3.8.98 \(\int \frac {(A+B x) (a+b x+c x^2)^3}{x^4} \, dx\)

Optimal. Leaf size=155 \[ -\frac {a^3 A}{3 x^3}-\frac {a^2 (a B+3 A b)}{2 x^2}+\frac {3}{2} c x^2 \left (a B c+A b c+b^2 B\right )-\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x}+x \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\log (x) \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{3} c^2 x^3 (A c+3 b B)+\frac {1}{4} B c^3 x^4 \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {765} \begin {gather*} -\frac {a^2 (a B+3 A b)}{2 x^2}-\frac {a^3 A}{3 x^3}+x \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {3}{2} c x^2 \left (a B c+A b c+b^2 B\right )-\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x}+\log (x) \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{3} c^2 x^3 (A c+3 b B)+\frac {1}{4} B c^3 x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^3)/x^4,x]

[Out]

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

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx &=\int \left (b^3 B \left (1+\frac {3 c \left (2 a b B+A \left (b^2+a c\right )\right )}{b^3 B}\right )+\frac {a^3 A}{x^4}+\frac {a^2 (3 A b+a B)}{x^3}+\frac {3 a \left (a b B+A \left (b^2+a c\right )\right )}{x^2}+\frac {3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )}{x}+3 c \left (b^2 B+A b c+a B c\right ) x+c^2 (3 b B+A c) x^2+B c^3 x^3\right ) \, dx\\ &=-\frac {a^3 A}{3 x^3}-\frac {a^2 (3 A b+a B)}{2 x^2}-\frac {3 a \left (a b B+A \left (b^2+a c\right )\right )}{x}+\left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x+\frac {3}{2} c \left (b^2 B+A b c+a B c\right ) x^2+\frac {1}{3} c^2 (3 b B+A c) x^3+\frac {1}{4} B c^3 x^4+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) \log (x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 160, normalized size = 1.03 \begin {gather*} \frac {-2 a^3 (2 A+3 B x)-18 a^2 x (A (b+2 c x)+2 b B x)+18 a x^2 \left (B c x^2 (4 b+c x)-2 A \left (b^2-c^2 x^2\right )\right )+12 x^3 \log (x) \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+x^4 \left (18 b^2 c (2 A+B x)+6 b c^2 x (3 A+2 B x)+c^3 x^2 (4 A+3 B x)+12 b^3 B\right )}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^3)/x^4,x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^3)/x^4,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^3)/x^4, x]

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 168, normalized size = 1.08 \begin {gather*} \frac {3 \, B c^{3} x^{7} + 4 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 18 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 12 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + 12 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} \log \relax (x) - 4 \, A a^{3} - 36 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^4,x, algorithm="fricas")

[Out]

1/12*(3*B*c^3*x^7 + 4*(3*B*b*c^2 + A*c^3)*x^6 + 18*(B*b^2*c + (B*a + A*b)*c^2)*x^5 + 12*(B*b^3 + 3*A*a*c^2 + 3
*(2*B*a*b + A*b^2)*c)*x^4 + 12*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3*log(x) - 4*A*a^3 - 36*(B*a^2*b
+ A*a*b^2 + A*a^2*c)*x^2 - 6*(B*a^3 + 3*A*a^2*b)*x)/x^3

________________________________________________________________________________________

giac [A]  time = 0.17, size = 169, normalized size = 1.09 \begin {gather*} \frac {1}{4} \, B c^{3} x^{4} + B b c^{2} x^{3} + \frac {1}{3} \, A c^{3} x^{3} + \frac {3}{2} \, B b^{2} c x^{2} + \frac {3}{2} \, B a c^{2} x^{2} + \frac {3}{2} \, A b c^{2} x^{2} + B b^{3} x + 6 \, B a b c x + 3 \, A b^{2} c x + 3 \, A a c^{2} x + {\left (3 \, B a b^{2} + A b^{3} + 3 \, B a^{2} c + 6 \, A a b c\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{3} + 18 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^4,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 179, normalized size = 1.15 \begin {gather*} \frac {B \,c^{3} x^{4}}{4}+\frac {A \,c^{3} x^{3}}{3}+B b \,c^{2} x^{3}+\frac {3 A b \,c^{2} x^{2}}{2}+\frac {3 B a \,c^{2} x^{2}}{2}+\frac {3 B \,b^{2} c \,x^{2}}{2}+6 A a b c \ln \relax (x )+3 A a \,c^{2} x +A \,b^{3} \ln \relax (x )+3 A \,b^{2} c x +3 B \,a^{2} c \ln \relax (x )+3 B a \,b^{2} \ln \relax (x )+6 B a b c x +B \,b^{3} x -\frac {3 A \,a^{2} c}{x}-\frac {3 A a \,b^{2}}{x}-\frac {3 B \,a^{2} b}{x}-\frac {3 A \,a^{2} b}{2 x^{2}}-\frac {B \,a^{3}}{2 x^{2}}-\frac {A \,a^{3}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^3/x^4,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 162, normalized size = 1.05 \begin {gather*} \frac {1}{4} \, B c^{3} x^{4} + \frac {1}{3} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + \frac {3}{2} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{2} + {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} \log \relax (x) - \frac {2 \, A a^{3} + 18 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 164, normalized size = 1.06 \begin {gather*} \ln \relax (x)\,\left (3\,B\,c\,a^2+3\,B\,a\,b^2+6\,A\,c\,a\,b+A\,b^3\right )-\frac {x\,\left (\frac {B\,a^3}{2}+\frac {3\,A\,b\,a^2}{2}\right )+\frac {A\,a^3}{3}+x^2\,\left (3\,B\,a^2\,b+3\,A\,c\,a^2+3\,A\,a\,b^2\right )}{x^3}+x^3\,\left (\frac {A\,c^3}{3}+B\,b\,c^2\right )+x^2\,\left (\frac {3\,B\,b^2\,c}{2}+\frac {3\,A\,b\,c^2}{2}+\frac {3\,B\,a\,c^2}{2}\right )+x\,\left (B\,b^3+3\,A\,b^2\,c+6\,B\,a\,b\,c+3\,A\,a\,c^2\right )+\frac {B\,c^3\,x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x + c*x^2)^3)/x^4,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.36, size = 187, normalized size = 1.21 \begin {gather*} \frac {B c^{3} x^{4}}{4} + x^{3} \left (\frac {A c^{3}}{3} + B b c^{2}\right ) + x^{2} \left (\frac {3 A b c^{2}}{2} + \frac {3 B a c^{2}}{2} + \frac {3 B b^{2} c}{2}\right ) + x \left (3 A a c^{2} + 3 A b^{2} c + 6 B a b c + B b^{3}\right ) + \left (6 A a b c + A b^{3} + 3 B a^{2} c + 3 B a b^{2}\right ) \log {\relax (x )} + \frac {- 2 A a^{3} + x^{2} \left (- 18 A a^{2} c - 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**3/x**4,x)

[Out]

B*c**3*x**4/4 + x**3*(A*c**3/3 + B*b*c**2) + x**2*(3*A*b*c**2/2 + 3*B*a*c**2/2 + 3*B*b**2*c/2) + x*(3*A*a*c**2
+ 3*A*b**2*c + 6*B*a*b*c + B*b**3) + (6*A*a*b*c + A*b**3 + 3*B*a**2*c + 3*B*a*b**2)*log(x) + (-2*A*a**3 + x**
2*(-18*A*a**2*c - 18*A*a*b**2 - 18*B*a**2*b) + x*(-9*A*a**2*b - 3*B*a**3))/(6*x**3)

________________________________________________________________________________________