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

Optimal. Leaf size=257 \[ -\frac {x^2 \left (B (c d-b e) \left (c d^2-e (b d-2 a e)\right )-A e \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{2 e^4}-\frac {x^3 \left (A c e (c d-2 b e)-B \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{3 e^3}-\frac {x \left (A e (c d-b e) \left (c d^2-e (b d-2 a e)\right )-B \left (c d^2-e (b d-a e)\right )^2\right )}{e^5}-\frac {(B d-A e) \log (d+e x) \left (a e^2-b d e+c d^2\right )^2}{e^6}-\frac {c x^4 (-A c e-2 b B e+B c d)}{4 e^2}+\frac {B c^2 x^5}{5 e} \]

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Rubi [A]  time = 0.54, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {771} \begin {gather*} -\frac {x^3 \left (A c e (c d-2 b e)-B \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{3 e^3}-\frac {x^2 \left (B (c d-b e) \left (c d^2-e (b d-2 a e)\right )-A e \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{2 e^4}-\frac {x \left (A e (c d-b e) \left (c d^2-e (b d-2 a e)\right )-B \left (c d^2-e (b d-a e)\right )^2\right )}{e^5}-\frac {(B d-A e) \log (d+e x) \left (a e^2-b d e+c d^2\right )^2}{e^6}-\frac {c x^4 (-A c e-2 b B e+B c d)}{4 e^2}+\frac {B c^2 x^5}{5 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{d+e x} \, dx &=\int \left (\frac {-A e (c d-b e) \left (c d^2-e (b d-2 a e)\right )+B \left (c d^2-e (b d-a e)\right )^2}{e^5}+\frac {\left (-B (c d-b e) \left (c d^2-e (b d-2 a e)\right )+A e \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right )\right ) x}{e^4}+\frac {\left (-A c e (c d-2 b e)+B \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right )\right ) x^2}{e^3}+\frac {c (-B c d+2 b B e+A c e) x^3}{e^2}+\frac {B c^2 x^4}{e}+\frac {(-B d+A e) \left (c d^2-b d e+a e^2\right )^2}{e^5 (d+e x)}\right ) \, dx\\ &=-\frac {\left (A e (c d-b e) \left (c d^2-e (b d-2 a e)\right )-B \left (c d^2-e (b d-a e)\right )^2\right ) x}{e^5}-\frac {\left (B (c d-b e) \left (c d^2-e (b d-2 a e)\right )-A e \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right )\right ) x^2}{2 e^4}-\frac {\left (A c e (c d-2 b e)-B \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right )\right ) x^3}{3 e^3}-\frac {c (B c d-2 b B e-A c e) x^4}{4 e^2}+\frac {B c^2 x^5}{5 e}-\frac {(B d-A e) \left (c d^2-b d e+a e^2\right )^2 \log (d+e x)}{e^6}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 298, normalized size = 1.16 \begin {gather*} \frac {e x \left (B \left (10 e^2 \left (6 a^2 e^2+6 a b e (e x-2 d)+b^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )+10 c e \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 )+c^2 \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )+5 A e \left (4 c 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 )+6 b e^2 (4 a e-2 b d+b e x)+c^2 \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )\right )-60 (B d-A e) \log (d+e x) \left (e (a e-b d)+c d^2\right )^2}{60 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(5*A*e*(6*b*e^2*(-2*b*d + 4*a*e + b*e*x) + c^2*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3) + 4*c*e*(3
*a*e*(-2*d + e*x) + b*(6*d^2 - 3*d*e*x + 2*e^2*x^2))) + B*(c^2*(60*d^4 - 30*d^3*e*x + 20*d^2*e^2*x^2 - 15*d*e^
3*x^3 + 12*e^4*x^4) + 10*e^2*(6*a^2*e^2 + 6*a*b*e*(-2*d + e*x) + b^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2)) + 10*c*e*(
