3.1041 \(\int \frac {(a+b x)^3 (A+B x)}{d+e x} \, dx\)

Optimal. Leaf size=124 \[ \frac {(b d-a e)^3 (B d-A e) \log (d+e x)}{e^5}-\frac {b x (b d-a e)^2 (B d-A e)}{e^4}+\frac {(a+b x)^2 (b d-a e) (B d-A e)}{2 e^3}-\frac {(a+b x)^3 (B d-A e)}{3 e^2}+\frac {B (a+b x)^4}{4 b e} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ -\frac {(a+b x)^3 (B d-A e)}{3 e^2}+\frac {(a+b x)^2 (b d-a e) (B d-A e)}{2 e^3}-\frac {b x (b d-a e)^2 (B d-A e)}{e^4}+\frac {(b d-a e)^3 (B d-A e) \log (d+e x)}{e^5}+\frac {B (a+b x)^4}{4 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, size = 169, normalized size = 1.36 \[ \frac {e x \left (12 a^3 B e^3+18 a^2 b e^2 (2 A e-2 B d+B e x)+6 a b^2 e \left (3 A e (e x-2 d)+B \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )+b^3 \left (2 A e \left (6 d^2-3 d e x+2 e^2 x^2\right )+B \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )\right )+12 (b d-a e)^3 (B d-A e) \log (d+e x)}{12 e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(12*a^3*B*e^3 + 18*a^2*b*e^2*(-2*B*d + 2*A*e + B*e*x) + 6*a*b^2*e*(3*A*e*(-2*d + e*x) + B*(6*d^2 - 3*d*e*
x + 2*e^2*x^2)) + b^3*(2*A*e*(6*d^2 - 3*d*e*x + 2*e^2*x^2) + B*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3)
)) + 12*(b*d - a*e)^3*(B*d - A*e)*Log[d + e*x])/(12*e^5)

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fricas [B]  time = 0.82, size = 260, normalized size = 2.10 \[ \frac {3 \, B b^{3} e^{4} x^{4} - 4 \, {\left (B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 6 \, {\left (B b^{3} d^{2} e^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} - 12 \, {\left (B b^{3} d^{3} e - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 12 \, {\left (B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3}\right )} \log \left (e x + d\right )}{12 \, e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*b^3*e^4*x^4 - 4*(B*b^3*d*e^3 - (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 6*(B*b^3*d^2*e^2 - (3*B*a*b^2 + A*b^3)
*d*e^3 + 3*(B*a^2*b + A*a*b^2)*e^4)*x^2 - 12*(B*b^3*d^3*e - (3*B*a*b^2 + A*b^3)*d^2*e^2 + 3*(B*a^2*b + A*a*b^2
)*d*e^3 - (B*a^3 + 3*A*a^2*b)*e^4)*x + 12*(B*b^3*d^4 + A*a^3*e^4 - (3*B*a*b^2 + A*b^3)*d^3*e + 3*(B*a^2*b + A*
a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3)*log(e*x + d))/e^5

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giac [B]  time = 1.33, size = 284, normalized size = 2.29 \[ {\left (B b^{3} d^{4} - 3 \, B a b^{2} d^{3} e - A b^{3} d^{3} e + 3 \, B a^{2} b d^{2} e^{2} + 3 \, A a b^{2} d^{2} e^{2} - B a^{3} d e^{3} - 3 \, A a^{2} b d e^{3} + A a^{3} e^{4}\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{12} \, {\left (3 \, B b^{3} x^{4} e^{3} - 4 \, B b^{3} d x^{3} e^{2} + 6 \, B b^{3} d^{2} x^{2} e - 12 \, B b^{3} d^{3} x + 12 \, B a b^{2} x^{3} e^{3} + 4 \, A b^{3} x^{3} e^{3} - 18 \, B a b^{2} d x^{2} e^{2} - 6 \, A b^{3} d x^{2} e^{2} + 36 \, B a b^{2} d^{2} x e + 12 \, A b^{3} d^{2} x e + 18 \, B a^{2} b x^{2} e^{3} + 18 \, A a b^{2} x^{2} e^{3} - 36 \, B a^{2} b d x e^{2} - 36 \, A a b^{2} d x e^{2} + 12 \, B a^{3} x e^{3} + 36 \, A a^{2} b x e^{3}\right )} e^{\left (-4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b^3*d^4 - 3*B*a*b^2*d^3*e - A*b^3*d^3*e + 3*B*a^2*b*d^2*e^2 + 3*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)*e^(-5)*log(abs(x*e + d)) + 1/12*(3*B*b^3*x^4*e^3 - 4*B*b^3*d*x^3*e^2 + 6*B*b^3*d^2*x^2*e - 1
2*B*b^3*d^3*x + 12*B*a*b^2*x^3*e^3 + 4*A*b^3*x^3*e^3 - 18*B*a*b^2*d*x^2*e^2 - 6*A*b^3*d*x^2*e^2 + 36*B*a*b^2*d
^2*x*e + 12*A*b^3*d^2*x*e + 18*B*a^2*b*x^2*e^3 + 18*A*a*b^2*x^2*e^3 - 36*B*a^2*b*d*x*e^2 - 36*A*a*b^2*d*x*e^2
+ 12*B*a^3*x*e^3 + 36*A*a^2*b*x*e^3)*e^(-4)

