\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^5} \, dx\) [594]

Optimal result

Integrand size = 26, antiderivative size = 250 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^5} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a b^4 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {b^5 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]

[Out]

-1/4*a^5*((b*x^2+a)^2)^(1/2)/x^4/(b*x^2+a)-5/2*a^4*b*((b*x^2+a)^2)^(1/2)/x^2/(b*x^2+a)+5*a^2*b^3*x^2*((b*x^2+a
)^2)^(1/2)/(b*x^2+a)+5/4*a*b^4*x^4*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/6*b^5*x^6*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+10*
a^3*b^2*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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^5} \, dx=\frac {b^5 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {5 a b^4 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {10 a^3 b^2 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^5} \, dx=\frac {\left (12 a^5+120 a^4 b x^2+57 a^3 b^2 x^4-240 a^2 b^3 x^6-60 a b^4 x^8-8 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{48 x^4 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )}-5 a^3 b^2 \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-5 \left (a^2\right )^{3/2} b^2 \log \left (x^2\right )+\frac {5}{2} \left (a^2\right )^{3/2} b^2 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+\frac {5}{2} \left (a^2\right )^{3/2} b^2 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right ) \]

[In]

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

[Out]

((12*a^5 + 120*a^4*b*x^2 + 57*a^3*b^2*x^4 - 240*a^2*b^3*x^6 - 60*a*b^4*x^8 - 8*b^5*x^10)*(Sqrt[a^2]*b*x^2 + a*
(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])))/(48*x^4*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2])) - 5*a^3*b^2*ArcTa
nh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 5*(a^2)^(3/2)*b^2*Log[x^2] + (5*(a^2)^(3/2)*b^2*Log[Sqrt[a^2]
- b*x^2 - Sqrt[(a + b*x^2)^2]])/2 + (5*(a^2)^(3/2)*b^2*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]])/2

Maple [C] (warning: unable to verify)

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

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.28

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-\frac {2 x^{10} b^{5}}{3}-5 a \,x^{8} b^{4}-20 a^{2} x^{6} b^{3}-20 a^{3} b^{2} \ln \left (x^{2}\right ) x^{4}+10 x^{2} a^{4} b +a^{5}\right )}{4 x^{4}}\)	\(70\)
default	\(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (2 x^{10} b^{5}+15 a \,x^{8} b^{4}+60 a^{2} x^{6} b^{3}+120 \ln \left (x \right ) x^{4} a^{3} b^{2}-30 x^{2} a^{4} b -3 a^{5}\right )}{12 \left (b \,x^{2}+a \right )^{5} x^{4}}\)	\(82\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{3} \left (\frac {1}{6} b^{2} x^{6}+\frac {5}{4} a b \,x^{4}+5 a^{2} x^{2}\right )}{b \,x^{2}+a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {5}{2} x^{2} a^{4} b -\frac {1}{4} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{4}}+\frac {10 a^{3} b^{2} \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(119\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^5} \, dx=\frac {2 \, b^{5} x^{10} + 15 \, a b^{4} x^{8} + 60 \, a^{2} b^{3} x^{6} + 120 \, a^{3} b^{2} x^{4} \log \left (x\right ) - 30 \, a^{4} b x^{2} - 3 \, a^{5}}{12 \, x^{4}} \]

[In]

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

[Out]

1/12*(2*b^5*x^10 + 15*a*b^4*x^8 + 60*a^2*b^3*x^6 + 120*a^3*b^2*x^4*log(x) - 30*a^4*b*x^2 - 3*a^5)/x^4

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^5} \, dx=\frac {1}{6} \, b^{5} x^{6} + \frac {5}{4} \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} \log \left (x\right ) - \frac {5 \, a^{4} b}{2 \, x^{2}} - \frac {a^{5}}{4 \, x^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^5} \, dx=\frac {1}{6} \, b^{5} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{4} \, a b^{4} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{2} b^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{3} b^{2} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {30 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 10 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, x^{4}} \]

[In]

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

[Out]

1/6*b^5*x^6*sgn(b*x^2 + a) + 5/4*a*b^4*x^4*sgn(b*x^2 + a) + 5*a^2*b^3*x^2*sgn(b*x^2 + a) + 5*a^3*b^2*log(x^2)*
sgn(b*x^2 + a) - 1/4*(30*a^3*b^2*x^4*sgn(b*x^2 + a) + 10*a^4*b*x^2*sgn(b*x^2 + a) + a^5*sgn(b*x^2 + a))/x^4

Mupad [F(-1)]

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

[In]

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

[Out]

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