3.10.77 \(\int \frac {(A+B x) (b x+c x^2)^2}{(d+e x)^4} \, dx\)

Optimal. Leaf size=238 \[ -\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{e^6 (d+e x)}-\frac {\log (d+e x) \left (2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )\right )}{e^6}+\frac {d^2 (B d-A e) (c d-b e)^2}{3 e^6 (d+e x)^3}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{2 e^6 (d+e x)^2}-\frac {c x (-A c e-2 b B e+4 B c d)}{e^5}+\frac {B c^2 x^2}{2 e^4} \]

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Rubi [A]  time = 0.30, antiderivative size = 238, 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 {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 )}{e^6 (d+e x)}-\frac {\log (d+e x) \left (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 )\right )}{e^6}+\frac {d^2 (B d-A e) (c d-b e)^2}{3 e^6 (d+e x)^3}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{2 e^6 (d+e x)^2}-\frac {c x (-A c e-2 b B e+4 B c d)}{e^5}+\frac {B c^2 x^2}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.16, size = 220, normalized size = 0.92 \begin {gather*} \frac {-\frac {6 \left (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 )\right )}{d+e x}+6 \log (d+e x) \left (2 A c e (b e-2 c d)+B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )\right )+\frac {2 d^2 (B d-A e) (c d-b e)^2}{(d+e x)^3}-\frac {3 d (c d-b e) (2 A e (b e-2 c d)+B d (5 c d-3 b e))}{(d+e x)^2}+6 c e x (A c e+2 b B e-4 B c d)+3 B c^2 e^2 x^2}{6 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*c*e*(-4*B*c*d + 2*b*B*e + A*c*e)*x + 3*B*c^2*e^2*x^2 + (2*d^2*(B*d - A*e)*(c*d - b*e)^2)/(d + e*x)^3 - (3*d
*(c*d - b*e)*(B*d*(5*c*d - 3*b*e) + 2*A*e*(-2*c*d + b*e)))/(d + e*x)^2 - (6*(B*d*(-10*c^2*d^2 + 12*b*c*d*e - 3
*b^2*e^2) + A*e*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2)))/(d + e*x) + 6*(2*A*c*e*(-2*c*d + b*e) + B*(10*c^2*d^2 - 8*
b*c*d*e + b^2*e^2))*Log[d + e*x])/(6*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 (b x+c x^2\right )^2}{(d+e x)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.39, size = 505, normalized size = 2.12 \begin {gather*} \frac {3 \, B c^{2} e^{5} x^{5} + 47 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 26 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + 11 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} - 3 \, {\left (5 \, B c^{2} d e^{4} - 2 \, {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} - 9 \, {\left (7 \, B c^{2} d^{2} e^{3} - 2 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4}\right )} x^{3} - 3 \, {\left (3 \, B c^{2} d^{3} e^{2} + 2 \, A b^{2} e^{5} + 6 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 6 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 3 \, {\left (27 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 18 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 9 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x + 6 \, {\left (10 \, B c^{2} d^{5} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + {\left (10 \, 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} + 3 \, {\left (10 \, B c^{2} d^{3} e^{2} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 3 \, {\left (10 \, B c^{2} d^{4} e - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\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((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/6*(3*B*c^2*e^5*x^5 + 47*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 26*(2*B*b*c + A*c^2)*d^4*e + 11*(B*b^2 + 2*A*b*c)*d^3*
e^2 - 3*(5*B*c^2*d*e^4 - 2*(2*B*b*c + A*c^2)*e^5)*x^4 - 9*(7*B*c^2*d^2*e^3 - 2*(2*B*b*c + A*c^2)*d*e^4)*x^3 -
3*(3*B*c^2*d^3*e^2 + 2*A*b^2*e^5 + 6*(2*B*b*c + A*c^2)*d^2*e^3 - 6*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 3*(27*B*c^2*
d^4*e - 2*A*b^2*d*e^4 - 18*(2*B*b*c + A*c^2)*d^3*e^2 + 9*(B*b^2 + 2*A*b*c)*d^2*e^3)*x + 6*(10*B*c^2*d^5 - 4*(2
*B*b*c + A*c^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2 + (10*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 + 3*(10*B*c^2*d^3*e^2 - 4*(2*B*b*c + A*c^2)*d^2*e^3 + (B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 3*(10*B*
c^2*d^4*e - 4*(2*B*b*c + A*c^2)*d^3*e^2 + (B*b^2 + 2*A*b*c)*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 = 299, normalized size = 1.26 \begin {gather*} {\left (10 \, B c^{2} d^{2} - 8 \, B b c d e - 4 \, A c^{2} d e + B b^{2} e^{2} + 2 \, A b c e^{2}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (B c^{2} x^{2} e^{4} - 8 \, B c^{2} d x e^{3} + 4 \, B b c x e^{4} + 2 \, A c^{2} x e^{4}\right )} e^{\left (-8\right )} + \frac {{\left (47 \, B c^{2} d^{5} - 52 \, B b c d^{4} e - 26 \, A c^{2} d^{4} e + 11 \, B b^{2} d^{3} e^{2} + 22 \, A b c d^{3} e^{2} - 2 \, A b^{2} d^{2} e^{3} + 6 \, {\left (10 \, B c^{2} d^{3} e^{2} - 12 \, B b c d^{2} e^{3} - 6 \, A c^{2} d^{2} e^{3} + 3 \, B b^{2} d e^{4} + 6 \, A b c d e^{4} - A b^{2} e^{5}\right )} x^{2} + 3 \, {\left (35 \, B c^{2} d^{4} e - 40 \, B b c d^{3} e^{2} - 20 \, A c^{2} d^{3} e^{2} + 9 \, B b^{2} d^{2} e^{3} + 18 \, A b c d^{2} e^{3} - 2 \, A b^{2} d e^{4}\right )} x\right )} e^{\left (-6\right )}}{6 \, {\left (x e + d\right )}^{3}} \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)^4,x, algorithm="giac")

