\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{17}} \, dx\) [600]

Optimal result

Integrand size = 26, antiderivative size = 128 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {\left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 a x^{16}}+\frac {b \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{56 a^2 x^{14}}-\frac {b^2 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{336 a^3 x^{12}} \]

[Out]

-1/16*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/a/x^16+1/56*b*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/a^2/x^14-1/336*b^2*(b*x^2+
a)^5*((b*x^2+a)^2)^(1/2)/a^3/x^12

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1125, 660, 47, 37} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{16 a x^{16}}+\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{56 a^2 x^{14}}-\frac {b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{336 a^3 x^{12}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^17,x]

[Out]

-1/16*((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a*x^16) + (b*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x
^4])/(56*a^2*x^14) - (b^2*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(336*a^3*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 47

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] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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^9} \, 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^9} \, 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}}{16 a x^{16}}-\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^8} \, dx,x,x^2\right )}{8 a b^3 \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}}{16 a x^{16}}+\frac {b \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{56 a^2 x^{14}}+\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 )}{56 a^2 b^2 \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}}{16 a x^{16}}+\frac {b \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{56 a^2 x^{14}}-\frac {b^2 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{336 a^3 x^{12}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.65 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (21 a^5+120 a^4 b x^2+280 a^3 b^2 x^4+336 a^2 b^3 x^6+210 a b^4 x^8+56 b^5 x^{10}\right )}{336 x^{16} \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^17,x]

[Out]

-1/336*(Sqrt[(a + b*x^2)^2]*(21*a^5 + 120*a^4*b*x^2 + 280*a^3*b^2*x^4 + 336*a^2*b^3*x^6 + 210*a*b^4*x^8 + 56*b
^5*x^10))/(x^16*(a + b*x^2))

Maple [C] (warning: unable to verify)

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

Time = 0.74 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52

method	result	size
pseudoelliptic	\(-\frac {\left (\frac {8}{3} x^{10} b^{5}+10 a \,x^{8} b^{4}+16 a^{2} x^{6} b^{3}+\frac {40}{3} a^{3} x^{4} b^{2}+\frac {40}{7} x^{2} a^{4} b +a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{16 x^{16}}\)	\(66\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{16} a^{5}-\frac {5}{14} x^{2} a^{4} b -\frac {5}{6} a^{3} x^{4} b^{2}-a^{2} x^{6} b^{3}-\frac {5}{8} a \,x^{8} b^{4}-\frac {1}{6} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{16}}\)	\(79\)
gosper	\(-\frac {\left (56 x^{10} b^{5}+210 a \,x^{8} b^{4}+336 a^{2} x^{6} b^{3}+280 a^{3} x^{4} b^{2}+120 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 x^{16} \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(-\frac {\left (56 x^{10} b^{5}+210 a \,x^{8} b^{4}+336 a^{2} x^{6} b^{3}+280 a^{3} x^{4} b^{2}+120 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 x^{16} \left (b \,x^{2}+a \right )^{5}}\)	\(80\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^17,x,method=_RETURNVERBOSE)

[Out]

-1/16*(8/3*x^10*b^5+10*a*x^8*b^4+16*a^2*x^6*b^3+40/3*a^3*x^4*b^2+40/7*x^2*a^4*b+a^5)*csgn(b*x^2+a)/x^16

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.46 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {56 \, b^{5} x^{10} + 210 \, a b^{4} x^{8} + 336 \, a^{2} b^{3} x^{6} + 280 \, a^{3} b^{2} x^{4} + 120 \, a^{4} b x^{2} + 21 \, a^{5}}{336 \, x^{16}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^17,x, algorithm="fricas")

[Out]

-1/336*(56*b^5*x^10 + 210*a*b^4*x^8 + 336*a^2*b^3*x^6 + 280*a^3*b^2*x^4 + 120*a^4*b*x^2 + 21*a^5)/x^16

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{17}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**17,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**17, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.45 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {b^{5}}{6 \, x^{6}} - \frac {5 \, a b^{4}}{8 \, x^{8}} - \frac {a^{2} b^{3}}{x^{10}} - \frac {5 \, a^{3} b^{2}}{6 \, x^{12}} - \frac {5 \, a^{4} b}{14 \, x^{14}} - \frac {a^{5}}{16 \, x^{16}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^17,x, algorithm="maxima")

[Out]

-1/6*b^5/x^6 - 5/8*a*b^4/x^8 - a^2*b^3/x^10 - 5/6*a^3*b^2/x^12 - 5/14*a^4*b/x^14 - 1/16*a^5/x^16

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {56 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 210 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 336 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 280 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 120 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 21 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{336 \, x^{16}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^17,x, algorithm="giac")

[Out]

-1/336*(56*b^5*x^10*sgn(b*x^2 + a) + 210*a*b^4*x^8*sgn(b*x^2 + a) + 336*a^2*b^3*x^6*sgn(b*x^2 + a) + 280*a^3*b
^2*x^4*sgn(b*x^2 + a) + 120*a^4*b*x^2*sgn(b*x^2 + a) + 21*a^5*sgn(b*x^2 + a))/x^16

Mupad [B] (verification not implemented)

Time = 13.60 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{16\,x^{16}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^6\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^{10}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^{12}\,\left (b\,x^2+a\right )} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^17,x)

[Out]

- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(16*x^16*(a + b*x^2)) - (b^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(6*x
^6*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(8*x^8*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 +
2*a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(x^10*(a + b*x^2)) - (5*
a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(6*x^12*(a + b*x^2))