3.1043 \(\int \frac{(a+b x)^3 (A+B x)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=145 \[ -\frac{b^2 x (-3 a B e-A b e+3 b B d)}{e^4}+\frac{(b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (d+e x)}-\frac{(b d-a e)^3 (B d-A e)}{2 e^5 (d+e x)^2}+\frac{3 b (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)}{e^5}+\frac{b^3 B x^2}{2 e^3} \]

[Out]

-((b^2*(3*b*B*d - A*b*e - 3*a*B*e)*x)/e^4) + (b^3*B*x^2)/(2*e^3) - ((b*d - a*e)^3*(B*d - A*e))/(2*e^5*(d + e*x
)^2) + ((b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(e^5*(d + e*x)) + (3*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e
)*Log[d + e*x])/e^5

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Rubi [A]  time = 0.152999, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {77} \[ -\frac{b^2 x (-3 a B e-A b e+3 b B d)}{e^4}+\frac{(b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (d+e x)}-\frac{(b d-a e)^3 (B d-A e)}{2 e^5 (d+e x)^2}+\frac{3 b (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)}{e^5}+\frac{b^3 B x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^3,x]

[Out]

-((b^2*(3*b*B*d - A*b*e - 3*a*B*e)*x)/e^4) + (b^3*B*x^2)/(2*e^3) - ((b*d - a*e)^3*(B*d - A*e))/(2*e^5*(d + e*x
)^2) + ((b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(e^5*(d + e*x)) + (3*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e
)*Log[d + e*x])/e^5

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.118814, size = 238, normalized size = 1.64 \[ \frac{-3 a^2 b e^2 (A e (d+2 e x)-B d (3 d+4 e x))-a^3 e^3 (A e+B (d+2 e x))+3 a b^2 e \left (A d e (3 d+4 e x)+B \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )\right )+6 b (d+e x)^2 (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)+b^3 \left (A e \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )+B \left (-11 d^2 e^2 x^2+2 d^3 e x+7 d^4-4 d e^3 x^3+e^4 x^4\right )\right )}{2 e^5 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^3*e^3*(A*e + B*(d + 2*e*x))) - 3*a^2*b*e^2*(A*e*(d + 2*e*x) - B*d*(3*d + 4*e*x)) + 3*a*b^2*e*(A*d*e*(3*d
+ 4*e*x) + B*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + b^3*(A*e*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*
e^3*x^3) + B*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4)) + 6*b*(b*d - a*e)*(2*b*B*d - A*b*e
- a*B*e)*(d + e*x)^2*Log[d + e*x])/(2*e^5*(d + e*x)^2)

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Maple [B]  time = 0.01, size = 404, normalized size = 2.8 \begin{align*}{\frac{{b}^{3}B{x}^{2}}{2\,{e}^{3}}}+{\frac{{b}^{3}Ax}{{e}^{3}}}+3\,{\frac{a{b}^{2}Bx}{{e}^{3}}}-3\,{\frac{{b}^{3}Bdx}{{e}^{4}}}-{\frac{{a}^{3}A}{2\,e \left ( ex+d \right ) ^{2}}}+{\frac{3\,Ad{a}^{2}b}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}-{\frac{3\,A{d}^{2}a{b}^{2}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{{b}^{3}A{d}^{3}}{2\,{e}^{4} \left ( ex+d \right ) ^{2}}}+{\frac{Bd{a}^{3}}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}-{\frac{3\,B{d}^{2}{a}^{2}b}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{3\,a{b}^{2}B{d}^{3}}{2\,{e}^{4} \left ( ex+d \right ) ^{2}}}-{\frac{{b}^{3}B{d}^{4}}{2\,{e}^{5} \left ( ex+d \right ) ^{2}}}-3\,{\frac{Ab{a}^{2}}{{e}^{2} \left ( ex+d \right ) }}+6\,{\frac{a{b}^{2}Ad}{{e}^{3} \left ( ex+d \right ) }}-3\,{\frac{{b}^{3}A{d}^{2}}{{e}^{4} \left ( ex+d \right ) }}-{\frac{B{a}^{3}}{{e}^{2} \left ( ex+d \right ) }}+6\,{\frac{{a}^{2}bBd}{{e}^{3} \left ( ex+d \right ) }}-9\,{\frac{a{b}^{2}B{d}^{2}}{{e}^{4} \left ( ex+d \right ) }}+4\,{\frac{{b}^{3}B{d}^{3}}{{e}^{5} \left ( ex+d \right ) }}+3\,{\frac{{b}^{2}\ln \left ( ex+d \right ) Aa}{{e}^{3}}}-3\,{\frac{{b}^{3}\ln \left ( ex+d \right ) Ad}{{e}^{4}}}+3\,{\frac{b\ln \left ( ex+d \right ) B{a}^{2}}{{e}^{3}}}-9\,{\frac{{b}^{2}\ln \left ( ex+d \right ) Bda}{{e}^{4}}}+6\,{\frac{{b}^{3}\ln \left ( ex+d \right ) B{d}^{2}}{{e}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*B*x^2/e^3+b^3/e^3*A*x+3*b^2/e^3*a*B*x-3*b^3/e^4*B*d*x-1/2/e/(e*x+d)^2*a^3*A+3/2/e^2/(e*x+d)^2*A*d*a^2*
b-3/2/e^3/(e*x+d)^2*A*d^2*a*b^2+1/2/e^4/(e*x+d)^2*A*b^3*d^3+1/2/e^2/(e*x+d)^2*B*d*a^3-3/2/e^3/(e*x+d)^2*B*d^2*
a^2*b+3/2/e^4/(e*x+d)^2*B*a*b^2*d^3-1/2/e^5/(e*x+d)^2*b^3*B*d^4-3/e^2/(e*x+d)*A*a^2*b+6/e^3/(e*x+d)*A*a*b^2*d-
3/e^4/(e*x+d)*A*b^3*d^2-1/e^2/(e*x+d)*B*a^3+6/e^3/(e*x+d)*B*a^2*b*d-9/e^4/(e*x+d)*B*a*b^2*d^2+4/e^5/(e*x+d)*b^
3*B*d^3+3*b^2/e^3*ln(e*x+d)*A*a-3*b^3/e^4*ln(e*x+d)*A*d+3*b/e^3*ln(e*x+d)*B*a^2-9*b^2/e^4*ln(e*x+d)*B*d*a+6*b^
3/e^5*ln(e*x+d)*B*d^2

