3.10.79 \(\int \frac {(A+B x) (b x+c x^2)^2}{(d+e x)^6} \, dx\)

Optimal. Leaf size=248 \[ \frac {2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )}{2 e^6 (d+e x)^2}-\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{3 e^6 (d+e x)^3}+\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}+\frac {c (-A c e-2 b B e+5 B c d)}{e^6 (d+e x)}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}+\frac {B c^2 \log (d+e x)}{e^6} \]

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Rubi [A]  time = 0.24, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {771} \begin {gather*} \frac {2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )}{2 e^6 (d+e x)^2}-\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{3 e^6 (d+e x)^3}+\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}+\frac {c (-A c e-2 b B e+5 B c d)}{e^6 (d+e x)}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}+\frac {B c^2 \log (d+e x)}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6,x]

[Out]

(d^2*(B*d - A*e)*(c*d - b*e)^2)/(5*e^6*(d + e*x)^5) - (d*(c*d - b*e)*(B*d*(5*c*d - 3*b*e) - 2*A*e*(2*c*d - b*e
)))/(4*e^6*(d + e*x)^4) - (A*e*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2) - B*d*(10*c^2*d^2 - 12*b*c*d*e + 3*b^2*e^2))/
(3*e^6*(d + e*x)^3) + (2*A*c*e*(2*c*d - b*e) - B*(10*c^2*d^2 - 8*b*c*d*e + b^2*e^2))/(2*e^6*(d + e*x)^2) + (c*
(5*B*c*d - 2*b*B*e - A*c*e))/(e^6*(d + e*x)) + (B*c^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 (b x+c x^2\right )^2}{(d+e x)^6} \, dx &=\int \left (-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)^6}+\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^5 (d+e x)^5}+\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{e^5 (d+e x)^4}+\frac {-2 A c e (2 c d-b e)+B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{e^5 (d+e x)^3}+\frac {c (-5 B c d+2 b B e+A c e)}{e^5 (d+e x)^2}+\frac {B c^2}{e^5 (d+e x)}\right ) \, dx\\ &=\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}-\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{3 e^6 (d+e x)^3}+\frac {2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{2 e^6 (d+e x)^2}+\frac {c (5 B c d-2 b B e-A c e)}{e^6 (d+e x)}+\frac {B c^2 \log (d+e x)}{e^6}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 269, normalized size = 1.08 \begin {gather*} \frac {-2 A e \left (b^2 e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+3 b c e \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+6 c^2 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )\right )+B \left (-3 b^2 e^2 \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )-24 b c e \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )+c^2 d \left (137 d^4+625 d^3 e x+1100 d^2 e^2 x^2+900 d e^3 x^3+300 e^4 x^4\right )\right )+60 B c^2 (d+e x)^5 \log (d+e x)}{60 e^6 (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6,x]

[Out]

(-2*A*e*(b^2*e^2*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*b*c*e*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + 6*c^2*
(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 + 5*e^4*x^4)) + B*(-3*b^2*e^2*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2
+ 10*e^3*x^3) - 24*b*c*e*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 + 5*e^4*x^4) + c^2*d*(137*d^4 + 625
*d^3*e*x + 1100*d^2*e^2*x^2 + 900*d*e^3*x^3 + 300*e^4*x^4)) + 60*B*c^2*(d + e*x)^5*Log[d + e*x])/(60*e^6*(d +
e*x)^5)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^6, x]

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fricas [A]  time = 0.41, size = 407, normalized size = 1.64 \begin {gather*} \frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x + 60 \, {\left (B c^{2} e^{5} x^{5} + 5 \, B c^{2} d e^{4} x^{4} + 10 \, B c^{2} d^{2} e^{3} x^{3} + 10 \, B c^{2} d^{3} e^{2} x^{2} + 5 \, B c^{2} d^{4} e x + B c^{2} d^{5}\right )} \log \left (e x + d\right )}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="fricas")

[Out]

1/60*(137*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 12*(2*B*b*c + A*c^2)*d^4*e - 3*(B*b^2 + 2*A*b*c)*d^3*e^2 + 60*(5*B*c^2
*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 30*(30*B*c^2*d^2*e^3 - 4*(2*B*b*c + A*c^2)*d*e^4 - (B*b^2 + 2*A*b*c)*e^5
)*x^3 + 10*(110*B*c^2*d^3*e^2 - 2*A*b^2*e^5 - 12*(2*B*b*c + A*c^2)*d^2*e^3 - 3*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 +
5*(125*B*c^2*d^4*e - 2*A*b^2*d*e^4 - 12*(2*B*b*c + A*c^2)*d^3*e^2 - 3*(B*b^2 + 2*A*b*c)*d^2*e^3)*x + 60*(B*c^2
*e^5*x^5 + 5*B*c^2*d*e^4*x^4 + 10*B*c^2*d^2*e^3*x^3 + 10*B*c^2*d^3*e^2*x^2 + 5*B*c^2*d^4*e*x + B*c^2*d^5)*log(
e*x + d))/(e^11*x^5 + 5*d*e^10*x^4 + 10*d^2*e^9*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5*e^6)

