3.14.20 \(\int \frac {(b+2 c x) (a+b x+c x^2)^2}{(d+e x)^4} \, dx\)

Optimal. Leaf size=217 \[ \frac {(2 c d-b e) \left (-2 c e (5 b d-3 a e)+b^2 e^2+10 c^2 d^2\right )}{e^6 (d+e x)}-\frac {\left (a e^2-b d e+c d^2\right ) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6 (d+e x)^2}+\frac {4 c \log (d+e x) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6}+\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )^2}{3 e^6 (d+e x)^3}-\frac {c^2 x (8 c d-5 b e)}{e^5}+\frac {c^3 x^2}{e^4} \]

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Rubi [A]  time = 0.23, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {771} \begin {gather*} \frac {(2 c d-b e) \left (-2 c e (5 b d-3 a e)+b^2 e^2+10 c^2 d^2\right )}{e^6 (d+e x)}-\frac {\left (a e^2-b d e+c d^2\right ) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6 (d+e x)^2}+\frac {4 c \log (d+e x) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6}+\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )^2}{3 e^6 (d+e x)^3}-\frac {c^2 x (8 c d-5 b e)}{e^5}+\frac {c^3 x^2}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(a + b*x + c*x^2)^2)/(d + e*x)^4,x]

[Out]

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

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Mathematica [A]  time = 0.13, size = 300, normalized size = 1.38 \begin {gather*} \frac {-c e^2 \left (a^2 e^2 (d+3 e x)+6 a b e \left (d^2+3 d e x+3 e^2 x^2\right )-2 b^2 d \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )-b e^3 \left (a^2 e^2+a b e (d+3 e x)+b^2 \left (d^2+3 d e x+3 e^2 x^2\right )\right )+12 c (d+e x)^3 \log (d+e x) \left (c e (a e-5 b d)+b^2 e^2+5 c^2 d^2\right )+c^2 e \left (2 a d e \left (11 d^2+27 d e x+18 e^2 x^2\right )-5 b \left (13 d^4+27 d^3 e x+9 d^2 e^2 x^2-9 d e^3 x^3-3 e^4 x^4\right )\right )+c^3 \left (47 d^5+81 d^4 e x-9 d^3 e^2 x^2-63 d^2 e^3 x^3-15 d e^4 x^4+3 e^5 x^5\right )}{3 e^6 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(a + b*x + c*x^2)^2)/(d + e*x)^4,x]

[Out]

(c^3*(47*d^5 + 81*d^4*e*x - 9*d^3*e^2*x^2 - 63*d^2*e^3*x^3 - 15*d*e^4*x^4 + 3*e^5*x^5) - b*e^3*(a^2*e^2 + a*b*
e*(d + 3*e*x) + b^2*(d^2 + 3*d*e*x + 3*e^2*x^2)) - c*e^2*(a^2*e^2*(d + 3*e*x) + 6*a*b*e*(d^2 + 3*d*e*x + 3*e^2
*x^2) - 2*b^2*d*(11*d^2 + 27*d*e*x + 18*e^2*x^2)) + c^2*e*(2*a*d*e*(11*d^2 + 27*d*e*x + 18*e^2*x^2) - 5*b*(13*
d^4 + 27*d^3*e*x + 9*d^2*e^2*x^2 - 9*d*e^3*x^3 - 3*e^4*x^4)) + 12*c*(5*c^2*d^2 + b^2*e^2 + c*e*(-5*b*d + a*e))
*(d + e*x)^3*Log[d + e*x])/(3*e^6*(d + e*x)^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^2}{(d+e x)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((b + 2*c*x)*(a + b*x + c*x^2)^2)/(d + e*x)^4,x]

[Out]

IntegrateAlgebraic[((b + 2*c*x)*(a + b*x + c*x^2)^2)/(d + e*x)^4, x]

