{6} B \ln \left (x \right )\)	\(142\)
risch	\(\frac {\left (-3 A a \,b^{5}-\frac {15}{2} B \,b^{4} a^{2}\right ) x^{5}+\left (-5 A \,a^{3} b^{3}-\frac {15}{4} B \,a^{4} b^{2}\right ) x^{3}+\left (-3 A \,a^{4} b^{2}-\frac {6}{5} B \,a^{5} b \right ) x^{2}+\left (-A \,a^{5} b -\frac {1}{6} B \,a^{6}\right ) x +\left (-5 A \,b^{4} a^{2}-\frac {20}{3} B \,a^{3} b^{3}\right ) x^{4}+\left (-A \,b^{6}-6 B a \,b^{5}\right ) x^{6}-\frac {A \,a^{6}}{7}}{x^{7}}+b^{6} B \ln \left (x \right )\)	\(142\)
parallelrisch	\(-\frac {-420 B \,b^{6} \ln \left (x \right ) x^{7}+420 A \,b^{6} x^{6}+2520 x^{6} B a \,b^{5}+1260 a A \,b^{5} x^{5}+3150 x^{5} B \,b^{4} a^{2}+2100 a^{2} A \,b^{4} x^{4}+2800 x^{4} B \,a^{3} b^{3}+2100 a^{3} A \,b^{3} x^{3}+1575 x^{3} B \,a^{4} b^{2}+1260 a^{4} A \,b^{2} x^{2}+504 x^{2} B \,a^{5} b +420 a^{5} A b x +70 x B \,a^{6}+60 A \,a^{6}}{420 x^{7}}\)	\(150\)

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/6*a^5*(6*A*b+B*a)/x^6-5/4*a^3*b^2*(4*A*b+3*B*a)/x^4-3/5*a^4*b*(5*A*b+2*B*a)/x^5+b^6*B*ln(x)-3/2*a*b^4*(2*A*
b+5*B*a)/x^2-1/7*A*a^6/x^7-b^5*(A*b+6*B*a)/x-5/3*a^2*b^3*(3*A*b+4*B*a)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.60 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.48 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=\frac {420 \, B b^{6} x^{7} \log \left (x\right ) - 60 \, A a^{6} - 420 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} - 630 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} - 700 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} - 525 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} - 252 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} - 70 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{420 \, x^{7}} \]

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^8,x, algorithm="fricas")

[Out]

1/420*(420*B*b^6*x^7*log(x) - 60*A*a^6 - 420*(6*B*a*b^5 + A*b^6)*x^6 - 630*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 - 700
*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 - 525*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 - 252*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 - 7
0*(B*a^6 + 6*A*a^5*b)*x)/x^7

Sympy [A] (verification not implemented)

Time = 2.57 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.54 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=B b^{6} \log {\left (x \right )} + \frac {- 60 A a^{6} + x^{6} \left (- 420 A b^{6} - 2520 B a b^{5}\right ) + x^{5} \left (- 1260 A a b^{5} - 3150 B a^{2} b^{4}\right ) + x^{4} \left (- 2100 A a^{2} b^{4} - 2800 B a^{3} b^{3}\right ) + x^{3} \left (- 2100 A a^{3} b^{3} - 1575 B a^{4} b^{2}\right ) + x^{2} \left (- 1260 A a^{4} b^{2} - 504 B a^{5} b\right ) + x \left (- 420 A a^{5} b - 70 B a^{6}\right )}{420 x^{7}} \]

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3/x**8,x)

[Out]

B*b**6*log(x) + (-60*A*a**6 + x**6*(-420*A*b**6 - 2520*B*a*b**5) + x**5*(-1260*A*a*b**5 - 3150*B*a**2*b**4) +
x**4*(-2100*A*a**2*b**4 - 2800*B*a**3*b**3) + x**3*(-2100*A*a**3*b**3 - 1575*B*a**4*b**2) + x**2*(-1260*A*a**4
*b**2 - 504*B*a**5*b) + x*(-420*A*a**5*b - 70*B*a**6))/(420*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.45 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=B b^{6} \log \left (x\right ) - \frac {60 \, A a^{6} + 420 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 630 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 700 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 525 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 252 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 70 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{420 \, x^{7}} \]

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^8,x, algorithm="maxima")

[Out]

B*b^6*log(x) - 1/420*(60*A*a^6 + 420*(6*B*a*b^5 + A*b^6)*x^6 + 630*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 700*(4*B*a^
3*b^3 + 3*A*a^2*b^4)*x^4 + 525*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 252*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 70*(B*a^6
+ 6*A*a^5*b)*x)/x^7

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.46 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=B b^{6} \log \left ({\left | x \right |}\right ) - \frac {60 \, A a^{6} + 420 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 630 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 700 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 525 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 252 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 70 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{420 \, x^{7}} \]

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^8,x, algorithm="giac")

[Out]

B*b^6*log(abs(x)) - 1/420*(60*A*a^6 + 420*(6*B*a*b^5 + A*b^6)*x^6 + 630*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 700*(4
*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 525*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 252*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 70*(
B*a^6 + 6*A*a^5*b)*x)/x^7

Mupad [B] (verification not implemented)

Time = 10.00 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^8} \, dx=B\,b^6\,\ln \left (x\right )-\frac {x\,\left (\frac {B\,a^6}{6}+A\,b\,a^5\right )+\frac {A\,a^6}{7}+x^2\,\left (\frac {6\,B\,a^5\,b}{5}+3\,A\,a^4\,b^2\right )+x^5\,\left (\frac {15\,B\,a^2\,b^4}{2}+3\,A\,a\,b^5\right )+x^6\,\left (A\,b^6+6\,B\,a\,b^5\right )+x^3\,\left (\frac {15\,B\,a^4\,b^2}{4}+5\,A\,a^3\,b^3\right )+x^4\,\left (\frac {20\,B\,a^3\,b^3}{3}+5\,A\,a^2\,b^4\right )}{x^7} \]

[In]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3)/x^8,x)

[Out]

B*b^6*log(x) - (x*((B*a^6)/6 + A*a^5*b) + (A*a^6)/7 + x^2*(3*A*a^4*b^2 + (6*B*a^5*b)/5) + x^5*((15*B*a^2*b^4)/
2 + 3*A*a*b^5) + x^6*(A*b^6 + 6*B*a*b^5) + x^3*(5*A*a^3*b^3 + (15*B*a^4*b^2)/4) + x^4*(5*A*a^2*b^4 + (20*B*a^3
*b^3)/3))/x^7