\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{16}} \, dx\) [48]

Optimal result

Integrand size = 26, antiderivative size = 84 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 a x^{15}}+\frac {b \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{60 a^2 x^{12}} \]

[Out]

-1/15*(b*x^3+a)^3*((b*x^3+a)^2)^(1/2)/a/x^15+1/60*b*(b*x^3+a)^3*((b*x^3+a)^2)^(1/2)/a^2/x^12

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1369, 272, 47, 37} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=\frac {b \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{60 a^2 x^{12}}-\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 a x^{15}} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^16,x]

[Out]

-1/15*((a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a*x^15) + (b*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x
^6])/(60*a^2*x^12)

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]]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (a b+b^2 x^3\right )^3}{x^{16}} \, dx}{b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^3}{x^6} \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )} \\ & = -\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 a x^{15}}-\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^3}{x^5} \, dx,x,x^3\right )}{15 a b \left (a b+b^2 x^3\right )} \\ & = -\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 a x^{15}}+\frac {b \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{60 a^2 x^{12}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (4 a^3+15 a^2 b x^3+20 a b^2 x^6+10 b^3 x^9\right )}{60 x^{15} \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^16,x]

[Out]

-1/60*(Sqrt[(a + b*x^3)^2]*(4*a^3 + 15*a^2*b*x^3 + 20*a*b^2*x^6 + 10*b^3*x^9))/(x^15*(a + b*x^3))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.52

method	result	size
pseudoelliptic	\(-\frac {\left (\frac {5}{2} b^{3} x^{9}+5 b^{2} x^{6} a +\frac {15}{4} a^{2} b \,x^{3}+a^{3}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{15 x^{15}}\)	\(44\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{15} a^{3}-\frac {1}{4} a^{2} b \,x^{3}-\frac {1}{3} b^{2} x^{6} a -\frac {1}{6} b^{3} x^{9}\right )}{\left (b \,x^{3}+a \right ) x^{15}}\)	\(57\)
gosper	\(-\frac {\left (10 b^{3} x^{9}+20 b^{2} x^{6} a +15 a^{2} b \,x^{3}+4 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{60 x^{15} \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
default	\(-\frac {\left (10 b^{3} x^{9}+20 b^{2} x^{6} a +15 a^{2} b \,x^{3}+4 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{60 x^{15} \left (b \,x^{3}+a \right )^{3}}\)	\(58\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x,method=_RETURNVERBOSE)

[Out]

-1/15*(5/2*b^3*x^9+5*b^2*x^6*a+15/4*a^2*b*x^3+a^3)*csgn(b*x^3+a)/x^15

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {10 \, b^{3} x^{9} + 20 \, a b^{2} x^{6} + 15 \, a^{2} b x^{3} + 4 \, a^{3}}{60 \, x^{15}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x, algorithm="fricas")

[Out]

-1/60*(10*b^3*x^9 + 20*a*b^2*x^6 + 15*a^2*b*x^3 + 4*a^3)/x^15

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{16}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**16,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**16, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (58) = 116\).

Time = 0.21 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.13 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{5}}{12 \, a^{5}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{4}}{12 \, a^{4} x^{3}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{3}}{12 \, a^{5} x^{6}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{2}}{12 \, a^{4} x^{9}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{12 \, a^{3} x^{12}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{15 \, a^{2} x^{15}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x, algorithm="maxima")

[Out]

-1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^5/a^5 - 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^4/(a^4*x^3) + 1/12*
(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^3/(a^5*x^6) - 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^2/(a^4*x^9) + 1/12*
(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b/(a^3*x^12) - 1/15*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^15)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {10 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 20 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 15 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 4 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{60 \, x^{15}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x, algorithm="giac")

[Out]

-1/60*(10*b^3*x^9*sgn(b*x^3 + a) + 20*a*b^2*x^6*sgn(b*x^3 + a) + 15*a^2*b*x^3*sgn(b*x^3 + a) + 4*a^3*sgn(b*x^3
+ a))/x^15

Mupad [B] (verification not implemented)

Time = 8.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{15\,x^{15}\,\left (b\,x^3+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,x^6\,\left (b\,x^3+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^9\,\left (b\,x^3+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{4\,x^{12}\,\left (b\,x^3+a\right )} \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^16,x)

[Out]

- (a^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(15*x^15*(a + b*x^3)) - (b^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(6*x
^6*(a + b*x^3)) - (a*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(3*x^9*(a + b*x^3)) - (a^2*b*(a^2 + b^2*x^6 + 2*a*
b*x^3)^(1/2))/(4*x^12*(a + b*x^3))