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Maxima [A]  time = 1.47091, size = 370, normalized size = 2.55 \begin{align*} \frac{7 \, B b^{3} d^{4} - A a^{3} e^{4} - 5 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 9 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} -{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 2 \,{\left (4 \, B b^{3} d^{3} e - 3 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 6 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} -{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{2 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} + \frac{B b^{3} e x^{2} - 2 \,{\left (3 \, B b^{3} d -{\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} x}{2 \, e^{4}} + \frac{3 \,{\left (2 \, B b^{3} d^{2} -{\left (3 \, B a b^{2} + A b^{3}\right )} d e +{\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(7*B*b^3*d^4 - A*a^3*e^4 - 5*(3*B*a*b^2 + A*b^3)*d^3*e + 9*(B*a^2*b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*
b)*d*e^3 + 2*(4*B*b^3*d^3*e - 3*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 6*(B*a^2*b + A*a*b^2)*d*e^3 - (B*a^3 + 3*A*a^2*b
)*e^4)*x)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5) + 1/2*(B*b^3*e*x^2 - 2*(3*B*b^3*d - (3*B*a*b^2 + A*b^3)*e)*x)/e^4 +
3*(2*B*b^3*d^2 - (3*B*a*b^2 + A*b^3)*d*e + (B*a^2*b + A*a*b^2)*e^2)*log(e*x + d)/e^5

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Fricas [B]  time = 1.73996, size = 849, normalized size = 5.86 \begin{align*} \frac{B b^{3} e^{4} x^{4} + 7 \, B b^{3} d^{4} - A a^{3} e^{4} - 5 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 9 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} -{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 2 \,{\left (2 \, B b^{3} d e^{3} -{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} -{\left (11 \, B b^{3} d^{2} e^{2} - 4 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3}\right )} x^{2} + 2 \,{\left (B b^{3} d^{3} e - 2 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 6 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} -{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 6 \,{\left (2 \, B b^{3} d^{4} -{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e +{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} +{\left (2 \, B b^{3} d^{2} e^{2} -{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} +{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 2 \,{\left (2 \, B b^{3} d^{3} e -{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} +{\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*b^3*e^4*x^4 + 7*B*b^3*d^4 - A*a^3*e^4 - 5*(3*B*a*b^2 + A*b^3)*d^3*e + 9*(B*a^2*b + A*a*b^2)*d^2*e^2 - (
B*a^3 + 3*A*a^2*b)*d*e^3 - 2*(2*B*b^3*d*e^3 - (3*B*a*b^2 + A*b^3)*e^4)*x^3 - (11*B*b^3*d^2*e^2 - 4*(3*B*a*b^2
+ A*b^3)*d*e^3)*x^2 + 2*(B*b^3*d^3*e - 2*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 6*(B*a^2*b + A*a*b^2)*d*e^3 - (B*a^3 +
3*A*a^2*b)*e^4)*x + 6*(2*B*b^3*d^4 - (3*B*a*b^2 + A*b^3)*d^3*e + (B*a^2*b + A*a*b^2)*d^2*e^2 + (2*B*b^3*d^2*e^
2 - (3*B*a*b^2 + A*b^3)*d*e^3 + (B*a^2*b + A*a*b^2)*e^4)*x^2 + 2*(2*B*b^3*d^3*e - (3*B*a*b^2 + A*b^3)*d^2*e^2
+ (B*a^2*b + A*a*b^2)*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)

