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

Optimal. Leaf size=194 \[ \frac {d^2 (B d-A e) (c d-b e)^2}{e^6 (d+e x)}+\frac {d (c d-b e) \log (d+e x) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^6}-\frac {x (c d-b e) (2 B d (2 c d-b e)-A e (3 c d-b e))}{e^5}+\frac {x^2 (c d-b e) (-2 A c e-b B e+3 B c d)}{2 e^4}-\frac {c x^3 (-A c e-2 b B e+2 B c d)}{3 e^3}+\frac {B c^2 x^4}{4 e^2} \]

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Rubi [A]  time = 0.28, antiderivative size = 194, 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 {d^2 (B d-A e) (c d-b e)^2}{e^6 (d+e x)}-\frac {c x^3 (-A c e-2 b B e+2 B c d)}{3 e^3}+\frac {x^2 (c d-b e) (-2 A c e-b B e+3 B c d)}{2 e^4}-\frac {x (c d-b e) (2 B d (2 c d-b e)-A e (3 c d-b e))}{e^5}+\frac {d (c d-b e) \log (d+e x) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^6}+\frac {B c^2 x^4}{4 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 184, normalized size = 0.95 \begin {gather*} \frac {\frac {12 d^2 (B d-A e) (c d-b e)^2}{d+e x}+4 c e^3 x^3 (A c e+2 b B e-2 B c d)+6 e^2 x^2 (b e-c d) (2 A c e+b B e-3 B c d)+12 e x (b e-c d) (A e (b e-3 c d)+2 B d (2 c d-b e))+12 d (c d-b e) \log (d+e x) (2 A e (b e-2 c d)+B d (5 c d-3 b e))+3 B c^2 e^4 x^4}{12 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 420, normalized size = 2.16 \begin {gather*} \frac {3 \, B c^{2} e^{5} x^{5} + 12 \, B c^{2} d^{5} - 12 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + 12 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} - {\left (5 \, B c^{2} d e^{4} - 4 \, {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 2 \, {\left (5 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} - 6 \, {\left (5 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 4 \, {\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} - 12 \, {\left (4 \, B c^{2} d^{4} e - A b^{2} d e^{4} - 3 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 2 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x + 12 \, {\left (5 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 4 \, {\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} + {\left (5 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 4 \, {\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\right )} \log \left (e x + d\right )}{12 \, {\left (e^{7} x + d 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)^2,x, algorithm="fricas")

[Out]

1/12*(3*B*c^2*e^5*x^5 + 12*B*c^2*d^5 - 12*A*b^2*d^2*e^3 - 12*(2*B*b*c + A*c^2)*d^4*e + 12*(B*b^2 + 2*A*b*c)*d^
3*e^2 - (5*B*c^2*d*e^4 - 4*(2*B*b*c + A*c^2)*e^5)*x^4 + 2*(5*B*c^2*d^2*e^3 - 4*(2*B*b*c + A*c^2)*d*e^4 + 3*(B*
b^2 + 2*A*b*c)*e^5)*x^3 - 6*(5*B*c^2*d^3*e^2 - 2*A*b^2*e^5 - 4*(2*B*b*c + A*c^2)*d^2*e^3 + 3*(B*b^2 + 2*A*b*c)
*d*e^4)*x^2 - 12*(4*B*c^2*d^4*e - A*b^2*d*e^4 - 3*(2*B*b*c + A*c^2)*d^3*e^2 + 2*(B*b^2 + 2*A*b*c)*d^2*e^3)*x +
12*(5*B*c^2*d^5 - 2*A*b^2*d^2*e^3 - 4*(2*B*b*c + A*c^2)*d^4*e + 3*(B*b^2 + 2*A*b*c)*d^3*e^2 + (5*B*c^2*d^4*e
- 2*A*b^2*d*e^4 - 4*(2*B*b*c + A*c^2)*d^3*e^2 + 3*(B*b^2 + 2*A*b*c)*d^2*e^3)*x)*log(e*x + d))/(e^7*x + d*e^6)

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giac [B]  time = 0.17, size = 380, normalized size = 1.96 \begin {gather*} \frac {1}{12} \, {\left (3 \, B c^{2} - \frac {4 \, {\left (5 \, B c^{2} d e - 2 \, B b c e^{2} - A c^{2} e^{2}\right )} e^{\left (-1\right )}}{x e + d} + \frac {6 \, {\left (10 \, 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 )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}} - \frac {12 \, {\left (10 \, B c^{2} d^{3} e^{3} - 12 \, B b c d^{2} e^{4} - 6 \, A c^{2} d^{2} e^{4} + 3 \, B b^{2} d e^{5} + 6 \, A b c d e^{5} - A b^{2} e^{6}\right )} e^{\left (-3\right )}}{{\left (x e + d\right )}^{3}}\right )} {\left (x e + d\right )}^{4} e^{\left (-6\right )} - {\left (5 \, B c^{2} d^{4} - 8 \, B b c d^{3} e - 4 \, 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 )} e^{\left (-6\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (\frac {B c^{2} d^{5} e^{4}}{x e + d} - \frac {2 \, B b c d^{4} e^{5}}{x e + d} - \frac {A c^{2} d^{4} e^{5}}{x e + d} + \frac {B b^{2} d^{3} e^{6}}{x e + d} + \frac {2 \, A b c d^{3} e^{6}}{x e + d} - \frac {A b^{2} d^{2} e^{7}}{x e + d}\right )} e^{\left (-10\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)^2,x, algorithm="giac")

