3.11.12 \(\int \frac {(A+B x) (d+e x)^2}{(b x+c x^2)^3} \, dx\)

Optimal. Leaf size=198 \[ -\frac {d (2 A b e-3 A c d+b B d)}{b^4 x}-\frac {(c d-b e) (A b e-3 A c d+2 b B d)}{b^4 (b+c x)}-\frac {(b B-A c) (c d-b e)^2}{2 b^3 c (b+c x)^2}-\frac {A d^2}{2 b^3 x^2}+\frac {\log (x) \left (b^2 e (A e+2 B d)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{b^5}-\frac {\log (b+c x) \left (b^2 e (A e+2 B d)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{b^5} \]

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 198, 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 {\log (x) \left (b^2 e (A e+2 B d)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{b^5}-\frac {\log (b+c x) \left (b^2 e (A e+2 B d)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{b^5}-\frac {d (2 A b e-3 A c d+b B d)}{b^4 x}-\frac {(c d-b e) (A b e-3 A c d+2 b B d)}{b^4 (b+c x)}-\frac {(b B-A c) (c d-b e)^2}{2 b^3 c (b+c x)^2}-\frac {A d^2}{2 b^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^3,x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.27, size = 190, normalized size = 0.96 \begin {gather*} -\frac {-2 \log (x) \left (b^2 e (A e+2 B d)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )+2 \log (b+c x) \left (b^2 e (A e+2 B d)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )+\frac {b^2 (b B-A c) (c d-b e)^2}{c (b+c x)^2}+\frac {A b^2 d^2}{x^2}+\frac {2 b d (2 A b e-3 A c d+b B d)}{x}-\frac {2 b (b e-c d) (A b e-3 A c d+2 b B d)}{b+c x}}{2 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^3,x]

[Out]

-1/2*((A*b^2*d^2)/x^2 + (2*b*d*(b*B*d - 3*A*c*d + 2*A*b*e))/x + (b^2*(b*B - A*c)*(c*d - b*e)^2)/(c*(b + c*x)^2
) - (2*b*(-(c*d) + b*e)*(2*b*B*d - 3*A*c*d + A*b*e))/(b + c*x) - 2*(6*A*c^2*d^2 + b^2*e*(2*B*d + A*e) - 3*b*c*
d*(B*d + 2*A*e))*Log[x] + 2*(6*A*c^2*d^2 + b^2*e*(2*B*d + A*e) - 3*b*c*d*(B*d + 2*A*e))*Log[b + c*x])/b^5

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) (d+e x)^2}{\left (b x+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^3,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^3, x]

