\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{11}} \, dx\) [597]

Optimal result

Integrand size = 26, antiderivative size = 251 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^6 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]

[Out]

-1/10*a^5*((b*x^2+a)^2)^(1/2)/x^10/(b*x^2+a)-5/8*a^4*b*((b*x^2+a)^2)^(1/2)/x^8/(b*x^2+a)-5/3*a^3*b^2*((b*x^2+a
)^2)^(1/2)/x^6/(b*x^2+a)-5/2*a^2*b^3*((b*x^2+a)^2)^(1/2)/x^4/(b*x^2+a)-5/2*a*b^4*((b*x^2+a)^2)^(1/2)/x^2/(b*x^
2+a)+b^5*ln(x)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 251, 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 = {1126, 272, 45} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\frac {b^5 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^6 \left (a+b x^2\right )} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^11,x]

[Out]

-1/10*(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^10*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*
x^8*(a + b*x^2)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*x^6*(a + b*x^2)) - (5*a^2*b^3*Sqrt[a^2 + 2*a
*b*x^2 + b^2*x^4])/(2*x^4*(a + b*x^2)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^2*(a + b*x^2)) + (b^5*
Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[x])/(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 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 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 )^5}{x^{11}} \, dx}{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 \frac {\left (a b+b^2 x\right )^5}{x^6} \, 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^6}+\frac {5 a^4 b^6}{x^5}+\frac {10 a^3 b^7}{x^4}+\frac {10 a^2 b^8}{x^3}+\frac {5 a b^9}{x^2}+\frac {b^{10}}{x}\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}}{10 x^{10} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^6 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\frac {1}{240} \left (-\frac {\sqrt {\left (a+b x^2\right )^2} \left (12 a^4+63 a^3 b x^2+137 a^2 b^2 x^4+163 a b^3 x^6+137 b^4 x^8\right )}{x^{10}}+\frac {\sqrt {a^2} \left (12 a^4+75 a^3 b x^2+200 a^2 b^2 x^4+300 a b^3 x^6+300 b^4 x^8\right )}{x^{10}}-120 b^5 \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-\frac {120 \sqrt {a^2} b^5 \log \left (x^2\right )}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{a}\right ) \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^11,x]

[Out]

(-((Sqrt[(a + b*x^2)^2]*(12*a^4 + 63*a^3*b*x^2 + 137*a^2*b^2*x^4 + 163*a*b^3*x^6 + 137*b^4*x^8))/x^10) + (Sqrt
[a^2]*(12*a^4 + 75*a^3*b*x^2 + 200*a^2*b^2*x^4 + 300*a*b^3*x^6 + 300*b^4*x^8))/x^10 - 120*b^5*ArcTanh[(b*x^2)/
(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - (120*Sqrt[a^2]*b^5*Log[x^2])/a + (60*Sqrt[a^2]*b^5*Log[a*(Sqrt[a^2] - b*x
^2 - Sqrt[(a + b*x^2)^2])])/a + (60*Sqrt[a^2]*b^5*Log[a*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/a)/240

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.28

method	result	size
pseudoelliptic	\(-\frac {\left (-5 b^{5} \ln \left (x^{2}\right ) x^{10}+a \left (25 b^{4} x^{8}+25 a \,b^{3} x^{6}+\frac {50}{3} a^{2} b^{2} x^{4}+\frac {25}{4} a^{3} b \,x^{2}+a^{4}\right )\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{10 x^{10}}\)	\(70\)
default	\(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (120 \ln \left (x \right ) x^{10} b^{5}-300 a \,x^{8} b^{4}-300 a^{2} x^{6} b^{3}-200 a^{3} x^{4} b^{2}-75 x^{2} a^{4} b -12 a^{5}\right )}{120 \left (b \,x^{2}+a \right )^{5} x^{10}}\)	\(82\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {5}{2} a \,x^{8} b^{4}-\frac {5}{2} a^{2} x^{6} b^{3}-\frac {5}{3} a^{3} x^{4} b^{2}-\frac {5}{8} x^{2} a^{4} b -\frac {1}{10} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{10}}+\frac {b^{5} \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(98\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^11,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\frac {120 \, b^{5} x^{10} \log \left (x\right ) - 300 \, a b^{4} x^{8} - 300 \, a^{2} b^{3} x^{6} - 200 \, a^{3} b^{2} x^{4} - 75 \, a^{4} b x^{2} - 12 \, a^{5}}{120 \, x^{10}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^11,x, algorithm="fricas")

[Out]

1/120*(120*b^5*x^10*log(x) - 300*a*b^4*x^8 - 300*a^2*b^3*x^6 - 200*a^3*b^2*x^4 - 75*a^4*b*x^2 - 12*a^5)/x^10

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{11}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**11,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**11, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=b^{5} \log \left (x\right ) - \frac {5 \, a b^{4}}{2 \, x^{2}} - \frac {5 \, a^{2} b^{3}}{2 \, x^{4}} - \frac {5 \, a^{3} b^{2}}{3 \, x^{6}} - \frac {5 \, a^{4} b}{8 \, x^{8}} - \frac {a^{5}}{10 \, x^{10}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^11,x, algorithm="maxima")

[Out]

b^5*log(x) - 5/2*a*b^4/x^2 - 5/2*a^2*b^3/x^4 - 5/3*a^3*b^2/x^6 - 5/8*a^4*b/x^8 - 1/10*a^5/x^10

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\frac {1}{2} \, b^{5} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {137 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 300 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 300 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 200 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 75 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 12 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{120 \, x^{10}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^11,x, algorithm="giac")

[Out]

1/2*b^5*log(x^2)*sgn(b*x^2 + a) - 1/120*(137*b^5*x^10*sgn(b*x^2 + a) + 300*a*b^4*x^8*sgn(b*x^2 + a) + 300*a^2*
b^3*x^6*sgn(b*x^2 + a) + 200*a^3*b^2*x^4*sgn(b*x^2 + a) + 75*a^4*b*x^2*sgn(b*x^2 + a) + 12*a^5*sgn(b*x^2 + a))
/x^10

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{11}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^{11}} \,d x \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^11,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^11, x)