\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^9} \, dx\) [596]

Optimal result

Integrand size = 26, antiderivative size = 250 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2 \left (a+b x^2\right )}+\frac {b^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 250, 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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {b^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2 \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^9,x]

[Out]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(696\) vs. \(2(250)=500\).

Time = 0.98 (sec) , antiderivative size = 696, normalized size of antiderivative = 2.78 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {48 a^6 \sqrt {a^2}+368 a^5 \sqrt {a^2} b x^2+1280 a^4 \sqrt {a^2} b^2 x^4+2880 a^3 \sqrt {a^2} b^3 x^6+2677 \left (a^2\right )^{3/2} b^4 x^8+565 a \sqrt {a^2} b^5 x^{10}-192 \sqrt {a^2} b^6 x^{12}-48 a^6 \sqrt {\left (a+b x^2\right )^2}-320 a^5 b x^2 \sqrt {\left (a+b x^2\right )^2}-960 a^4 b^2 x^4 \sqrt {\left (a+b x^2\right )^2}-1920 a^3 b^3 x^6 \sqrt {\left (a+b x^2\right )^2}-757 a^2 b^4 x^8 \sqrt {\left (a+b x^2\right )^2}+192 a b^5 x^{10} \sqrt {\left (a+b x^2\right )^2}-960 a b^4 x^8 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right ) \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-960 b^4 x^8 \left (\left (a^2\right )^{3/2}+a \sqrt {a^2} b x^2-a^2 \sqrt {\left (a+b x^2\right )^2}\right ) \log \left (x^2\right )+480 \left (a^2\right )^{3/2} b^4 x^8 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+480 a \sqrt {a^2} b^5 x^{10} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-480 a^2 b^4 x^8 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+480 \left (a^2\right )^{3/2} b^4 x^8 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+480 a \sqrt {a^2} b^5 x^{10} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-480 a^2 b^4 x^8 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )}{384 x^8 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^9,x]

[Out]

(48*a^6*Sqrt[a^2] + 368*a^5*Sqrt[a^2]*b*x^2 + 1280*a^4*Sqrt[a^2]*b^2*x^4 + 2880*a^3*Sqrt[a^2]*b^3*x^6 + 2677*(
a^2)^(3/2)*b^4*x^8 + 565*a*Sqrt[a^2]*b^5*x^10 - 192*Sqrt[a^2]*b^6*x^12 - 48*a^6*Sqrt[(a + b*x^2)^2] - 320*a^5*
b*x^2*Sqrt[(a + b*x^2)^2] - 960*a^4*b^2*x^4*Sqrt[(a + b*x^2)^2] - 1920*a^3*b^3*x^6*Sqrt[(a + b*x^2)^2] - 757*a
^2*b^4*x^8*Sqrt[(a + b*x^2)^2] + 192*a*b^5*x^10*Sqrt[(a + b*x^2)^2] - 960*a*b^4*x^8*(a^2 + a*b*x^2 - Sqrt[a^2]
*Sqrt[(a + b*x^2)^2])*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 960*b^4*x^8*((a^2)^(3/2) + a*Sqrt[a
^2]*b*x^2 - a^2*Sqrt[(a + b*x^2)^2])*Log[x^2] + 480*(a^2)^(3/2)*b^4*x^8*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^
2)^2]] + 480*a*Sqrt[a^2]*b^5*x^10*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] - 480*a^2*b^4*x^8*Sqrt[(a + b*x
^2)^2]*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + 480*(a^2)^(3/2)*b^4*x^8*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a
+ b*x^2)^2]] + 480*a*Sqrt[a^2]*b^5*x^10*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]] - 480*a^2*b^4*x^8*Sqrt[(a
+ b*x^2)^2]*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]])/(384*x^8*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2
)^2]))

Maple [C] (warning: unable to verify)

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

Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.28

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-4 x^{10} b^{5}-20 b^{4} a \ln \left (x^{2}\right ) x^{8}+40 a^{2} x^{6} b^{3}+20 a^{3} x^{4} b^{2}+\frac {20 x^{2} a^{4} b}{3}+a^{5}\right )}{8 x^{8}}\)	\(70\)
default	\(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (12 x^{10} b^{5}+120 \ln \left (x \right ) x^{8} a \,b^{4}-120 a^{2} x^{6} b^{3}-60 a^{3} x^{4} b^{2}-20 x^{2} a^{4} b -3 a^{5}\right )}{24 \left (b \,x^{2}+a \right )^{5} x^{8}}\)	\(82\)
risch	\(\frac {b^{5} x^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 b \,x^{2}+2 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-5 a^{2} x^{6} b^{3}-\frac {5}{2} a^{3} x^{4} b^{2}-\frac {5}{6} x^{2} a^{4} b -\frac {1}{8} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{8}}+\frac {5 a \,b^{4} \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(119\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/8*csgn(b*x^2+a)*(-4*x^10*b^5-20*b^4*a*ln(x^2)*x^8+40*a^2*x^6*b^3+20*a^3*x^4*b^2+20/3*x^2*a^4*b+a^5)/x^8

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {12 \, b^{5} x^{10} + 120 \, a b^{4} x^{8} \log \left (x\right ) - 120 \, a^{2} b^{3} x^{6} - 60 \, a^{3} b^{2} x^{4} - 20 \, a^{4} b x^{2} - 3 \, a^{5}}{24 \, x^{8}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x, algorithm="fricas")

[Out]

1/24*(12*b^5*x^10 + 120*a*b^4*x^8*log(x) - 120*a^2*b^3*x^6 - 60*a^3*b^2*x^4 - 20*a^4*b*x^2 - 3*a^5)/x^8

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{9}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**9,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**9, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {1}{2} \, b^{5} x^{2} + 5 \, a b^{4} \log \left (x\right ) - \frac {5 \, a^{2} b^{3}}{x^{2}} - \frac {5 \, a^{3} b^{2}}{2 \, x^{4}} - \frac {5 \, a^{4} b}{6 \, x^{6}} - \frac {a^{5}}{8 \, x^{8}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\frac {1}{2} \, b^{5} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a b^{4} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {125 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 120 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 60 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{24 \, x^{8}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^9,x, algorithm="giac")

[Out]

1/2*b^5*x^2*sgn(b*x^2 + a) + 5/2*a*b^4*log(x^2)*sgn(b*x^2 + a) - 1/24*(125*a*b^4*x^8*sgn(b*x^2 + a) + 120*a^2*
b^3*x^6*sgn(b*x^2 + a) + 60*a^3*b^2*x^4*sgn(b*x^2 + a) + 20*a^4*b*x^2*sgn(b*x^2 + a) + 3*a^5*sgn(b*x^2 + a))/x
^8

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^9} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^9} \,d x \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^9,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^9, x)