\(\int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx\) [184]

Optimal result

Integrand size = 16, antiderivative size = 113 \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=\frac {a^2 (3 A b-4 a B) x}{b^5}-\frac {a (2 A b-3 a B) x^2}{2 b^4}+\frac {(A b-2 a B) x^3}{3 b^3}+\frac {B x^4}{4 b^2}-\frac {a^4 (A b-a B)}{b^6 (a+b x)}-\frac {a^3 (4 A b-5 a B) \log (a+b x)}{b^6} \]

[Out]

a^2*(3*A*b-4*B*a)*x/b^5-1/2*a*(2*A*b-3*B*a)*x^2/b^4+1/3*(A*b-2*B*a)*x^3/b^3+1/4*B*x^4/b^2-a^4*(A*b-B*a)/b^6/(b
*x+a)-a^3*(4*A*b-5*B*a)*ln(b*x+a)/b^6

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=-\frac {a^4 (A b-a B)}{b^6 (a+b x)}-\frac {a^3 (4 A b-5 a B) \log (a+b x)}{b^6}+\frac {a^2 x (3 A b-4 a B)}{b^5}-\frac {a x^2 (2 A b-3 a B)}{2 b^4}+\frac {x^3 (A b-2 a B)}{3 b^3}+\frac {B x^4}{4 b^2} \]

[In]

Int[(x^4*(A + B*x))/(a + b*x)^2,x]

[Out]

(a^2*(3*A*b - 4*a*B)*x)/b^5 - (a*(2*A*b - 3*a*B)*x^2)/(2*b^4) + ((A*b - 2*a*B)*x^3)/(3*b^3) + (B*x^4)/(4*b^2)
- (a^4*(A*b - a*B))/(b^6*(a + b*x)) - (a^3*(4*A*b - 5*a*B)*Log[a + b*x])/b^6

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 {a^2 (-3 A b+4 a B)}{b^5}+\frac {a (-2 A b+3 a B) x}{b^4}+\frac {(A b-2 a B) x^2}{b^3}+\frac {B x^3}{b^2}-\frac {a^4 (-A b+a B)}{b^5 (a+b x)^2}+\frac {a^3 (-4 A b+5 a B)}{b^5 (a+b x)}\right ) \, dx \\ & = \frac {a^2 (3 A b-4 a B) x}{b^5}-\frac {a (2 A b-3 a B) x^2}{2 b^4}+\frac {(A b-2 a B) x^3}{3 b^3}+\frac {B x^4}{4 b^2}-\frac {a^4 (A b-a B)}{b^6 (a+b x)}-\frac {a^3 (4 A b-5 a B) \log (a+b x)}{b^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=\frac {-12 a^2 b (-3 A b+4 a B) x+6 a b^2 (-2 A b+3 a B) x^2+4 b^3 (A b-2 a B) x^3+3 b^4 B x^4+\frac {12 a^4 (-A b+a B)}{a+b x}+12 a^3 (-4 A b+5 a B) \log (a+b x)}{12 b^6} \]

[In]

Integrate[(x^4*(A + B*x))/(a + b*x)^2,x]

[Out]

