3.1042 \(\int \frac {(a+b x)^3 (A+B x)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=150 \[ -\frac {b^2 (d+e x)^2 (-3 a B e-A b e+4 b B d)}{2 e^5}-\frac {(b d-a e)^3 (B d-A e)}{e^5 (d+e x)}-\frac {(b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)}{e^5}+\frac {3 b x (b d-a e) (-a B e-A b e+2 b B d)}{e^4}+\frac {b^3 B (d+e x)^3}{3 e^5} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 150, 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} \[ -\frac {b^2 (d+e x)^2 (-3 a B e-A b e+4 b B d)}{2 e^5}-\frac {(b d-a e)^3 (B d-A e)}{e^5 (d+e x)}+\frac {3 b x (b d-a e) (-a B e-A b e+2 b B d)}{e^4}-\frac {(b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)}{e^5}+\frac {b^3 B (d+e x)^3}{3 e^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^2,x]

[Out]

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

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Mathematica [A]  time = 0.11, size = 244, normalized size = 1.63 \[ \frac {6 a^3 e^3 (B d-A e)+18 a^2 b e^2 \left (A d e+B \left (-d^2+d e x+e^2 x^2\right )\right )+9 a b^2 e \left (2 A e \left (-d^2+d e x+e^2 x^2\right )+B \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )\right )-6 (d+e x) (b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)+b^3 \left (3 A e \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+2 B \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )\right )}{6 e^5 (d+e x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^2,x]

[Out]

(6*a^3*e^3*(B*d - A*e) + 18*a^2*b*e^2*(A*d*e + B*(-d^2 + d*e*x + e^2*x^2)) + 9*a*b^2*e*(2*A*e*(-d^2 + d*e*x +
e^2*x^2) + B*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^3)) + b^3*(3*A*e*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*
x^3) + 2*B*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4)) - 6*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e
- a*B*e)*(d + e*x)*Log[d + e*x])/(6*e^5*(d + e*x))

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fricas [B]  time = 0.92, size = 396, normalized size = 2.64 \[ \frac {2 \, B b^{3} e^{4} x^{4} - 6 \, B b^{3} d^{4} - 6 \, A a^{3} e^{4} + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 18 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (4 \, B b^{3} d e^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \, {\left (4 \, B b^{3} d^{2} e^{2} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 6 \, {\left (3 \, B b^{3} d^{3} e - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x - 6 \, {\left (4 \, B b^{3} d^{4} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + {\left (4 \, B b^{3} d^{3} e - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{6} x + d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.23, size = 362, normalized size = 2.41 \[ \frac {1}{6} \, {\left (2 \, B b^{3} - \frac {3 \, {\left (4 \, B b^{3} d e - 3 \, B a b^{2} e^{2} - A b^{3} e^{2}\right )} e^{\left (-1\right )}}{x e + d} + \frac {18 \, {\left (2 \, B b^{3} d^{2} e^{2} - 3 \, B a b^{2} d e^{3} - A b^{3} d e^{3} + B a^{2} b e^{4} + A a b^{2} e^{4}\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}}\right )} {\left (x e + d\right )}^{3} e^{\left (-5\right )} + {\left (4 \, B b^{3} d^{3} - 9 \, B a b^{2} d^{2} e - 3 \, A b^{3} d^{2} e + 6 \, B a^{2} b d e^{2} + 6 \, A a b^{2} d e^{2} - B a^{3} e^{3} - 3 \, A a^{2} b e^{3}\right )} e^{\left (-5\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - {\left (\frac {B b^{3} d^{4} e^{3}}{x e + d} - \frac {3 \, B a b^{2} d^{3} e^{4}}{x e + d} - \frac {A b^{3} d^{3} e^{4}}{x e + d} + \frac {3 \, B a^{2} b d^{2} e^{5}}{x e + d} + \frac {3 \, A a b^{2} d^{2} e^{5}}{x e + d} - \frac {B a^{3} d e^{6}}{x e + d} - \frac {3 \, A a^{2} b d e^{6}}{x e + d} + \frac {A a^{3} e^{7}}{x e + d}\right )} e^{\left (-8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^2,x, algorithm="giac")

[Out]

1/6*(2*B*b^3 - 3*(4*B*b^3*d*e - 3*B*a*b^2*e^2 - A*b^3*e^2)*e^(-1)/(x*e + d) + 18*(2*B*b^3*d^2*e^2 - 3*B*a*b^2*
d*e^3 - A*b^3*d*e^3 + B*a^2*b*e^4 + A*a*b^2*e^4)*e^(-2)/(x*e + d)^2)*(x*e + d)^3*e^(-5) + (4*B*b^3*d^3 - 9*B*a
*b^2*d^2*e - 3*A*b^3*d^2*e + 6*B*a^2*b*d*e^2 + 6*A*a*b^2*d*e^2 - B*a^3*e^3 - 3*A*a^2*b*e^3)*e^(-5)*log(abs(x*e
+ d)*e^(-1)/(x*e + d)^2) - (B*b^3*d^4*e^3/(x*e + d) - 3*B*a*b^2*d^3*e^4/(x*e + d) - A*b^3*d^3*e^4/(x*e + d) +
3*B*a^2*b*d^2*e^5/(x*e + d) + 3*A*a*b^2*d^2*e^5/(x*e + d) - B*a^3*d*e^6/(x*e + d) - 3*A*a^2*b*d*e^6/(x*e + d)
+ A*a^3*e^7/(x*e + d))*e^(-8)