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)))) - 60*(B*d - A*e)*(c
*d^2 + e*(-(b*d) + a*e))^2*Log[d + e*x])/(60*e^6)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{d+e x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 379, normalized size = 1.47 \begin {gather*} \frac {12 \, B c^{2} e^{5} x^{5} - 15 \, {\left (B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 20 \, {\left (B c^{2} d^{2} e^{3} - {\left (2 \, B b c + A c^{2}\right )} d e^{4} + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} e^{5}\right )} x^{3} - 30 \, {\left (B c^{2} d^{3} e^{2} - {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d e^{4} - {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} e^{5}\right )} x^{2} + 60 \, {\left (B c^{2} d^{4} e - {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{2} e^{3} - {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d e^{4} + {\left (B a^{2} + 2 \, A a b\right )} e^{5}\right )} x - 60 \, {\left (B c^{2} d^{5} - A a^{2} e^{5} - {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{3} e^{2} - {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d^{2} e^{3} + {\left (B a^{2} + 2 \, A a b\right )} d e^{4}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*c^2*e^5*x^5 - 15*(B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 20*(B*c^2*d^2*e^3 - (2*B*b*c + A*c^2)*
d*e^4 + (B*b^2 + 2*(B*a + A*b)*c)*e^5)*x^3 - 30*(B*c^2*d^3*e^2 - (2*B*b*c + A*c^2)*d^2*e^3 + (B*b^2 + 2*(B*a +
A*b)*c)*d*e^4 - (2*B*a*b + A*b^2 + 2*A*a*c)*e^5)*x^2 + 60*(B*c^2*d^4*e - (2*B*b*c + A*c^2)*d^3*e^2 + (B*b^2 +
2*(B*a + A*b)*c)*d^2*e^3 - (2*B*a*b + A*b^2 + 2*A*a*c)*d*e^4 + (B*a^2 + 2*A*a*b)*e^5)*x - 60*(B*c^2*d^5 - A*a
^2*e^5 - (2*B*b*c + A*c^2)*d^4*e + (B*b^2 + 2*(B*a + A*b)*c)*d^3*e^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d^2*e^3 + (
B*a^2 + 2*A*a*b)*d*e^4)*log(e*x + d))/e^6

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giac [A]  time = 0.16, size = 463, normalized size = 1.80 \begin {gather*} -{\left (B c^{2} d^{5} - 2 \, B b c d^{4} e - A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 2 \, B a c d^{3} e^{2} + 2 \, A b c d^{3} e^{2} - 2 \, B a b d^{2} e^{3} - A b^{2} d^{2} e^{3} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + 2 \, A a b d e^{4} - A a^{2} e^{5}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{60} \, {\left (12 \, B c^{2} x^{5} e^{4} - 15 \, B c^{2} d x^{4} e^{3} + 20 \, B c^{2} d^{2} x^{3} e^{2} - 30 \, B c^{2} d^{3} x^{2} e + 60 \, B c^{2} d^{4} x + 30 \, B b c x^{4} e^{4} + 15 \, A c^{2} x^{4} e^{4} - 40 \, B b c d x^{3} e^{3} - 20 \, A c^{2} d x^{3} e^{3} + 60 \, B b c d^{2} x^{2} e^{2} + 30 \, A c^{2} d^{2} x^{2} e^{2} - 120 \, B b c d^{3} x e - 60 \, A c^{2} d^{3} x e + 20 \, B b^{2} x^{3} e^{4} + 40 \, B a c x^{3} e^{4} + 40 \, A b c x^{3} e^{4} - 30 \, B b^{2} d x^{2} e^{3} - 60 \, B a c d x^{2} e^{3} - 60 \, A b c d x^{2} e^{3} + 60 \, B b^{2} d^{2} x e^{2} + 120 \, B a c d^{2} x e^{2} + 120 \, A b c d^{2} x e^{2} + 60 \, B a b x^{2} e^{4} + 30 \, A b^{2} x^{2} e^{4} + 60 \, A a c x^{2} e^{4} - 120 \, B a b d x e^{3} - 60 \, A b^{2} d x e^{3} - 120 \, A a c d x e^{3} + 60 \, B a^{2} x e^{4} + 120 \, A a b x e^{4}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*c^2*d^5 - 2*B*b*c*d^4*e - A*c^2*d^4*e + B*b^2*d^3*e^2 + 2*B*a*c*d^3*e^2 + 2*A*b*c*d^3*e^2 - 2*B*a*b*d^2*e^
