3.2.14 \(\int \frac {(a+b x)^3 (A+B x)}{x^6} \, dx\) [114]

Optimal. Leaf size=44 \[ -\frac {A (a+b x)^4}{5 a x^5}+\frac {(A b-5 a B) (a+b x)^4}{20 a^2 x^4} \]

[Out]

-1/5*A*(b*x+a)^4/a/x^5+1/20*(A*b-5*B*a)*(b*x+a)^4/a^2/x^4

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Rubi [A]
time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {79, 37} \begin {gather*} \frac {(a+b x)^4 (A b-5 a B)}{20 a^2 x^4}-\frac {A (a+b x)^4}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/x^6,x]

[Out]

-1/5*(A*(a + b*x)^4)/(a*x^5) + ((A*b - 5*a*B)*(a + b*x)^4)/(20*a^2*x^4)

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)^3 (A+B x)}{x^6} \, dx &=-\frac {A (a+b x)^4}{5 a x^5}+\frac {(-A b+5 a B) \int \frac {(a+b x)^3}{x^5} \, dx}{5 a}\\ &=-\frac {A (a+b x)^4}{5 a x^5}+\frac {(A b-5 a B) (a+b x)^4}{20 a^2 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 66, normalized size = 1.50 \begin {gather*} -\frac {10 b^3 x^3 (A+2 B x)+10 a b^2 x^2 (2 A+3 B x)+5 a^2 b x (3 A+4 B x)+a^3 (4 A+5 B x)}{20 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*(A + B*x))/x^6,x]

[Out]

-1/20*(10*b^3*x^3*(A + 2*B*x) + 10*a*b^2*x^2*(2*A + 3*B*x) + 5*a^2*b*x*(3*A + 4*B*x) + a^3*(4*A + 5*B*x))/x^5

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Maple [A]
time = 0.06, size = 66, normalized size = 1.50

method	result	size
default	\(-\frac {b^{3} B}{x}-\frac {a^{3} A}{5 x^{5}}-\frac {a^{2} \left (3 A b +B a \right )}{4 x^{4}}-\frac {b^{2} \left (A b +3 B a \right )}{2 x^{2}}-\frac {a b \left (A b +B a \right )}{x^{3}}\)	\(66\)
norman	\(\frac {-b^{3} B \,x^{4}+\left (-\frac {1}{2} b^{3} A -\frac {3}{2} a \,b^{2} B \right ) x^{3}+\left (-a \,b^{2} A -a^{2} b B \right ) x^{2}+\left (-\frac {3}{4} a^{2} b A -\frac {1}{4} a^{3} B \right ) x -\frac {a^{3} A}{5}}{x^{5}}\)	\(74\)
risch	\(\frac {-b^{3} B \,x^{4}+\left (-\frac {1}{2} b^{3} A -\frac {3}{2} a \,b^{2} B \right ) x^{3}+\left (-a \,b^{2} A -a^{2} b B \right ) x^{2}+\left (-\frac {3}{4} a^{2} b A -\frac {1}{4} a^{3} B \right ) x -\frac {a^{3} A}{5}}{x^{5}}\)	\(74\)
gosper	\(-\frac {20 b^{3} B \,x^{4}+10 A \,b^{3} x^{3}+30 B a \,b^{2} x^{3}+20 a A \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+15 a^{2} A b x +5 a^{3} B x +4 a^{3} A}{20 x^{5}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A)/x^6,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.33, size = 73, normalized size = 1.66 \begin {gather*} -\frac {20 \, B b^{3} x^{4} + 4 \, A a^{3} + 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 20 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/x^6,x, algorithm="maxima")

[Out]

-1/20*(20*B*b^3*x^4 + 4*A*a^3 + 10*(3*B*a*b^2 + A*b^3)*x^3 + 20*(B*a^2*b + A*a*b^2)*x^2 + 5*(B*a^3 + 3*A*a^2*b
)*x)/x^5

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Fricas [A]
time = 0.76, size = 73, normalized size = 1.66 \begin {gather*} -\frac {20 \, B b^{3} x^{4} + 4 \, A a^{3} + 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 20 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/x^6,x, algorithm="fricas")

[Out]

-1/20*(20*B*b^3*x^4 + 4*A*a^3 + 10*(3*B*a*b^2 + A*b^3)*x^3 + 20*(B*a^2*b + A*a*b^2)*x^2 + 5*(B*a^3 + 3*A*a^2*b
)*x)/x^5

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (37) = 74\).
time = 0.80, size = 82, normalized size = 1.86 \begin {gather*} \frac {- 4 A a^{3} - 20 B b^{3} x^{4} + x^{3} \left (- 10 A b^{3} - 30 B a b^{2}\right ) + x^{2} \left (- 20 A a b^{2} - 20 B a^{2} b\right ) + x \left (- 15 A a^{2} b - 5 B a^{3}\right )}{20 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A)/x**6,x)

[Out]

(-4*A*a**3 - 20*B*b**3*x**4 + x**3*(-10*A*b**3 - 30*B*a*b**2) + x**2*(-20*A*a*b**2 - 20*B*a**2*b) + x*(-15*A*a
**2*b - 5*B*a**3))/(20*x**5)

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Giac [A]
time = 1.14, size = 75, normalized size = 1.70 \begin {gather*} -\frac {20 \, B b^{3} x^{4} + 30 \, B a b^{2} x^{3} + 10 \, A b^{3} x^{3} + 20 \, B a^{2} b x^{2} + 20 \, A a b^{2} x^{2} + 5 \, B a^{3} x + 15 \, A a^{2} b x + 4 \, A a^{3}}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/x^6,x, algorithm="giac")

[Out]

-1/20*(20*B*b^3*x^4 + 30*B*a*b^2*x^3 + 10*A*b^3*x^3 + 20*B*a^2*b*x^2 + 20*A*a*b^2*x^2 + 5*B*a^3*x + 15*A*a^2*b
*x + 4*A*a^3)/x^5

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Mupad [B]
time = 0.31, size = 71, normalized size = 1.61 \begin {gather*} -\frac {x^2\,\left (B\,a^2\,b+A\,a\,b^2\right )+x\,\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )+\frac {A\,a^3}{5}+x^3\,\left (\frac {A\,b^3}{2}+\frac {3\,B\,a\,b^2}{2}\right )+B\,b^3\,x^4}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^3)/x^6,x)

[Out]

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

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