\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^3}{x^{19}} \, dx\) [471]

Optimal result

Integrand size = 24, antiderivative size = 62 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=-\frac {\left (a+b x^2\right )^7}{18 a x^{18}}+\frac {b \left (a+b x^2\right )^7}{72 a^2 x^{16}}-\frac {b^2 \left (a+b x^2\right )^7}{504 a^3 x^{14}} \]

[Out]

-1/18*(b*x^2+a)^7/a/x^18+1/72*b*(b*x^2+a)^7/a^2/x^16-1/504*b^2*(b*x^2+a)^7/a^3/x^14

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 272, 47, 37} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=-\frac {b^2 \left (a+b x^2\right )^7}{504 a^3 x^{14}}+\frac {b \left (a+b x^2\right )^7}{72 a^2 x^{16}}-\frac {\left (a+b x^2\right )^7}{18 a x^{18}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^19,x]

[Out]

-1/18*(a + b*x^2)^7/(a*x^18) + (b*(a + b*x^2)^7)/(72*a^2*x^16) - (b^2*(a + b*x^2)^7)/(504*a^3*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^{19}} \, dx}{b^6} \\ & = \frac {\text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^{10}} \, dx,x,x^2\right )}{2 b^6} \\ & = -\frac {\left (a+b x^2\right )^7}{18 a x^{18}}-\frac {\text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^9} \, dx,x,x^2\right )}{9 a b^5} \\ & = -\frac {\left (a+b x^2\right )^7}{18 a x^{18}}+\frac {b \left (a+b x^2\right )^7}{72 a^2 x^{16}}+\frac {\text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^8} \, dx,x,x^2\right )}{72 a^2 b^4} \\ & = -\frac {\left (a+b x^2\right )^7}{18 a x^{18}}+\frac {b \left (a+b x^2\right )^7}{72 a^2 x^{16}}-\frac {b^2 \left (a+b x^2\right )^7}{504 a^3 x^{14}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=-\frac {a^6}{18 x^{18}}-\frac {3 a^5 b}{8 x^{16}}-\frac {15 a^4 b^2}{14 x^{14}}-\frac {5 a^3 b^3}{3 x^{12}}-\frac {3 a^2 b^4}{2 x^{10}}-\frac {3 a b^5}{4 x^8}-\frac {b^6}{6 x^6} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^19,x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11

method	result	size
default	\(-\frac {5 a^{3} b^{3}}{3 x^{12}}-\frac {15 a^{4} b^{2}}{14 x^{14}}-\frac {3 b^{5} a}{4 x^{8}}-\frac {b^{6}}{6 x^{6}}-\frac {a^{6}}{18 x^{18}}-\frac {3 a^{5} b}{8 x^{16}}-\frac {3 a^{2} b^{4}}{2 x^{10}}\)	\(69\)
norman	\(\frac {-\frac {1}{18} a^{6}-\frac {3}{8} a^{5} b \,x^{2}-\frac {15}{14} a^{4} b^{2} x^{4}-\frac {5}{3} a^{3} b^{3} x^{6}-\frac {3}{2} a^{2} b^{4} x^{8}-\frac {3}{4} a \,b^{5} x^{10}-\frac {1}{6} b^{6} x^{12}}{x^{18}}\)	\(70\)
risch	\(\frac {-\frac {1}{18} a^{6}-\frac {3}{8} a^{5} b \,x^{2}-\frac {15}{14} a^{4} b^{2} x^{4}-\frac {5}{3} a^{3} b^{3} x^{6}-\frac {3}{2} a^{2} b^{4} x^{8}-\frac {3}{4} a \,b^{5} x^{10}-\frac {1}{6} b^{6} x^{12}}{x^{18}}\)	\(70\)
gosper	\(-\frac {84 b^{6} x^{12}+378 a \,b^{5} x^{10}+756 a^{2} b^{4} x^{8}+840 a^{3} b^{3} x^{6}+540 a^{4} b^{2} x^{4}+189 a^{5} b \,x^{2}+28 a^{6}}{504 x^{18}}\)	\(71\)
parallelrisch	\(\frac {-84 b^{6} x^{12}-378 a \,b^{5} x^{10}-756 a^{2} b^{4} x^{8}-840 a^{3} b^{3} x^{6}-540 a^{4} b^{2} x^{4}-189 a^{5} b \,x^{2}-28 a^{6}}{504 x^{18}}\)	\(71\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^3/x^19,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=-\frac {84 \, b^{6} x^{12} + 378 \, a b^{5} x^{10} + 756 \, a^{2} b^{4} x^{8} + 840 \, a^{3} b^{3} x^{6} + 540 \, a^{4} b^{2} x^{4} + 189 \, a^{5} b x^{2} + 28 \, a^{6}}{504 \, x^{18}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^19,x, algorithm="fricas")

[Out]

-1/504*(84*b^6*x^12 + 378*a*b^5*x^10 + 756*a^2*b^4*x^8 + 840*a^3*b^3*x^6 + 540*a^4*b^2*x^4 + 189*a^5*b*x^2 + 2
8*a^6)/x^18

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=\frac {- 28 a^{6} - 189 a^{5} b x^{2} - 540 a^{4} b^{2} x^{4} - 840 a^{3} b^{3} x^{6} - 756 a^{2} b^{4} x^{8} - 378 a b^{5} x^{10} - 84 b^{6} x^{12}}{504 x^{18}} \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**3/x**19,x)

[Out]

(-28*a**6 - 189*a**5*b*x**2 - 540*a**4*b**2*x**4 - 840*a**3*b**3*x**6 - 756*a**2*b**4*x**8 - 378*a*b**5*x**10
- 84*b**6*x**12)/(504*x**18)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=-\frac {84 \, b^{6} x^{12} + 378 \, a b^{5} x^{10} + 756 \, a^{2} b^{4} x^{8} + 840 \, a^{3} b^{3} x^{6} + 540 \, a^{4} b^{2} x^{4} + 189 \, a^{5} b x^{2} + 28 \, a^{6}}{504 \, x^{18}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^19,x, algorithm="maxima")

[Out]

-1/504*(84*b^6*x^12 + 378*a*b^5*x^10 + 756*a^2*b^4*x^8 + 840*a^3*b^3*x^6 + 540*a^4*b^2*x^4 + 189*a^5*b*x^2 + 2
8*a^6)/x^18

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=-\frac {84 \, b^{6} x^{12} + 378 \, a b^{5} x^{10} + 756 \, a^{2} b^{4} x^{8} + 840 \, a^{3} b^{3} x^{6} + 540 \, a^{4} b^{2} x^{4} + 189 \, a^{5} b x^{2} + 28 \, a^{6}}{504 \, x^{18}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^19,x, algorithm="giac")

[Out]

-1/504*(84*b^6*x^12 + 378*a*b^5*x^10 + 756*a^2*b^4*x^8 + 840*a^3*b^3*x^6 + 540*a^4*b^2*x^4 + 189*a^5*b*x^2 + 2
8*a^6)/x^18

Mupad [B] (verification not implemented)

Time = 13.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^{19}} \, dx=-\frac {\frac {a^6}{18}+\frac {3\,a^5\,b\,x^2}{8}+\frac {15\,a^4\,b^2\,x^4}{14}+\frac {5\,a^3\,b^3\,x^6}{3}+\frac {3\,a^2\,b^4\,x^8}{2}+\frac {3\,a\,b^5\,x^{10}}{4}+\frac {b^6\,x^{12}}{6}}{x^{18}} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^3/x^19,x)

[Out]

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