3.6.20 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)}{x^6} \, dx\) [520]

Optimal. Leaf size=55 \[ -\frac {a^2 A}{5 x^5}-\frac {a (2 A b+a B)}{4 x^4}-\frac {b (A b+2 a B)}{3 x^3}-\frac {b^2 B}{2 x^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {27, 77} \begin {gather*} -\frac {a^2 A}{5 x^5}-\frac {a (a B+2 A b)}{4 x^4}-\frac {b (2 a B+A b)}{3 x^3}-\frac {b^2 B}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^6,x]

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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*} \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{x^6} \, dx &=\int \frac {(a+b x)^2 (A+B x)}{x^6} \, dx\\ &=\int \left (\frac {a^2 A}{x^6}+\frac {a (2 A b+a B)}{x^5}+\frac {b (A b+2 a B)}{x^4}+\frac {b^2 B}{x^3}\right ) \, dx\\ &=-\frac {a^2 A}{5 x^5}-\frac {a (2 A b+a B)}{4 x^4}-\frac {b (A b+2 a B)}{3 x^3}-\frac {b^2 B}{2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 50, normalized size = 0.91 \begin {gather*} -\frac {10 b^2 x^2 (2 A+3 B x)+10 a b x (3 A+4 B x)+3 a^2 (4 A+5 B x)}{60 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^6,x]

[Out]

-1/60*(10*b^2*x^2*(2*A + 3*B*x) + 10*a*b*x*(3*A + 4*B*x) + 3*a^2*(4*A + 5*B*x))/x^5

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Maple [A]
time = 0.04, size = 48, normalized size = 0.87

method	result	size
default	\(-\frac {a^{2} A}{5 x^{5}}-\frac {a \left (2 A b +B a \right )}{4 x^{4}}-\frac {b \left (A b +2 B a \right )}{3 x^{3}}-\frac {b^{2} B}{2 x^{2}}\)	\(48\)
norman	\(\frac {-\frac {b^{2} B \,x^{3}}{2}+\left (-\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) x^{2}+\left (-\frac {1}{2} a b A -\frac {1}{4} a^{2} B \right ) x -\frac {a^{2} A}{5}}{x^{5}}\)	\(51\)
risch	\(\frac {-\frac {b^{2} B \,x^{3}}{2}+\left (-\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) x^{2}+\left (-\frac {1}{2} a b A -\frac {1}{4} a^{2} B \right ) x -\frac {a^{2} A}{5}}{x^{5}}\)	\(51\)
gosper	\(-\frac {30 b^{2} B \,x^{3}+20 b^{2} A \,x^{2}+40 B a b \,x^{2}+30 a b A x +15 a^{2} B x +12 a^{2} A}{60 x^{5}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/x^6,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 51, normalized size = 0.93 \begin {gather*} -\frac {30 \, B b^{2} x^{3} + 12 \, A a^{2} + 20 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/x^6,x, algorithm="maxima")

[Out]

-1/60*(30*B*b^2*x^3 + 12*A*a^2 + 20*(2*B*a*b + A*b^2)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/x^5

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Fricas [A]
time = 1.74, size = 51, normalized size = 0.93 \begin {gather*} -\frac {30 \, B b^{2} x^{3} + 12 \, A a^{2} + 20 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/x^6,x, algorithm="fricas")

[Out]

-1/60*(30*B*b^2*x^3 + 12*A*a^2 + 20*(2*B*a*b + A*b^2)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/x^5

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Sympy [A]
time = 1.19, size = 56, normalized size = 1.02 \begin {gather*} \frac {- 12 A a^{2} - 30 B b^{2} x^{3} + x^{2} \left (- 20 A b^{2} - 40 B a b\right ) + x \left (- 30 A a b - 15 B a^{2}\right )}{60 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)/x**6,x)

[Out]

(-12*A*a**2 - 30*B*b**2*x**3 + x**2*(-20*A*b**2 - 40*B*a*b) + x*(-30*A*a*b - 15*B*a**2))/(60*x**5)

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Giac [A]
time = 1.39, size = 51, normalized size = 0.93 \begin {gather*} -\frac {30 \, B b^{2} x^{3} + 40 \, B a b x^{2} + 20 \, A b^{2} x^{2} + 15 \, B a^{2} x + 30 \, A a b x + 12 \, A a^{2}}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/x^6,x, algorithm="giac")

[Out]

-1/60*(30*B*b^2*x^3 + 40*B*a*b*x^2 + 20*A*b^2*x^2 + 15*B*a^2*x + 30*A*a*b*x + 12*A*a^2)/x^5

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Mupad [B]
time = 0.04, size = 51, normalized size = 0.93 \begin {gather*} -\frac {x^2\,\left (\frac {A\,b^2}{3}+\frac {2\,B\,a\,b}{3}\right )+\frac {A\,a^2}{5}+x\,\left (\frac {B\,a^2}{4}+\frac {A\,b\,a}{2}\right )+\frac {B\,b^2\,x^3}{2}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x))/x^6,x)

[Out]

-(x^2*((A*b^2)/3 + (2*B*a*b)/3) + (A*a^2)/5 + x*((B*a^2)/4 + (A*a*b)/2) + (B*b^2*x^3)/2)/x^5

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