[Out]

(10*B*c^2*d^2 - 8*B*b*c*d*e - 4*A*c^2*d*e + B*b^2*e^2 + 2*A*b*c*e^2)*e^(-6)*log(abs(x*e + d)) + 1/2*(B*c^2*x^2
*e^4 - 8*B*c^2*d*x*e^3 + 4*B*b*c*x*e^4 + 2*A*c^2*x*e^4)*e^(-8) + 1/6*(47*B*c^2*d^5 - 52*B*b*c*d^4*e - 26*A*c^2
*d^4*e + 11*B*b^2*d^3*e^2 + 22*A*b*c*d^3*e^2 - 2*A*b^2*d^2*e^3 + 6*(10*B*c^2*d^3*e^2 - 12*B*b*c*d^2*e^3 - 6*A*
c^2*d^2*e^3 + 3*B*b^2*d*e^4 + 6*A*b*c*d*e^4 - A*b^2*e^5)*x^2 + 3*(35*B*c^2*d^4*e - 40*B*b*c*d^3*e^2 - 20*A*c^2
*d^3*e^2 + 9*B*b^2*d^2*e^3 + 18*A*b*c*d^2*e^3 - 2*A*b^2*d*e^4)*x)*e^(-6)/(x*e + d)^3

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maple [A]  time = 0.07, size = 446, normalized size = 1.87 \begin {gather*} -\frac {A \,b^{2} d^{2}}{3 \left (e x +d \right )^{3} e^{3}}+\frac {2 A b c \,d^{3}}{3 \left (e x +d \right )^{3} e^{4}}-\frac {A \,c^{2} d^{4}}{3 \left (e x +d \right )^{3} e^{5}}+\frac {B \,b^{2} d^{3}}{3 \left (e x +d \right )^{3} e^{4}}-\frac {2 B b c \,d^{4}}{3 \left (e x +d \right )^{3} e^{5}}+\frac {B \,c^{2} d^{5}}{3 \left (e x +d \right )^{3} e^{6}}+\frac {A \,b^{2} d}{\left (e x +d \right )^{2} e^{3}}-\frac {3 A b c \,d^{2}}{\left (e x +d \right )^{2} e^{4}}+\frac {2 A \,c^{2} d^{3}}{\left (e x +d \right )^{2} e^{5}}-\frac {3 B \,b^{2} d^{2}}{2 \left (e x +d \right )^{2} e^{4}}+\frac {4 B b c \,d^{3}}{\left (e x +d \right )^{2} e^{5}}-\frac {5 B \,c^{2} d^{4}}{2 \left (e x +d \right )^{2} e^{6}}+\frac {B \,c^{2} x^{2}}{2 e^{4}}-\frac {A \,b^{2}}{\left (e x +d \right ) e^{3}}+\frac {6 A b c d}{\left (e x +d \right ) e^{4}}+\frac {2 A b c \ln \left (e x +d \right )}{e^{4}}-\frac {6 A \,c^{2} d^{2}}{\left (e x +d \right ) e^{5}}-\frac {4 A \,c^{2} d \ln \left (e x +d \right )}{e^{5}}+\frac {A \,c^{2} x}{e^{4}}+\frac {3 B \,b^{2} d}{\left (e x +d \right ) e^{4}}+\frac {B \,b^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {12 B b c \,d^{2}}{\left (e x +d \right ) e^{5}}-\frac {8 B b c d \ln \left (e x +d \right )}{e^{5}}+\frac {2 B b c x}{e^{4}}+\frac {10 B \,c^{2} d^{3}}{\left (e x +d \right ) e^{6}}+\frac {10 B \,c^{2} d^{2} \ln \left (e x +d \right )}{e^{6}}-\frac {4 B \,c^{2} d 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)^2/(e*x+d)^4,x)

