\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^2} \, dx\) [612]

Optimal result

Integrand size = 26, antiderivative size = 247 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {5 a^4 b x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {2 a^2 b^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a b^4 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {b^5 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )} \]

[Out]

-a^5*((b*x^2+a)^2)^(1/2)/x/(b*x^2+a)+5*a^4*b*x*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+10/3*a^3*b^2*x^3*((b*x^2+a)^2)^(1
/2)/(b*x^2+a)+2*a^2*b^3*x^5*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/7*a*b^4*x^7*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/9*b^5*
x^9*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 247, 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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {b^5 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {5 a b^4 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {2 a^2 b^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {5 a^4 b x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (-63 a^5+315 a^4 b x^2+210 a^3 b^2 x^4+126 a^2 b^3 x^6+45 a b^4 x^8+7 b^5 x^{10}\right )}{63 x \left (a+b x^2\right )} \]

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(-63*a^5 + 315*a^4*b*x^2 + 210*a^3*b^2*x^4 + 126*a^2*b^3*x^6 + 45*a*b^4*x^8 + 7*b^5*x^10)
)/(63*x*(a + b*x^2))

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32

method	result	size
gosper	\(-\frac {\left (-7 x^{10} b^{5}-45 a \,x^{8} b^{4}-126 a^{2} x^{6} b^{3}-210 a^{3} x^{4} b^{2}-315 x^{2} a^{4} b +63 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{63 x \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(-\frac {\left (-7 x^{10} b^{5}-45 a \,x^{8} b^{4}-126 a^{2} x^{6} b^{3}-210 a^{3} x^{4} b^{2}-315 x^{2} a^{4} b +63 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{63 x \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \left (\frac {1}{9} b^{4} x^{9}+\frac {5}{7} a \,b^{3} x^{7}+2 a^{2} b^{2} x^{5}+\frac {10}{3} a^{3} b \,x^{3}+5 a^{4} x \right )}{b \,x^{2}+a}-\frac {a^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x \left (b \,x^{2}+a \right )}\)	\(96\)

[In]

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

[Out]

-1/63*(-7*b^5*x^10-45*a*b^4*x^8-126*a^2*b^3*x^6-210*a^3*b^2*x^4-315*a^4*b*x^2+63*a^5)*((b*x^2+a)^2)^(5/2)/x/(b
*x^2+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {7 \, b^{5} x^{10} + 45 \, a b^{4} x^{8} + 126 \, a^{2} b^{3} x^{6} + 210 \, a^{3} b^{2} x^{4} + 315 \, a^{4} b x^{2} - 63 \, a^{5}}{63 \, x} \]

[In]

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

[Out]

1/63*(7*b^5*x^10 + 45*a*b^4*x^8 + 126*a^2*b^3*x^6 + 210*a^3*b^2*x^4 + 315*a^4*b*x^2 - 63*a^5)/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {1}{9} \, b^{5} x^{9} + \frac {5}{7} \, a b^{4} x^{7} + 2 \, a^{2} b^{3} x^{5} + \frac {10}{3} \, a^{3} b^{2} x^{3} + 5 \, a^{4} b x - \frac {a^{5}}{x} \]

[In]

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

[Out]

1/9*b^5*x^9 + 5/7*a*b^4*x^7 + 2*a^2*b^3*x^5 + 10/3*a^3*b^2*x^3 + 5*a^4*b*x - a^5/x

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {1}{9} \, b^{5} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a b^{4} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{3} \, a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{4} b x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{x} \]

[In]

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

[Out]

1/9*b^5*x^9*sgn(b*x^2 + a) + 5/7*a*b^4*x^7*sgn(b*x^2 + a) + 2*a^2*b^3*x^5*sgn(b*x^2 + a) + 10/3*a^3*b^2*x^3*sg
n(b*x^2 + a) + 5*a^4*b*x*sgn(b*x^2 + a) - a^5*sgn(b*x^2 + a)/x

Mupad [F(-1)]

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

[In]

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

[Out]

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