3.1017 \(\int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=63 \[ -\frac {(b d-a e) (B d-A e)}{e^3 (d+e x)}-\frac {\log (d+e x) (-a B e-A b e+2 b B d)}{e^3}+\frac {b B x}{e^2} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \[ -\frac {(b d-a e) (B d-A e)}{e^3 (d+e x)}-\frac {\log (d+e x) (-a B e-A b e+2 b B d)}{e^3}+\frac {b B x}{e^2} \]

Antiderivative was successfully verified.

[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 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 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*}

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Mathematica [A]  time = 0.05, size = 56, normalized size = 0.89 \[ \frac {-\frac {(b d-a e) (B d-A e)}{d+e x}+\log (d+e x) (a B e+A b e-2 b B d)+b B e x}{e^3} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.81, size = 102, normalized size = 1.62 \[ \frac {B b e^{2} x^{2} + B b d e x - B b d^{2} - A a e^{2} + {\left (B a + A b\right )} d e - {\left (2 \, B b d^{2} - {\left (B a + A b\right )} d e + {\left (2 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \]

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*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)

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giac [A]  time = 1.19, size = 116, normalized size = 1.84 \[ {\left (x e + d\right )} B b e^{\left (-3\right )} + {\left (2 \, B b d - B a e - A b e\right )} e^{\left (-3\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - {\left (\frac {B b d^{2} e}{x e + d} - \frac {B a d e^{2}}{x e + d} - \frac {A b d e^{2}}{x e + d} + \frac {A a e^{3}}{x e + d}\right )} e^{\left (-4\right )} \]

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]

(x*e + d)*B*b*e^(-3) + (2*B*b*d - B*a*e - A*b*e)*e^(-3)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) - (B*b*d^2*e/(x*e
+ d) - B*a*d*e^2/(x*e + d) - A*b*d*e^2/(x*e + d) + A*a*e^3/(x*e + d))*e^(-4)

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maple [A]  time = 0.01, size = 106, normalized size = 1.68 \[ -\frac {A a}{\left (e x +d \right ) e}+\frac {A b d}{\left (e x +d \right ) e^{2}}+\frac {A b \ln \left (e x +d \right )}{e^{2}}+\frac {B a d}{\left (e x +d \right ) e^{2}}+\frac {B a \ln \left (e x +d \right )}{e^{2}}-\frac {B b \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 B b d \ln \left (e x +d \right )}{e^{3}}+\frac {B b x}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 74, normalized size = 1.17 \[ \frac {B b x}{e^{2}} - \frac {B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e}{e^{4} x + d e^{3}} - \frac {{\left (2 \, B b d - {\left (B a + A b\right )} e\right )} \log \left (e x + d\right )}{e^{3}} \]

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*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

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mupad [B]  time = 1.09, size = 75, normalized size = 1.19 \[ \frac {\ln \left (d+e\,x\right )\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^3}-\frac {A\,a\,e^2+B\,b\,d^2-A\,b\,d\,e-B\,a\,d\,e}{e\,\left (x\,e^3+d\,e^2\right )}+\frac {B\,b\,x}{e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.46, size = 71, normalized size = 1.13 \[ \frac {B b x}{e^{2}} + \frac {- A a e^{2} + A b d e + B a d e - B b d^{2}}{d e^{3} + e^{4} x} + \frac {\left (A b e + B a e - 2 B b d\right ) \log {\left (d + e x \right )}}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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