3 - A*b^2*d^2*e^3 - 2*A*a*c*d^2*e^3 + B*a^2*d*e^4 + 2*A*a*b*d*e^4 - A*a^2*e^5)*e^(-6)*log(abs(x*e + d)) + 1/60
*(12*B*c^2*x^5*e^4 - 15*B*c^2*d*x^4*e^3 + 20*B*c^2*d^2*x^3*e^2 - 30*B*c^2*d^3*x^2*e + 60*B*c^2*d^4*x + 30*B*b*
c*x^4*e^4 + 15*A*c^2*x^4*e^4 - 40*B*b*c*d*x^3*e^3 - 20*A*c^2*d*x^3*e^3 + 60*B*b*c*d^2*x^2*e^2 + 30*A*c^2*d^2*x
^2*e^2 - 120*B*b*c*d^3*x*e - 60*A*c^2*d^3*x*e + 20*B*b^2*x^3*e^4 + 40*B*a*c*x^3*e^4 + 40*A*b*c*x^3*e^4 - 30*B*
b^2*d*x^2*e^3 - 60*B*a*c*d*x^2*e^3 - 60*A*b*c*d*x^2*e^3 + 60*B*b^2*d^2*x*e^2 + 120*B*a*c*d^2*x*e^2 + 120*A*b*c
*d^2*x*e^2 + 60*B*a*b*x^2*e^4 + 30*A*b^2*x^2*e^4 + 60*A*a*c*x^2*e^4 - 120*B*a*b*d*x*e^3 - 60*A*b^2*d*x*e^3 - 1
20*A*a*c*d*x*e^3 + 60*B*a^2*x*e^4 + 120*A*a*b*x*e^4)*e^(-5)

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maple [B]  time = 0.05, size = 558, normalized size = 2.17 \begin {gather*} \frac {B \,c^{2} x^{5}}{5 e}+\frac {A \,c^{2} x^{4}}{4 e}+\frac {B b c \,x^{4}}{2 e}-\frac {B \,c^{2} d \,x^{4}}{4 e^{2}}+\frac {2 A b c \,x^{3}}{3 e}-\frac {A \,c^{2} d \,x^{3}}{3 e^{2}}+\frac {2 B a c \,x^{3}}{3 e}+\frac {B \,b^{2} x^{3}}{3 e}-\frac {2 B b c d \,x^{3}}{3 e^{2}}+\frac {B \,c^{2} d^{2} x^{3}}{3 e^{3}}+\frac {A a c \,x^{2}}{e}+\frac {A \,b^{2} x^{2}}{2 e}-\frac {A b c d \,x^{2}}{e^{2}}+\frac {A \,c^{2} d^{2} x^{2}}{2 e^{3}}+\frac {B a b \,x^{2}}{e}-\frac {B a c d \,x^{2}}{e^{2}}-\frac {B \,b^{2} d \,x^{2}}{2 e^{2}}+\frac {B b c \,d^{2} x^{2}}{e^{3}}-\frac {B \,c^{2} d^{3} x^{2}}{2 e^{4}}+\frac {A \,a^{2} \ln \left (e x +d \right )}{e}-\frac {2 A a b d \ln \left (e x +d \right )}{e^{2}}+\frac {2 A a b x}{e}+\frac {2 A a c \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {2 A a c d x}{e^{2}}+\frac {A \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {A \,b^{2} d x}{e^{2}}-\frac {2 A b c \,d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {2 A b c \,d^{2} x}{e^{3}}+\frac {A \,c^{2} d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {A \,c^{2} d^{3} x}{e^{4}}-\frac {B \,a^{2} d \ln \left (e x +d \right )}{e^{2}}+\frac {B \,a^{2} x}{e}+\frac {2 B a b \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {2 B a b d x}{e^{2}}-\frac {2 B a c \,d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {2 B a c \,d^{2} x}{e^{3}}-\frac {B \,b^{2} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {B \,b^{2} d^{2} x}{e^{3}}+\frac {2 B b c \,d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {2 B b c \,d^{3} x}{e^{4}}-\frac {B \,c^{2} d^{5} \ln \left (e x +d \right )}{e^{6}}+\frac {B \,c^{2} d^{4} x}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.67, size = 378, normalized size = 1.47 \begin {gather*} \frac {12 \, B c^{2} e^{4} x^{5} - 15 \, {\left (B c^{2} d e^{3} - {\left (2 \, B b c + A c^{2}\right )} e^{4}\right )} x^{4} + 20 \, {\left (B c^{2} d^{2} e^{2} - {\left (2 \, B b c + A c^{2}\right )} d e^{3} + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} e^{4}\right )} x^{3} - 30 \, {\left (B c^{2} d^{3} e - {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{2} + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d e^{3} - {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} e^{4}\right )} x^{2} + 60 \, {\left (B c^{2} d^{4} - {\left (2 \, B b c + A c^{2}\right )} d^{3} e + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{2} e^{2} - {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d