\(\int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx\) [1126]

Optimal result

Integrand size = 18, antiderivative size = 60 \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=\frac {B e x}{b^2}-\frac {(A b-a B) (b d-a e)}{b^3 (a+b x)}+\frac {(b B d+A b e-2 a B e) \log (a+b x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=-\frac {(A b-a B) (b d-a e)}{b^3 (a+b x)}+\frac {\log (a+b x) (-2 a B e+A b e+b B d)}{b^3}+\frac {B e x}{b^2} \]

[In]

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

[Out]

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

Rule 78

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=\frac {b B e x-\frac {(A b-a B) (b d-a e)}{a+b x}+(b B d+A b e-2 a B e) \log (a+b x)}{b^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17

method	result	size
default	\(\frac {B e x}{b^{2}}+\frac {\left (A b e -2 B a e +B b d \right ) \ln \left (b x +a \right )}{b^{3}}-\frac {-A a b e +A \,b^{2} d +B \,a^{2} e -B a b d}{b^{3} \left (b x +a \right )}\)	\(70\)
norman	\(\frac {\frac {A a b e -A \,b^{2} d -2 B \,a^{2} e +B a b d}{b^{3}}+\frac {B e \,x^{2}}{b}}{b x +a}+\frac {\left (A b e -2 B a e +B b d \right ) \ln \left (b x +a \right )}{b^{3}}\)	\(73\)
risch	\(\frac {B e x}{b^{2}}+\frac {A a e}{b^{2} \left (b x +a \right )}-\frac {A d}{b \left (b x +a \right )}-\frac {B \,a^{2} e}{b^{3} \left (b x +a \right )}+\frac {B a d}{b^{2} \left (b x +a \right )}+\frac {\ln \left (b x +a \right ) A e}{b^{2}}-\frac {2 \ln \left (b x +a \right ) B a e}{b^{3}}+\frac {\ln \left (b x +a \right ) B d}{b^{2}}\)	\(106\)
parallelrisch	\(\frac {A \ln \left (b x +a \right ) x \,b^{2} e -2 B \ln \left (b x +a \right ) x a b e +B \ln \left (b x +a \right ) x \,b^{2} d +B e \,x^{2} b^{2}+A \ln \left (b x +a \right ) a b e -2 B \ln \left (b x +a \right ) a^{2} e +B \ln \left (b x +a \right ) a b d +A a b e -A \,b^{2} d -2 B \,a^{2} e +B a b d}{b^{3} \left (b x +a \right )}\)	\(120\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=\frac {B b^{2} e x^{2} + B a b e x + {\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e + {\left (B a b d - {\left (2 \, B a^{2} - A a b\right )} e + {\left (B b^{2} d - {\left (2 \, B a b - A b^{2}\right )} e\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=\frac {B e x}{b^{2}} + \frac {A a b e - A b^{2} d - B a^{2} e + B a b d}{a b^{3} + b^{4} x} - \frac {\left (- A b e + 2 B a e - B b d\right ) \log {\left (a + b x \right )}}{b^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=\frac {B e x}{b^{2}} + \frac {{\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e}{b^{4} x + a b^{3}} + \frac {{\left (B b d - {\left (2 \, B a - A b\right )} e\right )} \log \left (b x + a\right )}{b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.87 \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )} B e}{b^{3}} - \frac {{\left (B b d - 2 \, B a e + A b e\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {\frac {B a b^{2} d}{b x + a} - \frac {A b^{3} d}{b x + a} - \frac {B a^{2} b e}{b x + a} + \frac {A a b^{2} e}{b x + a}}{b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) (d+e x)}{(a+b x)^2} \, dx=\frac {\ln \left (a+b\,x\right )\,\left (A\,b\,e-2\,B\,a\,e+B\,b\,d\right )}{b^3}-\frac {A\,b^2\,d+B\,a^2\,e-A\,a\,b\,e-B\,a\,b\,d}{b\,\left (x\,b^3+a\,b^2\right )}+\frac {B\,e\,x}{b^2} \]

[In]

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

[Out]

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