\(\int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx\) [126]

Optimal result

Integrand size = 16, antiderivative size = 105 \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=-\frac {a^5 A}{x}+5 a^3 b (2 A b+a B) x+5 a^2 b^2 (A b+a B) x^2+\frac {5}{3} a b^3 (A b+2 a B) x^3+\frac {1}{4} b^4 (A b+5 a B) x^4+\frac {1}{5} b^5 B x^5+a^4 (5 A b+a B) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=-\frac {a^5 A}{x}+a^4 \log (x) (a B+5 A b)+5 a^3 b x (a B+2 A b)+5 a^2 b^2 x^2 (a B+A b)+\frac {1}{4} b^4 x^4 (5 a B+A b)+\frac {5}{3} a b^3 x^3 (2 a B+A b)+\frac {1}{5} b^5 B x^5 \]

[In]

Int[((a + b*x)^5*(A + B*x))/x^2,x]

[Out]

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

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \left (5 a^3 b (2 A b+a B)+\frac {a^5 A}{x^2}+\frac {a^4 (5 A b+a B)}{x}+10 a^2 b^2 (A b+a B) x+5 a b^3 (A b+2 a B) x^2+b^4 (A b+5 a B) x^3+b^5 B x^4\right ) \, dx \\ & = -\frac {a^5 A}{x}+5 a^3 b (2 A b+a B) x+5 a^2 b^2 (A b+a B) x^2+\frac {5}{3} a b^3 (A b+2 a B) x^3+\frac {1}{4} b^4 (A b+5 a B) x^4+\frac {1}{5} b^5 B x^5+a^4 (5 A b+a B) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=-\frac {a^5 A}{x}+5 a^3 b (2 A b+a B) x+5 a^2 b^2 (A b+a B) x^2+\frac {5}{3} a b^3 (A b+2 a B) x^3+\frac {1}{4} b^4 (A b+5 a B) x^4+\frac {1}{5} b^5 B x^5+\left (5 a^4 A b+a^5 B\right ) \log (x) \]

[In]

Integrate[((a + b*x)^5*(A + B*x))/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11

method	result	size
default	\(\frac {b^{5} B \,x^{5}}{5}+\frac {A \,b^{5} x^{4}}{4}+\frac {5 B a \,b^{4} x^{4}}{4}+\frac {5 A a \,b^{4} x^{3}}{3}+\frac {10 B \,a^{2} b^{3} x^{3}}{3}+5 A \,a^{2} b^{3} x^{2}+5 B \,a^{3} b^{2} x^{2}+10 a^{3} b^{2} A x +5 a^{4} b B x +a^{4} \left (5 A b +B a \right ) \ln \left (x \right )-\frac {a^{5} A}{x}\)	\(117\)
risch	\(\frac {b^{5} B \,x^{5}}{5}+\frac {A \,b^{5} x^{4}}{4}+\frac {5 B a \,b^{4} x^{4}}{4}+\frac {5 A a \,b^{4} x^{3}}{3}+\frac {10 B \,a^{2} b^{3} x^{3}}{3}+5 A \,a^{2} b^{3} x^{2}+5 B \,a^{3} b^{2} x^{2}+10 a^{3} b^{2} A x +5 a^{4} b B x -\frac {a^{5} A}{x}+5 A \ln \left (x \right ) a^{4} b +B \ln \left (x \right ) a^{5}\)	\(119\)
norman	\(\frac {\left (\frac {1}{4} b^{5} A +\frac {5}{4} a \,b^{4} B \right ) x^{5}+\left (\frac {5}{3} a \,b^{4} A +\frac {10}{3} a^{2} b^{3} B \right ) x^{4}+\left (5 a^{2} b^{3} A +5 a^{3} b^{2} B \right ) x^{3}+\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) x^{2}-a^{5} A +\frac {b^{5} B \,x^{6}}{5}}{x}+\left (5 a^{4} b A +a^{5} B \right ) \ln \left (x \right )\)	\(121\)
parallelrisch	\(\frac {12 b^{5} B \,x^{6}+15 A \,b^{5} x^{5}+75 B a \,b^{4} x^{5}+100 a A \,b^{4} x^{4}+200 B \,a^{2} b^{3} x^{4}+300 a^{2} A \,b^{3} x^{3}+300 B \,a^{3} b^{2} x^{3}+300 A \ln \left (x \right ) x \,a^{4} b +600 a^{3} A \,b^{2} x^{2}+60 B \ln \left (x \right ) x \,a^{5}+300 B \,a^{4} b \,x^{2}-60 a^{5} A}{60 x}\)	\(128\)

