\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{13}} \, dx\) [598]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=-\frac {\left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 a x^{12}} \]

[Out]

-1/12*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/a/x^12

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 37} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=-\frac {\left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 a x^{12}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^13,x]

[Out]

-1/12*((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a*x^12)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 )^{5/2}}{x^7} \, 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 )^5}{x^7} \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = -\frac {\left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 a x^{12}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (a^5+6 a^4 b x^2+15 a^3 b^2 x^4+20 a^2 b^3 x^6+15 a b^4 x^8+6 b^5 x^{10}\right )}{12 x^{12} \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^13,x]

[Out]

-1/12*(Sqrt[(a + b*x^2)^2]*(a^5 + 6*a^4*b*x^2 + 15*a^3*b^2*x^4 + 20*a^2*b^3*x^6 + 15*a*b^4*x^8 + 6*b^5*x^10))/
(x^12*(a + b*x^2))

Maple [C] (warning: unable to verify)

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

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.41

method	result	size
pseudoelliptic	\(-\frac {\left (2 b \,x^{2}+a \right ) \left (3 b^{2} x^{4}+3 a b \,x^{2}+a^{2}\right ) \left (b^{2} x^{4}+a b \,x^{2}+a^{2}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{12 x^{12}}\)	\(58\)
gosper	\(-\frac {\left (6 x^{10} b^{5}+15 a \,x^{8} b^{4}+20 a^{2} x^{6} b^{3}+15 a^{3} x^{4} b^{2}+6 x^{2} a^{4} b +a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{12 x^{12} \left (b \,x^{2}+a \right )^{5}}\)	\(78\)
default	\(-\frac {\left (6 x^{10} b^{5}+15 a \,x^{8} b^{4}+20 a^{2} x^{6} b^{3}+15 a^{3} x^{4} b^{2}+6 x^{2} a^{4} b +a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{12 x^{12} \left (b \,x^{2}+a \right )^{5}}\)	\(78\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{2} x^{10} b^{5}-\frac {5}{4} a \,x^{8} b^{4}-\frac {5}{3} a^{2} x^{6} b^{3}-\frac {5}{4} a^{3} x^{4} b^{2}-\frac {1}{2} x^{2} a^{4} b -\frac {1}{12} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{12}}\)	\(79\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^13,x,method=_RETURNVERBOSE)

[Out]

-1/12*(2*b*x^2+a)*(3*b^2*x^4+3*a*b*x^2+a^2)*(b^2*x^4+a*b*x^2+a^2)*csgn(b*x^2+a)/x^12

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=-\frac {6 \, b^{5} x^{10} + 15 \, a b^{4} x^{8} + 20 \, a^{2} b^{3} x^{6} + 15 \, a^{3} b^{2} x^{4} + 6 \, a^{4} b x^{2} + a^{5}}{12 \, x^{12}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^13,x, algorithm="fricas")

[Out]

-1/12*(6*b^5*x^10 + 15*a*b^4*x^8 + 20*a^2*b^3*x^6 + 15*a^3*b^2*x^4 + 6*a^4*b*x^2 + a^5)/x^12

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{13}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**13,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**13, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=-\frac {b^{5}}{2 \, x^{2}} - \frac {5 \, a b^{4}}{4 \, x^{4}} - \frac {5 \, a^{2} b^{3}}{3 \, x^{6}} - \frac {5 \, a^{3} b^{2}}{4 \, x^{8}} - \frac {a^{4} b}{2 \, x^{10}} - \frac {a^{5}}{12 \, x^{12}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^13,x, algorithm="maxima")

[Out]

-1/2*b^5/x^2 - 5/4*a*b^4/x^4 - 5/3*a^2*b^3/x^6 - 5/4*a^3*b^2/x^8 - 1/2*a^4*b/x^10 - 1/12*a^5/x^12

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (28) = 56\).

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.59 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=-\frac {6 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{12 \, x^{12}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^13,x, algorithm="giac")

[Out]

-1/12*(6*b^5*x^10*sgn(b*x^2 + a) + 15*a*b^4*x^8*sgn(b*x^2 + a) + 20*a^2*b^3*x^6*sgn(b*x^2 + a) + 15*a^3*b^2*x^
4*sgn(b*x^2 + a) + 6*a^4*b*x^2*sgn(b*x^2 + a) + a^5*sgn(b*x^2 + a))/x^12

Mupad [B] (verification not implemented)

Time = 13.15 (sec) , antiderivative size = 231, normalized size of antiderivative = 5.63 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{13}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{12\,x^{12}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^2\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^4\,\left (b\,x^2+a\right )}-\frac {a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^{10}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^6\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^8\,\left (b\,x^2+a\right )} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^13,x)

[Out]

- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(12*x^12*(a + b*x^2)) - (b^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(2*x
^2*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(4*x^4*(a + b*x^2)) - (a^4*b*(a^2 + b^2*x^4 + 2*
a*b*x^2)^(1/2))/(2*x^10*(a + b*x^2)) - (5*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(3*x^6*(a + b*x^2)) - (5*
a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(4*x^8*(a + b*x^2))