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fricas [B]  time = 0.41, size = 487, normalized size = 2.24 \begin {gather*} \frac {3 \, c^{3} e^{5} x^{5} + 47 \, c^{3} d^{5} - 65 \, b c^{2} d^{4} e - a^{2} b e^{5} + 22 \, {\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} - {\left (b^{3} + 6 \, a b c\right )} d^{2} e^{3} - {\left (a b^{2} + a^{2} c\right )} d e^{4} - 15 \, {\left (c^{3} d e^{4} - b c^{2} e^{5}\right )} x^{4} - 9 \, {\left (7 \, c^{3} d^{2} e^{3} - 5 \, b c^{2} d e^{4}\right )} x^{3} - 3 \, {\left (3 \, c^{3} d^{3} e^{2} + 15 \, b c^{2} d^{2} e^{3} - 12 \, {\left (b^{2} c + a c^{2}\right )} d e^{4} + {\left (b^{3} + 6 \, a b c\right )} e^{5}\right )} x^{2} + 3 \, {\left (27 \, c^{3} d^{4} e - 45 \, b c^{2} d^{3} e^{2} + 18 \, {\left (b^{2} c + a c^{2}\right )} d^{2} e^{3} - {\left (b^{3} + 6 \, a b c\right )} d e^{4} - {\left (a b^{2} + a^{2} c\right )} e^{5}\right )} x + 12 \, {\left (5 \, c^{3} d^{5} - 5 \, b c^{2} d^{4} e + {\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} + {\left (5 \, c^{3} d^{2} e^{3} - 5 \, b c^{2} d e^{4} + {\left (b^{2} c + a c^{2}\right )} e^{5}\right )} x^{3} + 3 \, {\left (5 \, c^{3} d^{3} e^{2} - 5 \, b c^{2} d^{2} e^{3} + {\left (b^{2} c + a c^{2}\right )} d e^{4}\right )} x^{2} + 3 \, {\left (5 \, c^{3} d^{4} e - 5 \, b c^{2} d^{3} e^{2} + {\left (b^{2} c + a c^{2}\right )} d^{2} e^{3}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x^2+b*x+a)^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/3*(3*c^3*e^5*x^5 + 47*c^3*d^5 - 65*b*c^2*d^4*e - a^2*b*e^5 + 22*(b^2*c + a*c^2)*d^3*e^2 - (b^3 + 6*a*b*c)*d^
2*e^3 - (a*b^2 + a^2*c)*d*e^4 - 15*(c^3*d*e^4 - b*c^2*e^5)*x^4 - 9*(7*c^3*d^2*e^3 - 5*b*c^2*d*e^4)*x^3 - 3*(3*
c^3*d^3*e^2 + 15*b*c^2*d^2*e^3 - 12*(b^2*c + a*c^2)*d*e^4 + (b^3 + 6*a*b*c)*e^5)*x^2 + 3*(27*c^3*d^4*e - 45*b*
c^2*d^3*e^2 + 18*(b^2*c + a*c^2)*d^2*e^3 - (b^3 + 6*a*b*c)*d*e^4 - (a*b^2 + a^2*c)*e^5)*x + 12*(5*c^3*d^5 - 5*
b*c^2*d^4*e + (b^2*c + a*c^2)*d^3*e^2 + (5*c^3*d^2*e^3 - 5*b*c^2*d*e^4 + (b^2*c + a*c^2)*e^5)*x^3 + 3*(5*c^3*d
^3*e^2 - 5*b*c^2*d^2*e^3 + (b^2*c + a*c^2)*d*e^4)*x^2 + 3*(5*c^3*d^4*e - 5*b*c^2*d^3*e^2 + (b^2*c + a*c^2)*d^2
*e^3)*x)*log(e*x + d))/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6)

