\(\int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx\) [111]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (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} \[ \int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx=-\frac {a^3 A}{2 x^2}-\frac {a^2 (a B+3 A b)}{x}+b^2 x (3 a B+A b)+3 a b \log (x) (a B+A b)+\frac {1}{2} b^3 B x^2 \]

[In]

Int[((a + b*x)^3*(A + B*x))/x^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx=\frac {1}{2} \left (-\frac {6 a^2 A b}{x}+6 a b^2 B x+b^3 x (2 A+B x)-\frac {a^3 (A+2 B x)}{x^2}+6 a b (A b+a B) \log (x)\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {b^{3} B \,x^{2}}{2}+A \,b^{3} x +3 B a \,b^{2} x +3 a b \left (A b +B a \right ) \ln \left (x \right )-\frac {a^{2} \left (3 A b +B a \right )}{x}-\frac {a^{3} A}{2 x^{2}}\)	\(63\)
risch	\(\frac {b^{3} B \,x^{2}}{2}+A \,b^{3} x +3 B a \,b^{2} x +\frac {\left (-3 a^{2} b A -a^{3} B \right ) x -\frac {a^{3} A}{2}}{x^{2}}+3 A \ln \left (x \right ) a \,b^{2}+3 B \ln \left (x \right ) a^{2} b\)	\(70\)
norman	\(\frac {\left (b^{3} A +3 a \,b^{2} B \right ) x^{3}+\left (-3 a^{2} b A -a^{3} B \right ) x -\frac {a^{3} A}{2}+\frac {b^{3} B \,x^{4}}{2}}{x^{2}}+\left (3 a \,b^{2} A +3 a^{2} b B \right ) \ln \left (x \right )\)	\(73\)
parallelrisch	\(\frac {b^{3} B \,x^{4}+6 A \ln \left (x \right ) x^{2} a \,b^{2}+2 A \,b^{3} x^{3}+6 B \ln \left (x \right ) x^{2} a^{2} b +6 B a \,b^{2} x^{3}-6 a^{2} A b x -2 a^{3} B x -a^{3} A}{2 x^{2}}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx=\frac {B b^{3} x^{4} - A a^{3} + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} \log \left (x\right ) - 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx=\frac {B b^{3} x^{2}}{2} + 3 a b \left (A b + B a\right ) \log {\left (x \right )} + x \left (A b^{3} + 3 B a b^{2}\right ) + \frac {- A a^{3} + x \left (- 6 A a^{2} b - 2 B a^{3}\right )}{2 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx=\frac {1}{2} \, B b^{3} x^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} x + 3 \, {\left (B a^{2} b + A a b^{2}\right )} \log \left (x\right ) - \frac {A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx=\frac {1}{2} \, B b^{3} x^{2} + 3 \, B a b^{2} x + A b^{3} x + 3 \, {\left (B a^{2} b + A a b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2 \, x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^3 (A+B x)}{x^3} \, dx=\ln \left (x\right )\,\left (3\,B\,a^2\,b+3\,A\,a\,b^2\right )-\frac {x\,\left (B\,a^3+3\,A\,b\,a^2\right )+\frac {A\,a^3}{2}}{x^2}+x\,\left (A\,b^3+3\,B\,a\,b^2\right )+\frac {B\,b^3\,x^2}{2} \]

[In]

int(((A + B*x)*(a + b*x)^3)/x^3,x)

[Out]

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