[In]

int((b*x+a)^5*(B*x+A)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/5*b^5*B*x^5+1/4*A*b^5*x^4+5/4*B*a*b^4*x^4+5/3*A*a*b^4*x^3+10/3*B*a^2*b^3*x^3+5*A*a^2*b^3*x^2+5*B*a^3*b^2*x^2
+10*a^3*b^2*A*x+5*a^4*b*B*x+a^4*(5*A*b+B*a)*ln(x)-a^5*A/x

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=\frac {12 \, B b^{5} x^{6} - 60 \, A a^{5} + 15 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 100 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 300 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 60 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x \log \left (x\right )}{60 \, x} \]

[In]

integrate((b*x+a)^5*(B*x+A)/x^2,x, algorithm="fricas")

[Out]

1/60*(12*B*b^5*x^6 - 60*A*a^5 + 15*(5*B*a*b^4 + A*b^5)*x^5 + 100*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 300*(B*a^3*b^2
+ A*a^2*b^3)*x^3 + 300*(B*a^4*b + 2*A*a^3*b^2)*x^2 + 60*(B*a^5 + 5*A*a^4*b)*x*log(x))/x

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=- \frac {A a^{5}}{x} + \frac {B b^{5} x^{5}}{5} + a^{4} \cdot \left (5 A b + B a\right ) \log {\left (x \right )} + x^{4} \left (\frac {A b^{5}}{4} + \frac {5 B a b^{4}}{4}\right ) + x^{3} \cdot \left (\frac {5 A a b^{4}}{3} + \frac {10 B a^{2} b^{3}}{3}\right ) + x^{2} \cdot \left (5 A a^{2} b^{3} + 5 B a^{3} b^{2}\right ) + x \left (10 A a^{3} b^{2} + 5 B a^{4} b\right ) \]

[In]

integrate((b*x+a)**5*(B*x+A)/x**2,x)

[Out]

-A*a**5/x + B*b**5*x**5/5 + a**4*(5*A*b + B*a)*log(x) + x**4*(A*b**5/4 + 5*B*a*b**4/4) + x**3*(5*A*a*b**4/3 +
10*B*a**2*b**3/3) + x**2*(5*A*a**2*b**3 + 5*B*a**3*b**2) + x*(10*A*a**3*b**2 + 5*B*a**4*b)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=\frac {1}{5} \, B b^{5} x^{5} - \frac {A a^{5}}{x} + \frac {1}{4} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{4} + \frac {5}{3} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{3} + 5 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{2} + 5 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x + {\left (B a^{5} + 5 \, A a^{4} b\right )} \log \left (x\right ) \]

[In]

integrate((b*x+a)^5*(B*x+A)/x^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=\frac {1}{5} \, B b^{5} x^{5} + \frac {5}{4} \, B a b^{4} x^{4} + \frac {1}{4} \, A b^{5} x^{4} + \frac {10}{3} \, B a^{2} b^{3} x^{3} + \frac {5}{3} \, A a b^{4} x^{3} + 5 \, B a^{3} b^{2} x^{2} + 5 \, A a^{2} b^{3} x^{2} + 5 \, B a^{4} b x + 10 \, A a^{3} b^{2} x - \frac {A a^{5}}{x} + {\left (B a^{5} + 5 \, A a^{4} b\right )} \log \left ({\left | x \right |}\right ) \]

[In]

integrate((b*x+a)^5*(B*x+A)/x^2,x, algorithm="giac")

[Out]

1/5*B*b^5*x^5 + 5/4*B*a*b^4*x^4 + 1/4*A*b^5*x^4 + 10/3*B*a^2*b^3*x^3 + 5/3*A*a*b^4*x^3 + 5*B*a^3*b^2*x^2 + 5*A
*a^2*b^3*x^2 + 5*B*a^4*b*x + 10*A*a^3*b^2*x - A*a^5/x + (B*a^5 + 5*A*a^4*b)*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^5 (A+B x)}{x^2} \, dx=x^4\,\left (\frac {A\,b^5}{4}+\frac {5\,B\,a\,b^4}{4}\right )+\ln \left (x\right )\,\left (B\,a^5+5\,A\,b\,a^4\right )-\frac {A\,a^5}{x}+\frac {B\,b^5\,x^5}{5}+5\,a^2\,b^2\,x^2\,\left (A\,b+B\,a\right )+5\,a^3\,b\,x\,\left (2\,A\,b+B\,a\right )+\frac {5\,a\,b^3\,x^3\,\left (A\,b+2\,B\,a\right )}{3} \]

[In]

int(((A + B*x)*(a + b*x)^5)/x^2,x)

[Out]

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