\(\int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx\) [1020]

Optimal result

Integrand size = 18, antiderivative size = 77 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=-\frac {(b d-a e) (B d-A e)}{4 e^3 (d+e x)^4}+\frac {2 b B d-A b e-a B e}{3 e^3 (d+e x)^3}-\frac {b B}{2 e^3 (d+e x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, 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} \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=\frac {-a B e-A b e+2 b B d}{3 e^3 (d+e x)^3}-\frac {(b d-a e) (B d-A e)}{4 e^3 (d+e x)^4}-\frac {b B}{2 e^3 (d+e x)^2} \]

[In]

Int[((a + b*x)*(A + B*x))/(d + e*x)^5,x]

[Out]

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

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 d+a e) (-B d+A e)}{e^2 (d+e x)^5}+\frac {-2 b B d+A b e+a B e}{e^2 (d+e x)^4}+\frac {b B}{e^2 (d+e x)^3}\right ) \, dx \\ & = -\frac {(b d-a e) (B d-A e)}{4 e^3 (d+e x)^4}+\frac {2 b B d-A b e-a B e}{3 e^3 (d+e x)^3}-\frac {b B}{2 e^3 (d+e x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=-\frac {a e (3 A e+B (d+4 e x))+b \left (A e (d+4 e x)+B \left (d^2+4 d e x+6 e^2 x^2\right )\right )}{12 e^3 (d+e x)^4} \]

[In]

Integrate[((a + b*x)*(A + B*x))/(d + e*x)^5,x]

[Out]

-1/12*(a*e*(3*A*e + B*(d + 4*e*x)) + b*(A*e*(d + 4*e*x) + B*(d^2 + 4*d*e*x + 6*e^2*x^2)))/(e^3*(d + e*x)^4)

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {-\frac {b B \,x^{2}}{2 e}-\frac {\left (A b e +B a e +B b d \right ) x}{3 e^{2}}-\frac {3 A a \,e^{2}+A b d e +B a d e +b B \,d^{2}}{12 e^{3}}}{\left (e x +d \right )^{4}}\)	\(68\)
gosper	\(-\frac {6 b B \,x^{2} e^{2}+4 A x b \,e^{2}+4 B x a \,e^{2}+4 B x b d e +3 A a \,e^{2}+A b d e +B a d e +b B \,d^{2}}{12 e^{3} \left (e x +d \right )^{4}}\)	\(70\)
parallelrisch	\(-\frac {6 b B \,x^{2} e^{3}+4 A b \,e^{3} x +4 B a \,e^{3} x +4 B b d \,e^{2} x +3 A a \,e^{3}+A b d \,e^{2}+B a d \,e^{2}+b B \,d^{2} e}{12 e^{4} \left (e x +d \right )^{4}}\)	\(77\)
norman	\(\frac {-\frac {b B \,x^{2}}{2 e}-\frac {\left (A b \,e^{2}+B a \,e^{2}+b B d e \right ) x}{3 e^{3}}-\frac {3 A a \,e^{3}+A b d \,e^{2}+B a d \,e^{2}+b B \,d^{2} e}{12 e^{4}}}{\left (e x +d \right )^{4}}\)	\(78\)
default	\(-\frac {A b e +B a e -2 B b d}{3 e^{3} \left (e x +d \right )^{3}}-\frac {b B}{2 e^{3} \left (e x +d \right )^{2}}-\frac {A a \,e^{2}-A b d e -B a d e +b B \,d^{2}}{4 e^{3} \left (e x +d \right )^{4}}\)	\(79\)

[In]

int((b*x+a)*(B*x+A)/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

(-1/2*b*B/e*x^2-1/3/e^2*(A*b*e+B*a*e+B*b*d)*x-1/12/e^3*(3*A*a*e^2+A*b*d*e+B*a*d*e+B*b*d^2))/(e*x+d)^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=-\frac {6 \, B b e^{2} x^{2} + B b d^{2} + 3 \, A a e^{2} + {\left (B a + A b\right )} d e + 4 \, {\left (B b d e + {\left (B a + A b\right )} e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)^5,x, algorithm="fricas")

