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

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

[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]

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Rubi [A]  time = 0.0188923, antiderivative size = 59, normalized size of antiderivative = 1., 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} \[ -\frac{3 a^2 b B}{2 x^2}-\frac{a^3 B}{3 x^3}-\frac{A (a+b x)^4}{4 a x^4}-\frac{3 a b^2 B}{x}+b^3 B \log (x) \]

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 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]))))

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])

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*}

Mathematica [A]  time = 0.0238631, size = 70, normalized size = 1.19 \[ -\frac{6 a^2 b x (2 A+3 B x)+a^3 (3 A+4 B x)+18 a b^2 x^2 (A+2 B x)+12 A b^3 x^3-12 b^3 B x^4 \log (x)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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*Log[x])
/(12*x^4)

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Maple [A]  time = 0.006, size = 76, normalized size = 1.3 \begin{align*}{b}^{3}B\ln \left ( x \right ) -{\frac{{a}^{2}bA}{{x}^{3}}}-{\frac{{a}^{3}B}{3\,{x}^{3}}}-{\frac{A{a}^{3}}{4\,{x}^{4}}}-{\frac{3\,a{b}^{2}A}{2\,{x}^{2}}}-{\frac{3\,{a}^{2}bB}{2\,{x}^{2}}}-{\frac{{b}^{3}A}{x}}-3\,{\frac{a{b}^{2}B}{x}} \end{align*}

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

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Maxima [A]  time = 1.03966, size = 97, normalized size = 1.64 \begin{align*} B b^{3} \log \left (x\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{align*}

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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Fricas [A]  time = 1.6708, size = 170, normalized size = 2.88 \begin{align*} \frac{12 \, B b^{3} x^{4} \log \left (x\right ) - 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{align*}

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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Sympy [A]  time = 1.2649, size = 75, normalized size = 1.27 \begin{align*} B b^{3} \log{\left (x \right )} - \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{align*}

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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Giac [A]  time = 1.16198, size = 99, normalized size = 1.68 \begin{align*} 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{align*}

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