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maple [B]  time = 0.01, size = 376, normalized size = 2.51 \[ \frac {B \,b^{3} x^{3}}{3 e^{2}}+\frac {A \,b^{3} x^{2}}{2 e^{2}}+\frac {3 B a \,b^{2} x^{2}}{2 e^{2}}-\frac {B \,b^{3} d \,x^{2}}{e^{3}}-\frac {A \,a^{3}}{\left (e x +d \right ) e}+\frac {3 A \,a^{2} b d}{\left (e x +d \right ) e^{2}}+\frac {3 A \,a^{2} b \ln \left (e x +d \right )}{e^{2}}-\frac {3 A a \,b^{2} d^{2}}{\left (e x +d \right ) e^{3}}-\frac {6 A a \,b^{2} d \ln \left (e x +d \right )}{e^{3}}+\frac {3 A a \,b^{2} x}{e^{2}}+\frac {A \,b^{3} d^{3}}{\left (e x +d \right ) e^{4}}+\frac {3 A \,b^{3} d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {2 A \,b^{3} d x}{e^{3}}+\frac {B \,a^{3} d}{\left (e x +d \right ) e^{2}}+\frac {B \,a^{3} \ln \left (e x +d \right )}{e^{2}}-\frac {3 B \,a^{2} b \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {6 B \,a^{2} b d \ln \left (e x +d \right )}{e^{3}}+\frac {3 B \,a^{2} b x}{e^{2}}+\frac {3 B a \,b^{2} d^{3}}{\left (e x +d \right ) e^{4}}+\frac {9 B a \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {6 B a \,b^{2} d x}{e^{3}}-\frac {B \,b^{3} d^{4}}{\left (e x +d \right ) e^{5}}-\frac {4 B \,b^{3} d^{3} \ln \left (e x +d \right )}{e^{5}}+\frac {3 B \,b^{3} d^{2} x}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A)/(e*x+d)^2,x)

[Out]

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

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maxima [A]  time = 0.51, size = 267, normalized size = 1.78 \[ -\frac {B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3}}{e^{6} x + d e^{5}} + \frac {2 \, B b^{3} e^{2} x^{3} - 3 \, {\left (2 \, B b^{3} d e - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 6 \, {\left (3 \, B b^{3} d^{2} - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )} x}{6 \, e^{4}} - \frac {{\left (4 \, B b^{3} d^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{3}\right )} \log \left (e x + d\right )}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.11, size = 293, normalized size = 1.95 \[ x^2\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{2\,e^2}-\frac {B\,b^3\,d}{e^3}\right )-x\,\left (\frac {2\,d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{e^2}-\frac {2\,B\,b^3\,d}{e^3}\right )}{e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{e^2}+\frac {B\,b^3\,d^2}{e^4}\right )+\frac {\ln \left (d+e\,x\right )\,\left (B\,a^3\,e^3-6\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+9\,B\,a\,b^2\,d^2\,e-6\,A\,a\,b^2\,d\,e^2-4\,B\,b^3\,d^3+3\,A\,b^3\,d^2\,e\right )}{e^5}-\frac {-B\,a^3\,d\,e^3+A\,a^3\,e^4+3\,B\,a^2\,b\,d^2\,e^2-3\,A\,a^2\,b\,d\,e^3-3\,B\,a\,b^2\,d^3\,e+3\,A\,a\,b^2\,d^2\,e^2+B\,b^3\,d^4-A\,b^3\,d^3\,e}{e\,\left (x\,e^5+d\,e^4\right )}+\frac {B\,b^3\,x^3}{3\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^3)/(d + e*x)^2,x)

[Out]

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

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sympy [A]  time = 1.58, size = 257, normalized size = 1.71 \[ \frac {B b^{3} x^{3}}{3 e^{2}} + x^{2} \left (\frac {A b^{3}}{2 e^{2}} + \frac {3 B a b^{2}}{2 e^{2}} - \frac {B b^{3} d}{e^{3}}\right ) + x \left (\frac {3 A a b^{2}}{e^{2}} - \frac {2 A b^{3} d}{e^{3}} + \frac {3 B a^{2} b}{e^{2}} - \frac {6 B a b^{2} d}{e^{3}} + \frac {3 B b^{3} d^{2}}{e^{4}}\right ) + \frac {- A a^{3} e^{4} + 3 A a^{2} b d e^{3} - 3 A a b^{2} d^{2} e^{2} + A b^{3} d^{3} e + B a^{3} d e^{3} - 3 B a^{2} b d^{2} e^{2} + 3 B a b^{2} d^{3} e - B b^{3} d^{4}}{d e^{5} + e^{6} x} + \frac {\left (a e - b d\right )^{2} \left (3 A b e + B a e - 4 B b d\right ) \log {\left (d + e x \right )}}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**2,x)

[Out]

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

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