3.571 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{17}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 x^{16} \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \[ -\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 x^{16} \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^17,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*x^16*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(14*x
^14*(a + b*x^2)) - (a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^12*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4])/(10*x^10*(a + b*x^2))

Rule 43

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 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 )^{3/2}}{x^{17}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{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 )^3}{x^9} \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \left (\frac {a^3 b^3}{x^9}+\frac {3 a^2 b^4}{x^8}+\frac {3 a b^5}{x^7}+\frac {b^6}{x^6}\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 x^{16} \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 61, normalized size = 0.37 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (35 a^3+120 a^2 b x^2+140 a b^2 x^4+56 b^3 x^6\right )}{560 x^{16} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^17,x]

[Out]

-1/560*(Sqrt[(a + b*x^2)^2]*(35*a^3 + 120*a^2*b*x^2 + 140*a*b^2*x^4 + 56*b^3*x^6))/(x^16*(a + b*x^2))

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fricas [A]  time = 0.98, size = 37, normalized size = 0.22 \[ -\frac {56 \, b^{3} x^{6} + 140 \, a b^{2} x^{4} + 120 \, a^{2} b x^{2} + 35 \, a^{3}}{560 \, x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^17,x, algorithm="fricas")

[Out]

-1/560*(56*b^3*x^6 + 140*a*b^2*x^4 + 120*a^2*b*x^2 + 35*a^3)/x^16

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giac [A]  time = 0.18, size = 69, normalized size = 0.41 \[ -\frac {56 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 140 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 120 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 35 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{560 \, x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^17,x, algorithm="giac")

[Out]

-1/560*(56*b^3*x^6*sgn(b*x^2 + a) + 140*a*b^2*x^4*sgn(b*x^2 + a) + 120*a^2*b*x^2*sgn(b*x^2 + a) + 35*a^3*sgn(b
*x^2 + a))/x^16

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \[ -\frac {\left (56 b^{3} x^{6}+140 a \,b^{2} x^{4}+120 a^{2} b \,x^{2}+35 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{560 \left (b \,x^{2}+a \right )^{3} x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^17,x)

[Out]

-1/560*(56*b^3*x^6+140*a*b^2*x^4+120*a^2*b*x^2+35*a^3)*((b*x^2+a)^2)^(3/2)/x^16/(b*x^2+a)^3

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maxima [A]  time = 1.32, size = 35, normalized size = 0.21 \[ -\frac {b^{3}}{10 \, x^{10}} - \frac {a b^{2}}{4 \, x^{12}} - \frac {3 \, a^{2} b}{14 \, x^{14}} - \frac {a^{3}}{16 \, x^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^17,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.23, size = 151, normalized size = 0.90 \[ -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{16\,x^{16}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^{12}\,\left (b\,x^2+a\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^17,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{17}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**17,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**17, x)

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