(-12*a^2*b*(-3*A*b + 4*a*B)*x + 6*a*b^2*(-2*A*b + 3*a*B)*x^2 + 4*b^3*(A*b - 2*a*B)*x^3 + 3*b^4*B*x^4 + (12*a^4
*(-(A*b) + a*B))/(a + b*x) + 12*a^3*(-4*A*b + 5*a*B)*Log[a + b*x])/(12*b^6)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {\frac {1}{4} b^{3} B \,x^{4}+\frac {1}{3} A \,b^{3} x^{3}-\frac {2}{3} B a \,b^{2} x^{3}-a A \,b^{2} x^{2}+\frac {3}{2} B \,a^{2} b \,x^{2}+3 a^{2} A b x -4 a^{3} B x}{b^{5}}-\frac {a^{3} \left (4 A b -5 B a \right ) \ln \left (b x +a \right )}{b^{6}}-\frac {a^{4} \left (A b -B a \right )}{b^{6} \left (b x +a \right )}\)	\(116\)
norman	\(\frac {\frac {B \,x^{5}}{4 b}-\frac {a \left (4 A \,a^{3} b -5 B \,a^{4}\right )}{b^{6}}+\frac {\left (4 A b -5 B a \right ) x^{4}}{12 b^{2}}-\frac {a \left (4 A b -5 B a \right ) x^{3}}{6 b^{3}}+\frac {a^{2} \left (4 A b -5 B a \right ) x^{2}}{2 b^{4}}}{b x +a}-\frac {a^{3} \left (4 A b -5 B a \right ) \ln \left (b x +a \right )}{b^{6}}\)	\(118\)
risch	\(\frac {B \,x^{4}}{4 b^{2}}+\frac {A \,x^{3}}{3 b^{2}}-\frac {2 B a \,x^{3}}{3 b^{3}}-\frac {a A \,x^{2}}{b^{3}}+\frac {3 B \,a^{2} x^{2}}{2 b^{4}}+\frac {3 a^{2} A x}{b^{4}}-\frac {4 a^{3} B x}{b^{5}}-\frac {4 a^{3} \ln \left (b x +a \right ) A}{b^{5}}+\frac {5 a^{4} \ln \left (b x +a \right ) B}{b^{6}}-\frac {a^{4} A}{b^{5} \left (b x +a \right )}+\frac {a^{5} B}{b^{6} \left (b x +a \right )}\)	\(133\)
parallelrisch	\(-\frac {-3 b^{5} B \,x^{5}-4 A \,b^{5} x^{4}+5 B a \,b^{4} x^{4}+8 A a \,b^{4} x^{3}-10 B \,a^{2} b^{3} x^{3}+48 A \ln \left (b x +a \right ) x \,a^{3} b^{2}-24 A \,a^{2} b^{3} x^{2}-60 B \ln \left (b x +a \right ) x \,a^{4} b +30 B \,a^{3} b^{2} x^{2}+48 A \ln \left (b x +a \right ) a^{4} b -60 B \ln \left (b x +a \right ) a^{5}+48 a^{4} b A -60 a^{5} B}{12 b^{6} \left (b x +a \right )}\)	\(156\)

[In]

int(x^4*(B*x+A)/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b^5*(1/4*b^3*B*x^4+1/3*A*b^3*x^3-2/3*B*a*b^2*x^3-a*A*b^2*x^2+3/2*B*a^2*b*x^2+3*a^2*A*b*x-4*a^3*B*x)-a^3*(4*A
*b-5*B*a)*ln(b*x+a)/b^6-a^4*(A*b-B*a)/b^6/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.45 \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=\frac {3 \, B b^{5} x^{5} + 12 \, B a^{5} - 12 \, A a^{4} b - {\left (5 \, B a b^{4} - 4 \, A b^{5}\right )} x^{4} + 2 \, {\left (5 \, B a^{2} b^{3} - 4 \, A a b^{4}\right )} x^{3} - 6 \, {\left (5 \, B a^{3} b^{2} - 4 \, A a^{2} b^{3}\right )} x^{2} - 12 \, {\left (4 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x + 12 \, {\left (5 \, B a^{5} - 4 \, A a^{4} b + {\left (5 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x\right )} \log \left (b x + a\right )}{12 \, {\left (b^{7} x + a b^{6}\right )}} \]

[In]

integrate(x^4*(B*x+A)/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.05 \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=\frac {B x^{4}}{4 b^{2}} + \frac {a^{3} \left (- 4 A b + 5 B a\right ) \log {\left (a + b x \right )}}{b^{6}} + x^{3} \left (\frac {A}{3 b^{2}} - \frac {2 B a}{3 b^{3}}\right ) + x^{2} \left (- \frac {A a}{b^{3}} + \frac {3 B a^{2}}{2 b^{4}}\right ) + x \left (\frac {3 A a^{2}}{b^{4}} - \frac {4 B a^{3}}{b^{5}}\right ) + \frac {- A a^{4} b + B a^{5}}{a b^{6} + b^{7} x} \]