________________________________________________________________________________________

fricas [B]  time = 0.42, size = 558, normalized size = 2.82 \begin {gather*} -\frac {A b^{4} c d^{2} - 2 \, {\left (A b^{3} c^{2} e^{2} - 3 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{2} + 2 \, {\left (B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} d e\right )} x^{3} + {\left (9 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{2} - 6 \, {\left (B b^{4} c - 3 \, A b^{3} c^{2}\right )} d e + {\left (B b^{5} - 3 \, A b^{4} c\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, A b^{4} c d e + {\left (B b^{4} c - 2 \, A b^{3} c^{2}\right )} d^{2}\right )} x + 2 \, {\left ({\left (A b^{2} c^{3} e^{2} - 3 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{2} + 2 \, {\left (B b^{2} c^{3} - 3 \, A b c^{4}\right )} d e\right )} x^{4} + 2 \, {\left (A b^{3} c^{2} e^{2} - 3 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{2} + 2 \, {\left (B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} d e\right )} x^{3} + {\left (A b^{4} c e^{2} - 3 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{2} + 2 \, {\left (B b^{4} c - 3 \, A b^{3} c^{2}\right )} d e\right )} x^{2}\right )} \log \left (c x + b\right ) - 2 \, {\left ({\left (A b^{2} c^{3} e^{2} - 3 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{2} + 2 \, {\left (B b^{2} c^{3} - 3 \, A b c^{4}\right )} d e\right )} x^{4} + 2 \, {\left (A b^{3} c^{2} e^{2} - 3 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{2} + 2 \, {\left (B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} d e\right )} x^{3} + {\left (A b^{4} c e^{2} - 3 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{2} + 2 \, {\left (B b^{4} c - 3 \, A b^{3} c^{2}\right )} d e\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left (b^{5} c^{3} x^{4} + 2 \, b^{6} c^{2} x^{3} + b^{7} c x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.19, size = 324, normalized size = 1.64 \begin {gather*} -\frac {{\left (3 \, B b c d^{2} - 6 \, A c^{2} d^{2} - 2 \, B b^{2} d e + 6 \, A b c d e - A b^{2} e^{2}\right )} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {{\left (3 \, B b c^{2} d^{2} - 6 \, A c^{3} d^{2} - 2 \, B b^{2} c d e + 6 \, A b c^{2} d e - A b^{2} c e^{2}\right )} \log \left ({\left | c x + b \right |}\right )}{b^{5} c} - \frac {6 \, B b c^{3} d^{2} x^{3} - 12 \, A c^{4} d^{2} x^{3} - 4 \, B b^{2} c^{2} d x^{3} e + 12 \, A b c^{3} d x^{3} e + 9 \, B b^{2} c^{2} d^{2} x^{2} - 18 \, A b c^{3} d^{2} x^{2} - 2 \, A b^{2} c^{2} x^{3} e^{2} - 6 \, B b^{3} c d x^{2} e + 18 \, A b^{2} c^{2} d x^{2} e + 2 \, B b^{3} c d^{2} x - 4 \, A b^{2} c^{2} d^{2} x + B b^{4} x^{2} e^{2} - 3 \, A b^{3} c x^{2} e^{2} + 4 \, A b^{3} c d x e + A b^{3} c d^{2}}{2 \, {\left (c x^{2} + b x\right )}^{2} b^{4} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