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giac [A]  time = 0.16, size = 301, normalized size = 1.21 \begin {gather*} B c^{2} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {{\left (60 \, {\left (5 \, B c^{2} d e^{3} - 2 \, B b c e^{4} - A c^{2} e^{4}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{2} - 8 \, B b c d e^{3} - 4 \, A c^{2} d e^{3} - B b^{2} e^{4} - 2 \, A b c e^{4}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e - 24 \, B b c d^{2} e^{2} - 12 \, A c^{2} d^{2} e^{2} - 3 \, B b^{2} d e^{3} - 6 \, A b c d e^{3} - 2 \, A b^{2} e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} - 24 \, B b c d^{3} e - 12 \, A c^{2} d^{3} e - 3 \, B b^{2} d^{2} e^{2} - 6 \, A b c d^{2} e^{2} - 2 \, A b^{2} d e^{3}\right )} x + {\left (137 \, B c^{2} d^{5} - 24 \, B b c d^{4} e - 12 \, A c^{2} d^{4} e - 3 \, B b^{2} d^{3} e^{2} - 6 \, A b c d^{3} e^{2} - 2 \, A b^{2} d^{2} e^{3}\right )} e^{\left (-1\right )}\right )} e^{\left (-5\right )}}{60 \, {\left (x e + d\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="giac")

[Out]

B*c^2*e^(-6)*log(abs(x*e + d)) + 1/60*(60*(5*B*c^2*d*e^3 - 2*B*b*c*e^4 - A*c^2*e^4)*x^4 + 30*(30*B*c^2*d^2*e^2
- 8*B*b*c*d*e^3 - 4*A*c^2*d*e^3 - B*b^2*e^4 - 2*A*b*c*e^4)*x^3 + 10*(110*B*c^2*d^3*e - 24*B*b*c*d^2*e^2 - 12*
A*c^2*d^2*e^2 - 3*B*b^2*d*e^3 - 6*A*b*c*d*e^3 - 2*A*b^2*e^4)*x^2 + 5*(125*B*c^2*d^4 - 24*B*b*c*d^3*e - 12*A*c^
2*d^3*e - 3*B*b^2*d^2*e^2 - 6*A*b*c*d^2*e^2 - 2*A*b^2*d*e^3)*x + (137*B*c^2*d^5 - 24*B*b*c*d^4*e - 12*A*c^2*d^
4*e - 3*B*b^2*d^3*e^2 - 6*A*b*c*d^3*e^2 - 2*A*b^2*d^2*e^3)*e^(-1))*e^(-5)/(x*e + d)^5

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maple [A]  time = 0.06, size = 472, normalized size = 1.90 \begin {gather*} -\frac {A \,b^{2} d^{2}}{5 \left (e x +d \right )^{5} e^{3}}+\frac {2 A b c \,d^{3}}{5 \left (e x +d \right )^{5} e^{4}}-\frac {A \,c^{2} d^{4}}{5 \left (e x +d \right )^{5} e^{5}}+\frac {B \,b^{2} d^{3}}{5 \left (e x +d \right )^{5} e^{4}}-\frac {2 B b c \,d^{4}}{5 \left (e x +d \right )^{5} e^{5}}+\frac {B \,c^{2} d^{5}}{5 \left (e x +d \right )^{5} e^{6}}+\frac {A \,b^{2} d}{2 \left (e x +d \right )^{4} e^{3}}-\frac {3 A b c \,d^{2}}{2 \left (e x +d \right )^{4} e^{4}}+\frac {A \,c^{2} d^{3}}{\left (e x +d \right )^{4} e^{5}}-\frac {3 B \,b^{2} d^{2}}{4 \left (e x +d \right )^{4} e^{4}}+\frac {2 B b c \,d^{3}}{\left (e x +d \right )^{4} e^{5}}-\frac {5 B \,c^{2} d^{4}}{4 \left (e x +d \right )^{4} e^{6}}-\frac {A \,b^{2}}{3 \left (e x +d \right )^{3} e^{3}}+\frac {2 A b c d}{\left (e x +d \right )^{3} e^{4}}-\frac {2 A \,c^{2} d^{2}}{\left (e x +d \right )^{3} e^{5}}+\frac {B \,b^{2} d}{\left (e x +d \right )^{3} e^{4}}-\frac {4 B b c \,d^{2}}{\left (e x +d \right )^{3} e^{5}}+\frac {10 B \,c^{2} d^{3}}{3 \left (e x +d \right )^{3} e^{6}}-\frac {A b c}{\left (e x +d \right )^{2} e^{4}}+\frac {2 A \,c^{2} d}{\left (e x +d \right )^{2} e^{5}}-\frac {B \,b^{2}}{2 \left (e x +d \right )^{2} e^{4}}+\frac {4 B b c d}{\left (e x +d \right )^{2} e^{5}}-\frac {5 B \,c^{2} d^{2}}{\left (e x +d \right )^{2} e^{6}}-\frac {A \,c^{2}}{\left (e x +d \right ) e^{5}}-\frac {2 B b c}{\left (e x +d \right ) e^{5}}+\frac {5 B \,c^{2} d}{\left (e x +d \right ) e^{6}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^6,x)