[In]

integrate(x**4*(B*x+A)/(b*x+a)**2,x)

[Out]

B*x**4/(4*b**2) + a**3*(-4*A*b + 5*B*a)*log(a + b*x)/b**6 + x**3*(A/(3*b**2) - 2*B*a/(3*b**3)) + x**2*(-A*a/b*
*3 + 3*B*a**2/(2*b**4)) + x*(3*A*a**2/b**4 - 4*B*a**3/b**5) + (-A*a**4*b + B*a**5)/(a*b**6 + b**7*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=\frac {B a^{5} - A a^{4} b}{b^{7} x + a b^{6}} + \frac {3 \, B b^{3} x^{4} - 4 \, {\left (2 \, B a b^{2} - A b^{3}\right )} x^{3} + 6 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{2} - 12 \, {\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} x}{12 \, b^{5}} + \frac {{\left (5 \, B a^{4} - 4 \, A a^{3} b\right )} \log \left (b x + a\right )}{b^{6}} \]

[In]

integrate(x^4*(B*x+A)/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.55 \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )}^{4} {\left (3 \, B - \frac {4 \, {\left (5 \, B a b - A b^{2}\right )}}{{\left (b x + a\right )} b} + \frac {12 \, {\left (5 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )}}{{\left (b x + a\right )}^{2} b^{2}} - \frac {24 \, {\left (5 \, B a^{3} b^{3} - 3 \, A a^{2} b^{4}\right )}}{{\left (b x + a\right )}^{3} b^{3}}\right )}}{12 \, b^{6}} - \frac {{\left (5 \, B a^{4} - 4 \, A a^{3} b\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{6}} + \frac {\frac {B a^{5} b^{4}}{b x + a} - \frac {A a^{4} b^{5}}{b x + a}}{b^{10}} \]

[In]

integrate(x^4*(B*x+A)/(b*x+a)^2,x, algorithm="giac")

[Out]

1/12*(b*x + a)^4*(3*B - 4*(5*B*a*b - A*b^2)/((b*x + a)*b) + 12*(5*B*a^2*b^2 - 2*A*a*b^3)/((b*x + a)^2*b^2) - 2
4*(5*B*a^3*b^3 - 3*A*a^2*b^4)/((b*x + a)^3*b^3))/b^6 - (5*B*a^4 - 4*A*a^3*b)*log(abs(b*x + a)/((b*x + a)^2*abs
(b)))/b^6 + (B*a^5*b^4/(b*x + a) - A*a^4*b^5/(b*x + a))/b^10

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.53 \[ \int \frac {x^4 (A+B x)}{(a+b x)^2} \, dx=x\,\left (\frac {2\,a\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )}{b}-\frac {a^2\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b^2}\right )+x^3\,\left (\frac {A}{3\,b^2}-\frac {2\,B\,a}{3\,b^3}\right )-x^2\,\left (\frac {a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{2\,b^4}\right )+\frac {\ln \left (a+b\,x\right )\,\left (5\,B\,a^4-4\,A\,a^3\,b\right )}{b^6}+\frac {B\,x^4}{4\,b^2}+\frac {B\,a^5-A\,a^4\,b}{b\,\left (x\,b^6+a\,b^5\right )} \]

[In]

int((x^4*(A + B*x))/(a + b*x)^2,x)

[Out]

x*((2*a*((2*a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^2)/b^4))/b - (a^2*(A/b^2 - (2*B*a)/b^3))/b^2) + x^3*(A/(3*b^2) -
(2*B*a)/(3*b^3)) - x^2*((a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^2)/(2*b^4)) + (log(a + b*x)*(5*B*a^4 - 4*A*a^3*b))
/b^6 + (B*x^4)/(4*b^2) + (B*a^5 - A*a^4*b)/(b*(a*b^5 + b^6*x))