\(\int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx\) [113]

Optimal result

Integrand size = 16, antiderivative size = 59 \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=-\frac {a^3 B}{3 x^3}-\frac {3 a^2 b B}{2 x^2}-\frac {3 a b^2 B}{x}-\frac {A (a+b x)^4}{4 a x^4}+b^3 B \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {79, 45} \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=-\frac {a^3 B}{3 x^3}-\frac {3 a^2 b B}{2 x^2}-\frac {A (a+b x)^4}{4 a x^4}-\frac {3 a b^2 B}{x}+b^3 B \log (x) \]

[In]

Int[((a + b*x)^3*(A + B*x))/x^5,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=-\frac {12 A b^3 x^3+18 a b^2 x^2 (A+2 B x)+6 a^2 b x (2 A+3 B x)+a^3 (3 A+4 B x)-12 b^3 B x^4 \log (x)}{12 x^4} \]

[In]

Integrate[((a + b*x)^3*(A + B*x))/x^5,x]

[Out]

-1/12*(12*A*b^3*x^3 + 18*a*b^2*x^2*(A + 2*B*x) + 6*a^2*b*x*(2*A + 3*B*x) + a^3*(3*A + 4*B*x) - 12*b^3*B*x^4*Lo
g[x])/x^4

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08

method	result	size
default	\(b^{3} B \ln \left (x \right )-\frac {a^{2} \left (3 A b +B a \right )}{3 x^{3}}-\frac {b^{2} \left (A b +3 B a \right )}{x}-\frac {3 a b \left (A b +B a \right )}{2 x^{2}}-\frac {a^{3} A}{4 x^{4}}\)	\(64\)
norman	\(\frac {\left (-\frac {3}{2} a \,b^{2} A -\frac {3}{2} a^{2} b B \right ) x^{2}+\left (-a^{2} b A -\frac {1}{3} a^{3} B \right ) x +\left (-b^{3} A -3 a \,b^{2} B \right ) x^{3}-\frac {a^{3} A}{4}}{x^{4}}+b^{3} B \ln \left (x \right )\)	\(73\)
risch	\(\frac {\left (-\frac {3}{2} a \,b^{2} A -\frac {3}{2} a^{2} b B \right ) x^{2}+\left (-a^{2} b A -\frac {1}{3} a^{3} B \right ) x +\left (-b^{3} A -3 a \,b^{2} B \right ) x^{3}-\frac {a^{3} A}{4}}{x^{4}}+b^{3} B \ln \left (x \right )\)	\(73\)
parallelrisch	\(-\frac {-12 b^{3} B \ln \left (x \right ) x^{4}+12 A \,b^{3} x^{3}+36 B a \,b^{2} x^{3}+18 a A \,b^{2} x^{2}+18 B \,a^{2} b \,x^{2}+12 a^{2} A b x +4 a^{3} B x +3 a^{3} A}{12 x^{4}}\)	\(78\)

[In]

int((b*x+a)^3*(B*x+A)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=\frac {12 \, B b^{3} x^{4} \log \left (x\right ) - 3 \, A a^{3} - 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^5,x, algorithm="fricas")

[Out]

1/12*(12*B*b^3*x^4*log(x) - 3*A*a^3 - 12*(3*B*a*b^2 + A*b^3)*x^3 - 18*(B*a^2*b + A*a*b^2)*x^2 - 4*(B*a^3 + 3*A
*a^2*b)*x)/x^4

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=B b^{3} \log {\left (x \right )} + \frac {- 3 A a^{3} + x^{3} \left (- 12 A b^{3} - 36 B a b^{2}\right ) + x^{2} \left (- 18 A a b^{2} - 18 B a^{2} b\right ) + x \left (- 12 A a^{2} b - 4 B a^{3}\right )}{12 x^{4}} \]

[In]

integrate((b*x+a)**3*(B*x+A)/x**5,x)

[Out]

B*b**3*log(x) + (-3*A*a**3 + x**3*(-12*A*b**3 - 36*B*a*b**2) + x**2*(-18*A*a*b**2 - 18*B*a**2*b) + x*(-12*A*a*
*2*b - 4*B*a**3))/(12*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.22 \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=B b^{3} \log \left (x\right ) - \frac {3 \, A a^{3} + 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^5,x, algorithm="maxima")

[Out]

B*b^3*log(x) - 1/12*(3*A*a^3 + 12*(3*B*a*b^2 + A*b^3)*x^3 + 18*(B*a^2*b + A*a*b^2)*x^2 + 4*(B*a^3 + 3*A*a^2*b)
*x)/x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=B b^{3} \log \left ({\left | x \right |}\right ) - \frac {3 \, A a^{3} + 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^5,x, algorithm="giac")

[Out]

B*b^3*log(abs(x)) - 1/12*(3*A*a^3 + 12*(3*B*a*b^2 + A*b^3)*x^3 + 18*(B*a^2*b + A*a*b^2)*x^2 + 4*(B*a^3 + 3*A*a
^2*b)*x)/x^4

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx=B\,b^3\,\ln \left (x\right )-\frac {x^2\,\left (\frac {3\,B\,a^2\,b}{2}+\frac {3\,A\,a\,b^2}{2}\right )+x\,\left (\frac {B\,a^3}{3}+A\,b\,a^2\right )+\frac {A\,a^3}{4}+x^3\,\left (A\,b^3+3\,B\,a\,b^2\right )}{x^4} \]

[In]

int(((A + B*x)*(a + b*x)^3)/x^5,x)

[Out]

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