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

Optimal. Leaf size=168 \[ -\frac {A b e-3 A c d+b B d}{b^4 x}-\frac {(b B-A c) (c d-b e)}{2 b^3 (b+c x)^2}-\frac {A d}{2 b^3 x^2}+\frac {\log (x) \left (-3 b c (A e+B d)+6 A c^2 d+b^2 B e\right )}{b^5}-\frac {\log (b+c x) \left (-3 b c (A e+B d)+6 A c^2 d+b^2 B e\right )}{b^5}+\frac {-2 b c (A e+B d)+3 A c^2 d+b^2 B e}{b^4 (b+c x)} \]

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Rubi [A]  time = 0.21, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {771} \begin {gather*} \frac {-2 b c (A e+B d)+3 A c^2 d+b^2 B e}{b^4 (b+c x)}+\frac {\log (x) \left (-3 b c (A e+B d)+6 A c^2 d+b^2 B e\right )}{b^5}-\frac {\log (b+c x) \left (-3 b c (A e+B d)+6 A c^2 d+b^2 B e\right )}{b^5}-\frac {A b e-3 A c d+b B d}{b^4 x}-\frac {(b B-A c) (c d-b e)}{2 b^3 (b+c x)^2}-\frac {A d}{2 b^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.14, size = 162, normalized size = 0.96 \begin {gather*} \frac {\frac {2 b \left (-2 b c (A e+B d)+3 A c^2 d+b^2 B e\right )}{b+c x}+2 \log (x) \left (-3 b c (A e+B d)+6 A c^2 d+b^2 B e\right )-2 \log (b+c x) \left (-3 b c (A e+B d)+6 A c^2 d+b^2 B e\right )+\frac {b^2 (b B-A c) (b e-c d)}{(b+c x)^2}-\frac {A b^2 d}{x^2}-\frac {2 b (A b e-3 A c d+b B d)}{x}}{2 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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fricas [B]  time = 0.42, size = 410, normalized size = 2.44 \begin {gather*} -\frac {A b^{4} d + 2 \, {\left (3 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d - {\left (B b^{3} c - 3 \, A b^{2} c^{2}\right )} e\right )} x^{3} + 3 \, {\left (3 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d - {\left (B b^{4} - 3 \, A b^{3} c\right )} e\right )} x^{2} + 2 \, {\left (A b^{4} e + {\left (B b^{4} - 2 \, A b^{3} c\right )} d\right )} x - 2 \, {\left ({\left (3 \, {\left (B b c^{3} - 2 \, A c^{4}\right )} d - {\left (B b^{2} c^{2} - 3 \, A b c^{3}\right )} e\right )} x^{4} + 2 \, {\left (3 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d - {\left (B b^{3} c - 3 \, A b^{2} c^{2}\right )} e\right )} x^{3} + {\left (3 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d - {\left (B b^{4} - 3 \, A b^{3} c\right )} e\right )} x^{2}\right )} \log \left (c x + b\right ) + 2 \, {\left ({\left (3 \, {\left (B b c^{3} - 2 \, A c^{4}\right )} d - {\left (B b^{2} c^{2} - 3 \, A b c^{3}\right )} e\right )} x^{4} + 2 \, {\left (3 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d - {\left (B b^{3} c - 3 \, A b^{2} c^{2}\right )} e\right )} x^{3} + {\left (3 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d - {\left (B b^{4} - 3 \, A b^{3} c\right )} e\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left (b^{5} c^{2} x^{4} + 2 \, b^{6} c x^{3} + b^{7} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 225, normalized size = 1.34 \begin {gather*} -\frac {{\left (3 \, B b c d - 6 \, A c^{2} d - B b^{2} e + 3 \, A b c e\right )} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {{\left (3 \, B b c^{2} d - 6 \, A c^{3} d - B b^{2} c e + 3 \, A b c^{2} e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{5} c} - \frac {6 \, B b c^{2} d x^{3} - 12 \, A c^{3} d x^{3} - 2 \, B b^{2} c x^{3} e + 6 \, A b c^{2} x^{3} e + 9 \, B b^{2} c d x^{2} - 18 \, A b c^{2} d x^{2} - 3 \, B b^{3} x^{2} e + 9 \, A b^{2} c x^{2} e + 2 \, B b^{3} d x - 4 \, A b^{2} c d x + 2 \, A b^{3} x e + A b^{3} d}{2 \, {\left (c x^{2} + b x\right )}^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 261, normalized size = 1.55 \begin {gather*} -\frac {A c e}{2 \left (c x +b \right )^{2} b^{2}}+\frac {A \,c^{2} d}{2 \left (c x +b \right )^{2} b^{3}}+\frac {B e}{2 \left (c x +b \right )^{2} b}-\frac {B c d}{2 \left (c x +b \right )^{2} b^{2}}-\frac {2 A c e}{\left (c x +b \right ) b^{3}}+\frac {3 A \,c^{2} d}{\left (c x +b \right ) b^{4}}-\frac {3 A c e \ln \relax (x )}{b^{4}}+\frac {3 A c e \ln \left (c x +b \right )}{b^{4}}+\frac {6 A \,c^{2} d \ln \relax (x )}{b^{5}}-\frac {6 A \,c^{2} d \ln \left (c x +b \right )}{b^{5}}+\frac {B e}{\left (c x +b \right ) b^{2}}-\frac {2 B c d}{\left (c x +b \right ) b^{3}}+\frac {B e \ln \relax (x )}{b^{3}}-\frac {B e \ln \left (c x +b \right )}{b^{3}}-\frac {3 B c d \ln \relax (x )}{b^{4}}+\frac {3 B c d \ln \left (c x +b \right )}{b^{4}}-\frac {A e}{b^{3} x}+\frac {3 A c d}{b^{4} x}-\frac {B d}{b^{3} x}-\frac {A d}{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)/(c*x^2+b*x)^3,x)

