3.1.99 \(\int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx\)

Optimal. Leaf size=59 \[ -\frac {a^3 B}{3 x^3}-\frac {3 a^2 b B}{2 x^2}-\frac {A (a+b x)^4}{4 a x^4}-\frac {3 a b^2 B}{x}+b^3 B \log (x) \]

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Rubi [A]  time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {78, 43} \begin {gather*} -\frac {3 a^2 b B}{2 x^2}-\frac {a^3 B}{3 x^3}-\frac {A (a+b x)^4}{4 a x^4}-\frac {3 a b^2 B}{x}+b^3 B \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/x^5,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

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] || LtQ
[p, n]))))

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 70, normalized size = 1.19 \begin {gather*} -\frac {a^3 (3 A+4 B x)+6 a^2 b x (2 A+3 B x)+18 a b^2 x^2 (A+2 B x)+12 A b^3 x^3-12 b^3 B x^4 \log (x)}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*(A + B*x))/x^5,x]

[Out]

-1/12*(12*A*b^3*x^3 + 18*a*b^2*x^2*(A + 2*B*x) + 6*a^2*b*x*(2*A + 3*B*x) + a^3*(3*A + 4*B*x) - 12*b^3*B*x^4*Lo
g[x])/x^4

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x)^3 (A+B x)}{x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x)^3*(A + B*x))/x^5,x]

[Out]

IntegrateAlgebraic[((a + b*x)^3*(A + B*x))/x^5, x]

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fricas [A]  time = 1.14, size = 75, normalized size = 1.27 \begin {gather*} \frac {12 \, B b^{3} x^{4} \log \relax (x) - 3 \, A a^{3} - 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/x^5,x, algorithm="fricas")

[Out]

1/12*(12*B*b^3*x^4*log(x) - 3*A*a^3 - 12*(3*B*a*b^2 + A*b^3)*x^3 - 18*(B*a^2*b + A*a*b^2)*x^2 - 4*(B*a^3 + 3*A
*a^2*b)*x)/x^4

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giac [A]  time = 1.23, size = 73, normalized size = 1.24 \begin {gather*} B b^{3} \log \left ({\left | x \right |}\right ) - \frac {3 \, A a^{3} + 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/x^5,x, algorithm="giac")

[Out]

B*b^3*log(abs(x)) - 1/12*(3*A*a^3 + 12*(3*B*a*b^2 + A*b^3)*x^3 + 18*(B*a^2*b + A*a*b^2)*x^2 + 4*(B*a^3 + 3*A*a
^2*b)*x)/x^4

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maple [A]  time = 0.01, size = 76, normalized size = 1.29 \begin {gather*} B \,b^{3} \ln \relax (x )-\frac {A \,b^{3}}{x}-\frac {3 B a \,b^{2}}{x}-\frac {3 A a \,b^{2}}{2 x^{2}}-\frac {3 B \,a^{2} b}{2 x^{2}}-\frac {A \,a^{2} b}{x^{3}}-\frac {B \,a^{3}}{3 x^{3}}-\frac {A \,a^{3}}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A)/x^5,x)

[Out]

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

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maxima [A]  time = 1.04, size = 72, normalized size = 1.22 \begin {gather*} B b^{3} \log \relax (x) - \frac {3 \, A a^{3} + 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/x^5,x, algorithm="maxima")

[Out]

B*b^3*log(x) - 1/12*(3*A*a^3 + 12*(3*B*a*b^2 + A*b^3)*x^3 + 18*(B*a^2*b + A*a*b^2)*x^2 + 4*(B*a^3 + 3*A*a^2*b)
*x)/x^4

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mupad [B]  time = 0.35, size = 71, normalized size = 1.20 \begin {gather*} B\,b^3\,\ln \relax (x)-\frac {x^2\,\left (\frac {3\,B\,a^2\,b}{2}+\frac {3\,A\,a\,b^2}{2}\right )+x\,\left (\frac {B\,a^3}{3}+A\,b\,a^2\right )+\frac {A\,a^3}{4}+x^3\,\left (A\,b^3+3\,B\,a\,b^2\right )}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^3)/x^5,x)

[Out]

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

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sympy [A]  time = 1.43, size = 80, normalized size = 1.36 \begin {gather*} B b^{3} \log {\relax (x )} + \frac {- 3 A a^{3} + x^{3} \left (- 12 A b^{3} - 36 B a b^{2}\right ) + x^{2} \left (- 18 A a b^{2} - 18 B a^{2} b\right ) + x \left (- 12 A a^{2} b - 4 B a^{3}\right )}{12 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A)/x**5,x)

[Out]

B*b**3*log(x) + (-3*A*a**3 + x**3*(-12*A*b**3 - 36*B*a*b**2) + x**2*(-18*A*a*b**2 - 18*B*a**2*b) + x*(-12*A*a*
*2*b - 4*B*a**3))/(12*x**4)

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