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giac [A]  time = 0.16, size = 314, normalized size = 1.45 \begin {gather*} 4 \, {\left (5 \, c^{3} d^{2} - 5 \, b c^{2} d e + b^{2} c e^{2} + a c^{2} e^{2}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + {\left (c^{3} x^{2} e^{4} - 8 \, c^{3} d x e^{3} + 5 \, b c^{2} x e^{4}\right )} e^{\left (-8\right )} + \frac {{\left (47 \, c^{3} d^{5} - 65 \, b c^{2} d^{4} e + 22 \, b^{2} c d^{3} e^{2} + 22 \, a c^{2} d^{3} e^{2} - b^{3} d^{2} e^{3} - 6 \, a b c d^{2} e^{3} - a b^{2} d e^{4} - a^{2} c d e^{4} - a^{2} b e^{5} + 3 \, {\left (20 \, c^{3} d^{3} e^{2} - 30 \, b c^{2} d^{2} e^{3} + 12 \, b^{2} c d e^{4} + 12 \, a c^{2} d e^{4} - b^{3} e^{5} - 6 \, a b c e^{5}\right )} x^{2} + 3 \, {\left (35 \, c^{3} d^{4} e - 50 \, b c^{2} d^{3} e^{2} + 18 \, b^{2} c d^{2} e^{3} + 18 \, a c^{2} d^{2} e^{3} - b^{3} d e^{4} - 6 \, a b c d e^{4} - a b^{2} e^{5} - a^{2} c e^{5}\right )} x\right )} e^{\left (-6\right )}}{3 \, {\left (x e + d\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x^2+b*x+a)^2/(e*x+d)^4,x, algorithm="giac")

[Out]

4*(5*c^3*d^2 - 5*b*c^2*d*e + b^2*c*e^2 + a*c^2*e^2)*e^(-6)*log(abs(x*e + d)) + (c^3*x^2*e^4 - 8*c^3*d*x*e^3 +
5*b*c^2*x*e^4)*e^(-8) + 1/3*(47*c^3*d^5 - 65*b*c^2*d^4*e + 22*b^2*c*d^3*e^2 + 22*a*c^2*d^3*e^2 - b^3*d^2*e^3 -
6*a*b*c*d^2*e^3 - a*b^2*d*e^4 - a^2*c*d*e^4 - a^2*b*e^5 + 3*(20*c^3*d^3*e^2 - 30*b*c^2*d^2*e^3 + 12*b^2*c*d*e
^4 + 12*a*c^2*d*e^4 - b^3*e^5 - 6*a*b*c*e^5)*x^2 + 3*(35*c^3*d^4*e - 50*b*c^2*d^3*e^2 + 18*b^2*c*d^2*e^3 + 18*
a*c^2*d^2*e^3 - b^3*d*e^4 - 6*a*b*c*d*e^4 - a*b^2*e^5 - a^2*c*e^5)*x)*e^(-6)/(x*e + d)^3