e^{3} + {\left (B a^{2} + 2 \, A a b\right )} e^{4}\right )} x}{60 \, e^{5}} - \frac {{\left (B c^{2} d^{5} - A a^{2} e^{5} - {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{3} e^{2} - {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d^{2} e^{3} + {\left (B a^{2} + 2 \, A a b\right )} d e^{4}\right )} \log \left (e x + d\right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*c^2*e^4*x^5 - 15*(B*c^2*d*e^3 - (2*B*b*c + A*c^2)*e^4)*x^4 + 20*(B*c^2*d^2*e^2 - (2*B*b*c + A*c^2)*
d*e^3 + (B*b^2 + 2*(B*a + A*b)*c)*e^4)*x^3 - 30*(B*c^2*d^3*e - (2*B*b*c + A*c^2)*d^2*e^2 + (B*b^2 + 2*(B*a + A
*b)*c)*d*e^3 - (2*B*a*b + A*b^2 + 2*A*a*c)*e^4)*x^2 + 60*(B*c^2*d^4 - (2*B*b*c + A*c^2)*d^3*e + (B*b^2 + 2*(B*
a + A*b)*c)*d^2*e^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d*e^3 + (B*a^2 + 2*A*a*b)*e^4)*x)/e^5 - (B*c^2*d^5 - A*a^2*e
^5 - (2*B*b*c + A*c^2)*d^4*e + (B*b^2 + 2*(B*a + A*b)*c)*d^3*e^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d^2*e^3 + (B*a^
2 + 2*A*a*b)*d*e^4)*log(e*x + d)/e^6

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mupad [B]  time = 0.08, size = 423, normalized size = 1.65 \begin {gather*} x^3\,\left (\frac {B\,b^2+2\,A\,c\,b+2\,B\,a\,c}{3\,e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{3\,e}\right )+x\,\left (\frac {B\,a^2+2\,A\,b\,a}{e}-\frac {d\,\left (\frac {A\,b^2+2\,B\,a\,b+2\,A\,a\,c}{e}-\frac {d\,\left (\frac {B\,b^2+2\,A\,c\,b+2\,B\,a\,c}{e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}\right )}{e}\right )}{e}\right )+x^4\,\left (\frac {A\,c^2+2\,B\,b\,c}{4\,e}-\frac {B\,c^2\,d}{4\,e^2}\right )+x^2\,\left (\frac {A\,b^2+2\,B\,a\,b+2\,A\,a\,c}{2\,e}-\frac {d\,\left (\frac {B\,b^2+2\,A\,c\,b+2\,B\,a\,c}{e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}\right )}{2\,e}\right )+\frac {\ln \left (d+e\,x\right )\,\left (-B\,a^2\,d\,e^4+A\,a^2\,e^5+2\,B\,a\,b\,d^2\,e^3-2\,A\,a\,b\,d\,e^4-2\,B\,a\,c\,d^3\,e^2+2\,A\,a\,c\,d^2\,e^3-B\,b^2\,d^3\,e^2+A\,b^2\,d^2\,e^3+2\,B\,b\,c\,d^4\,e-2\,A\,b\,c\,d^3\,e^2-B\,c^2\,d^5+A\,c^2\,d^4\,e\right )}{e^6}+\frac {B\,c^2\,x^5}{5\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.14, size = 372, normalized size = 1.45 \begin {gather*} \frac {B c^{2} x^{5}}{5 e} + x^{4} \left (\frac {A c^{2}}{4 e} + \frac {B b c}{2 e} - \frac {B c^{2} d}{4 e^{2}}\right ) + x^{3} \left (\frac {2 A b c}{3 e} - \frac {A c^{2} d}{3 e^{2}} + \frac {2 B a c}{3 e} + \frac {B b^{2}}{3 e} - \frac {2 B b c d}{3 e^{2}} + \frac {B c^{2} d^{2}}{3 e^{3}}\right ) + x^{2} \left (\frac {A a c}{e} + \frac {A b^{2}}{2 e} - \frac {A b c d}{e^{2}} + \frac {A c^{2} d^{2}}{2 e^{3}} + \frac {B a b}{e} - \frac {B a c d}{e^{2}} - \frac {B b^{2} d}{2 e^{2}} + \frac {B b c d^{2}}{e^{3}} - \frac {B c^{2} d^{3}}{2 e^{4}}\right ) + x \left (\frac {2 A a b}{e} - \frac {2 A a c d}{e^{2}} - \frac {A b^{2} d}{e^{2}} + \frac {2 A b c d^{2}}{e^{3}} - \frac {A c^{2} d^{3}}{e^{4}} + \frac {B a^{2}}{e} - \frac {2 B a b d}{e^{2}} + \frac {2 B a c d^{2}}{e^{3}} + \frac {B b^{2} d^{2}}{e^{3}} - \frac {2 B b c d^{3}}{e^{4}} + \frac {B c^{2} d^{4}}{e^{5}}\right ) - \frac {\left (- A e + B d\right ) \left (a e^{2} - b d e + c d^{2}\right )^{2} \log {\left (d + e x \right )}}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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