\(\int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx\) [110]

Optimal result

Integrand size = 16, antiderivative size = 65 \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=-\frac {a^3 A}{x}+3 a b (A b+a B) x+\frac {1}{2} b^2 (A b+3 a B) x^2+\frac {1}{3} b^3 B x^3+a^2 (3 A b+a B) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, 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} \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=-\frac {a^3 A}{x}+a^2 \log (x) (a B+3 A b)+\frac {1}{2} b^2 x^2 (3 a B+A b)+3 a b x (a B+A b)+\frac {1}{3} b^3 B x^3 \]

[In]

Int[((a + b*x)^3*(A + B*x))/x^2,x]

[Out]

-((a^3*A)/x) + 3*a*b*(A*b + a*B)*x + (b^2*(A*b + 3*a*B)*x^2)/2 + (b^3*B*x^3)/3 + a^2*(3*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 (3 a b (A b+a B)+\frac {a^3 A}{x^2}+\frac {a^2 (3 A b+a B)}{x}+b^2 (A b+3 a B) x+b^3 B x^2\right ) \, dx \\ & = -\frac {a^3 A}{x}+3 a b (A b+a B) x+\frac {1}{2} b^2 (A b+3 a B) x^2+\frac {1}{3} b^3 B x^3+a^2 (3 A b+a B) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=-\frac {a^3 A}{x}+3 a b (A b+a B) x+\frac {1}{2} b^2 (A b+3 a B) x^2+\frac {1}{3} b^3 B x^3+\left (3 a^2 A b+a^3 B\right ) \log (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {b^{3} B \,x^{3}}{3}+\frac {A \,b^{3} x^{2}}{2}+\frac {3 B a \,b^{2} x^{2}}{2}+3 a \,b^{2} A x +3 a^{2} b B x +a^{2} \left (3 A b +B a \right ) \ln \left (x \right )-\frac {a^{3} A}{x}\)	\(69\)
risch	\(\frac {b^{3} B \,x^{3}}{3}+\frac {A \,b^{3} x^{2}}{2}+\frac {3 B a \,b^{2} x^{2}}{2}+3 a \,b^{2} A x +3 a^{2} b B x -\frac {a^{3} A}{x}+3 A \ln \left (x \right ) a^{2} b +B \ln \left (x \right ) a^{3}\)	\(71\)
norman	\(\frac {\left (\frac {1}{2} b^{3} A +\frac {3}{2} a \,b^{2} B \right ) x^{3}+\left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{2}-a^{3} A +\frac {b^{3} B \,x^{4}}{3}}{x}+\left (3 a^{2} b A +a^{3} B \right ) \ln \left (x \right )\)	\(75\)
parallelrisch	\(\frac {2 b^{3} B \,x^{4}+3 A \,b^{3} x^{3}+9 B a \,b^{2} x^{3}+18 A \ln \left (x \right ) x \,a^{2} b +18 a A \,b^{2} x^{2}+6 B \ln \left (x \right ) x \,a^{3}+18 B \,a^{2} b \,x^{2}-6 a^{3} A}{6 x}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=\frac {2 \, B b^{3} x^{4} - 6 \, A a^{3} + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x \log \left (x\right )}{6 \, x} \]

[In]

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

[Out]

1/6*(2*B*b^3*x^4 - 6*A*a^3 + 3*(3*B*a*b^2 + A*b^3)*x^3 + 18*(B*a^2*b + A*a*b^2)*x^2 + 6*(B*a^3 + 3*A*a^2*b)*x*
log(x))/x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=- \frac {A a^{3}}{x} + \frac {B b^{3} x^{3}}{3} + a^{2} \cdot \left (3 A b + B a\right ) \log {\left (x \right )} + x^{2} \left (\frac {A b^{3}}{2} + \frac {3 B a b^{2}}{2}\right ) + x \left (3 A a b^{2} + 3 B a^{2} b\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=\frac {1}{3} \, B b^{3} x^{3} - \frac {A a^{3}}{x} + \frac {1}{2} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} x + {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=\frac {1}{3} \, B b^{3} x^{3} + \frac {3}{2} \, B a b^{2} x^{2} + \frac {1}{2} \, A b^{3} x^{2} + 3 \, B a^{2} b x + 3 \, A a b^{2} x - \frac {A a^{3}}{x} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/3*B*b^3*x^3 + 3/2*B*a*b^2*x^2 + 1/2*A*b^3*x^2 + 3*B*a^2*b*x + 3*A*a*b^2*x - A*a^3/x + (B*a^3 + 3*A*a^2*b)*lo
g(abs(x))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^3 (A+B x)}{x^2} \, dx=x^2\,\left (\frac {A\,b^3}{2}+\frac {3\,B\,a\,b^2}{2}\right )+\ln \left (x\right )\,\left (B\,a^3+3\,A\,b\,a^2\right )-\frac {A\,a^3}{x}+\frac {B\,b^3\,x^3}{3}+3\,a\,b\,x\,\left (A\,b+B\,a\right ) \]

[In]

int(((A + B*x)*(a + b*x)^3)/x^2,x)

[Out]

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