\(\int x^2 (a+b x)^5 \, dx\) [81]

Optimal result

Integrand size = 11, antiderivative size = 47 \[ \int x^2 (a+b x)^5 \, dx=\frac {a^2 (a+b x)^6}{6 b^3}-\frac {2 a (a+b x)^7}{7 b^3}+\frac {(a+b x)^8}{8 b^3} \]

[Out]

1/6*a^2*(b*x+a)^6/b^3-2/7*a*(b*x+a)^7/b^3+1/8*(b*x+a)^8/b^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, 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} \[ \int x^2 (a+b x)^5 \, dx=\frac {a^2 (a+b x)^6}{6 b^3}+\frac {(a+b x)^8}{8 b^3}-\frac {2 a (a+b x)^7}{7 b^3} \]

[In]

Int[x^2*(a + b*x)^5,x]

[Out]

(a^2*(a + b*x)^6)/(6*b^3) - (2*a*(a + b*x)^7)/(7*b^3) + (a + b*x)^8/(8*b^3)

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^2 (a+b x)^5}{b^2}-\frac {2 a (a+b x)^6}{b^2}+\frac {(a+b x)^7}{b^2}\right ) \, dx \\ & = \frac {a^2 (a+b x)^6}{6 b^3}-\frac {2 a (a+b x)^7}{7 b^3}+\frac {(a+b x)^8}{8 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.43 \[ \int x^2 (a+b x)^5 \, dx=\frac {a^5 x^3}{3}+\frac {5}{4} a^4 b x^4+2 a^3 b^2 x^5+\frac {5}{3} a^2 b^3 x^6+\frac {5}{7} a b^4 x^7+\frac {b^5 x^8}{8} \]

[In]

Integrate[x^2*(a + b*x)^5,x]

[Out]

(a^5*x^3)/3 + (5*a^4*b*x^4)/4 + 2*a^3*b^2*x^5 + (5*a^2*b^3*x^6)/3 + (5*a*b^4*x^7)/7 + (b^5*x^8)/8

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23

method	result	size
gosper	\(\frac {1}{8} b^{5} x^{8}+\frac {5}{7} a \,b^{4} x^{7}+\frac {5}{3} a^{2} b^{3} x^{6}+2 a^{3} b^{2} x^{5}+\frac {5}{4} a^{4} b \,x^{4}+\frac {1}{3} a^{5} x^{3}\)	\(58\)
default	\(\frac {1}{8} b^{5} x^{8}+\frac {5}{7} a \,b^{4} x^{7}+\frac {5}{3} a^{2} b^{3} x^{6}+2 a^{3} b^{2} x^{5}+\frac {5}{4} a^{4} b \,x^{4}+\frac {1}{3} a^{5} x^{3}\)	\(58\)
norman	\(\frac {1}{8} b^{5} x^{8}+\frac {5}{7} a \,b^{4} x^{7}+\frac {5}{3} a^{2} b^{3} x^{6}+2 a^{3} b^{2} x^{5}+\frac {5}{4} a^{4} b \,x^{4}+\frac {1}{3} a^{5} x^{3}\)	\(58\)
risch	\(\frac {1}{8} b^{5} x^{8}+\frac {5}{7} a \,b^{4} x^{7}+\frac {5}{3} a^{2} b^{3} x^{6}+2 a^{3} b^{2} x^{5}+\frac {5}{4} a^{4} b \,x^{4}+\frac {1}{3} a^{5} x^{3}\)	\(58\)
parallelrisch	\(\frac {1}{8} b^{5} x^{8}+\frac {5}{7} a \,b^{4} x^{7}+\frac {5}{3} a^{2} b^{3} x^{6}+2 a^{3} b^{2} x^{5}+\frac {5}{4} a^{4} b \,x^{4}+\frac {1}{3} a^{5} x^{3}\)	\(58\)

[In]

int(x^2*(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/8*b^5*x^8+5/7*a*b^4*x^7+5/3*a^2*b^3*x^6+2*a^3*b^2*x^5+5/4*a^4*b*x^4+1/3*a^5*x^3

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int x^2 (a+b x)^5 \, dx=\frac {1}{8} \, b^{5} x^{8} + \frac {5}{7} \, a b^{4} x^{7} + \frac {5}{3} \, a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{4} \, a^{4} b x^{4} + \frac {1}{3} \, a^{5} x^{3} \]

[In]

integrate(x^2*(b*x+a)^5,x, algorithm="fricas")

[Out]

1/8*b^5*x^8 + 5/7*a*b^4*x^7 + 5/3*a^2*b^3*x^6 + 2*a^3*b^2*x^5 + 5/4*a^4*b*x^4 + 1/3*a^5*x^3

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int x^2 (a+b x)^5 \, dx=\frac {a^{5} x^{3}}{3} + \frac {5 a^{4} b x^{4}}{4} + 2 a^{3} b^{2} x^{5} + \frac {5 a^{2} b^{3} x^{6}}{3} + \frac {5 a b^{4} x^{7}}{7} + \frac {b^{5} x^{8}}{8} \]

[In]

integrate(x**2*(b*x+a)**5,x)

[Out]

a**5*x**3/3 + 5*a**4*b*x**4/4 + 2*a**3*b**2*x**5 + 5*a**2*b**3*x**6/3 + 5*a*b**4*x**7/7 + b**5*x**8/8

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int x^2 (a+b x)^5 \, dx=\frac {1}{8} \, b^{5} x^{8} + \frac {5}{7} \, a b^{4} x^{7} + \frac {5}{3} \, a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{4} \, a^{4} b x^{4} + \frac {1}{3} \, a^{5} x^{3} \]

[In]

integrate(x^2*(b*x+a)^5,x, algorithm="maxima")

[Out]

1/8*b^5*x^8 + 5/7*a*b^4*x^7 + 5/3*a^2*b^3*x^6 + 2*a^3*b^2*x^5 + 5/4*a^4*b*x^4 + 1/3*a^5*x^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int x^2 (a+b x)^5 \, dx=\frac {1}{8} \, b^{5} x^{8} + \frac {5}{7} \, a b^{4} x^{7} + \frac {5}{3} \, a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{4} \, a^{4} b x^{4} + \frac {1}{3} \, a^{5} x^{3} \]

[In]

integrate(x^2*(b*x+a)^5,x, algorithm="giac")

[Out]

1/8*b^5*x^8 + 5/7*a*b^4*x^7 + 5/3*a^2*b^3*x^6 + 2*a^3*b^2*x^5 + 5/4*a^4*b*x^4 + 1/3*a^5*x^3

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int x^2 (a+b x)^5 \, dx=\frac {a^5\,x^3}{3}+\frac {5\,a^4\,b\,x^4}{4}+2\,a^3\,b^2\,x^5+\frac {5\,a^2\,b^3\,x^6}{3}+\frac {5\,a\,b^4\,x^7}{7}+\frac {b^5\,x^8}{8} \]

[In]

int(x^2*(a + b*x)^5,x)

[Out]

(a^5*x^3)/3 + (b^5*x^8)/8 + (5*a^4*b*x^4)/4 + (5*a*b^4*x^7)/7 + 2*a^3*b^2*x^5 + (5*a^2*b^3*x^6)/3