\(\int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx\) [1017]

Optimal result

Integrand size = 18, antiderivative size = 63 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=\frac {b B x}{e^2}-\frac {(b d-a e) (B d-A e)}{e^3 (d+e x)}-\frac {(2 b B d-A b e-a B e) \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, 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} \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=-\frac {(b d-a e) (B d-A e)}{e^3 (d+e x)}-\frac {\log (d+e x) (-a B e-A b e+2 b B d)}{e^3}+\frac {b B x}{e^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=\frac {b B e x-\frac {(b d-a e) (B d-A e)}{d+e x}+(-2 b B d+A b e+a B e) \log (d+e x)}{e^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11

method	result	size
default	\(\frac {b B x}{e^{2}}-\frac {A a \,e^{2}-A b d e -B a d e +b B \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {\left (A b e +B a e -2 B b d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(70\)
norman	\(\frac {\frac {b B \,x^{2}}{e}-\frac {A a \,e^{2}-A b d e -B a d e +2 b B \,d^{2}}{e^{3}}}{e x +d}+\frac {\left (A b e +B a e -2 B b d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(75\)
risch	\(\frac {b B x}{e^{2}}-\frac {A a}{e \left (e x +d \right )}+\frac {A b d}{e^{2} \left (e x +d \right )}+\frac {B a d}{e^{2} \left (e x +d \right )}-\frac {b B \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {\ln \left (e x +d \right ) A b}{e^{2}}+\frac {\ln \left (e x +d \right ) B a}{e^{2}}-\frac {2 \ln \left (e x +d \right ) B b d}{e^{3}}\)	\(106\)
parallelrisch	\(\frac {A \ln \left (e x +d \right ) x b \,e^{2}+B \ln \left (e x +d \right ) x a \,e^{2}-2 B \ln \left (e x +d \right ) x b d e +b B \,x^{2} e^{2}+A \ln \left (e x +d \right ) b d e +B \ln \left (e x +d \right ) a d e -2 B \ln \left (e x +d \right ) b \,d^{2}-A a \,e^{2}+A b d e +B a d e -2 b B \,d^{2}}{e^{3} \left (e x +d \right )}\)	\(120\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=\frac {B b e^{2} x^{2} + B b d e x - B b d^{2} - A a e^{2} + {\left (B a + A b\right )} d e - {\left (2 \, B b d^{2} - {\left (B a + A b\right )} d e + {\left (2 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=\frac {B b x}{e^{2}} + \frac {- A a e^{2} + A b d e + B a d e - B b d^{2}}{d e^{3} + e^{4} x} + \frac {\left (A b e + B a e - 2 B b d\right ) \log {\left (d + e x \right )}}{e^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=\frac {B b x}{e^{2}} - \frac {B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e}{e^{4} x + d e^{3}} - \frac {{\left (2 \, B b d - {\left (B a + A b\right )} e\right )} \log \left (e x + d\right )}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.81 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=\frac {{\left (e x + d\right )} B b}{e^{3}} + \frac {{\left (2 \, B b d - B a e - A b e\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{3}} - \frac {\frac {B b d^{2} e}{e x + d} - \frac {B a d e^{2}}{e x + d} - \frac {A b d e^{2}}{e x + d} + \frac {A a e^{3}}{e x + d}}{e^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^3}-\frac {A\,a\,e^2+B\,b\,d^2-A\,b\,d\,e-B\,a\,d\,e}{e\,\left (x\,e^3+d\,e^2\right )}+\frac {B\,b\,x}{e^2} \]

[In]

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

[Out]

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