\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{21}} \, dx\) [602]

Optimal result

Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{20 x^{20} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 x^{18} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^{14} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^{14} \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{20 x^{20} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 x^{18} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^21,x]

[Out]

-1/20*(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^20*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(18
*x^18*(a + b*x^2)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*x^16*(a + b*x^2)) - (5*a^2*b^3*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])/(7*x^14*(a + b*x^2)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*x^12*(a + b*x^2)) -
(b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*x^10*(a + b*x^2))

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])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^5}{x^{11}} \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (\frac {a^5 b^5}{x^{11}}+\frac {5 a^4 b^6}{x^{10}}+\frac {10 a^3 b^7}{x^9}+\frac {10 a^2 b^8}{x^8}+\frac {5 a b^9}{x^7}+\frac {b^{10}}{x^6}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{20 x^{20} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 x^{18} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^{14} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (126 a^5+700 a^4 b x^2+1575 a^3 b^2 x^4+1800 a^2 b^3 x^6+1050 a b^4 x^8+252 b^5 x^{10}\right )}{2520 x^{20} \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^21,x]

[Out]

-1/2520*(Sqrt[(a + b*x^2)^2]*(126*a^5 + 700*a^4*b*x^2 + 1575*a^3*b^2*x^4 + 1800*a^2*b^3*x^6 + 1050*a*b^4*x^8 +
252*b^5*x^10))/(x^20*(a + b*x^2))

Maple [C] (warning: unable to verify)

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

Time = 1.83 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (2 x^{10} b^{5}+\frac {25}{3} a \,x^{8} b^{4}+\frac {100}{7} a^{2} x^{6} b^{3}+\frac {25}{2} a^{3} x^{4} b^{2}+\frac {50}{9} x^{2} a^{4} b +a^{5}\right )}{20 x^{20}}\)	\(66\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{20} a^{5}-\frac {5}{18} x^{2} a^{4} b -\frac {5}{8} a^{3} x^{4} b^{2}-\frac {5}{7} a^{2} x^{6} b^{3}-\frac {5}{12} a \,x^{8} b^{4}-\frac {1}{10} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{20}}\)	\(79\)
gosper	\(-\frac {\left (252 x^{10} b^{5}+1050 a \,x^{8} b^{4}+1800 a^{2} x^{6} b^{3}+1575 a^{3} x^{4} b^{2}+700 x^{2} a^{4} b +126 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{2520 x^{20} \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(-\frac {\left (252 x^{10} b^{5}+1050 a \,x^{8} b^{4}+1800 a^{2} x^{6} b^{3}+1575 a^{3} x^{4} b^{2}+700 x^{2} a^{4} b +126 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{2520 x^{20} \left (b \,x^{2}+a \right )^{5}}\)	\(80\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^21,x,method=_RETURNVERBOSE)

[Out]

-1/20*csgn(b*x^2+a)*(2*x^10*b^5+25/3*a*x^8*b^4+100/7*a^2*x^6*b^3+25/2*a^3*x^4*b^2+50/9*x^2*a^4*b+a^5)/x^20

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=-\frac {252 \, b^{5} x^{10} + 1050 \, a b^{4} x^{8} + 1800 \, a^{2} b^{3} x^{6} + 1575 \, a^{3} b^{2} x^{4} + 700 \, a^{4} b x^{2} + 126 \, a^{5}}{2520 \, x^{20}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^21,x, algorithm="fricas")

[Out]

-1/2520*(252*b^5*x^10 + 1050*a*b^4*x^8 + 1800*a^2*b^3*x^6 + 1575*a^3*b^2*x^4 + 700*a^4*b*x^2 + 126*a^5)/x^20

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{21}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**21,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**21, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=-\frac {b^{5}}{10 \, x^{10}} - \frac {5 \, a b^{4}}{12 \, x^{12}} - \frac {5 \, a^{2} b^{3}}{7 \, x^{14}} - \frac {5 \, a^{3} b^{2}}{8 \, x^{16}} - \frac {5 \, a^{4} b}{18 \, x^{18}} - \frac {a^{5}}{20 \, x^{20}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^21,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=-\frac {252 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 1050 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 1800 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 1575 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 700 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 126 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{2520 \, x^{20}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^21,x, algorithm="giac")

[Out]

-1/2520*(252*b^5*x^10*sgn(b*x^2 + a) + 1050*a*b^4*x^8*sgn(b*x^2 + a) + 1800*a^2*b^3*x^6*sgn(b*x^2 + a) + 1575*
a^3*b^2*x^4*sgn(b*x^2 + a) + 700*a^4*b*x^2*sgn(b*x^2 + a) + 126*a^5*sgn(b*x^2 + a))/x^20

Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{21}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{20\,x^{20}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{12\,x^{12}\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{18\,x^{18}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^{14}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^{16}\,\left (b\,x^2+a\right )} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^21,x)

[Out]

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