\(\int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx\) [112]

Optimal result

Integrand size = 16, antiderivative size = 64 \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=-\frac {a^3 A}{3 x^3}-\frac {a^2 (3 A b+a B)}{2 x^2}-\frac {3 a b (A b+a B)}{x}+b^3 B x+b^2 (A b+3 a B) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, 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} \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=-\frac {a^3 A}{3 x^3}-\frac {a^2 (a B+3 A b)}{2 x^2}+b^2 \log (x) (3 a B+A b)-\frac {3 a b (a B+A b)}{x}+b^3 B x \]

[In]

Int[((a + b*x)^3*(A + B*x))/x^4,x]

[Out]

-1/3*(a^3*A)/x^3 - (a^2*(3*A*b + a*B))/(2*x^2) - (3*a*b*(A*b + a*B))/x + b^3*B*x + b^2*(A*b + 3*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 (b^3 B+\frac {a^3 A}{x^4}+\frac {a^2 (3 A b+a B)}{x^3}+\frac {3 a b (A b+a B)}{x^2}+\frac {b^2 (A b+3 a B)}{x}\right ) \, dx \\ & = -\frac {a^3 A}{3 x^3}-\frac {a^2 (3 A b+a B)}{2 x^2}-\frac {3 a b (A b+a B)}{x}+b^3 B x+b^2 (A b+3 a B) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=-\frac {18 a A b^2 x^2-6 b^3 B x^4+9 a^2 b x (A+2 B x)+a^3 (2 A+3 B x)}{6 x^3}+b^2 (A b+3 a B) \log (x) \]

[In]

Integrate[((a + b*x)^3*(A + B*x))/x^4,x]

[Out]

-1/6*(18*a*A*b^2*x^2 - 6*b^3*B*x^4 + 9*a^2*b*x*(A + 2*B*x) + a^3*(2*A + 3*B*x))/x^3 + b^2*(A*b + 3*a*B)*Log[x]

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {a^{3} A}{3 x^{3}}-\frac {a^{2} \left (3 A b +B a \right )}{2 x^{2}}-\frac {3 a b \left (A b +B a \right )}{x}+b^{3} B x +b^{2} \left (A b +3 B a \right ) \ln \left (x \right )\)	\(61\)
risch	\(b^{3} B x +\frac {\left (-3 a \,b^{2} A -3 a^{2} b B \right ) x^{2}+\left (-\frac {3}{2} a^{2} b A -\frac {1}{2} a^{3} B \right ) x -\frac {a^{3} A}{3}}{x^{3}}+A \ln \left (x \right ) b^{3}+3 B \ln \left (x \right ) a \,b^{2}\)	\(70\)
norman	\(\frac {\left (-\frac {3}{2} a^{2} b A -\frac {1}{2} a^{3} B \right ) x +\left (-3 a \,b^{2} A -3 a^{2} b B \right ) x^{2}+b^{3} B \,x^{4}-\frac {a^{3} A}{3}}{x^{3}}+\left (b^{3} A +3 a \,b^{2} B \right ) \ln \left (x \right )\)	\(72\)
parallelrisch	\(\frac {6 A \ln \left (x \right ) x^{3} b^{3}+18 B \ln \left (x \right ) x^{3} a \,b^{2}+6 b^{3} B \,x^{4}-18 a A \,b^{2} x^{2}-18 B \,a^{2} b \,x^{2}-9 a^{2} A b x -3 a^{3} B x -2 a^{3} A}{6 x^{3}}\)	\(80\)

[In]

int((b*x+a)^3*(B*x+A)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=\frac {6 \, B b^{3} x^{4} + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} \log \left (x\right ) - 2 \, A a^{3} - 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=B b^{3} x + b^{2} \left (A b + 3 B a\right ) \log {\left (x \right )} + \frac {- 2 A a^{3} + x^{2} \left (- 18 A a b^{2} - 18 B a^{2} b\right ) + x \left (- 9 A a^{2} b - 3 B a^{3}\right )}{6 x^{3}} \]

[In]

integrate((b*x+a)**3*(B*x+A)/x**4,x)

[Out]

B*b**3*x + b**2*(A*b + 3*B*a)*log(x) + (-2*A*a**3 + x**2*(-18*A*a*b**2 - 18*B*a**2*b) + x*(-9*A*a**2*b - 3*B*a
**3))/(6*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=B b^{3} x + {\left (3 \, B a b^{2} + A b^{3}\right )} \log \left (x\right ) - \frac {2 \, A a^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=B b^{3} x + {\left (3 \, B a b^{2} + A b^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^3 (A+B x)}{x^4} \, dx=\ln \left (x\right )\,\left (A\,b^3+3\,B\,a\,b^2\right )-\frac {x^2\,\left (3\,B\,a^2\,b+3\,A\,a\,b^2\right )+x\,\left (\frac {B\,a^3}{2}+\frac {3\,A\,b\,a^2}{2}\right )+\frac {A\,a^3}{3}}{x^3}+B\,b^3\,x \]

[In]

int(((A + B*x)*(a + b*x)^3)/x^4,x)

[Out]

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