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maple [B]  time = 0.05, size = 495, normalized size = 2.28 \begin {gather*} -\frac {a^{2} b}{3 \left (e x +d \right )^{3} e}+\frac {2 a^{2} c d}{3 \left (e x +d \right )^{3} e^{2}}+\frac {2 a \,b^{2} d}{3 \left (e x +d \right )^{3} e^{2}}-\frac {2 a b c \,d^{2}}{\left (e x +d \right )^{3} e^{3}}+\frac {4 a \,c^{2} d^{3}}{3 \left (e x +d \right )^{3} e^{4}}-\frac {b^{3} d^{2}}{3 \left (e x +d \right )^{3} e^{3}}+\frac {4 b^{2} c \,d^{3}}{3 \left (e x +d \right )^{3} e^{4}}-\frac {5 b \,c^{2} d^{4}}{3 \left (e x +d \right )^{3} e^{5}}+\frac {2 c^{3} d^{5}}{3 \left (e x +d \right )^{3} e^{6}}-\frac {a^{2} c}{\left (e x +d \right )^{2} e^{2}}-\frac {a \,b^{2}}{\left (e x +d \right )^{2} e^{2}}+\frac {6 a b c d}{\left (e x +d \right )^{2} e^{3}}-\frac {6 a \,c^{2} d^{2}}{\left (e x +d \right )^{2} e^{4}}+\frac {b^{3} d}{\left (e x +d \right )^{2} e^{3}}-\frac {6 b^{2} c \,d^{2}}{\left (e x +d \right )^{2} e^{4}}+\frac {10 b \,c^{2} d^{3}}{\left (e x +d \right )^{2} e^{5}}-\frac {5 c^{3} d^{4}}{\left (e x +d \right )^{2} e^{6}}+\frac {c^{3} x^{2}}{e^{4}}-\frac {6 a b c}{\left (e x +d \right ) e^{3}}+\frac {12 a \,c^{2} d}{\left (e x +d \right ) e^{4}}+\frac {4 a \,c^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {b^{3}}{\left (e x +d \right ) e^{3}}+\frac {12 b^{2} c d}{\left (e x +d \right ) e^{4}}+\frac {4 b^{2} c \ln \left (e x +d \right )}{e^{4}}-\frac {30 b \,c^{2} d^{2}}{\left (e x +d \right ) e^{5}}-\frac {20 b \,c^{2} d \ln \left (e x +d \right )}{e^{5}}+\frac {5 b \,c^{2} x}{e^{4}}+\frac {20 c^{3} d^{3}}{\left (e x +d \right ) e^{6}}+\frac {20 c^{3} d^{2} \ln \left (e x +d \right )}{e^{6}}-\frac {8 c^{3} d x}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(c*x^2+b*x+a)^2/(e*x+d)^4,x)

[Out]

6/e^3/(e*x+d)^2*a*b*c*d-2/e^3/(e*x+d)^3*d^2*a*b*c-1/e^3/(e*x+d)*b^3-1/3/e/(e*x+d)^3*a^2*b-1/3/e^3/(e*x+d)^3*d^
2*b^3+2/3/e^6/(e*x+d)^3*c^3*d^5+20/e^6/(e*x+d)*c^3*d^3+5*c^2/e^4*b*x-8*c^3/e^5*x*d+4*c^2/e^4*ln(e*x+d)*a+4*c/e
^4*ln(e*x+d)*b^2+20*c^3/e^6*ln(e*x+d)*d^2-1/e^2/(e*x+d)^2*a*b^2+1/e^3/(e*x+d)^2*b^3*d-5/e^6/(e*x+d)^2*c^3*d^4-
1/e^2/(e*x+d)^2*a^2*c-6/e^4/(e*x+d)^2*a*c^2*d^2-6/e^4/(e*x+d)^2*b^2*c*d^2-30/e^5/(e*x+d)*b*c^2*d^2-5/3/e^5/(e*
x+d)^3*b*c^2*d^4+10/e^5/(e*x+d)^2*b*c^2*d^3+2/3/e^2/(e*x+d)^3*a^2*c*d+2/3/e^2/(e*x+d)^3*d*a*b^2+4/3/e^4/(e*x+d
)^3*a*c^2*d^3+4/3/e^4/(e*x+d)^3*d^3*b^2*c-20*c^2/e^5*ln(e*x+d)*b*d-6/e^3/(e*x+d)*a*b*c+12/e^4/(e*x+d)*c^2*a*d+
12/e^4/(e*x+d)*b^2*c*d+c^3*x^2/e^4

