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

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

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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} \begin {gather*} -\frac {B d-A e}{e (d+e x) (b d-a e)}+\frac {(A b-a B) \log (a+b x)}{(b d-a e)^2}-\frac {(A b-a B) \log (d+e x)}{(b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.11, size = 80, normalized size = 0.98 \begin {gather*} \frac {B d-A e}{e (d+e x) (a e-b d)}+\frac {(A b-a B) \log (a+b x)}{(b d-a e)^2}+\frac {(a B-A b) \log (d+e x)}{(b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.97, size = 148, normalized size = 1.80 \begin {gather*} -\frac {B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e + {\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \log \left (b x + a\right ) - {\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \log \left (e x + d\right )}{b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3} + {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.39, size = 108, normalized size = 1.32 \begin {gather*} -\frac {{\left (B a e - A b e\right )} \log \left ({\left | b - \frac {b d}{x e + d} + \frac {a e}{x e + d} \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} - \frac {\frac {B d}{x e + d} - \frac {A e}{x e + d}}{b d e - a e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 119, normalized size = 1.45 \begin {gather*} -\frac {{\left (B a - A b\right )} \log \left (b x + a\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} + \frac {{\left (B a - A b\right )} \log \left (e x + d\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} - \frac {B d - A e}{b d^{2} e - a d e^{2} + {\left (b d e^{2} - a e^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.19, size = 95, normalized size = 1.16 \begin {gather*} \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\,b-B\,a\right )}{{\left (a\,e-b\,d\right )}^2}-\frac {A\,e-B\,d}{e\,\left (a\,e-b\,d\right )\,\left (d+e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.22, size = 355, normalized size = 4.33 \begin {gather*} \frac {\left (- A b + B a\right ) \log {\left (x + \frac {- A a b e - A b^{2} d + B a^{2} e + B a b d - \frac {a^{3} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac {3 a^{2} b d e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac {3 a b^{2} d^{2} e \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac {b^{3} d^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}}}{- 2 A b^{2} e + 2 B a b e} \right )}}{\left (a e - b d\right )^{2}} - \frac {\left (- A b + B a\right ) \log {\left (x + \frac {- A a b e - A b^{2} d + B a^{2} e + B a b d + \frac {a^{3} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac {3 a^{2} b d e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac {3 a b^{2} d^{2} e \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac {b^{3} d^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}}}{- 2 A b^{2} e + 2 B a b e} \right )}}{\left (a e - b d\right )^{2}} + \frac {- A e + B d}{a d e^{2} - b d^{2} e + x \left (a e^{3} - b d e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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