\(\int x (a+b x)^5 \, dx\) [82]

Optimal result

Integrand size = 9, antiderivative size = 30 \[ \int x (a+b x)^5 \, dx=-\frac {a (a+b x)^6}{6 b^2}+\frac {(a+b x)^7}{7 b^2} \]

[Out]

-1/6*a*(b*x+a)^6/b^2+1/7*(b*x+a)^7/b^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {45} \[ \int x (a+b x)^5 \, dx=\frac {(a+b x)^7}{7 b^2}-\frac {a (a+b x)^6}{6 b^2} \]

[In]

Int[x*(a + b*x)^5,x]

[Out]

-1/6*(a*(a + b*x)^6)/b^2 + (a + b*x)^7/(7*b^2)

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 (a+b x)^5}{b}+\frac {(a+b x)^6}{b}\right ) \, dx \\ & = -\frac {a (a+b x)^6}{6 b^2}+\frac {(a+b x)^7}{7 b^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(30)=60\).

Time = 0.00 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.23 \[ \int x (a+b x)^5 \, dx=\frac {a^5 x^2}{2}+\frac {5}{3} a^4 b x^3+\frac {5}{2} a^3 b^2 x^4+2 a^2 b^3 x^5+\frac {5}{6} a b^4 x^6+\frac {b^5 x^7}{7} \]

[In]

Integrate[x*(a + b*x)^5,x]

[Out]

(a^5*x^2)/2 + (5*a^4*b*x^3)/3 + (5*a^3*b^2*x^4)/2 + 2*a^2*b^3*x^5 + (5*a*b^4*x^6)/6 + (b^5*x^7)/7

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(57\) vs. \(2(26)=52\).

Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93

method	result	size
gosper	\(\frac {1}{7} b^{5} x^{7}+\frac {5}{6} a \,b^{4} x^{6}+2 a^{2} b^{3} x^{5}+\frac {5}{2} a^{3} b^{2} x^{4}+\frac {5}{3} a^{4} b \,x^{3}+\frac {1}{2} a^{5} x^{2}\)	\(58\)
default	\(\frac {1}{7} b^{5} x^{7}+\frac {5}{6} a \,b^{4} x^{6}+2 a^{2} b^{3} x^{5}+\frac {5}{2} a^{3} b^{2} x^{4}+\frac {5}{3} a^{4} b \,x^{3}+\frac {1}{2} a^{5} x^{2}\)	\(58\)
norman	\(\frac {1}{7} b^{5} x^{7}+\frac {5}{6} a \,b^{4} x^{6}+2 a^{2} b^{3} x^{5}+\frac {5}{2} a^{3} b^{2} x^{4}+\frac {5}{3} a^{4} b \,x^{3}+\frac {1}{2} a^{5} x^{2}\)	\(58\)
risch	\(\frac {1}{7} b^{5} x^{7}+\frac {5}{6} a \,b^{4} x^{6}+2 a^{2} b^{3} x^{5}+\frac {5}{2} a^{3} b^{2} x^{4}+\frac {5}{3} a^{4} b \,x^{3}+\frac {1}{2} a^{5} x^{2}\)	\(58\)
parallelrisch	\(\frac {1}{7} b^{5} x^{7}+\frac {5}{6} a \,b^{4} x^{6}+2 a^{2} b^{3} x^{5}+\frac {5}{2} a^{3} b^{2} x^{4}+\frac {5}{3} a^{4} b \,x^{3}+\frac {1}{2} a^{5} x^{2}\)	\(58\)

[In]

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

[Out]

1/7*b^5*x^7+5/6*a*b^4*x^6+2*a^2*b^3*x^5+5/2*a^3*b^2*x^4+5/3*a^4*b*x^3+1/2*a^5*x^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (26) = 52\).

Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int x (a+b x)^5 \, dx=\frac {1}{7} \, b^{5} x^{7} + \frac {5}{6} \, a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{5} + \frac {5}{2} \, a^{3} b^{2} x^{4} + \frac {5}{3} \, a^{4} b x^{3} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

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

[Out]

1/7*b^5*x^7 + 5/6*a*b^4*x^6 + 2*a^2*b^3*x^5 + 5/2*a^3*b^2*x^4 + 5/3*a^4*b*x^3 + 1/2*a^5*x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (24) = 48\).

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int x (a+b x)^5 \, dx=\frac {a^{5} x^{2}}{2} + \frac {5 a^{4} b x^{3}}{3} + \frac {5 a^{3} b^{2} x^{4}}{2} + 2 a^{2} b^{3} x^{5} + \frac {5 a b^{4} x^{6}}{6} + \frac {b^{5} x^{7}}{7} \]

[In]

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

[Out]

a**5*x**2/2 + 5*a**4*b*x**3/3 + 5*a**3*b**2*x**4/2 + 2*a**2*b**3*x**5 + 5*a*b**4*x**6/6 + b**5*x**7/7

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (26) = 52\).

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int x (a+b x)^5 \, dx=\frac {1}{7} \, b^{5} x^{7} + \frac {5}{6} \, a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{5} + \frac {5}{2} \, a^{3} b^{2} x^{4} + \frac {5}{3} \, a^{4} b x^{3} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

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

[Out]

1/7*b^5*x^7 + 5/6*a*b^4*x^6 + 2*a^2*b^3*x^5 + 5/2*a^3*b^2*x^4 + 5/3*a^4*b*x^3 + 1/2*a^5*x^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (26) = 52\).

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int x (a+b x)^5 \, dx=\frac {1}{7} \, b^{5} x^{7} + \frac {5}{6} \, a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{5} + \frac {5}{2} \, a^{3} b^{2} x^{4} + \frac {5}{3} \, a^{4} b x^{3} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

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

[Out]

1/7*b^5*x^7 + 5/6*a*b^4*x^6 + 2*a^2*b^3*x^5 + 5/2*a^3*b^2*x^4 + 5/3*a^4*b*x^3 + 1/2*a^5*x^2

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int x (a+b x)^5 \, dx=\frac {a^5\,x^2}{2}+\frac {5\,a^4\,b\,x^3}{3}+\frac {5\,a^3\,b^2\,x^4}{2}+2\,a^2\,b^3\,x^5+\frac {5\,a\,b^4\,x^6}{6}+\frac {b^5\,x^7}{7} \]

[In]

int(x*(a + b*x)^5,x)

[Out]

(a^5*x^2)/2 + (b^5*x^7)/7 + (5*a^4*b*x^3)/3 + (5*a*b^4*x^6)/6 + (5*a^3*b^2*x^4)/2 + 2*a^2*b^3*x^5