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maxima [A]  time = 0.63, size = 327, normalized size = 1.51 \begin {gather*} \frac {47 \, c^{3} d^{5} - 65 \, b c^{2} d^{4} e - a^{2} b e^{5} + 22 \, {\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} - {\left (b^{3} + 6 \, a b c\right )} d^{2} e^{3} - {\left (a b^{2} + a^{2} c\right )} d e^{4} + 3 \, {\left (20 \, c^{3} d^{3} e^{2} - 30 \, b c^{2} d^{2} e^{3} + 12 \, {\left (b^{2} c + a c^{2}\right )} d e^{4} - {\left (b^{3} + 6 \, a b c\right )} e^{5}\right )} x^{2} + 3 \, {\left (35 \, c^{3} d^{4} e - 50 \, b c^{2} d^{3} e^{2} + 18 \, {\left (b^{2} c + a c^{2}\right )} d^{2} e^{3} - {\left (b^{3} + 6 \, a b c\right )} d e^{4} - {\left (a b^{2} + a^{2} c\right )} e^{5}\right )} x}{3 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} + \frac {c^{3} e x^{2} - {\left (8 \, c^{3} d - 5 \, b c^{2} e\right )} x}{e^{5}} + \frac {4 \, {\left (5 \, c^{3} d^{2} - 5 \, b c^{2} d e + {\left (b^{2} c + a c^{2}\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x^2+b*x+a)^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/3*(47*c^3*d^5 - 65*b*c^2*d^4*e - a^2*b*e^5 + 22*(b^2*c + a*c^2)*d^3*e^2 - (b^3 + 6*a*b*c)*d^2*e^3 - (a*b^2 +
a^2*c)*d*e^4 + 3*(20*c^3*d^3*e^2 - 30*b*c^2*d^2*e^3 + 12*(b^2*c + a*c^2)*d*e^4 - (b^3 + 6*a*b*c)*e^5)*x^2 + 3
*(35*c^3*d^4*e - 50*b*c^2*d^3*e^2 + 18*(b^2*c + a*c^2)*d^2*e^3 - (b^3 + 6*a*b*c)*d*e^4 - (a*b^2 + a^2*c)*e^5)*
x)/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6) + (c^3*e*x^2 - (8*c^3*d - 5*b*c^2*e)*x)/e^5 + 4*(5*c^3*d^2
- 5*b*c^2*d*e + (b^2*c + a*c^2)*e^2)*log(e*x + d)/e^6

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mupad [B]  time = 1.86, size = 349, normalized size = 1.61 \begin {gather*} x\,\left (\frac {5\,b\,c^2}{e^4}-\frac {8\,c^3\,d}{e^5}\right )-\frac {x\,\left (a^2\,c\,e^4+a\,b^2\,e^4+6\,a\,b\,c\,d\,e^3-18\,a\,c^2\,d^2\,e^2+b^3\,d\,e^3-18\,b^2\,c\,d^2\,e^2+50\,b\,c^2\,d^3\,e-35\,c^3\,d^4\right )+x^2\,\left (b^3\,e^4-12\,b^2\,c\,d\,e^3+30\,b\,c^2\,d^2\,e^2+6\,a\,b\,c\,e^4-20\,c^3\,d^3\,e-12\,a\,c^2\,d\,e^3\right )+\frac {a^2\,b\,e^5+a^2\,c\,d\,e^4+a\,b^2\,d\,e^4+6\,a\,b\,c\,d^2\,e^3-22\,a\,c^2\,d^3\,e^2+b^3\,d^2\,e^3-22\,b^2\,c\,d^3\,e^2+65\,b\,c^2\,d^4\,e-47\,c^3\,d^5}{3\,e}}{d^3\,e^5+3\,d^2\,e^6\,x+3\,d\,e^7\,x^2+e^8\,x^3}+\frac {\ln \left (d+e\,x\right )\,\left (4\,b^2\,c\,e^2-20\,b\,c^2\,d\,e+20\,c^3\,d^2+4\,a\,c^2\,e^2\right )}{e^6}+\frac {c^3\,x^2}{e^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b + 2*c*x)*(a + b*x + c*x^2)^2)/(d + e*x)^4,x)

