\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{15}} \, dx\) [570]

Optimal result

Integrand size = 26, antiderivative size = 167 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )} \]

[Out]

-1/14*a^3*((b*x^2+a)^2)^(1/2)/x^14/(b*x^2+a)-1/4*a^2*b*((b*x^2+a)^2)^(1/2)/x^12/(b*x^2+a)-3/10*a*b^2*((b*x^2+a
)^2)^(1/2)/x^10/(b*x^2+a)-1/8*b^3*((b*x^2+a)^2)^(1/2)/x^8/(b*x^2+a)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 167, 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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^15,x]

[Out]

-1/14*(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^14*(a + b*x^2)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^
12*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*x^10*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(8*x^8*(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 )^{3/2}}{x^8} \, 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 )^3}{x^8} \, dx,x,x^2\right )}{2 b^2 \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^3 b^3}{x^8}+\frac {3 a^2 b^4}{x^7}+\frac {3 a b^5}{x^6}+\frac {b^6}{x^5}\right ) \, dx,x,x^2\right )}{2 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}}{14 x^{14} \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (20 a^3+70 a^2 b x^2+84 a b^2 x^4+35 b^3 x^6\right )}{280 x^{14} \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^15,x]

[Out]

-1/280*(Sqrt[(a + b*x^2)^2]*(20*a^3 + 70*a^2*b*x^2 + 84*a*b^2*x^4 + 35*b^3*x^6))/(x^14*(a + b*x^2))

Maple [C] (warning: unable to verify)

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

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.28

method	result	size
pseudoelliptic	\(-\frac {\left (35 b^{3} x^{6}+84 b^{2} x^{4} a +70 a^{2} b \,x^{2}+20 a^{3}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{280 x^{14}}\)	\(46\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{14} a^{3}-\frac {1}{4} a^{2} b \,x^{2}-\frac {3}{10} b^{2} x^{4} a -\frac {1}{8} b^{3} x^{6}\right )}{\left (b \,x^{2}+a \right ) x^{14}}\)	\(57\)
gosper	\(-\frac {\left (35 b^{3} x^{6}+84 b^{2} x^{4} a +70 a^{2} b \,x^{2}+20 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{280 x^{14} \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
default	\(-\frac {\left (35 b^{3} x^{6}+84 b^{2} x^{4} a +70 a^{2} b \,x^{2}+20 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{280 x^{14} \left (b \,x^{2}+a \right )^{3}}\)	\(58\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x,method=_RETURNVERBOSE)

[Out]

-1/280*(35*b^3*x^6+84*a*b^2*x^4+70*a^2*b*x^2+20*a^3)*csgn(b*x^2+a)/x^14

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {35 \, b^{3} x^{6} + 84 \, a b^{2} x^{4} + 70 \, a^{2} b x^{2} + 20 \, a^{3}}{280 \, x^{14}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x, algorithm="fricas")

[Out]

-1/280*(35*b^3*x^6 + 84*a*b^2*x^4 + 70*a^2*b*x^2 + 20*a^3)/x^14

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{15}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**15,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**15, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {b^{3}}{8 \, x^{8}} - \frac {3 \, a b^{2}}{10 \, x^{10}} - \frac {a^{2} b}{4 \, x^{12}} - \frac {a^{3}}{14 \, x^{14}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x, algorithm="maxima")

[Out]

-1/8*b^3/x^8 - 3/10*a*b^2/x^10 - 1/4*a^2*b/x^12 - 1/14*a^3/x^14

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {35 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 84 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 70 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{280 \, x^{14}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x, algorithm="giac")

[Out]

-1/280*(35*b^3*x^6*sgn(b*x^2 + a) + 84*a*b^2*x^4*sgn(b*x^2 + a) + 70*a^2*b*x^2*sgn(b*x^2 + a) + 20*a^3*sgn(b*x
^2 + a))/x^14

Mupad [B] (verification not implemented)

Time = 13.24 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^{12}\,\left (b\,x^2+a\right )} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^15,x)

[Out]

- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(8*x
^8*(a + b*x^2)) - (3*a*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(10*x^10*(a + b*x^2)) - (a^2*b*(a^2 + b^2*x^4 +
2*a*b*x^2)^(1/2))/(4*x^12*(a + b*x^2))