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

Optimal. Leaf size=128 \[ -\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{16 a x^{16}}+\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{56 a^2 x^{14}}-\frac {b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{336 a^3 x^{12}} \]

[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

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Rubi [A]  time = 0.09, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1111, 646, 45, 37} \[ -\frac {b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{336 a^3 x^{12}}+\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{56 a^2 x^{14}}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{16 a x^{16}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*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 45

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 646

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 1111

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*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{17}} \, dx &=\frac {1}{2} \operatorname {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} \operatorname {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} \operatorname {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} \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.65 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (21 a^5+120 a^4 b x^2+280 a^3 b^2 x^4+336 a^2 b^3 x^6+210 a b^4 x^8+56 b^5 x^{10}\right )}{336 x^{16} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 1.21, size = 59, normalized size = 0.46 \[ -\frac {56 \, b^{5} x^{10} + 210 \, a b^{4} x^{8} + 336 \, a^{2} b^{3} x^{6} + 280 \, a^{3} b^{2} x^{4} + 120 \, a^{4} b x^{2} + 21 \, a^{5}}{336 \, x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 107, normalized size = 0.84 \[ -\frac {56 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 210 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 336 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 280 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 120 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 21 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{336 \, x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.01, size = 80, normalized size = 0.62 \[ -\frac {\left (56 b^{5} x^{10}+210 a \,b^{4} x^{8}+336 a^{2} b^{3} x^{6}+280 a^{3} b^{2} x^{4}+120 a^{4} b \,x^{2}+21 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{336 \left (b \,x^{2}+a \right )^{5} x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*((b*x^2+a)^2)^(5/2)/x^
16/(b*x^2+a)^5

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maxima [A]  time = 1.35, size = 57, normalized size = 0.45 \[ -\frac {b^{5}}{6 \, x^{6}} - \frac {5 \, a b^{4}}{8 \, x^{8}} - \frac {a^{2} b^{3}}{x^{10}} - \frac {5 \, a^{3} b^{2}}{6 \, x^{12}} - \frac {5 \, a^{4} b}{14 \, x^{14}} - \frac {a^{5}}{16 \, x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.24, size = 231, normalized size = 1.80 \[ -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{16\,x^{16}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^6\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^{10}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^{12}\,\left (b\,x^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{17}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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