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Sympy [B]  time = 6.15213, size = 296, normalized size = 2.04 \begin{align*} \frac{B b^{3} x^{2}}{2 e^{3}} + \frac{3 b \left (a e - b d\right ) \left (A b e + B a e - 2 B b d\right ) \log{\left (d + e x \right )}}{e^{5}} - \frac{A a^{3} e^{4} + 3 A a^{2} b d e^{3} - 9 A a b^{2} d^{2} e^{2} + 5 A b^{3} d^{3} e + B a^{3} d e^{3} - 9 B a^{2} b d^{2} e^{2} + 15 B a b^{2} d^{3} e - 7 B b^{3} d^{4} + x \left (6 A a^{2} b e^{4} - 12 A a b^{2} d e^{3} + 6 A b^{3} d^{2} e^{2} + 2 B a^{3} e^{4} - 12 B a^{2} b d e^{3} + 18 B a b^{2} d^{2} e^{2} - 8 B b^{3} d^{3} e\right )}{2 d^{2} e^{5} + 4 d e^{6} x + 2 e^{7} x^{2}} + \frac{x \left (A b^{3} e + 3 B a b^{2} e - 3 B b^{3} d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b**3*x**2/(2*e**3) + 3*b*(a*e - b*d)*(A*b*e + B*a*e - 2*B*b*d)*log(d + e*x)/e**5 - (A*a**3*e**4 + 3*A*a**2*b
*d*e**3 - 9*A*a*b**2*d**2*e**2 + 5*A*b**3*d**3*e + B*a**3*d*e**3 - 9*B*a**2*b*d**2*e**2 + 15*B*a*b**2*d**3*e -
7*B*b**3*d**4 + x*(6*A*a**2*b*e**4 - 12*A*a*b**2*d*e**3 + 6*A*b**3*d**2*e**2 + 2*B*a**3*e**4 - 12*B*a**2*b*d*
e**3 + 18*B*a*b**2*d**2*e**2 - 8*B*b**3*d**3*e))/(2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + x*(A*b**3*e + 3*B*
a*b**2*e - 3*B*b**3*d)/e**4

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Giac [A]  time = 3.08768, size = 369, normalized size = 2.54 \begin{align*} 3 \,{\left (2 \, B b^{3} d^{2} - 3 \, B a b^{2} d e - A b^{3} d e + B a^{2} b e^{2} + A a b^{2} e^{2}\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{2} \,{\left (B b^{3} x^{2} e^{3} - 6 \, B b^{3} d x e^{2} + 6 \, B a b^{2} x e^{3} + 2 \, A b^{3} x e^{3}\right )} e^{\left (-6\right )} + \frac{{\left (7 \, B b^{3} d^{4} - 15 \, B a b^{2} d^{3} e - 5 \, A b^{3} d^{3} e + 9 \, B a^{2} b d^{2} e^{2} + 9 \, A a b^{2} d^{2} e^{2} - B a^{3} d e^{3} - 3 \, A a^{2} b d e^{3} - A a^{3} e^{4} + 2 \,{\left (4 \, B b^{3} d^{3} e - 9 \, B a b^{2} d^{2} e^{2} - 3 \, A b^{3} d^{2} e^{2} + 6 \, B a^{2} b d e^{3} + 6 \, A a b^{2} d e^{3} - B a^{3} e^{4} - 3 \, A a^{2} b e^{4}\right )} x\right )} e^{\left (-5\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(2*B*b^3*d^2 - 3*B*a*b^2*d*e - A*b^3*d*e + B*a^2*b*e^2 + A*a*b^2*e^2)*e^(-5)*log(abs(x*e + d)) + 1/2*(B*b^3*
x^2*e^3 - 6*B*b^3*d*x*e^2 + 6*B*a*b^2*x*e^3 + 2*A*b^3*x*e^3)*e^(-6) + 1/2*(7*B*b^3*d^4 - 15*B*a*b^2*d^3*e - 5*
A*b^3*d^3*e + 9*B*a^2*b*d^2*e^2 + 9*A*a*b^2*d^2*e^2 - B*a^3*d*e^3 - 3*A*a^2*b*d*e^3 - A*a^3*e^4 + 2*(4*B*b^3*d
^3*e - 9*B*a*b^2*d^2*e^2 - 3*A*b^3*d^2*e^2 + 6*B*a^2*b*d*e^3 + 6*A*a*b^2*d*e^3 - B*a^3*e^4 - 3*A*a^2*b*e^4)*x)
*e^(-5)/(x*e + d)^2