\(\int \frac {(a+b x)^3 (A+B x)}{x} \, dx\) [109]

Optimal result

Integrand size = 16, antiderivative size = 54 \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=3 a^2 A b x+\frac {3}{2} a A b^2 x^2+\frac {1}{3} A b^3 x^3+\frac {B (a+b x)^4}{4 b}+a^3 A \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, 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 = {81, 45} \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=a^3 A \log (x)+3 a^2 A b x+\frac {3}{2} a A b^2 x^2+\frac {B (a+b x)^4}{4 b}+\frac {1}{3} A b^3 x^3 \]

[In]

Int[((a + b*x)^3*(A + B*x))/x,x]

[Out]

3*a^2*A*b*x + (3*a*A*b^2*x^2)/2 + (A*b^3*x^3)/3 + (B*(a + b*x)^4)/(4*b) + a^3*A*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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=\frac {1}{12} x \left (12 a^3 B+18 a^2 b (2 A+B x)+6 a b^2 x (3 A+2 B x)+b^3 x^2 (4 A+3 B x)\right )+a^3 A \log (x) \]

[In]

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

[Out]

(x*(12*a^3*B + 18*a^2*b*(2*A + B*x) + 6*a*b^2*x*(3*A + 2*B*x) + b^3*x^2*(4*A + 3*B*x)))/12 + a^3*A*Log[x]

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.28

method	result	size
norman	\(\left (\frac {1}{3} b^{3} A +a \,b^{2} B \right ) x^{3}+\left (\frac {3}{2} a \,b^{2} A +\frac {3}{2} a^{2} b B \right ) x^{2}+\left (3 a^{2} b A +a^{3} B \right ) x +\frac {b^{3} B \,x^{4}}{4}+a^{3} A \ln \left (x \right )\)	\(69\)
default	\(\frac {b^{3} B \,x^{4}}{4}+\frac {A \,b^{3} x^{3}}{3}+B a \,b^{2} x^{3}+\frac {3 a A \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}+3 a^{2} A b x +a^{3} B x +a^{3} A \ln \left (x \right )\)	\(70\)
risch	\(\frac {b^{3} B \,x^{4}}{4}+\frac {A \,b^{3} x^{3}}{3}+B a \,b^{2} x^{3}+\frac {3 a A \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}+3 a^{2} A b x +a^{3} B x +a^{3} A \ln \left (x \right )\)	\(70\)
parallelrisch	\(\frac {b^{3} B \,x^{4}}{4}+\frac {A \,b^{3} x^{3}}{3}+B a \,b^{2} x^{3}+\frac {3 a A \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}+3 a^{2} A b x +a^{3} B x +a^{3} A \ln \left (x \right )\)	\(70\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=\frac {1}{4} \, B b^{3} x^{4} + A a^{3} \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + \frac {3}{2} \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=A a^{3} \log {\left (x \right )} + \frac {B b^{3} x^{4}}{4} + x^{3} \left (\frac {A b^{3}}{3} + B a b^{2}\right ) + x^{2} \cdot \left (\frac {3 A a b^{2}}{2} + \frac {3 B a^{2} b}{2}\right ) + x \left (3 A a^{2} b + B a^{3}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=\frac {1}{4} \, B b^{3} x^{4} + A a^{3} \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + \frac {3}{2} \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=\frac {1}{4} \, B b^{3} x^{4} + B a b^{2} x^{3} + \frac {1}{3} \, A b^{3} x^{3} + \frac {3}{2} \, B a^{2} b x^{2} + \frac {3}{2} \, A a b^{2} x^{2} + B a^{3} x + 3 \, A a^{2} b x + A a^{3} \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x)^3 (A+B x)}{x} \, dx=x\,\left (B\,a^3+3\,A\,b\,a^2\right )+x^3\,\left (\frac {A\,b^3}{3}+B\,a\,b^2\right )+\frac {B\,b^3\,x^4}{4}+A\,a^3\,\ln \left (x\right )+\frac {3\,a\,b\,x^2\,\left (A\,b+B\,a\right )}{2} \]

[In]

int(((A + B*x)*(a + b*x)^3)/x,x)

[Out]

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