\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^3}{x^{21}} \, dx\) [473]

Optimal result

Integrand size = 24, antiderivative size = 84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {\left (a+b x^2\right )^7}{20 a x^{20}}+\frac {b \left (a+b x^2\right )^7}{60 a^2 x^{18}}-\frac {b^2 \left (a+b x^2\right )^7}{240 a^3 x^{16}}+\frac {b^3 \left (a+b x^2\right )^7}{1680 a^4 x^{14}} \]

[Out]

-1/20*(b*x^2+a)^7/a/x^20+1/60*b*(b*x^2+a)^7/a^2/x^18-1/240*b^2*(b*x^2+a)^7/a^3/x^16+1/1680*b^3*(b*x^2+a)^7/a^4
/x^14

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 272, 47, 37} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=\frac {b^3 \left (a+b x^2\right )^7}{1680 a^4 x^{14}}-\frac {b^2 \left (a+b x^2\right )^7}{240 a^3 x^{16}}+\frac {b \left (a+b x^2\right )^7}{60 a^2 x^{18}}-\frac {\left (a+b x^2\right )^7}{20 a x^{20}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^21,x]

[Out]

-1/20*(a + b*x^2)^7/(a*x^20) + (b*(a + b*x^2)^7)/(60*a^2*x^18) - (b^2*(a + b*x^2)^7)/(240*a^3*x^16) + (b^3*(a
+ b*x^2)^7)/(1680*a^4*x^14)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\left (a b+b^2 x^2\right )^6}{x^{21}} \, dx}{b^6} \\ & = \frac {\text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^{11}} \, dx,x,x^2\right )}{2 b^6} \\ & = -\frac {\left (a+b x^2\right )^7}{20 a x^{20}}-\frac {3 \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^{10}} \, dx,x,x^2\right )}{20 a b^5} \\ & = -\frac {\left (a+b x^2\right )^7}{20 a x^{20}}+\frac {b \left (a+b x^2\right )^7}{60 a^2 x^{18}}+\frac {\text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^9} \, dx,x,x^2\right )}{30 a^2 b^4} \\ & = -\frac {\left (a+b x^2\right )^7}{20 a x^{20}}+\frac {b \left (a+b x^2\right )^7}{60 a^2 x^{18}}-\frac {b^2 \left (a+b x^2\right )^7}{240 a^3 x^{16}}-\frac {\text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^8} \, dx,x,x^2\right )}{240 a^3 b^3} \\ & = -\frac {\left (a+b x^2\right )^7}{20 a x^{20}}+\frac {b \left (a+b x^2\right )^7}{60 a^2 x^{18}}-\frac {b^2 \left (a+b x^2\right )^7}{240 a^3 x^{16}}+\frac {b^3 \left (a+b x^2\right )^7}{1680 a^4 x^{14}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {a^6}{20 x^{20}}-\frac {a^5 b}{3 x^{18}}-\frac {15 a^4 b^2}{16 x^{16}}-\frac {10 a^3 b^3}{7 x^{14}}-\frac {5 a^2 b^4}{4 x^{12}}-\frac {3 a b^5}{5 x^{10}}-\frac {b^6}{8 x^8} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^21,x]

[Out]

-1/20*a^6/x^20 - (a^5*b)/(3*x^18) - (15*a^4*b^2)/(16*x^16) - (10*a^3*b^3)/(7*x^14) - (5*a^2*b^4)/(4*x^12) - (3
*a*b^5)/(5*x^10) - b^6/(8*x^8)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {5 a^{2} b^{4}}{4 x^{12}}-\frac {10 a^{3} b^{3}}{7 x^{14}}-\frac {b^{6}}{8 x^{8}}-\frac {a^{6}}{20 x^{20}}-\frac {a^{5} b}{3 x^{18}}-\frac {15 a^{4} b^{2}}{16 x^{16}}-\frac {3 b^{5} a}{5 x^{10}}\)	\(69\)
norman	\(\frac {-\frac {1}{20} a^{6}-\frac {1}{3} a^{5} b \,x^{2}-\frac {15}{16} a^{4} b^{2} x^{4}-\frac {10}{7} a^{3} b^{3} x^{6}-\frac {5}{4} a^{2} b^{4} x^{8}-\frac {3}{5} a \,b^{5} x^{10}-\frac {1}{8} b^{6} x^{12}}{x^{20}}\)	\(70\)
risch	\(\frac {-\frac {1}{20} a^{6}-\frac {1}{3} a^{5} b \,x^{2}-\frac {15}{16} a^{4} b^{2} x^{4}-\frac {10}{7} a^{3} b^{3} x^{6}-\frac {5}{4} a^{2} b^{4} x^{8}-\frac {3}{5} a \,b^{5} x^{10}-\frac {1}{8} b^{6} x^{12}}{x^{20}}\)	\(70\)
gosper	\(-\frac {210 b^{6} x^{12}+1008 a \,b^{5} x^{10}+2100 a^{2} b^{4} x^{8}+2400 a^{3} b^{3} x^{6}+1575 a^{4} b^{2} x^{4}+560 a^{5} b \,x^{2}+84 a^{6}}{1680 x^{20}}\)	\(71\)
parallelrisch	\(\frac {-210 b^{6} x^{12}-1008 a \,b^{5} x^{10}-2100 a^{2} b^{4} x^{8}-2400 a^{3} b^{3} x^{6}-1575 a^{4} b^{2} x^{4}-560 a^{5} b \,x^{2}-84 a^{6}}{1680 x^{20}}\)	\(71\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^3/x^21,x,method=_RETURNVERBOSE)

