3.1151 \(\int \frac {A+B x}{(d+e x) (b x+c x^2)^2} \, dx\)

Optimal. Leaf size=147 \[ \frac {\log (x) (-A b e-2 A c d+b B d)}{b^3 d^2}+\frac {c (b B-A c)}{b^2 (b+c x) (c d-b e)}-\frac {A}{b^2 d x}+\frac {c \log (b+c x) \left (-b c (3 A e+B d)+2 A c^2 d+2 b^2 B e\right )}{b^3 (c d-b e)^2}-\frac {e^2 (B d-A e) \log (d+e x)}{d^2 (c d-b e)^2} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 147, 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} \[ \frac {c \log (b+c x) \left (-b c (3 A e+B d)+2 A c^2 d+2 b^2 B e\right )}{b^3 (c d-b e)^2}+\frac {\log (x) (-A b e-2 A c d+b B d)}{b^3 d^2}+\frac {c (b B-A c)}{b^2 (b+c x) (c d-b e)}-\frac {A}{b^2 d x}-\frac {e^2 (B d-A e) \log (d+e x)}{d^2 (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.16, size = 146, normalized size = 0.99 \[ \frac {\log (x) (-A b e-2 A c d+b B d)}{b^3 d^2}+\frac {c (A c-b B)}{b^2 (b+c x) (b e-c d)}-\frac {A}{b^2 d x}+\frac {c \log (b+c x) \left (-b c (3 A e+B d)+2 A c^2 d+2 b^2 B e\right )}{b^3 (c d-b e)^2}+\frac {e^2 (A e-B d) \log (d+e x)}{d^2 (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 67.63, size = 450, normalized size = 3.06 \[ -\frac {A b^{2} c^{2} d^{3} - 2 \, A b^{3} c d^{2} e + A b^{4} d e^{2} + {\left (A b^{3} c d e^{2} - {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d^{3} + {\left (B b^{3} c - 3 \, A b^{2} c^{2}\right )} d^{2} e\right )} x + {\left ({\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} - {\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} d^{2} e\right )} x^{2} + {\left ({\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d^{3} - {\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} d^{2} e\right )} x\right )} \log \left (c x + b\right ) + {\left ({\left (B b^{3} c d e^{2} - A b^{3} c e^{3}\right )} x^{2} + {\left (B b^{4} d e^{2} - A b^{4} e^{3}\right )} x\right )} \log \left (e x + d\right ) - {\left ({\left (B b^{3} c d e^{2} - A b^{3} c e^{3} + {\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} - {\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} d^{2} e\right )} x^{2} + {\left (B b^{4} d e^{2} - A b^{4} e^{3} + {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d^{3} - {\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} d^{2} e\right )} x\right )} \log \relax (x)}{{\left (b^{3} c^{3} d^{4} - 2 \, b^{4} c^{2} d^{3} e + b^{5} c d^{2} e^{2}\right )} x^{2} + {\left (b^{4} c^{2} d^{4} - 2 \, b^{5} c d^{3} e + b^{6} d^{2} e^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 268, normalized size = 1.82 \[ -\frac {{\left (B b c^{3} d - 2 \, A c^{4} d - 2 \, B b^{2} c^{2} e + 3 \, A b c^{3} e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c^{3} d^{2} - 2 \, b^{4} c^{2} d e + b^{5} c e^{2}} - \frac {{\left (B d e^{3} - A e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}} + \frac {{\left (B b d - 2 \, A c d - A b e\right )} \log \left ({\left | x \right |}\right )}{b^{3} d^{2}} - \frac {A b c^{2} d^{3} - 2 \, A b^{2} c d^{2} e + A b^{3} d e^{2} - {\left (B b c^{2} d^{3} - 2 \, A c^{3} d^{3} - B b^{2} c d^{2} e + 3 \, A b c^{2} d^{2} e - A b^{2} c d e^{2}\right )} x}{{\left (c d - b e\right )}^{2} {\left (c x + b\right )} b^{2} d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 248, normalized size = 1.69 \[ -\frac {3 A \,c^{2} e \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b^{2}}+\frac {2 A \,c^{3} d \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b^{3}}+\frac {A \,e^{3} \ln \left (e x +d \right )}{\left (b e -c d \right )^{2} d^{2}}+\frac {2 B c e \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b}-\frac {B \,c^{2} d \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b^{2}}-\frac {B \,e^{2} \ln \left (e x +d \right )}{\left (b e -c d \right )^{2} d}+\frac {A \,c^{2}}{\left (b e -c d \right ) \left (c x +b \right ) b^{2}}-\frac {B c}{\left (b e -c d \right ) \left (c x +b \right ) b}-\frac {A e \ln \relax (x )}{b^{2} d^{2}}-\frac {2 A c \ln \relax (x )}{b^{3} d}+\frac {B \ln \relax (x )}{b^{2} d}-\frac {A}{b^{2} d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 227, normalized size = 1.54 \[ -\frac {{\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} e\right )} \log \left (c x + b\right )}{b^{3} c^{2} d^{2} - 2 \, b^{4} c d e + b^{5} e^{2}} - \frac {{\left (B d e^{2} - A e^{3}\right )} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} - \frac {A b c d - A b^{2} e - {\left (A b c e + {\left (B b c - 2 \, A c^{2}\right )} d\right )} x}{{\left (b^{2} c^{2} d^{2} - b^{3} c d e\right )} x^{2} + {\left (b^{3} c d^{2} - b^{4} d e\right )} x} - \frac {{\left (A b e - {\left (B b - 2 \, A c\right )} d\right )} \log \relax (x)}{b^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.04, size = 201, normalized size = 1.37 \[ \frac {\ln \left (b+c\,x\right )\,\left (d\,\left (2\,A\,c^3-B\,b\,c^2\right )-3\,A\,b\,c^2\,e+2\,B\,b^2\,c\,e\right )}{b^5\,e^2-2\,b^4\,c\,d\,e+b^3\,c^2\,d^2}-\frac {\frac {A}{b\,d}+\frac {x\,\left (A\,b\,c\,e-2\,A\,c^2\,d+B\,b\,c\,d\right )}{b^2\,d\,\left (b\,e-c\,d\right )}}{c\,x^2+b\,x}+\frac {\ln \left (d+e\,x\right )\,\left (A\,e^3-B\,d\,e^2\right )}{b^2\,d^2\,e^2-2\,b\,c\,d^3\,e+c^2\,d^4}-\frac {\ln \relax (x)\,\left (b\,\left (A\,e-B\,d\right )+2\,A\,c\,d\right )}{b^3\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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