-(3*B*b*c*d^2 - 6*A*c^2*d^2 - 2*B*b^2*d*e + 6*A*b*c*d*e - A*b^2*e^2)*log(abs(x))/b^5 + (3*B*b*c^2*d^2 - 6*A*c^
3*d^2 - 2*B*b^2*c*d*e + 6*A*b*c^2*d*e - A*b^2*c*e^2)*log(abs(c*x + b))/(b^5*c) - 1/2*(6*B*b*c^3*d^2*x^3 - 12*A
*c^4*d^2*x^3 - 4*B*b^2*c^2*d*x^3*e + 12*A*b*c^3*d*x^3*e + 9*B*b^2*c^2*d^2*x^2 - 18*A*b*c^3*d^2*x^2 - 2*A*b^2*c
^2*x^3*e^2 - 6*B*b^3*c*d*x^2*e + 18*A*b^2*c^2*d*x^2*e + 2*B*b^3*c*d^2*x - 4*A*b^2*c^2*d^2*x + B*b^4*x^2*e^2 -
3*A*b^3*c*x^2*e^2 + 4*A*b^3*c*d*x*e + A*b^3*c*d^2)/((c*x^2 + b*x)^2*b^4*c)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 365, normalized size = 1.84 \begin {gather*} \frac {A \,e^{2}}{2 \left (c x +b \right )^{2} b}-\frac {A c d e}{\left (c x +b \right )^{2} b^{2}}+\frac {A \,c^{2} d^{2}}{2 \left (c x +b \right )^{2} b^{3}}+\frac {B d e}{\left (c x +b \right )^{2} b}-\frac {B c \,d^{2}}{2 \left (c x +b \right )^{2} b^{2}}-\frac {B \,e^{2}}{2 \left (c x +b \right )^{2} c}+\frac {A \,e^{2}}{\left (c x +b \right ) b^{2}}-\frac {4 A c d e}{\left (c x +b \right ) b^{3}}+\frac {A \,e^{2} \ln \relax (x )}{b^{3}}-\frac {A \,e^{2} \ln \left (c x +b \right )}{b^{3}}+\frac {3 A \,c^{2} d^{2}}{\left (c x +b \right ) b^{4}}-\frac {6 A c d e \ln \relax (x )}{b^{4}}+\frac {6 A c d e \ln \left (c x +b \right )}{b^{4}}+\frac {6 A \,c^{2} d^{2} \ln \relax (x )}{b^{5}}-\frac {6 A \,c^{2} d^{2} \ln \left (c x +b \right )}{b^{5}}+\frac {2 B d e}{\left (c x +b \right ) b^{2}}-\frac {2 B c \,d^{2}}{\left (c x +b \right ) b^{3}}+\frac {2 B d e \ln \relax (x )}{b^{3}}-\frac {2 B d e \ln \left (c x +b \right )}{b^{3}}-\frac {3 B c \,d^{2} \ln \relax (x )}{b^{4}}+\frac {3 B c \,d^{2} \ln \left (c x +b \right )}{b^{4}}-\frac {2 A d e}{b^{3} x}+\frac {3 A c \,d^{2}}{b^{4} x}-\frac {B \,d^{2}}{b^{3} x}-\frac {A \,d^{2}}{2 b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^2/(c*x^2+b*x)^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.59, size = 293, normalized size = 1.48 \begin {gather*} -\frac {A b^{3} c d^{2} - 2 \, {\left (A b^{2} c^{2} e^{2} - 3 \, {\left (B b c^{3} - 2 \, A c^{4}\right )} d^{2} + 2 \, {\left (B b^{2} c^{2} - 3 \, A b c^{3}\right )} d e\right )} x^{3} + {\left (9 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d^{2} - 6 \, {\left (B b^{3} c - 3 \, A b^{2} c^{2}\right )} d e + {\left (B b^{4} - 3 \, A b^{3} c\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, A b^{3} c d e + {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d^{2}\right )} x}{2 \, {\left (b^{4} c^{3} x^{4} + 2 \, b^{5} c^{2} x^{3} + b^{6} c x^{2}\right )}} - \frac {{\left (A b^{2} e^{2} - 3 \, {\left (B b c - 2 \, A c^{2}\right )} d^{2} + 2 \, {\left (B b^{2} - 3 \, A b c\right )} d e\right )} \log \left (c x + b\right )}{b^{5}} + \frac {{\left (A b^{2} e^{2} - 3 \, {\left (B b c - 2 \, A c^{2}\right )} d^{2} + 2 \, {\left (B b^{2} - 3 \, A b c\right )} d e\right )} \log \relax (x)}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

-1/2*(A*b^3*c*d^2 - 2*(A*b^2*c^2*e^2 - 3*(B*b*c^3 - 2*A*c^4)*d^2 + 2*(B*b^2*c^2 - 3*A*b*c^3)*d*e)*x^3 + (9*(B*
b^2*c^2 - 2*A*b*c^3)*d^2 - 6*(B*b^3*c - 3*A*b^2*c^2)*d*e + (B*b^4 - 3*A*b^3*c)*e^2)*x^2 + 2*(2*A*b^3*c*d*e + (
B*b^3*c - 2*A*b^2*c^2)*d^2)*x)/(b^4*c^3*x^4 + 2*b^5*c^2*x^3 + b^6*c*x^2) - (A*b^2*e^2 - 3*(B*b*c - 2*A*c^2)*d^
2 + 2*(B*b^2 - 3*A*b*c)*d*e)*log(c*x + b)/b^5 + (A*b^2*e^2 - 3*(B*b*c - 2*A*c^2)*d^2 + 2*(B*b^2 - 3*A*b*c)*d*e
)*log(x)/b^5

