\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{11}} \, dx\) [568]

Optimal result

Integrand size = 26, antiderivative size = 72 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 a x^{10}}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{40 a^2 x^{10}} \]

[Out]

-1/8*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/a/x^10+1/40*(b^2*x^4+2*a*b*x^2+a^2)^(5/2)/a^2/x^10

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1124} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{40 a^2 x^{10}}-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 a x^{10}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^11,x]

[Out]

-1/8*((a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(a*x^10) + (a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(40*a^2*x^10
)

Rule 1124

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^2
+ c*x^4)^(p + 1)/(4*a*d*(p + 1)*(2*p + 1))), x] - Simp[(d*x)^(m + 1)*(2*a + b*x^2)*((a + b*x^2 + c*x^4)^p/(4*a
*d*(2*p + 1))), x] /; FreeQ[{a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[m + 4*p + 5,
0] && NeQ[p, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 a x^{10}}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{40 a^2 x^{10}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (4 a^3+15 a^2 b x^2+20 a b^2 x^4+10 b^3 x^6\right )}{40 x^{10} \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^11,x]

[Out]

-1/40*(Sqrt[(a + b*x^2)^2]*(4*a^3 + 15*a^2*b*x^2 + 20*a*b^2*x^4 + 10*b^3*x^6))/(x^10*(a + b*x^2))

Maple [C] (warning: unable to verify)

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

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\frac {5}{2} b^{3} x^{6}+5 b^{2} x^{4} a +\frac {15}{4} a^{2} b \,x^{2}+a^{3}\right )}{10 x^{10}}\)	\(44\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{4} b^{3} x^{6}-\frac {1}{2} b^{2} x^{4} a -\frac {3}{8} a^{2} b \,x^{2}-\frac {1}{10} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{10}}\)	\(57\)
gosper	\(-\frac {\left (10 b^{3} x^{6}+20 b^{2} x^{4} a +15 a^{2} b \,x^{2}+4 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{40 x^{10} \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
default	\(-\frac {\left (10 b^{3} x^{6}+20 b^{2} x^{4} a +15 a^{2} b \,x^{2}+4 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{40 x^{10} \left (b \,x^{2}+a \right )^{3}}\)	\(58\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x,method=_RETURNVERBOSE)

[Out]

-1/10*csgn(b*x^2+a)*(5/2*b^3*x^6+5*b^2*x^4*a+15/4*a^2*b*x^2+a^3)/x^10

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {10 \, b^{3} x^{6} + 20 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 4 \, a^{3}}{40 \, x^{10}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x, algorithm="fricas")

[Out]

-1/40*(10*b^3*x^6 + 20*a*b^2*x^4 + 15*a^2*b*x^2 + 4*a^3)/x^10

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{11}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**11,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**11, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {b^{3}}{4 \, x^{4}} - \frac {a b^{2}}{2 \, x^{6}} - \frac {3 \, a^{2} b}{8 \, x^{8}} - \frac {a^{3}}{10 \, x^{10}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x, algorithm="maxima")

[Out]

-1/4*b^3/x^4 - 1/2*a*b^2/x^6 - 3/8*a^2*b/x^8 - 1/10*a^3/x^10

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {10 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{40 \, x^{10}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^11,x, algorithm="giac")

[Out]

-1/40*(10*b^3*x^6*sgn(b*x^2 + a) + 20*a*b^2*x^4*sgn(b*x^2 + a) + 15*a^2*b*x^2*sgn(b*x^2 + a) + 4*a^3*sgn(b*x^2
+ a))/x^10

Mupad [B] (verification not implemented)

Time = 13.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.10 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^4\,\left (b\,x^2+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^6\,\left (b\,x^2+a\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^11,x)

[Out]

- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(10*x^10*(a + b*x^2)) - (b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(4*x
^4*(a + b*x^2)) - (a*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(2*x^6*(a + b*x^2)) - (3*a^2*b*(a^2 + b^2*x^4 + 2*
a*b*x^2)^(1/2))/(8*x^8*(a + b*x^2))