[Out]

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

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maxima [A]  time = 0.50, size = 217, normalized size = 1.29 \begin {gather*} -\frac {A b^{3} d + 2 \, {\left (3 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c - 3 \, A b c^{2}\right )} e\right )} x^{3} + 3 \, {\left (3 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} d - {\left (B b^{3} - 3 \, A b^{2} c\right )} e\right )} x^{2} + 2 \, {\left (A b^{3} e + {\left (B b^{3} - 2 \, A b^{2} c\right )} d\right )} x}{2 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} + \frac {{\left (3 \, {\left (B b c - 2 \, A c^{2}\right )} d - {\left (B b^{2} - 3 \, A b c\right )} e\right )} \log \left (c x + b\right )}{b^{5}} - \frac {{\left (3 \, {\left (B b c - 2 \, A c^{2}\right )} d - {\left (B b^{2} - 3 \, A b c\right )} 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)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.53, size = 223, normalized size = 1.33 \begin {gather*} -\frac {\frac {x\,\left (A\,b\,e-2\,A\,c\,d+B\,b\,d\right )}{b^2}-\frac {3\,x^2\,\left (6\,A\,c^2\,d+B\,b^2\,e-3\,A\,b\,c\,e-3\,B\,b\,c\,d\right )}{2\,b^3}+\frac {A\,d}{2\,b}-\frac {c\,x^3\,\left (6\,A\,c^2\,d+B\,b^2\,e-3\,A\,b\,c\,e-3\,B\,b\,c\,d\right )}{b^4}}{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 (6\,A\,c^2\,d-b\,\left (3\,A\,c\,e+3\,B\,c\,d\right )+B\,b^2\,e\right )}{b\,\left (6\,A\,c^2\,d+B\,b^2\,e-3\,A\,b\,c\,e-3\,B\,b\,c\,d\right )}\right )\,\left (6\,A\,c^2\,d-b\,\left (3\,A\,c\,e+3\,B\,c\,d\right )+B\,b^2\,e\right )}{b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.87, size = 449, normalized size = 2.67 \begin {gather*} \frac {- A b^{3} d + x^{3} \left (- 6 A b c^{2} e + 12 A c^{3} d + 2 B b^{2} c e - 6 B b c^{2} d\right ) + x^{2} \left (- 9 A b^{2} c e + 18 A b c^{2} d + 3 B b^{3} e - 9 B b^{2} c d\right ) + x \left (- 2 A b^{3} e + 4 A b^{2} c d - 2 B b^{3} d\right )}{2 b^{6} x^{2} + 4 b^{5} c x^{3} + 2 b^{4} c^{2} x^{4}} + \frac {\left (- 3 A b c e + 6 A c^{2} d + B b^{2} e - 3 B b c d\right ) \log {\left (x + \frac {- 3 A b^{2} c e + 6 A b c^{2} d + B b^{3} e - 3 B b^{2} c d - b \left (- 3 A b c e + 6 A c^{2} d + B b^{2} e - 3 B b c d\right )}{- 6 A b c^{2} e + 12 A c^{3} d + 2 B b^{2} c e - 6 B b c^{2} d} \right )}}{b^{5}} - \frac {\left (- 3 A b c e + 6 A c^{2} d + B b^{2} e - 3 B b c d\right ) \log {\left (x + \frac {- 3 A b^{2} c e + 6 A b c^{2} d + B b^{3} e - 3 B b^{2} c d + b \left (- 3 A b c e + 6 A c^{2} d + B b^{2} e - 3 B b c d\right )}{- 6 A b c^{2} e + 12 A c^{3} d + 2 B b^{2} c e - 6 B b c^{2} d} \right )}}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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