________________________________________________________________________________________

mupad [B]  time = 0.26, size = 319, normalized size = 1.61 \begin {gather*} -\frac {\frac {A\,d^2}{2\,b}-\frac {c\,x^3\,\left (2\,B\,b^2\,d\,e+A\,b^2\,e^2-3\,B\,b\,c\,d^2-6\,A\,b\,c\,d\,e+6\,A\,c^2\,d^2\right )}{b^4}+\frac {d\,x\,\left (2\,A\,b\,e-2\,A\,c\,d+B\,b\,d\right )}{b^2}-\frac {x^2\,\left (-B\,b^3\,e^2+6\,B\,b^2\,c\,d\,e+3\,A\,b^2\,c\,e^2-9\,B\,b\,c^2\,d^2-18\,A\,b\,c^2\,d\,e+18\,A\,c^3\,d^2\right )}{2\,b^3\,c}}{b^2\,x^2+2\,b\,c\,x^3+c^2\,x^4}-\frac {2\,\mathrm {atanh}\left (\frac {\left (b+2\,c\,x\right )\,\left (b^2\,\left (A\,e^2+2\,B\,d\,e\right )-b\,\left (3\,B\,c\,d^2+6\,A\,c\,e\,d\right )+6\,A\,c^2\,d^2\right )}{b\,\left (2\,B\,b^2\,d\,e+A\,b^2\,e^2-3\,B\,b\,c\,d^2-6\,A\,b\,c\,d\,e+6\,A\,c^2\,d^2\right )}\right )\,\left (b^2\,\left (A\,e^2+2\,B\,d\,e\right )-b\,\left (3\,B\,c\,d^2+6\,A\,c\,e\,d\right )+6\,A\,c^2\,d^2\right )}{b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^3,x)

[Out]

- ((A*d^2)/(2*b) - (c*x^3*(A*b^2*e^2 + 6*A*c^2*d^2 - 3*B*b*c*d^2 + 2*B*b^2*d*e - 6*A*b*c*d*e))/b^4 + (d*x*(2*A
*b*e - 2*A*c*d + B*b*d))/b^2 - (x^2*(18*A*c^3*d^2 - B*b^3*e^2 + 3*A*b^2*c*e^2 - 9*B*b*c^2*d^2 - 18*A*b*c^2*d*e
+ 6*B*b^2*c*d*e))/(2*b^3*c))/(b^2*x^2 + c^2*x^4 + 2*b*c*x^3) - (2*atanh(((b + 2*c*x)*(b^2*(A*e^2 + 2*B*d*e) -
b*(3*B*c*d^2 + 6*A*c*d*e) + 6*A*c^2*d^2))/(b*(A*b^2*e^2 + 6*A*c^2*d^2 - 3*B*b*c*d^2 + 2*B*b^2*d*e - 6*A*b*c*d
*e)))*(b^2*(A*e^2 + 2*B*d*e) - b*(3*B*c*d^2 + 6*A*c*d*e) + 6*A*c^2*d^2))/b^5