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maple [B]  time = 0.01, size = 341, normalized size = 2.75 \[ \frac {B \,b^{3} x^{4}}{4 e}+\frac {A \,b^{3} x^{3}}{3 e}+\frac {B a \,b^{2} x^{3}}{e}-\frac {B \,b^{3} d \,x^{3}}{3 e^{2}}+\frac {3 A a \,b^{2} x^{2}}{2 e}-\frac {A \,b^{3} d \,x^{2}}{2 e^{2}}+\frac {3 B \,a^{2} b \,x^{2}}{2 e}-\frac {3 B a \,b^{2} d \,x^{2}}{2 e^{2}}+\frac {B \,b^{3} d^{2} x^{2}}{2 e^{3}}+\frac {A \,a^{3} \ln \left (e x +d \right )}{e}-\frac {3 A \,a^{2} b d \ln \left (e x +d \right )}{e^{2}}+\frac {3 A \,a^{2} b x}{e}+\frac {3 A a \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {3 A a \,b^{2} d x}{e^{2}}-\frac {A \,b^{3} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {A \,b^{3} d^{2} x}{e^{3}}-\frac {B \,a^{3} d \ln \left (e x +d \right )}{e^{2}}+\frac {B \,a^{3} x}{e}+\frac {3 B \,a^{2} b \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {3 B \,a^{2} b d x}{e^{2}}-\frac {3 B a \,b^{2} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {3 B a \,b^{2} d^{2} x}{e^{3}}+\frac {B \,b^{3} d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {B \,b^{3} d^{3} x}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.50, size = 258, normalized size = 2.08 \[ \frac {3 \, B b^{3} e^{3} x^{4} - 4 \, {\left (B b^{3} d e^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{3}\right )} x^{3} + 6 \, {\left (B b^{3} d^{2} e - {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{3}\right )} x^{2} - 12 \, {\left (B b^{3} d^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{3}\right )} x}{12 \, e^{4}} + \frac {{\left (B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3}\right )} \log \left (e x + d\right )}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*b^3*e^3*x^4 - 4*(B*b^3*d*e^2 - (3*B*a*b^2 + A*b^3)*e^3)*x^3 + 6*(B*b^3*d^2*e - (3*B*a*b^2 + A*b^3)*d
*e^2 + 3*(B*a^2*b + A*a*b^2)*e^3)*x^2 - 12*(B*b^3*d^3 - (3*B*a*b^2 + A*b^3)*d^2*e + 3*(B*a^2*b + A*a*b^2)*d*e^
2 - (B*a^3 + 3*A*a^2*b)*e^3)*x)/e^4 + (B*b^3*d^4 + A*a^3*e^4 - (3*B*a*b^2 + A*b^3)*d^3*e + 3*(B*a^2*b + A*a*b^
2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3)*log(e*x + d)/e^5

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mupad [B]  time = 1.08, size = 268, normalized size = 2.16 \[ x\,\left (\frac {B\,a^3+3\,A\,b\,a^2}{e}+\frac {d\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{e}-\frac {B\,b^3\,d}{e^2}\right )}{e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{e}\right )}{e}\right )+x^3\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{3\,e}-\frac {B\,b^3\,d}{3\,e^2}\right )-x^2\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{e}-\frac {B\,b^3\,d}{e^2}\right )}{2\,e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{2\,e}\right )+\frac {\ln \left (d+e\,x\right )\,\left (-B\,a^3\,d\,e^3+A\,a^3\,e^4+3\,B\,a^2\,b\,d^2\,e^2-3\,A\,a^2\,b\,d\,e^3-3\,B\,a\,b^2\,d^3\,e+3\,A\,a\,b^2\,d^2\,e^2+B\,b^3\,d^4-A\,b^3\,d^3\,e\right )}{e^5}+\frac {B\,b^3\,x^4}{4\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.68, size = 221, normalized size = 1.78 \[ \frac {B b^{3} x^{4}}{4 e} + x^{3} \left (\frac {A b^{3}}{3 e} + \frac {B a b^{2}}{e} - \frac {B b^{3} d}{3 e^{2}}\right ) + x^{2} \left (\frac {3 A a b^{2}}{2 e} - \frac {A b^{3} d}{2 e^{2}} + \frac {3 B a^{2} b}{2 e} - \frac {3 B a b^{2} d}{2 e^{2}} + \frac {B b^{3} d^{2}}{2 e^{3}}\right ) + x \left (\frac {3 A a^{2} b}{e} - \frac {3 A a b^{2} d}{e^{2}} + \frac {A b^{3} d^{2}}{e^{3}} + \frac {B a^{3}}{e} - \frac {3 B a^{2} b d}{e^{2}} + \frac {3 B a b^{2} d^{2}}{e^{3}} - \frac {B b^{3} d^{3}}{e^{4}}\right ) - \frac {\left (- A e + B d\right ) \left (a e - b d\right )^{3} \log {\left (d + e x \right )}}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b**3*x**4/(4*e) + x**3*(A*b**3/(3*e) + B*a*b**2/e - B*b**3*d/(3*e**2)) + x**2*(3*A*a*b**2/(2*e) - A*b**3*d/(
2*e**2) + 3*B*a**2*b/(2*e) - 3*B*a*b**2*d/(2*e**2) + B*b**3*d**2/(2*e**3)) + x*(3*A*a**2*b/e - 3*A*a*b**2*d/e*
*2 + A*b**3*d**2/e**3 + B*a**3/e - 3*B*a**2*b*d/e**2 + 3*B*a*b**2*d**2/e**3 - B*b**3*d**3/e**4) - (-A*e + B*d)
*(a*e - b*d)**3*log(d + e*x)/e**5

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