[Out]

x*((5*b*c^2)/e^4 - (8*c^3*d)/e^5) - (x*(a*b^2*e^4 - 35*c^3*d^4 + a^2*c*e^4 + b^3*d*e^3 - 18*a*c^2*d^2*e^2 - 18
*b^2*c*d^2*e^2 + 50*b*c^2*d^3*e + 6*a*b*c*d*e^3) + x^2*(b^3*e^4 - 20*c^3*d^3*e + 30*b*c^2*d^2*e^2 + 6*a*b*c*e^
4 - 12*a*c^2*d*e^3 - 12*b^2*c*d*e^3) + (a^2*b*e^5 - 47*c^3*d^5 + b^3*d^2*e^3 - 22*a*c^2*d^3*e^2 - 22*b^2*c*d^3
*e^2 + a*b^2*d*e^4 + a^2*c*d*e^4 + 65*b*c^2*d^4*e + 6*a*b*c*d^2*e^3)/(3*e))/(d^3*e^5 + e^8*x^3 + 3*d^2*e^6*x +
3*d*e^7*x^2) + (log(d + e*x)*(20*c^3*d^2 + 4*a*c^2*e^2 + 4*b^2*c*e^2 - 20*b*c^2*d*e))/e^6 + (c^3*x^2)/e^4

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sympy [A]  time = 21.03, size = 379, normalized size = 1.75 \begin {gather*} \frac {c^{3} x^{2}}{e^{4}} + \frac {4 c \left (a c e^{2} + b^{2} e^{2} - 5 b c d e + 5 c^{2} d^{2}\right ) \log {\left (d + e x \right )}}{e^{6}} + x \left (\frac {5 b c^{2}}{e^{4}} - \frac {8 c^{3} d}{e^{5}}\right ) + \frac {- a^{2} b e^{5} - a^{2} c d e^{4} - a b^{2} d e^{4} - 6 a b c d^{2} e^{3} + 22 a c^{2} d^{3} e^{2} - b^{3} d^{2} e^{3} + 22 b^{2} c d^{3} e^{2} - 65 b c^{2} d^{4} e + 47 c^{3} d^{5} + x^{2} \left (- 18 a b c e^{5} + 36 a c^{2} d e^{4} - 3 b^{3} e^{5} + 36 b^{2} c d e^{4} - 90 b c^{2} d^{2} e^{3} + 60 c^{3} d^{3} e^{2}\right ) + x \left (- 3 a^{2} c e^{5} - 3 a b^{2} e^{5} - 18 a b c d e^{4} + 54 a c^{2} d^{2} e^{3} - 3 b^{3} d e^{4} + 54 b^{2} c d^{2} e^{3} - 150 b c^{2} d^{3} e^{2} + 105 c^{3} d^{4} e\right )}{3 d^{3} e^{6} + 9 d^{2} e^{7} x + 9 d e^{8} x^{2} + 3 e^{9} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x**2+b*x+a)**2/(e*x+d)**4,x)

[Out]

c**3*x**2/e**4 + 4*c*(a*c*e**2 + b**2*e**2 - 5*b*c*d*e + 5*c**2*d**2)*log(d + e*x)/e**6 + x*(5*b*c**2/e**4 - 8
*c**3*d/e**5) + (-a**2*b*e**5 - a**2*c*d*e**4 - a*b**2*d*e**4 - 6*a*b*c*d**2*e**3 + 22*a*c**2*d**3*e**2 - b**3
*d**2*e**3 + 22*b**2*c*d**3*e**2 - 65*b*c**2*d**4*e + 47*c**3*d**5 + x**2*(-18*a*b*c*e**5 + 36*a*c**2*d*e**4 -
3*b**3*e**5 + 36*b**2*c*d*e**4 - 90*b*c**2*d**2*e**3 + 60*c**3*d**3*e**2) + x*(-3*a**2*c*e**5 - 3*a*b**2*e**5
- 18*a*b*c*d*e**4 + 54*a*c**2*d**2*e**3 - 3*b**3*d*e**4 + 54*b**2*c*d**2*e**3 - 150*b*c**2*d**3*e**2 + 105*c*
*3*d**4*e))/(3*d**3*e**6 + 9*d**2*e**7*x + 9*d*e**8*x**2 + 3*e**9*x**3)

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