[Out]

1/12*(3*B*c^2 - 4*(5*B*c^2*d*e - 2*B*b*c*e^2 - A*c^2*e^2)*e^(-1)/(x*e + d) + 6*(10*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)*e^(-2)/(x*e + d)^2 - 12*(10*B*c^2*d^3*e^3 - 12*B*b*c*d^2*e^4 - 6
*A*c^2*d^2*e^4 + 3*B*b^2*d*e^5 + 6*A*b*c*d*e^5 - A*b^2*e^6)*e^(-3)/(x*e + d)^3)*(x*e + d)^4*e^(-6) - (5*B*c^2*
d^4 - 8*B*b*c*d^3*e - 4*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)*e^(-6)*log(abs(x*e +
d)*e^(-1)/(x*e + d)^2) + (B*c^2*d^5*e^4/(x*e + d) - 2*B*b*c*d^4*e^5/(x*e + d) - A*c^2*d^4*e^5/(x*e + d) + B*b^
2*d^3*e^6/(x*e + d) + 2*A*b*c*d^3*e^6/(x*e + d) - A*b^2*d^2*e^7/(x*e + d))*e^(-10)

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maple [B]  time = 0.05, size = 394, normalized size = 2.03 \begin {gather*} \frac {B \,c^{2} x^{4}}{4 e^{2}}+\frac {A \,c^{2} x^{3}}{3 e^{2}}+\frac {2 B b c \,x^{3}}{3 e^{2}}-\frac {2 B \,c^{2} d \,x^{3}}{3 e^{3}}+\frac {A b c \,x^{2}}{e^{2}}-\frac {A \,c^{2} d \,x^{2}}{e^{3}}+\frac {B \,b^{2} x^{2}}{2 e^{2}}-\frac {2 B b c d \,x^{2}}{e^{3}}+\frac {3 B \,c^{2} d^{2} x^{2}}{2 e^{4}}-\frac {A \,b^{2} d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 A \,b^{2} d \ln \left (e x +d \right )}{e^{3}}+\frac {A \,b^{2} x}{e^{2}}+\frac {2 A b c \,d^{3}}{\left (e x +d \right ) e^{4}}+\frac {6 A b c \,d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {4 A b c d x}{e^{3}}-\frac {A \,c^{2} d^{4}}{\left (e x +d \right ) e^{5}}-\frac {4 A \,c^{2} d^{3} \ln \left (e x +d \right )}{e^{5}}+\frac {3 A \,c^{2} d^{2} x}{e^{4}}+\frac {B \,b^{2} d^{3}}{\left (e x +d \right ) e^{4}}+\frac {3 B \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {2 B \,b^{2} d x}{e^{3}}-\frac {2 B b c \,d^{4}}{\left (e x +d \right ) e^{5}}-\frac {8 B b c \,d^{3} \ln \left (e x +d \right )}{e^{5}}+\frac {6 B b c \,d^{2} x}{e^{4}}+\frac {B \,c^{2} d^{5}}{\left (e x +d \right ) e^{6}}+\frac {5 B \,c^{2} d^{4} \ln \left (e x +d \right )}{e^{6}}-\frac {4 B \,c^{2} d^{3} 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)^2,x)

[Out]

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

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maxima [A]  time = 0.53, size = 291, normalized size = 1.50 \begin {gather*} \frac {B c^{2} d^{5} - A b^{2} d^{2} e^{3} - {\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}}{e^{7} x + d e^{6}} + \frac {3 \, B c^{2} e^{3} x^{4} - 4 \, {\left (2 \, B c^{2} d e^{2} - {\left (2 \, B b c + A c^{2}\right )} e^{3}\right )} x^{3} + 6 \, {\left (3 \, B c^{2} d^{2} e - 2 \, {\left (2 \, B b c + A c^{2}\right )} d e^{2} + {\left (B b^{2} + 2 \, A b c\right )} e^{3}\right )} x^{2} - 12 \, {\left (4 \, B c^{2} d^{3} - A b^{2} e^{3} - 3 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e + 2 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{2}\right )} x}{12 \, e^{5}} + \frac {{\left (5 \, B c^{2} d^{4} - 2 \, A b^{2} d e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} 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)^2,x, algorithm="maxima")

[Out]

