\(\int \frac {x^5}{a+b x} \, dx\) [157]

Optimal result

Integrand size = 11, antiderivative size = 70 \[ \int \frac {x^5}{a+b x} \, dx=\frac {a^4 x}{b^5}-\frac {a^3 x^2}{2 b^4}+\frac {a^2 x^3}{3 b^3}-\frac {a x^4}{4 b^2}+\frac {x^5}{5 b}-\frac {a^5 \log (a+b x)}{b^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {x^5}{a+b x} \, dx=-\frac {a^5 \log (a+b x)}{b^6}+\frac {a^4 x}{b^5}-\frac {a^3 x^2}{2 b^4}+\frac {a^2 x^3}{3 b^3}-\frac {a x^4}{4 b^2}+\frac {x^5}{5 b} \]

[In]

Int[x^5/(a + b*x),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^4}{b^5}-\frac {a^3 x}{b^4}+\frac {a^2 x^2}{b^3}-\frac {a x^3}{b^2}+\frac {x^4}{b}-\frac {a^5}{b^5 (a+b x)}\right ) \, dx \\ & = \frac {a^4 x}{b^5}-\frac {a^3 x^2}{2 b^4}+\frac {a^2 x^3}{3 b^3}-\frac {a x^4}{4 b^2}+\frac {x^5}{5 b}-\frac {a^5 \log (a+b x)}{b^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{a+b x} \, dx=\frac {a^4 x}{b^5}-\frac {a^3 x^2}{2 b^4}+\frac {a^2 x^3}{3 b^3}-\frac {a x^4}{4 b^2}+\frac {x^5}{5 b}-\frac {a^5 \log (a+b x)}{b^6} \]

[In]

Integrate[x^5/(a + b*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {\frac {1}{5} b^{4} x^{5}-\frac {1}{4} a \,b^{3} x^{4}+\frac {1}{3} a^{2} b^{2} x^{3}-\frac {1}{2} a^{3} b \,x^{2}+a^{4} x}{b^{5}}-\frac {a^{5} \ln \left (b x +a \right )}{b^{6}}\)	\(63\)
norman	\(\frac {a^{4} x}{b^{5}}-\frac {a^{3} x^{2}}{2 b^{4}}+\frac {a^{2} x^{3}}{3 b^{3}}-\frac {a \,x^{4}}{4 b^{2}}+\frac {x^{5}}{5 b}-\frac {a^{5} \ln \left (b x +a \right )}{b^{6}}\)	\(63\)
risch	\(\frac {a^{4} x}{b^{5}}-\frac {a^{3} x^{2}}{2 b^{4}}+\frac {a^{2} x^{3}}{3 b^{3}}-\frac {a \,x^{4}}{4 b^{2}}+\frac {x^{5}}{5 b}-\frac {a^{5} \ln \left (b x +a \right )}{b^{6}}\)	\(63\)
parallelrisch	\(-\frac {-12 b^{5} x^{5}+15 a \,b^{4} x^{4}-20 a^{2} b^{3} x^{3}+30 a^{3} b^{2} x^{2}+60 a^{5} \ln \left (b x +a \right )-60 a^{4} b x}{60 b^{6}}\)	\(64\)

[In]

int(x^5/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {x^5}{a+b x} \, dx=\frac {12 \, b^{5} x^{5} - 15 \, a b^{4} x^{4} + 20 \, a^{2} b^{3} x^{3} - 30 \, a^{3} b^{2} x^{2} + 60 \, a^{4} b x - 60 \, a^{5} \log \left (b x + a\right )}{60 \, b^{6}} \]

[In]

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

[Out]

1/60*(12*b^5*x^5 - 15*a*b^4*x^4 + 20*a^2*b^3*x^3 - 30*a^3*b^2*x^2 + 60*a^4*b*x - 60*a^5*log(b*x + a))/b^6

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87 \[ \int \frac {x^5}{a+b x} \, dx=- \frac {a^{5} \log {\left (a + b x \right )}}{b^{6}} + \frac {a^{4} x}{b^{5}} - \frac {a^{3} x^{2}}{2 b^{4}} + \frac {a^{2} x^{3}}{3 b^{3}} - \frac {a x^{4}}{4 b^{2}} + \frac {x^{5}}{5 b} \]

[In]

integrate(x**5/(b*x+a),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.91 \[ \int \frac {x^5}{a+b x} \, dx=-\frac {a^{5} \log \left (b x + a\right )}{b^{6}} + \frac {12 \, b^{4} x^{5} - 15 \, a b^{3} x^{4} + 20 \, a^{2} b^{2} x^{3} - 30 \, a^{3} b x^{2} + 60 \, a^{4} x}{60 \, b^{5}} \]

[In]

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

[Out]

-a^5*log(b*x + a)/b^6 + 1/60*(12*b^4*x^5 - 15*a*b^3*x^4 + 20*a^2*b^2*x^3 - 30*a^3*b*x^2 + 60*a^4*x)/b^5

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {x^5}{a+b x} \, dx=-\frac {a^{5} \log \left ({\left | b x + a \right |}\right )}{b^{6}} + \frac {12 \, b^{4} x^{5} - 15 \, a b^{3} x^{4} + 20 \, a^{2} b^{2} x^{3} - 30 \, a^{3} b x^{2} + 60 \, a^{4} x}{60 \, b^{5}} \]

[In]

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

[Out]

-a^5*log(abs(b*x + a))/b^6 + 1/60*(12*b^4*x^5 - 15*a*b^3*x^4 + 20*a^2*b^2*x^3 - 30*a^3*b*x^2 + 60*a^4*x)/b^5

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89 \[ \int \frac {x^5}{a+b x} \, dx=\frac {x^5}{5\,b}-\frac {a^5\,\ln \left (a+b\,x\right )}{b^6}-\frac {a\,x^4}{4\,b^2}+\frac {a^4\,x}{b^5}+\frac {a^2\,x^3}{3\,b^3}-\frac {a^3\,x^2}{2\,b^4} \]

[In]

int(x^5/(a + b*x),x)

[Out]

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