\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{12}} \, dx\) [582]

Optimal result

Integrand size = 26, antiderivative size = 167 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^9 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )} \]

[Out]

-1/11*a^3*((b*x^2+a)^2)^(1/2)/x^11/(b*x^2+a)-1/3*a^2*b*((b*x^2+a)^2)^(1/2)/x^9/(b*x^2+a)-3/7*a*b^2*((b*x^2+a)^
2)^(1/2)/x^7/(b*x^2+a)-1/5*b^3*((b*x^2+a)^2)^(1/2)/x^5/(b*x^2+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1126, 276} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^9 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^12,x]

[Out]

-1/11*(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^11*(a + b*x^2)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*x^
9*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*x^7*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4])/(5*x^5*(a + b*x^2))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && 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^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^3}{x^{12}} \, dx}{b^2 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^3 b^3}{x^{12}}+\frac {3 a^2 b^4}{x^{10}}+\frac {3 a b^5}{x^8}+\frac {b^6}{x^6}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )} \\ & = -\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^9 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (105 a^3+385 a^2 b x^2+495 a b^2 x^4+231 b^3 x^6\right )}{1155 x^{11} \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^12,x]

[Out]

-1/1155*(Sqrt[(a + b*x^2)^2]*(105*a^3 + 385*a^2*b*x^2 + 495*a*b^2*x^4 + 231*b^3*x^6))/(x^11*(a + b*x^2))

Maple [A] (verified)

Time = 5.47 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.34

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{5} b^{3} x^{6}-\frac {3}{7} b^{2} x^{4} a -\frac {1}{3} a^{2} b \,x^{2}-\frac {1}{11} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{11}}\)	\(57\)
gosper	\(-\frac {\left (231 b^{3} x^{6}+495 b^{2} x^{4} a +385 a^{2} b \,x^{2}+105 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{1155 x^{11} \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
default	\(-\frac {\left (231 b^{3} x^{6}+495 b^{2} x^{4} a +385 a^{2} b \,x^{2}+105 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{1155 x^{11} \left (b \,x^{2}+a \right )^{3}}\)	\(58\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^12,x,method=_RETURNVERBOSE)

[Out]

((b*x^2+a)^2)^(1/2)/(b*x^2+a)*(-1/5*b^3*x^6-3/7*b^2*x^4*a-1/3*a^2*b*x^2-1/11*a^3)/x^11

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {231 \, b^{3} x^{6} + 495 \, a b^{2} x^{4} + 385 \, a^{2} b x^{2} + 105 \, a^{3}}{1155 \, x^{11}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^12,x, algorithm="fricas")

[Out]

-1/1155*(231*b^3*x^6 + 495*a*b^2*x^4 + 385*a^2*b*x^2 + 105*a^3)/x^11

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{12}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**12,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**12, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {b^{3}}{5 \, x^{5}} - \frac {3 \, a b^{2}}{7 \, x^{7}} - \frac {a^{2} b}{3 \, x^{9}} - \frac {a^{3}}{11 \, x^{11}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^12,x, algorithm="maxima")

[Out]

-1/5*b^3/x^5 - 3/7*a*b^2/x^7 - 1/3*a^2*b/x^9 - 1/11*a^3/x^11

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {231 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 495 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 385 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{1155 \, x^{11}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^12,x, algorithm="giac")

[Out]

-1/1155*(231*b^3*x^6*sgn(b*x^2 + a) + 495*a*b^2*x^4*sgn(b*x^2 + a) + 385*a^2*b*x^2*sgn(b*x^2 + a) + 105*a^3*sg
n(b*x^2 + a))/x^11

Mupad [B] (verification not implemented)

Time = 13.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{5\,x^5\,\left (b\,x^2+a\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^7\,\left (b\,x^2+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^9\,\left (b\,x^2+a\right )} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^12,x)

[Out]

- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(11*x^11*(a + b*x^2)) - (b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(5*x
^5*(a + b*x^2)) - (3*a*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(7*x^7*(a + b*x^2)) - (a^2*b*(a^2 + b^2*x^4 + 2*
a*b*x^2)^(1/2))/(3*x^9*(a + b*x^2))