3.1128 \(\int \frac {A+B x}{(a+b x)^2 (d+e x)} \, dx\)

Optimal. Leaf size=82 \[ -\frac {A b-a B}{b (a+b x) (b d-a e)}+\frac {\log (a+b x) (B d-A e)}{(b d-a e)^2}-\frac {(B d-A e) \log (d+e x)}{(b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ -\frac {A b-a B}{b (a+b x) (b d-a e)}+\frac {\log (a+b x) (B d-A e)}{(b d-a e)^2}-\frac {(B d-A e) \log (d+e x)}{(b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 69, normalized size = 0.84 \[ \frac {\frac {(a B-A b) (b d-a e)}{b (a+b x)}+\log (a+b x) (B d-A e)+(A e-B d) \log (d+e x)}{(b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 157, normalized size = 1.91 \[ \frac {{\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e + {\left (B a b d - A a b e + {\left (B b^{2} d - A b^{2} e\right )} x\right )} \log \left (b x + a\right ) - {\left (B a b d - A a b e + {\left (B b^{2} d - A b^{2} e\right )} x\right )} \log \left (e x + d\right )}{a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.24, size = 104, normalized size = 1.27 \[ -\frac {{\left (B b d - A b e\right )} \log \left ({\left | \frac {b d}{b x + a} - \frac {a e}{b x + a} + e \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} + \frac {\frac {B a}{b x + a} - \frac {A b}{b x + a}}{b^{2} d - a b e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 123, normalized size = 1.50 \[ -\frac {A e \ln \left (b x +a \right )}{\left (a e -b d \right )^{2}}+\frac {A e \ln \left (e x +d \right )}{\left (a e -b d \right )^{2}}+\frac {B d \ln \left (b x +a \right )}{\left (a e -b d \right )^{2}}-\frac {B d \ln \left (e x +d \right )}{\left (a e -b d \right )^{2}}+\frac {A}{\left (a e -b d \right ) \left (b x +a \right )}-\frac {B a}{\left (a e -b d \right ) \left (b x +a \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 118, normalized size = 1.44 \[ \frac {{\left (B d - A e\right )} \log \left (b x + a\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} - \frac {{\left (B d - A e\right )} \log \left (e x + d\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} + \frac {B a - A b}{a b^{2} d - a^{2} b e + {\left (b^{3} d - a b^{2} e\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.15, size = 94, normalized size = 1.15 \[ \frac {A\,b-B\,a}{b\,\left (a\,e-b\,d\right )\,\left (a+b\,x\right )}-\frac {2\,\mathrm {atanh}\left (\frac {a^2\,e^2-b^2\,d^2}{{\left (a\,e-b\,d\right )}^2}+\frac {2\,b\,e\,x}{a\,e-b\,d}\right )\,\left (A\,e-B\,d\right )}{{\left (a\,e-b\,d\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*b - B*a)/(b*(a*e - b*d)*(a + b*x)) - (2*atanh((a^2*e^2 - b^2*d^2)/(a*e - b*d)^2 + (2*b*e*x)/(a*e - b*d))*(A
*e - B*d))/(a*e - b*d)^2

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sympy [B]  time = 1.27, size = 355, normalized size = 4.33 \[ \frac {A b - B a}{a^{2} b e - a b^{2} d + x \left (a b^{2} e - b^{3} d\right )} - \frac {\left (- A e + B d\right ) \log {\left (x + \frac {- A a e^{2} - A b d e + B a d e + B b d^{2} - \frac {a^{3} e^{3} \left (- A e + B d\right )}{\left (a e - b d\right )^{2}} + \frac {3 a^{2} b d e^{2} \left (- A e + B d\right )}{\left (a e - b d\right )^{2}} - \frac {3 a b^{2} d^{2} e \left (- A e + B d\right )}{\left (a e - b d\right )^{2}} + \frac {b^{3} d^{3} \left (- A e + B d\right )}{\left (a e - b d\right )^{2}}}{- 2 A b e^{2} + 2 B b d e} \right )}}{\left (a e - b d\right )^{2}} + \frac {\left (- A e + B d\right ) \log {\left (x + \frac {- A a e^{2} - A b d e + B a d e + B b d^{2} + \frac {a^{3} e^{3} \left (- A e + B d\right )}{\left (a e - b d\right )^{2}} - \frac {3 a^{2} b d e^{2} \left (- A e + B d\right )}{\left (a e - b d\right )^{2}} + \frac {3 a b^{2} d^{2} e \left (- A e + B d\right )}{\left (a e - b d\right )^{2}} - \frac {b^{3} d^{3} \left (- A e + B d\right )}{\left (a e - b d\right )^{2}}}{- 2 A b e^{2} + 2 B b d e} \right )}}{\left (a e - b d\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*b - B*a)/(a**2*b*e - a*b**2*d + x*(a*b**2*e - b**3*d)) - (-A*e + B*d)*log(x + (-A*a*e**2 - A*b*d*e + B*a*d*
e + B*b*d**2 - a**3*e**3*(-A*e + B*d)/(a*e - b*d)**2 + 3*a**2*b*d*e**2*(-A*e + B*d)/(a*e - b*d)**2 - 3*a*b**2*
d**2*e*(-A*e + B*d)/(a*e - b*d)**2 + b**3*d**3*(-A*e + B*d)/(a*e - b*d)**2)/(-2*A*b*e**2 + 2*B*b*d*e))/(a*e -
b*d)**2 + (-A*e + B*d)*log(x + (-A*a*e**2 - A*b*d*e + B*a*d*e + B*b*d**2 + a**3*e**3*(-A*e + B*d)/(a*e - b*d)*
*2 - 3*a**2*b*d*e**2*(-A*e + B*d)/(a*e - b*d)**2 + 3*a*b**2*d**2*e*(-A*e + B*d)/(a*e - b*d)**2 - b**3*d**3*(-A
*e + B*d)/(a*e - b*d)**2)/(-2*A*b*e**2 + 2*B*b*d*e))/(a*e - b*d)**2

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