\(\int \frac {(A+B x) (d+e x)}{a+b x} \, dx\) [1115]

Optimal result

Integrand size = 18, antiderivative size = 59 \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=\frac {B (b d-a e) x}{b^2}+\frac {e (A+B x)^2}{2 b B}+\frac {(A b-a B) (b d-a e) \log (a+b x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, 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} \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=\frac {(A b-a B) (b d-a e) \log (a+b x)}{b^3}+\frac {B x (b d-a e)}{b^2}+\frac {e (A+B x)^2}{2 b B} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=\frac {b x (-2 a B e+b (2 B d+2 A e+B e x))+2 (A b-a B) (b d-a e) \log (a+b x)}{2 b^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {\frac {1}{2} B b e \,x^{2}+A b e x -B a e x +B b d x}{b^{2}}+\frac {\left (-A a b e +A \,b^{2} d +B \,a^{2} e -B a b d \right ) \ln \left (b x +a \right )}{b^{3}}\)	\(66\)
norman	\(\frac {\left (A b e -B a e +B b d \right ) x}{b^{2}}+\frac {B e \,x^{2}}{2 b}-\frac {\left (A a b e -A \,b^{2} d -B \,a^{2} e +B a b d \right ) \ln \left (b x +a \right )}{b^{3}}\)	\(67\)
risch	\(\frac {B e \,x^{2}}{2 b}+\frac {A e x}{b}-\frac {B a e x}{b^{2}}+\frac {B d x}{b}-\frac {\ln \left (b x +a \right ) A a e}{b^{2}}+\frac {\ln \left (b x +a \right ) A d}{b}+\frac {\ln \left (b x +a \right ) B \,a^{2} e}{b^{3}}-\frac {\ln \left (b x +a \right ) B a d}{b^{2}}\)	\(90\)
parallelrisch	\(-\frac {-B e \,x^{2} b^{2}+2 A \ln \left (b x +a \right ) a b e -2 A \ln \left (b x +a \right ) b^{2} d -2 A x \,b^{2} e -2 B \ln \left (b x +a \right ) a^{2} e +2 B \ln \left (b x +a \right ) a b d +2 B x a b e -2 B x \,b^{2} d}{2 b^{3}}\)	\(90\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=\frac {B b^{2} e x^{2} + 2 \, {\left (B b^{2} d - {\left (B a b - A b^{2}\right )} e\right )} x - 2 \, {\left ({\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e\right )} \log \left (b x + a\right )}{2 \, b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=\frac {B e x^{2}}{2 b} + x \left (\frac {A e}{b} - \frac {B a e}{b^{2}} + \frac {B d}{b}\right ) + \frac {\left (- A b + B a\right ) \left (a e - b d\right ) \log {\left (a + b x \right )}}{b^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.22 \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=\frac {B b e x^{2} + 2 \, {\left (B b d - {\left (B a - A b\right )} e\right )} x}{2 \, b^{2}} - \frac {{\left ({\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e\right )} \log \left (b x + a\right )}{b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=\frac {B b e x^{2} + 2 \, B b d x - 2 \, B a e x + 2 \, A b e x}{2 \, b^{2}} - \frac {{\left (B a b d - A b^{2} d - B a^{2} e + A a b e\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) (d+e x)}{a+b x} \, dx=x\,\left (\frac {A\,e+B\,d}{b}-\frac {B\,a\,e}{b^2}\right )+\frac {\ln \left (a+b\,x\right )\,\left (A\,b^2\,d+B\,a^2\,e-A\,a\,b\,e-B\,a\,b\,d\right )}{b^3}+\frac {B\,e\,x^2}{2\,b} \]

[In]

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

[Out]

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