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

Optimal. Leaf size=41 \[ -\frac {\left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 a x^{12}} \]

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 37} \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*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 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^{13}} \, dx &=\frac {1}{2} \operatorname {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} \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 81, normalized size = 1.98 \begin {gather*} -\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 )} \end {gather*}

Antiderivative was successfully verified.

[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))

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IntegrateAlgebraic [B]  time = 2.63, size = 442, normalized size = 10.78 \begin {gather*} \frac {8 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-a^{10} b-11 a^9 b^2 x^2-55 a^8 b^3 x^4-165 a^7 b^4 x^6-330 a^6 b^5 x^8-462 a^5 b^6 x^{10}-461 a^4 b^7 x^{12}-325 a^3 b^8 x^{14}-155 a^2 b^9 x^{16}-45 a b^{10} x^{18}-6 b^{11} x^{20}\right )+8 \sqrt {b^2} b^5 \left (a^{11}+12 a^{10} b x^2+66 a^9 b^2 x^4+220 a^8 b^3 x^6+495 a^7 b^4 x^8+792 a^6 b^5 x^{10}+923 a^5 b^6 x^{12}+786 a^4 b^7 x^{14}+480 a^3 b^8 x^{16}+200 a^2 b^9 x^{18}+51 a b^{10} x^{20}+6 b^{11} x^{22}\right )}{3 \sqrt {b^2} x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-32 a^5 b^5-160 a^4 b^6 x^2-320 a^3 b^7 x^4-320 a^2 b^8 x^6-160 a b^9 x^8-32 b^{10} x^{10}\right )+3 x^{12} \left (32 a^6 b^6+192 a^5 b^7 x^2+480 a^4 b^8 x^4+640 a^3 b^9 x^6+480 a^2 b^{10} x^8+192 a b^{11} x^{10}+32 b^{12} x^{12}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-(a^10*b) - 11*a^9*b^2*x^2 - 55*a^8*b^3*x^4 - 165*a^7*b^4*x^6 - 330*a^
6*b^5*x^8 - 462*a^5*b^6*x^10 - 461*a^4*b^7*x^12 - 325*a^3*b^8*x^14 - 155*a^2*b^9*x^16 - 45*a*b^10*x^18 - 6*b^1
1*x^20) + 8*b^5*Sqrt[b^2]*(a^11 + 12*a^10*b*x^2 + 66*a^9*b^2*x^4 + 220*a^8*b^3*x^6 + 495*a^7*b^4*x^8 + 792*a^6
*b^5*x^10 + 923*a^5*b^6*x^12 + 786*a^4*b^7*x^14 + 480*a^3*b^8*x^16 + 200*a^2*b^9*x^18 + 51*a*b^10*x^20 + 6*b^1
1*x^22))/(3*Sqrt[b^2]*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-32*a^5*b^5 - 160*a^4*b^6*x^2 - 320*a^3*b^7*x^4 -
320*a^2*b^8*x^6 - 160*a*b^9*x^8 - 32*b^10*x^10) + 3*x^12*(32*a^6*b^6 + 192*a^5*b^7*x^2 + 480*a^4*b^8*x^4 + 640
*a^3*b^9*x^6 + 480*a^2*b^10*x^8 + 192*a*b^11*x^10 + 32*b^12*x^12))

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fricas [B]  time = 0.87, size = 57, normalized size = 1.39 \begin {gather*} -\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}} \end {gather*}

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^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

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giac [B]  time = 0.16, size = 106, normalized size = 2.59 \begin {gather*} -\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}} \end {gather*}

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^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

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maple [B]  time = 0.01, size = 78, normalized size = 1.90 \begin {gather*} -\frac {\left (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}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{12 \left (b \,x^{2}+a \right )^{5} x^{12}} \end {gather*}

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^13,x)

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

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maxima [B]  time = 1.40, size = 57, normalized size = 1.39 \begin {gather*} -\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}} \end {gather*}

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^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

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mupad [B]  time = 4.18, size = 231, normalized size = 5.63 \begin {gather*} -\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 )} \end {gather*}

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^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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{13}}\, dx \end {gather*}

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**13,x)

[Out]

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

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