3.9.62 \(\int \frac {(A+B x) (a+b x+c x^2)^2}{x^5} \, dx\) [862]

Optimal. Leaf size=90 \[ -\frac {a^2 A}{4 x^4}-\frac {a (2 A b+a B)}{3 x^3}-\frac {2 a b B+A \left (b^2+2 a c\right )}{2 x^2}-\frac {b^2 B+2 A b c+2 a B c}{x}+B c^2 x+c (2 b B+A c) \log (x) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 90, 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 = {779} \begin {gather*} -\frac {a^2 A}{4 x^4}-\frac {A \left (2 a c+b^2\right )+2 a b B}{2 x^2}-\frac {2 a B c+2 A b c+b^2 B}{x}-\frac {a (a B+2 A b)}{3 x^3}+c \log (x) (A c+2 b B)+B c^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^5,x]

[Out]

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

Rule 779

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 )^2}{x^5} \, dx &=\int \left (B c^2+\frac {a^2 A}{x^5}+\frac {a (2 A b+a B)}{x^4}+\frac {2 a b B+A \left (b^2+2 a c\right )}{x^3}+\frac {b^2 B+2 A b c+2 a B c}{x^2}+\frac {c (2 b B+A c)}{x}\right ) \, dx\\ &=-\frac {a^2 A}{4 x^4}-\frac {a (2 A b+a B)}{3 x^3}-\frac {2 a b B+A \left (b^2+2 a c\right )}{2 x^2}-\frac {b^2 B+2 A b c+2 a B c}{x}+B c^2 x+c (2 b B+A c) \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 92, normalized size = 1.02 \begin {gather*} -\frac {a^2 (3 A+4 B x)+4 a x (3 B x (b+2 c x)+A (2 b+3 c x))+6 x^2 \left (A b (b+4 c x)+2 B x \left (b^2-c^2 x^2\right )\right )-12 c (2 b B+A c) x^4 \log (x)}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^5,x]

[Out]

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

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Maple [A]
time = 0.83, size = 85, normalized size = 0.94

method	result	size
default	\(B \,c^{2} x -\frac {a^{2} A}{4 x^{4}}-\frac {2 A a c +b^{2} A +2 a b B}{2 x^{2}}-\frac {a \left (2 A b +B a \right )}{3 x^{3}}+c \left (A c +2 B b \right ) \ln \left (x \right )-\frac {2 A b c +2 a B c +b^{2} B}{x}\)	\(85\)
risch	\(B \,c^{2} x +\frac {\left (-2 A b c -2 a B c -b^{2} B \right ) x^{3}+\left (-A a c -\frac {1}{2} b^{2} A -a b B \right ) x^{2}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} B \right ) x -\frac {a^{2} A}{4}}{x^{4}}+A \ln \left (x \right ) c^{2}+2 B \ln \left (x \right ) b c\)	\(89\)
norman	\(\frac {\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} B \right ) x +\left (-A a c -\frac {1}{2} b^{2} A -a b B \right ) x^{2}+\left (-2 A b c -2 a B c -b^{2} B \right ) x^{3}+B \,c^{2} x^{5}-\frac {a^{2} A}{4}}{x^{4}}+\left (A \,c^{2}+2 b B c \right ) \ln \left (x \right )\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^2/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 89, normalized size = 0.99 \begin {gather*} B c^{2} x + {\left (2 \, B b c + A c^{2}\right )} \log \left (x\right ) - \frac {12 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + 3 \, A a^{2} + 6 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 4 \, {\left (B a^{2} + 2 \, A a b\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.39, size = 95, normalized size = 1.06 \begin {gather*} \frac {12 \, B c^{2} x^{5} + 12 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} \log \left (x\right ) - 12 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} - 3 \, A a^{2} - 6 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 4 \, {\left (B a^{2} + 2 \, A a b\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 1.61, size = 99, normalized size = 1.10 \begin {gather*} B c^{2} x + c \left (A c + 2 B b\right ) \log {\left (x \right )} + \frac {- 3 A a^{2} + x^{3} \left (- 24 A b c - 24 B a c - 12 B b^{2}\right ) + x^{2} \left (- 12 A a c - 6 A b^{2} - 12 B a b\right ) + x \left (- 8 A a b - 4 B a^{2}\right )}{12 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**2/x**5,x)

[Out]

B*c**2*x + c*(A*c + 2*B*b)*log(x) + (-3*A*a**2 + x**3*(-24*A*b*c - 24*B*a*c - 12*B*b**2) + x**2*(-12*A*a*c - 6
*A*b**2 - 12*B*a*b) + x*(-8*A*a*b - 4*B*a**2))/(12*x**4)

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Giac [A]
time = 0.86, size = 90, normalized size = 1.00 \begin {gather*} B c^{2} x + {\left (2 \, B b c + A c^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, {\left (B b^{2} + 2 \, B a c + 2 \, A b c\right )} x^{3} + 3 \, A a^{2} + 6 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 4 \, {\left (B a^{2} + 2 \, A a b\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^5,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.07, size = 86, normalized size = 0.96 \begin {gather*} \ln \left (x\right )\,\left (A\,c^2+2\,B\,b\,c\right )-\frac {\frac {A\,a^2}{4}+x^2\,\left (\frac {A\,b^2}{2}+B\,a\,b+A\,a\,c\right )+x^3\,\left (B\,b^2+2\,A\,c\,b+2\,B\,a\,c\right )+x\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )}{x^4}+B\,c^2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x + c*x^2)^2)/x^5,x)

[Out]

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

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