\(\int x^3 (a+b x)^5 \, dx\) [80]

Optimal result

Integrand size = 11, antiderivative size = 64 \[ \int x^3 (a+b x)^5 \, dx=-\frac {a^3 (a+b x)^6}{6 b^4}+\frac {3 a^2 (a+b x)^7}{7 b^4}-\frac {3 a (a+b x)^8}{8 b^4}+\frac {(a+b x)^9}{9 b^4} \]

[Out]

-1/6*a^3*(b*x+a)^6/b^4+3/7*a^2*(b*x+a)^7/b^4-3/8*a*(b*x+a)^8/b^4+1/9*(b*x+a)^9/b^4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, 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} \[ \int x^3 (a+b x)^5 \, dx=-\frac {a^3 (a+b x)^6}{6 b^4}+\frac {3 a^2 (a+b x)^7}{7 b^4}+\frac {(a+b x)^9}{9 b^4}-\frac {3 a (a+b x)^8}{8 b^4} \]

[In]

Int[x^3*(a + b*x)^5,x]

[Out]

-1/6*(a^3*(a + b*x)^6)/b^4 + (3*a^2*(a + b*x)^7)/(7*b^4) - (3*a*(a + b*x)^8)/(8*b^4) + (a + b*x)^9/(9*b^4)

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^3 (a+b x)^5}{b^3}+\frac {3 a^2 (a+b x)^6}{b^3}-\frac {3 a (a+b x)^7}{b^3}+\frac {(a+b x)^8}{b^3}\right ) \, dx \\ & = -\frac {a^3 (a+b x)^6}{6 b^4}+\frac {3 a^2 (a+b x)^7}{7 b^4}-\frac {3 a (a+b x)^8}{8 b^4}+\frac {(a+b x)^9}{9 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int x^3 (a+b x)^5 \, dx=\frac {a^5 x^4}{4}+a^4 b x^5+\frac {5}{3} a^3 b^2 x^6+\frac {10}{7} a^2 b^3 x^7+\frac {5}{8} a b^4 x^8+\frac {b^5 x^9}{9} \]

[In]

Integrate[x^3*(a + b*x)^5,x]

[Out]

(a^5*x^4)/4 + a^4*b*x^5 + (5*a^3*b^2*x^6)/3 + (10*a^2*b^3*x^7)/7 + (5*a*b^4*x^8)/8 + (b^5*x^9)/9

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89

method	result	size
gosper	\(\frac {1}{9} b^{5} x^{9}+\frac {5}{8} a \,b^{4} x^{8}+\frac {10}{7} a^{2} b^{3} x^{7}+\frac {5}{3} a^{3} b^{2} x^{6}+a^{4} b \,x^{5}+\frac {1}{4} a^{5} x^{4}\)	\(57\)
default	\(\frac {1}{9} b^{5} x^{9}+\frac {5}{8} a \,b^{4} x^{8}+\frac {10}{7} a^{2} b^{3} x^{7}+\frac {5}{3} a^{3} b^{2} x^{6}+a^{4} b \,x^{5}+\frac {1}{4} a^{5} x^{4}\)	\(57\)
norman	\(\frac {1}{9} b^{5} x^{9}+\frac {5}{8} a \,b^{4} x^{8}+\frac {10}{7} a^{2} b^{3} x^{7}+\frac {5}{3} a^{3} b^{2} x^{6}+a^{4} b \,x^{5}+\frac {1}{4} a^{5} x^{4}\)	\(57\)
risch	\(\frac {1}{9} b^{5} x^{9}+\frac {5}{8} a \,b^{4} x^{8}+\frac {10}{7} a^{2} b^{3} x^{7}+\frac {5}{3} a^{3} b^{2} x^{6}+a^{4} b \,x^{5}+\frac {1}{4} a^{5} x^{4}\)	\(57\)
parallelrisch	\(\frac {1}{9} b^{5} x^{9}+\frac {5}{8} a \,b^{4} x^{8}+\frac {10}{7} a^{2} b^{3} x^{7}+\frac {5}{3} a^{3} b^{2} x^{6}+a^{4} b \,x^{5}+\frac {1}{4} a^{5} x^{4}\)	\(57\)