[Out]

-5/4*a^2*b^4/x^12-10/7*a^3*b^3/x^14-1/8*b^6/x^8-1/20*a^6/x^20-1/3*a^5*b/x^18-15/16*a^4*b^2/x^16-3/5*b^5*a/x^10

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {210 \, b^{6} x^{12} + 1008 \, a b^{5} x^{10} + 2100 \, a^{2} b^{4} x^{8} + 2400 \, a^{3} b^{3} x^{6} + 1575 \, a^{4} b^{2} x^{4} + 560 \, a^{5} b x^{2} + 84 \, a^{6}}{1680 \, x^{20}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^21,x, algorithm="fricas")

[Out]

-1/1680*(210*b^6*x^12 + 1008*a*b^5*x^10 + 2100*a^2*b^4*x^8 + 2400*a^3*b^3*x^6 + 1575*a^4*b^2*x^4 + 560*a^5*b*x
^2 + 84*a^6)/x^20

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=\frac {- 84 a^{6} - 560 a^{5} b x^{2} - 1575 a^{4} b^{2} x^{4} - 2400 a^{3} b^{3} x^{6} - 2100 a^{2} b^{4} x^{8} - 1008 a b^{5} x^{10} - 210 b^{6} x^{12}}{1680 x^{20}} \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**3/x**21,x)

[Out]

(-84*a**6 - 560*a**5*b*x**2 - 1575*a**4*b**2*x**4 - 2400*a**3*b**3*x**6 - 2100*a**2*b**4*x**8 - 1008*a*b**5*x*
*10 - 210*b**6*x**12)/(1680*x**20)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {210 \, b^{6} x^{12} + 1008 \, a b^{5} x^{10} + 2100 \, a^{2} b^{4} x^{8} + 2400 \, a^{3} b^{3} x^{6} + 1575 \, a^{4} b^{2} x^{4} + 560 \, a^{5} b x^{2} + 84 \, a^{6}}{1680 \, x^{20}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^21,x, algorithm="maxima")

[Out]

-1/1680*(210*b^6*x^12 + 1008*a*b^5*x^10 + 2100*a^2*b^4*x^8 + 2400*a^3*b^3*x^6 + 1575*a^4*b^2*x^4 + 560*a^5*b*x
^2 + 84*a^6)/x^20

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {210 \, b^{6} x^{12} + 1008 \, a b^{5} x^{10} + 2100 \, a^{2} b^{4} x^{8} + 2400 \, a^{3} b^{3} x^{6} + 1575 \, a^{4} b^{2} x^{4} + 560 \, a^{5} b x^{2} + 84 \, a^{6}}{1680 \, x^{20}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^21,x, algorithm="giac")

[Out]

-1/1680*(210*b^6*x^12 + 1008*a*b^5*x^10 + 2100*a^2*b^4*x^8 + 2400*a^3*b^3*x^6 + 1575*a^4*b^2*x^4 + 560*a^5*b*x
^2 + 84*a^6)/x^20

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{21}} \, dx=-\frac {\frac {a^6}{20}+\frac {a^5\,b\,x^2}{3}+\frac {15\,a^4\,b^2\,x^4}{16}+\frac {10\,a^3\,b^3\,x^6}{7}+\frac {5\,a^2\,b^4\,x^8}{4}+\frac {3\,a\,b^5\,x^{10}}{5}+\frac {b^6\,x^{12}}{8}}{x^{20}} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^3/x^21,x)

[Out]

-(a^6/20 + (b^6*x^12)/8 + (a^5*b*x^2)/3 + (3*a*b^5*x^10)/5 + (15*a^4*b^2*x^4)/16 + (10*a^3*b^3*x^6)/7 + (5*a^2
*b^4*x^8)/4)/x^20