[Out]

-1/3/e^3/(e*x+d)^3*A*b^2-c^2/e^5/(e*x+d)*A-2/5*d^4/e^5/(e*x+d)^5*B*b*c+2*d^3/e^5/(e*x+d)^4*B*b*c-3/2*d^2/e^4/(
e*x+d)^4*A*b*c-4/e^5/(e*x+d)^3*B*d^2*b*c+2/e^4/(e*x+d)^3*A*b*c*d+4/e^5/(e*x+d)^2*B*d*b*c+2/5*d^3/e^4/(e*x+d)^5
*A*b*c-1/5*d^4/e^5/(e*x+d)^5*A*c^2+1/5*d^3/e^4/(e*x+d)^5*B*b^2-5/4*d^4/e^6/(e*x+d)^4*B*c^2-1/5*d^2/e^3/(e*x+d)
^5*A*b^2+d^3/e^5/(e*x+d)^4*A*c^2-3/4*d^2/e^4/(e*x+d)^4*B*b^2+5*c^2/e^6/(e*x+d)*B*d+1/2*d/e^3/(e*x+d)^4*A*b^2+1
/5*d^5/e^6/(e*x+d)^5*B*c^2+10/3/e^6/(e*x+d)^3*B*c^2*d^3-1/e^4/(e*x+d)^2*A*b*c+2/e^5/(e*x+d)^2*A*c^2*d-5/e^6/(e
*x+d)^2*B*d^2*c^2-2*c/e^5/(e*x+d)*B*b-2/e^5/(e*x+d)^3*A*c^2*d^2+1/e^4/(e*x+d)^3*B*d*b^2+B*c^2*ln(e*x+d)/e^6-1/
2*b^2*B/e^4/(e*x+d)^2

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maxima [A]  time = 0.67, size = 340, normalized size = 1.37 \begin {gather*} \frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} + \frac {B c^{2} \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)^2/(e*x+d)^6,x, algorithm="maxima")

[Out]

1/60*(137*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 12*(2*B*b*c + A*c^2)*d^4*e - 3*(B*b^2 + 2*A*b*c)*d^3*e^2 + 60*(5*B*c^2
*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 30*(30*B*c^2*d^2*e^3 - 4*(2*B*b*c + A*c^2)*d*e^4 - (B*b^2 + 2*A*b*c)*e^5
)*x^3 + 10*(110*B*c^2*d^3*e^2 - 2*A*b^2*e^5 - 12*(2*B*b*c + A*c^2)*d^2*e^3 - 3*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 +
5*(125*B*c^2*d^4*e - 2*A*b^2*d*e^4 - 12*(2*B*b*c + A*c^2)*d^3*e^2 - 3*(B*b^2 + 2*A*b*c)*d^2*e^3)*x)/(e^11*x^5
+ 5*d*e^10*x^4 + 10*d^2*e^9*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5*e^6) + B*c^2*log(e*x + d)/e^6

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mupad [B]  time = 1.46, size = 343, normalized size = 1.38 \begin {gather*} \frac {B\,c^2\,\ln \left (d+e\,x\right )}{e^6}-\frac {\frac {3\,B\,b^2\,d^3\,e^2+2\,A\,b^2\,d^2\,e^3+24\,B\,b\,c\,d^4\,e+6\,A\,b\,c\,d^3\,e^2-137\,B\,c^2\,d^5+12\,A\,c^2\,d^4\,e}{60\,e^6}+\frac {x^3\,\left (B\,b^2\,e^2+8\,B\,b\,c\,d\,e+2\,A\,b\,c\,e^2-30\,B\,c^2\,d^2+4\,A\,c^2\,d\,e\right )}{2\,e^3}+\frac {x\,\left (3\,B\,b^2\,d^2\,e^2+2\,A\,b^2\,d\,e^3+24\,B\,b\,c\,d^3\,e+6\,A\,b\,c\,d^2\,e^2-125\,B\,c^2\,d^4+12\,A\,c^2\,d^3\,e\right )}{12\,e^5}+\frac {x^2\,\left (3\,B\,b^2\,d\,e^2+2\,A\,b^2\,e^3+24\,B\,b\,c\,d^2\,e+6\,A\,b\,c\,d\,e^2-110\,B\,c^2\,d^3+12\,A\,c^2\,d^2\,e\right )}{6\,e^4}+\frac {c\,x^4\,\left (A\,c\,e+2\,B\,b\,e-5\,B\,c\,d\right )}{e^2}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^2*(A + B*x))/(d + e*x)^6,x)