________________________________________________________________________________________

sympy [B]  time = 10.67, size = 660, normalized size = 3.33 \begin {gather*} \frac {- A b^{3} c d^{2} + x^{3} \left (2 A b^{2} c^{2} e^{2} - 12 A b c^{3} d e + 12 A c^{4} d^{2} + 4 B b^{2} c^{2} d e - 6 B b c^{3} d^{2}\right ) + x^{2} \left (3 A b^{3} c e^{2} - 18 A b^{2} c^{2} d e + 18 A b c^{3} d^{2} - B b^{4} e^{2} + 6 B b^{3} c d e - 9 B b^{2} c^{2} d^{2}\right ) + x \left (- 4 A b^{3} c d e + 4 A b^{2} c^{2} d^{2} - 2 B b^{3} c d^{2}\right )}{2 b^{6} c x^{2} + 4 b^{5} c^{2} x^{3} + 2 b^{4} c^{3} x^{4}} + \frac {\left (A b^{2} e^{2} - 6 A b c d e + 6 A c^{2} d^{2} + 2 B b^{2} d e - 3 B b c d^{2}\right ) \log {\left (x + \frac {A b^{3} e^{2} - 6 A b^{2} c d e + 6 A b c^{2} d^{2} + 2 B b^{3} d e - 3 B b^{2} c d^{2} - b \left (A b^{2} e^{2} - 6 A b c d e + 6 A c^{2} d^{2} + 2 B b^{2} d e - 3 B b c d^{2}\right )}{2 A b^{2} c e^{2} - 12 A b c^{2} d e + 12 A c^{3} d^{2} + 4 B b^{2} c d e - 6 B b c^{2} d^{2}} \right )}}{b^{5}} - \frac {\left (A b^{2} e^{2} - 6 A b c d e + 6 A c^{2} d^{2} + 2 B b^{2} d e - 3 B b c d^{2}\right ) \log {\left (x + \frac {A b^{3} e^{2} - 6 A b^{2} c d e + 6 A b c^{2} d^{2} + 2 B b^{3} d e - 3 B b^{2} c d^{2} + b \left (A b^{2} e^{2} - 6 A b c d e + 6 A c^{2} d^{2} + 2 B b^{2} d e - 3 B b c d^{2}\right )}{2 A b^{2} c e^{2} - 12 A b c^{2} d e + 12 A c^{3} d^{2} + 4 B b^{2} c d e - 6 B b c^{2} d^{2}} \right )}}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**2/(c*x**2+b*x)**3,x)

[Out]

(-A*b**3*c*d**2 + x**3*(2*A*b**2*c**2*e**2 - 12*A*b*c**3*d*e + 12*A*c**4*d**2 + 4*B*b**2*c**2*d*e - 6*B*b*c**3
*d**2) + x**2*(3*A*b**3*c*e**2 - 18*A*b**2*c**2*d*e + 18*A*b*c**3*d**2 - B*b**4*e**2 + 6*B*b**3*c*d*e - 9*B*b*
*2*c**2*d**2) + x*(-4*A*b**3*c*d*e + 4*A*b**2*c**2*d**2 - 2*B*b**3*c*d**2))/(2*b**6*c*x**2 + 4*b**5*c**2*x**3
+ 2*b**4*c**3*x**4) + (A*b**2*e**2 - 6*A*b*c*d*e + 6*A*c**2*d**2 + 2*B*b**2*d*e - 3*B*b*c*d**2)*log(x + (A*b**
3*e**2 - 6*A*b**2*c*d*e + 6*A*b*c**2*d**2 + 2*B*b**3*d*e - 3*B*b**2*c*d**2 - b*(A*b**2*e**2 - 6*A*b*c*d*e + 6*
A*c**2*d**2 + 2*B*b**2*d*e - 3*B*b*c*d**2))/(2*A*b**2*c*e**2 - 12*A*b*c**2*d*e + 12*A*c**3*d**2 + 4*B*b**2*c*d
*e - 6*B*b*c**2*d**2))/b**5 - (A*b**2*e**2 - 6*A*b*c*d*e + 6*A*c**2*d**2 + 2*B*b**2*d*e - 3*B*b*c*d**2)*log(x
+ (A*b**3*e**2 - 6*A*b**2*c*d*e + 6*A*b*c**2*d**2 + 2*B*b**3*d*e - 3*B*b**2*c*d**2 + b*(A*b**2*e**2 - 6*A*b*c*
d*e + 6*A*c**2*d**2 + 2*B*b**2*d*e - 3*B*b*c*d**2))/(2*A*b**2*c*e**2 - 12*A*b*c**2*d*e + 12*A*c**3*d**2 + 4*B*
b**2*c*d*e - 6*B*b*c**2*d**2))/b**5

________________________________________________________________________________________