[In]

int(x^3*(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/9*b^5*x^9+5/8*a*b^4*x^8+10/7*a^2*b^3*x^7+5/3*a^3*b^2*x^6+a^4*b*x^5+1/4*a^5*x^4

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int x^3 (a+b x)^5 \, dx=\frac {1}{9} \, b^{5} x^{9} + \frac {5}{8} \, a b^{4} x^{8} + \frac {10}{7} \, a^{2} b^{3} x^{7} + \frac {5}{3} \, a^{3} b^{2} x^{6} + a^{4} b x^{5} + \frac {1}{4} \, a^{5} x^{4} \]

[In]

integrate(x^3*(b*x+a)^5,x, algorithm="fricas")

[Out]

1/9*b^5*x^9 + 5/8*a*b^4*x^8 + 10/7*a^2*b^3*x^7 + 5/3*a^3*b^2*x^6 + a^4*b*x^5 + 1/4*a^5*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int x^3 (a+b x)^5 \, dx=\frac {a^{5} x^{4}}{4} + a^{4} b x^{5} + \frac {5 a^{3} b^{2} x^{6}}{3} + \frac {10 a^{2} b^{3} x^{7}}{7} + \frac {5 a b^{4} x^{8}}{8} + \frac {b^{5} x^{9}}{9} \]

[In]

integrate(x**3*(b*x+a)**5,x)

[Out]

a**5*x**4/4 + a**4*b*x**5 + 5*a**3*b**2*x**6/3 + 10*a**2*b**3*x**7/7 + 5*a*b**4*x**8/8 + b**5*x**9/9

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int x^3 (a+b x)^5 \, dx=\frac {1}{9} \, b^{5} x^{9} + \frac {5}{8} \, a b^{4} x^{8} + \frac {10}{7} \, a^{2} b^{3} x^{7} + \frac {5}{3} \, a^{3} b^{2} x^{6} + a^{4} b x^{5} + \frac {1}{4} \, a^{5} x^{4} \]

[In]

integrate(x^3*(b*x+a)^5,x, algorithm="maxima")

[Out]

1/9*b^5*x^9 + 5/8*a*b^4*x^8 + 10/7*a^2*b^3*x^7 + 5/3*a^3*b^2*x^6 + a^4*b*x^5 + 1/4*a^5*x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int x^3 (a+b x)^5 \, dx=\frac {1}{9} \, b^{5} x^{9} + \frac {5}{8} \, a b^{4} x^{8} + \frac {10}{7} \, a^{2} b^{3} x^{7} + \frac {5}{3} \, a^{3} b^{2} x^{6} + a^{4} b x^{5} + \frac {1}{4} \, a^{5} x^{4} \]

[In]

integrate(x^3*(b*x+a)^5,x, algorithm="giac")

[Out]

1/9*b^5*x^9 + 5/8*a*b^4*x^8 + 10/7*a^2*b^3*x^7 + 5/3*a^3*b^2*x^6 + a^4*b*x^5 + 1/4*a^5*x^4

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int x^3 (a+b x)^5 \, dx=\frac {a^5\,x^4}{4}+a^4\,b\,x^5+\frac {5\,a^3\,b^2\,x^6}{3}+\frac {10\,a^2\,b^3\,x^7}{7}+\frac {5\,a\,b^4\,x^8}{8}+\frac {b^5\,x^9}{9} \]

[In]

int(x^3*(a + b*x)^5,x)

[Out]

(a^5*x^4)/4 + (b^5*x^9)/9 + a^4*b*x^5 + (5*a*b^4*x^8)/8 + (5*a^3*b^2*x^6)/3 + (10*a^2*b^3*x^7)/7