[Out]

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

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maxima [A]  time = 0.54, size = 311, normalized size = 1.31 \begin {gather*} \frac {47 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 26 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + 11 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 6 \, {\left (10 \, B c^{2} d^{3} e^{2} - A b^{2} e^{5} - 6 \, {\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} + 3 \, {\left (35 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 20 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 9 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} + \frac {B c^{2} e x^{2} - 2 \, {\left (4 \, B c^{2} d - {\left (2 \, B b c + A c^{2}\right )} e\right )} x}{2 \, e^{5}} + \frac {{\left (10 \, B c^{2} d^{2} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e + {\left (B b^{2} + 2 \, A b c\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((B*x+A)*(c*x^2+b*x)^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*(47*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 26*(2*B*b*c + A*c^2)*d^4*e + 11*(B*b^2 + 2*A*b*c)*d^3*e^2 + 6*(10*B*c^2*
d^3*e^2 - A*b^2*e^5 - 6*(2*B*b*c + A*c^2)*d^2*e^3 + 3*(B*b^2 + 2*A*b*c)*d*e^4)*x^2 + 3*(35*B*c^2*d^4*e - 2*A*b
^2*d*e^4 - 20*(2*B*b*c + A*c^2)*d^3*e^2 + 9*(B*b^2 + 2*A*b*c)*d^2*e^3)*x)/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x
+ d^3*e^6) + 1/2*(B*c^2*e*x^2 - 2*(4*B*c^2*d - (2*B*b*c + A*c^2)*e)*x)/e^5 + (10*B*c^2*d^2 - 4*(2*B*b*c + A*c
^2)*d*e + (B*b^2 + 2*A*b*c)*e^2)*log(e*x + d)/e^6

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mupad [B]  time = 0.14, size = 328, normalized size = 1.38 \begin {gather*} x\,\left (\frac {A\,c^2+2\,B\,b\,c}{e^4}-\frac {4\,B\,c^2\,d}{e^5}\right )+\frac {\frac {11\,B\,b^2\,d^3\,e^2-2\,A\,b^2\,d^2\,e^3-52\,B\,b\,c\,d^4\,e+22\,A\,b\,c\,d^3\,e^2+47\,B\,c^2\,d^5-26\,A\,c^2\,d^4\,e}{6\,e}-x^2\,\left (-3\,B\,b^2\,d\,e^3+A\,b^2\,e^4+12\,B\,b\,c\,d^2\,e^2-6\,A\,b\,c\,d\,e^3-10\,B\,c^2\,d^3\,e+6\,A\,c^2\,d^2\,e^2\right )+x\,\left (\frac {9\,B\,b^2\,d^2\,e^2}{2}-A\,b^2\,d\,e^3-20\,B\,b\,c\,d^3\,e+9\,A\,b\,c\,d^2\,e^2+\frac {35\,B\,c^2\,d^4}{2}-10\,A\,c^2\,d^3\,e\right )}{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 (B\,b^2\,e^2-8\,B\,b\,c\,d\,e+2\,A\,b\,c\,e^2+10\,B\,c^2\,d^2-4\,A\,c^2\,d\,e\right )}{e^6}+\frac {B\,c^2\,x^2}{2\,e^4} \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)^4,x)

