3.6.19 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)}{x^5} \, dx\) [519]

Optimal. Leaf size=44 \[ -\frac {A (a+b x)^3}{4 a x^4}+\frac {(A b-4 a B) (a+b x)^3}{12 a^2 x^3} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {27, 79, 37} \begin {gather*} \frac {(a+b x)^3 (A b-4 a B)}{12 a^2 x^3}-\frac {A (a+b x)^3}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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*} \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{x^5} \, dx &=\int \frac {(a+b x)^2 (A+B x)}{x^5} \, dx\\ &=-\frac {A (a+b x)^3}{4 a x^4}+\frac {(-A b+4 a B) \int \frac {(a+b x)^2}{x^4} \, dx}{4 a}\\ &=-\frac {A (a+b x)^3}{4 a x^4}+\frac {(A b-4 a B) (a+b x)^3}{12 a^2 x^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 47, normalized size = 1.07 \begin {gather*} -\frac {6 b^2 x^2 (A+2 B x)+4 a b x (2 A+3 B x)+a^2 (3 A+4 B x)}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 48, normalized size = 1.09

method	result	size
default	\(-\frac {a^{2} A}{4 x^{4}}-\frac {b \left (A b +2 B a \right )}{2 x^{2}}-\frac {a \left (2 A b +B a \right )}{3 x^{3}}-\frac {b^{2} B}{x}\)	\(48\)
norman	\(\frac {-b^{2} B \,x^{3}+\left (-\frac {1}{2} b^{2} A -a b B \right ) x^{2}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} B \right ) x -\frac {a^{2} A}{4}}{x^{4}}\)	\(51\)
risch	\(\frac {-b^{2} B \,x^{3}+\left (-\frac {1}{2} b^{2} A -a b B \right ) x^{2}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} B \right ) x -\frac {a^{2} A}{4}}{x^{4}}\)	\(51\)
gosper	\(-\frac {12 b^{2} B \,x^{3}+6 b^{2} A \,x^{2}+12 B a b \,x^{2}+8 a b A x +4 a^{2} B x +3 a^{2} A}{12 x^{4}}\)	\(52\)

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^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 51, normalized size = 1.16 \begin {gather*} -\frac {12 \, B b^{2} x^{3} + 3 \, A a^{2} + 6 \, {\left (2 \, B a b + A b^{2}\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)*(b^2*x^2+2*a*b*x+a^2)/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.61, size = 51, normalized size = 1.16 \begin {gather*} -\frac {12 \, B b^{2} x^{3} + 3 \, A a^{2} + 6 \, {\left (2 \, B a b + A b^{2}\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)*(b^2*x^2+2*a*b*x+a^2)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.37, size = 56, normalized size = 1.27 \begin {gather*} \frac {- 3 A a^{2} - 12 B b^{2} x^{3} + x^{2} \left (- 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)*(b**2*x**2+2*a*b*x+a**2)/x**5,x)

[Out]

(-3*A*a**2 - 12*B*b**2*x**3 + x**2*(-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 = 1.04, size = 51, normalized size = 1.16 \begin {gather*} -\frac {12 \, B b^{2} x^{3} + 12 \, B a b x^{2} + 6 \, A b^{2} x^{2} + 4 \, B a^{2} x + 8 \, A a b x + 3 \, A a^{2}}{12 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

-1/12*(12*B*b^2*x^3 + 12*B*a*b*x^2 + 6*A*b^2*x^2 + 4*B*a^2*x + 8*A*a*b*x + 3*A*a^2)/x^4

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Mupad [B]
time = 0.04, size = 49, normalized size = 1.11 \begin {gather*} -\frac {x^2\,\left (\frac {A\,b^2}{2}+B\,a\,b\right )+\frac {A\,a^2}{4}+x\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+B\,b^2\,x^3}{x^4} \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^5,x)

[Out]

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

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