[Out]

-1/12*(6*B*b*e^2*x^2 + B*b*d^2 + 3*A*a*e^2 + (B*a + A*b)*d*e + 4*(B*b*d*e + (B*a + A*b)*e^2)*x)/(e^7*x^4 + 4*d
*e^6*x^3 + 6*d^2*e^5*x^2 + 4*d^3*e^4*x + d^4*e^3)

Sympy [A] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.52 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=\frac {- 3 A a e^{2} - A b d e - B a d e - B b d^{2} - 6 B b e^{2} x^{2} + x \left (- 4 A b e^{2} - 4 B a e^{2} - 4 B b d e\right )}{12 d^{4} e^{3} + 48 d^{3} e^{4} x + 72 d^{2} e^{5} x^{2} + 48 d e^{6} x^{3} + 12 e^{7} x^{4}} \]

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)**5,x)

[Out]

(-3*A*a*e**2 - A*b*d*e - B*a*d*e - B*b*d**2 - 6*B*b*e**2*x**2 + x*(-4*A*b*e**2 - 4*B*a*e**2 - 4*B*b*d*e))/(12*
d**4*e**3 + 48*d**3*e**4*x + 72*d**2*e**5*x**2 + 48*d*e**6*x**3 + 12*e**7*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=-\frac {6 \, B b e^{2} x^{2} + B b d^{2} + 3 \, A a e^{2} + {\left (B a + A b\right )} d e + 4 \, {\left (B b d e + {\left (B a + A b\right )} e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)^5,x, algorithm="maxima")

[Out]

-1/12*(6*B*b*e^2*x^2 + B*b*d^2 + 3*A*a*e^2 + (B*a + A*b)*d*e + 4*(B*b*d*e + (B*a + A*b)*e^2)*x)/(e^7*x^4 + 4*d
*e^6*x^3 + 6*d^2*e^5*x^2 + 4*d^3*e^4*x + d^4*e^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=-\frac {\frac {3 \, A a}{{\left (e x + d\right )}^{4}} + \frac {6 \, B b}{{\left (e x + d\right )}^{2} e^{2}} - \frac {8 \, B b d}{{\left (e x + d\right )}^{3} e^{2}} + \frac {3 \, B b d^{2}}{{\left (e x + d\right )}^{4} e^{2}} + \frac {4 \, B a}{{\left (e x + d\right )}^{3} e} + \frac {4 \, A b}{{\left (e x + d\right )}^{3} e} - \frac {3 \, B a d}{{\left (e x + d\right )}^{4} e} - \frac {3 \, A b d}{{\left (e x + d\right )}^{4} e}}{12 \, e} \]

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)^5,x, algorithm="giac")

[Out]

-1/12*(3*A*a/(e*x + d)^4 + 6*B*b/((e*x + d)^2*e^2) - 8*B*b*d/((e*x + d)^3*e^2) + 3*B*b*d^2/((e*x + d)^4*e^2) +
4*B*a/((e*x + d)^3*e) + 4*A*b/((e*x + d)^3*e) - 3*B*a*d/((e*x + d)^4*e) - 3*A*b*d/((e*x + d)^4*e))/e

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^5} \, dx=-\frac {\frac {3\,A\,a\,e^2+B\,b\,d^2+A\,b\,d\,e+B\,a\,d\,e}{12\,e^3}+\frac {x\,\left (A\,b\,e+B\,a\,e+B\,b\,d\right )}{3\,e^2}+\frac {B\,b\,x^2}{2\,e}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]

[In]

int(((A + B*x)*(a + b*x))/(d + e*x)^5,x)

[Out]

-((3*A*a*e^2 + B*b*d^2 + A*b*d*e + B*a*d*e)/(12*e^3) + (x*(A*b*e + B*a*e + B*b*d))/(3*e^2) + (B*b*x^2)/(2*e))/
(d^4 + e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x)