\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{22}} \, dx\) [85]

Optimal result

Integrand size = 26, antiderivative size = 84 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=-\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 a x^{21}}+\frac {b \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{126 a^2 x^{18}} \]

[Out]

-1/21*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/a/x^21+1/126*b*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/a^2/x^18

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} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=\frac {b \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{126 a^2 x^{18}}-\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 a x^{21}} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^22,x]

[Out]

-1/21*((a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a*x^21) + (b*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x
^6])/(126*a^2*x^18)

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 )^5}{x^{22}} \, dx}{b^4 \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 )^5}{x^8} \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )} \\ & = -\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 a x^{21}}-\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 )^5}{x^7} \, dx,x,x^3\right )}{21 a b^3 \left (a b+b^2 x^3\right )} \\ & = -\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 a x^{21}}+\frac {b \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{126 a^2 x^{18}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (6 a^5+35 a^4 b x^3+84 a^3 b^2 x^6+105 a^2 b^3 x^9+70 a b^4 x^{12}+21 b^5 x^{15}\right )}{126 x^{21} \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^22,x]

[Out]

-1/126*(Sqrt[(a + b*x^3)^2]*(6*a^5 + 35*a^4*b*x^3 + 84*a^3*b^2*x^6 + 105*a^2*b^3*x^9 + 70*a*b^4*x^12 + 21*b^5*
x^15))/(x^21*(a + b*x^3))

Maple [C] (warning: unable to verify)

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

Time = 2.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (21 b^{5} x^{15}+70 a \,b^{4} x^{12}+105 a^{2} b^{3} x^{9}+84 a^{3} b^{2} x^{6}+35 a^{4} b \,x^{3}+6 a^{5}\right )}{126 x^{21}}\)	\(68\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{21} a^{5}-\frac {5}{18} a^{4} b \,x^{3}-\frac {2}{3} a^{3} b^{2} x^{6}-\frac {5}{6} a^{2} b^{3} x^{9}-\frac {5}{9} a \,b^{4} x^{12}-\frac {1}{6} b^{5} x^{15}\right )}{\left (b \,x^{3}+a \right ) x^{21}}\)	\(79\)
gosper	\(-\frac {\left (21 b^{5} x^{15}+70 a \,b^{4} x^{12}+105 a^{2} b^{3} x^{9}+84 a^{3} b^{2} x^{6}+35 a^{4} b \,x^{3}+6 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{126 x^{21} \left (b \,x^{3}+a \right )^{5}}\)	\(80\)
default	\(-\frac {\left (21 b^{5} x^{15}+70 a \,b^{4} x^{12}+105 a^{2} b^{3} x^{9}+84 a^{3} b^{2} x^{6}+35 a^{4} b \,x^{3}+6 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{126 x^{21} \left (b \,x^{3}+a \right )^{5}}\)	\(80\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^22,x,method=_RETURNVERBOSE)

[Out]

-1/126*csgn(b*x^3+a)*(21*b^5*x^15+70*a*b^4*x^12+105*a^2*b^3*x^9+84*a^3*b^2*x^6+35*a^4*b*x^3+6*a^5)/x^21

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.70 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=-\frac {21 \, b^{5} x^{15} + 70 \, a b^{4} x^{12} + 105 \, a^{2} b^{3} x^{9} + 84 \, a^{3} b^{2} x^{6} + 35 \, a^{4} b x^{3} + 6 \, a^{5}}{126 \, x^{21}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^22,x, algorithm="fricas")

[Out]

-1/126*(21*b^5*x^15 + 70*a*b^4*x^12 + 105*a^2*b^3*x^9 + 84*a^3*b^2*x^6 + 35*a^4*b*x^3 + 6*a^5)/x^21

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{22}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x**22,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**22, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (58) = 116\).

Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.87 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=-\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{7}}{18 \, a^{7}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{6}}{18 \, a^{6} x^{3}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{5}}{18 \, a^{7} x^{6}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{4}}{18 \, a^{6} x^{9}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{3}}{18 \, a^{5} x^{12}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{2}}{18 \, a^{4} x^{15}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b}{18 \, a^{3} x^{18}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{21 \, a^{2} x^{21}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^22,x, algorithm="maxima")

[Out]

-1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^7/a^7 - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^6/(a^6*x^3) + 1/18*
(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^5/(a^7*x^6) - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^4/(a^6*x^9) + 1/18*
(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^3/(a^5*x^12) - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^2/(a^4*x^15) + 1/1
8*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b/(a^3*x^18) - 1/21*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)/(a^2*x^21)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=-\frac {21 \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + 70 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 105 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 84 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 35 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 6 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{126 \, x^{21}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^22,x, algorithm="giac")

[Out]

-1/126*(21*b^5*x^15*sgn(b*x^3 + a) + 70*a*b^4*x^12*sgn(b*x^3 + a) + 105*a^2*b^3*x^9*sgn(b*x^3 + a) + 84*a^3*b^
2*x^6*sgn(b*x^3 + a) + 35*a^4*b*x^3*sgn(b*x^3 + a) + 6*a^5*sgn(b*x^3 + a))/x^21

Mupad [B] (verification not implemented)

Time = 8.36 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.75 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{22}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{21\,x^{21}\,\left (b\,x^3+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,x^6\,\left (b\,x^3+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{9\,x^9\,\left (b\,x^3+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{18\,x^{18}\,\left (b\,x^3+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,x^{12}\,\left (b\,x^3+a\right )}-\frac {2\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^{15}\,\left (b\,x^3+a\right )} \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^22,x)

[Out]

- (a^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(21*x^21*(a + b*x^3)) - (b^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(6*x
^6*(a + b*x^3)) - (5*a*b^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(9*x^9*(a + b*x^3)) - (5*a^4*b*(a^2 + b^2*x^6 +
2*a*b*x^3)^(1/2))/(18*x^18*(a + b*x^3)) - (5*a^2*b^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(6*x^12*(a + b*x^3)) -
(2*a^3*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(3*x^15*(a + b*x^3))