[Out]

(B*c^2*log(d + e*x))/e^6 - ((12*A*c^2*d^4*e - 137*B*c^2*d^5 + 2*A*b^2*d^2*e^3 + 3*B*b^2*d^3*e^2 + 24*B*b*c*d^4
*e + 6*A*b*c*d^3*e^2)/(60*e^6) + (x^3*(B*b^2*e^2 - 30*B*c^2*d^2 + 2*A*b*c*e^2 + 4*A*c^2*d*e + 8*B*b*c*d*e))/(2
*e^3) + (x*(2*A*b^2*d*e^3 - 125*B*c^2*d^4 + 12*A*c^2*d^3*e + 3*B*b^2*d^2*e^2 + 24*B*b*c*d^3*e + 6*A*b*c*d^2*e^
2))/(12*e^5) + (x^2*(2*A*b^2*e^3 - 110*B*c^2*d^3 + 12*A*c^2*d^2*e + 3*B*b^2*d*e^2 + 6*A*b*c*d*e^2 + 24*B*b*c*d
^2*e))/(6*e^4) + (c*x^4*(A*c*e + 2*B*b*e - 5*B*c*d))/e^2)/(d^5 + e^5*x^5 + 5*d*e^4*x^4 + 10*d^3*e^2*x^2 + 10*d
^2*e^3*x^3 + 5*d^4*e*x)

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sympy [A]  time = 94.46, size = 405, normalized size = 1.63 \begin {gather*} \frac {B c^{2} \log {\left (d + e x \right )}}{e^{6}} + \frac {- 2 A b^{2} d^{2} e^{3} - 6 A b c d^{3} e^{2} - 12 A c^{2} d^{4} e - 3 B b^{2} d^{3} e^{2} - 24 B b c d^{4} e + 137 B c^{2} d^{5} + x^{4} \left (- 60 A c^{2} e^{5} - 120 B b c e^{5} + 300 B c^{2} d e^{4}\right ) + x^{3} \left (- 60 A b c e^{5} - 120 A c^{2} d e^{4} - 30 B b^{2} e^{5} - 240 B b c d e^{4} + 900 B c^{2} d^{2} e^{3}\right ) + x^{2} \left (- 20 A b^{2} e^{5} - 60 A b c d e^{4} - 120 A c^{2} d^{2} e^{3} - 30 B b^{2} d e^{4} - 240 B b c d^{2} e^{3} + 1100 B c^{2} d^{3} e^{2}\right ) + x \left (- 10 A b^{2} d e^{4} - 30 A b c d^{2} e^{3} - 60 A c^{2} d^{3} e^{2} - 15 B b^{2} d^{2} e^{3} - 120 B b c d^{3} e^{2} + 625 B c^{2} d^{4} e\right )}{60 d^{5} e^{6} + 300 d^{4} e^{7} x + 600 d^{3} e^{8} x^{2} + 600 d^{2} e^{9} x^{3} + 300 d e^{10} x^{4} + 60 e^{11} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**2/(e*x+d)**6,x)

[Out]

B*c**2*log(d + e*x)/e**6 + (-2*A*b**2*d**2*e**3 - 6*A*b*c*d**3*e**2 - 12*A*c**2*d**4*e - 3*B*b**2*d**3*e**2 -
24*B*b*c*d**4*e + 137*B*c**2*d**5 + x**4*(-60*A*c**2*e**5 - 120*B*b*c*e**5 + 300*B*c**2*d*e**4) + x**3*(-60*A*
b*c*e**5 - 120*A*c**2*d*e**4 - 30*B*b**2*e**5 - 240*B*b*c*d*e**4 + 900*B*c**2*d**2*e**3) + x**2*(-20*A*b**2*e*
*5 - 60*A*b*c*d*e**4 - 120*A*c**2*d**2*e**3 - 30*B*b**2*d*e**4 - 240*B*b*c*d**2*e**3 + 1100*B*c**2*d**3*e**2)
+ x*(-10*A*b**2*d*e**4 - 30*A*b*c*d**2*e**3 - 60*A*c**2*d**3*e**2 - 15*B*b**2*d**2*e**3 - 120*B*b*c*d**3*e**2
+ 625*B*c**2*d**4*e))/(60*d**5*e**6 + 300*d**4*e**7*x + 600*d**3*e**8*x**2 + 600*d**2*e**9*x**3 + 300*d*e**10*
x**4 + 60*e**11*x**5)

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