[Out]

x*((A*c^2 + 2*B*b*c)/e^4 - (4*B*c^2*d)/e^5) + ((47*B*c^2*d^5 - 26*A*c^2*d^4*e - 2*A*b^2*d^2*e^3 + 11*B*b^2*d^3
*e^2 - 52*B*b*c*d^4*e + 22*A*b*c*d^3*e^2)/(6*e) - x^2*(A*b^2*e^4 - 3*B*b^2*d*e^3 - 10*B*c^2*d^3*e + 6*A*c^2*d^
2*e^2 - 6*A*b*c*d*e^3 + 12*B*b*c*d^2*e^2) + x*((35*B*c^2*d^4)/2 - A*b^2*d*e^3 - 10*A*c^2*d^3*e + (9*B*b^2*d^2*
e^2)/2 - 20*B*b*c*d^3*e + 9*A*b*c*d^2*e^2))/(d^3*e^5 + e^8*x^3 + 3*d^2*e^6*x + 3*d*e^7*x^2) + (log(d + e*x)*(B
*b^2*e^2 + 10*B*c^2*d^2 + 2*A*b*c*e^2 - 4*A*c^2*d*e - 8*B*b*c*d*e))/e^6 + (B*c^2*x^2)/(2*e^4)

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sympy [A]  time = 11.73, size = 374, normalized size = 1.57 \begin {gather*} \frac {B c^{2} x^{2}}{2 e^{4}} + x \left (\frac {A c^{2}}{e^{4}} + \frac {2 B b c}{e^{4}} - \frac {4 B c^{2} d}{e^{5}}\right ) + \frac {- 2 A b^{2} d^{2} e^{3} + 22 A b c d^{3} e^{2} - 26 A c^{2} d^{4} e + 11 B b^{2} d^{3} e^{2} - 52 B b c d^{4} e + 47 B c^{2} d^{5} + x^{2} \left (- 6 A b^{2} e^{5} + 36 A b c d e^{4} - 36 A c^{2} d^{2} e^{3} + 18 B b^{2} d e^{4} - 72 B b c d^{2} e^{3} + 60 B c^{2} d^{3} e^{2}\right ) + x \left (- 6 A b^{2} d e^{4} + 54 A b c d^{2} e^{3} - 60 A c^{2} d^{3} e^{2} + 27 B b^{2} d^{2} e^{3} - 120 B b c d^{3} e^{2} + 105 B c^{2} d^{4} e\right )}{6 d^{3} e^{6} + 18 d^{2} e^{7} x + 18 d e^{8} x^{2} + 6 e^{9} x^{3}} + \frac {\left (2 A b c e^{2} - 4 A c^{2} d e + B b^{2} e^{2} - 8 B b c d e + 10 B c^{2} d^{2}\right ) \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)**2/(e*x+d)**4,x)

[Out]

B*c**2*x**2/(2*e**4) + x*(A*c**2/e**4 + 2*B*b*c/e**4 - 4*B*c**2*d/e**5) + (-2*A*b**2*d**2*e**3 + 22*A*b*c*d**3
*e**2 - 26*A*c**2*d**4*e + 11*B*b**2*d**3*e**2 - 52*B*b*c*d**4*e + 47*B*c**2*d**5 + x**2*(-6*A*b**2*e**5 + 36*
A*b*c*d*e**4 - 36*A*c**2*d**2*e**3 + 18*B*b**2*d*e**4 - 72*B*b*c*d**2*e**3 + 60*B*c**2*d**3*e**2) + x*(-6*A*b*
*2*d*e**4 + 54*A*b*c*d**2*e**3 - 60*A*c**2*d**3*e**2 + 27*B*b**2*d**2*e**3 - 120*B*b*c*d**3*e**2 + 105*B*c**2*
d**4*e))/(6*d**3*e**6 + 18*d**2*e**7*x + 18*d*e**8*x**2 + 6*e**9*x**3) + (2*A*b*c*e**2 - 4*A*c**2*d*e + B*b**2
*e**2 - 8*B*b*c*d*e + 10*B*c**2*d**2)*log(d + e*x)/e**6

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