(B*c^2*d^5 - A*b^2*d^2*e^3 - (2*B*b*c + A*c^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2)/(e^7*x + d*e^6) + 1/12*(3*B*
c^2*e^3*x^4 - 4*(2*B*c^2*d*e^2 - (2*B*b*c + A*c^2)*e^3)*x^3 + 6*(3*B*c^2*d^2*e - 2*(2*B*b*c + A*c^2)*d*e^2 + (
B*b^2 + 2*A*b*c)*e^3)*x^2 - 12*(4*B*c^2*d^3 - A*b^2*e^3 - 3*(2*B*b*c + A*c^2)*d^2*e + 2*(B*b^2 + 2*A*b*c)*d*e^
2)*x)/e^5 + (5*B*c^2*d^4 - 2*A*b^2*d*e^3 - 4*(2*B*b*c + A*c^2)*d^3*e + 3*(B*b^2 + 2*A*b*c)*d^2*e^2)*log(e*x +
d)/e^6

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mupad [B]  time = 1.43, size = 371, normalized size = 1.91 \begin {gather*} x\,\left (\frac {A\,b^2}{e^2}-\frac {d^2\,\left (\frac {A\,c^2+2\,B\,b\,c}{e^2}-\frac {2\,B\,c^2\,d}{e^3}\right )}{e^2}+\frac {2\,d\,\left (\frac {2\,d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e^2}-\frac {2\,B\,c^2\,d}{e^3}\right )}{e}-\frac {B\,b^2+2\,A\,c\,b}{e^2}+\frac {B\,c^2\,d^2}{e^4}\right )}{e}\right )+x^3\,\left (\frac {A\,c^2+2\,B\,b\,c}{3\,e^2}-\frac {2\,B\,c^2\,d}{3\,e^3}\right )-x^2\,\left (\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e^2}-\frac {2\,B\,c^2\,d}{e^3}\right )}{e}-\frac {B\,b^2+2\,A\,c\,b}{2\,e^2}+\frac {B\,c^2\,d^2}{2\,e^4}\right )+\frac {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}{e\,\left (x\,e^6+d\,e^5\right )}+\frac {\ln \left (d+e\,x\right )\,\left (3\,B\,b^2\,d^2\,e^2-2\,A\,b^2\,d\,e^3-8\,B\,b\,c\,d^3\,e+6\,A\,b\,c\,d^2\,e^2+5\,B\,c^2\,d^4-4\,A\,c^2\,d^3\,e\right )}{e^6}+\frac {B\,c^2\,x^4}{4\,e^2} \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)^2,x)

[Out]

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

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sympy [A]  time = 1.39, size = 316, normalized size = 1.63 \begin {gather*} \frac {B c^{2} x^{4}}{4 e^{2}} + \frac {d \left (b e - c d\right ) \left (- 2 A b e^{2} + 4 A c d e + 3 B b d e - 5 B c d^{2}\right ) \log {\left (d + e x \right )}}{e^{6}} + x^{3} \left (\frac {A c^{2}}{3 e^{2}} + \frac {2 B b c}{3 e^{2}} - \frac {2 B c^{2} d}{3 e^{3}}\right ) + x^{2} \left (\frac {A b c}{e^{2}} - \frac {A c^{2} d}{e^{3}} + \frac {B b^{2}}{2 e^{2}} - \frac {2 B b c d}{e^{3}} + \frac {3 B c^{2} d^{2}}{2 e^{4}}\right ) + x \left (\frac {A b^{2}}{e^{2}} - \frac {4 A b c d}{e^{3}} + \frac {3 A c^{2} d^{2}}{e^{4}} - \frac {2 B b^{2} d}{e^{3}} + \frac {6 B b c d^{2}}{e^{4}} - \frac {4 B c^{2} d^{3}}{e^{5}}\right ) + \frac {- A b^{2} d^{2} e^{3} + 2 A b c d^{3} e^{2} - A c^{2} d^{4} e + B b^{2} d^{3} e^{2} - 2 B b c d^{4} e + B c^{2} d^{5}}{d e^{6} + e^{7} x} \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)**2,x)

[Out]

B*c**2*x**4/(4*e**2) + d*(b*e - c*d)*(-2*A*b*e**2 + 4*A*c*d*e + 3*B*b*d*e - 5*B*c*d**2)*log(d + e*x)/e**6 + x*
*3*(A*c**2/(3*e**2) + 2*B*b*c/(3*e**2) - 2*B*c**2*d/(3*e**3)) + x**2*(A*b*c/e**2 - A*c**2*d/e**3 + B*b**2/(2*e
**2) - 2*B*b*c*d/e**3 + 3*B*c**2*d**2/(2*e**4)) + x*(A*b**2/e**2 - 4*A*b*c*d/e**3 + 3*A*c**2*d**2/e**4 - 2*B*b
**2*d/e**3 + 6*B*b*c*d**2/e**4 - 4*B*c**2*d**3/e**5) + (-A*b**2*d**2*e**3 + 2*A*b*c*d**3*e**2 - A*c**2*d**4*e
+ B*b**2*d**3*e**2 - 2*B*b*c*d**4*e + B*c**2*d**5)/(d*e**6 + e**7*x)

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