3.4.96 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{11}} \, dx\)

Optimal. Leaf size=72 \[ \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{40 a^2 x^{10}}-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 a x^{10}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1110} \begin {gather*} \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{40 a^2 x^{10}}-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 a x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^11,x]

[Out]

-((a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(8*a*x^10) + (a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(40*a^2*x^10)

Rule 1110

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*x^2
+ c*x^4)^(p + 1))/(4*a*d*(p + 1)*(2*p + 1)), x] - Simp[((d*x)^(m + 1)*(2*a + b*x^2)*(a + b*x^2 + c*x^4)^p)/(4*
a*d*(2*p + 1)), x] /; FreeQ[{a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[m + 4*p + 5,
0] && NeQ[p, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{11}} \, dx &=-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 a x^{10}}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{40 a^2 x^{10}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 61, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (4 a^3+15 a^2 b x^2+20 a b^2 x^4+10 b^3 x^6\right )}{40 x^{10} \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)^(3/2)/x^11,x]

[Out]

-1/40*(Sqrt[(a + b*x^2)^2]*(4*a^3 + 15*a^2*b*x^2 + 20*a*b^2*x^4 + 10*b^3*x^6))/(x^10*(a + b*x^2))

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 0.95, size = 356, normalized size = 4.94 \begin {gather*} \frac {2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-4 a^7 b-31 a^6 b^2 x^2-104 a^5 b^3 x^4-196 a^4 b^4 x^6-224 a^3 b^5 x^8-155 a^2 b^6 x^{10}-60 a b^7 x^{12}-10 b^8 x^{14}\right )+2 \sqrt {b^2} b^4 \left (4 a^8+35 a^7 b x^2+135 a^6 b^2 x^4+300 a^5 b^3 x^6+420 a^4 b^4 x^8+379 a^3 b^5 x^{10}+215 a^2 b^6 x^{12}+70 a b^7 x^{14}+10 b^8 x^{16}\right )}{5 \sqrt {b^2} x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-16 a^4 b^4-64 a^3 b^5 x^2-96 a^2 b^6 x^4-64 a b^7 x^6-16 b^8 x^8\right )+5 x^{10} \left (16 a^5 b^5+80 a^4 b^6 x^2+160 a^3 b^7 x^4+160 a^2 b^8 x^6+80 a b^9 x^8+16 b^{10} x^{10}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^11,x]

[Out]

(2*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-4*a^7*b - 31*a^6*b^2*x^2 - 104*a^5*b^3*x^4 - 196*a^4*b^4*x^6 - 224*a^
3*b^5*x^8 - 155*a^2*b^6*x^10 - 60*a*b^7*x^12 - 10*b^8*x^14) + 2*b^4*Sqrt[b^2]*(4*a^8 + 35*a^7*b*x^2 + 135*a^6*
b^2*x^4 + 300*a^5*b^3*x^6 + 420*a^4*b^4*x^8 + 379*a^3*b^5*x^10 + 215*a^2*b^6*x^12 + 70*a*b^7*x^14 + 10*b^8*x^1
6))/(5*Sqrt[b^2]*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-16*a^4*b^4 - 64*a^3*b^5*x^2 - 96*a^2*b^6*x^4 - 64*a*b^
7*x^6 - 16*b^8*x^8) + 5*x^10*(16*a^5*b^5 + 80*a^4*b^6*x^2 + 160*a^3*b^7*x^4 + 160*a^2*b^8*x^6 + 80*a*b^9*x^8 +
16*b^10*x^10))

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 37, normalized size = 0.51 \begin {gather*} -\frac {10 \, b^{3} x^{6} + 20 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 4 \, a^{3}}{40 \, x^{10}} \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)^(3/2)/x^11,x, algorithm="fricas")

[Out]

-1/40*(10*b^3*x^6 + 20*a*b^2*x^4 + 15*a^2*b*x^2 + 4*a^3)/x^10

________________________________________________________________________________________

giac [A]  time = 0.21, size = 69, normalized size = 0.96 \begin {gather*} -\frac {10 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{40 \, x^{10}} \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)^(3/2)/x^11,x, algorithm="giac")

[Out]

-1/40*(10*b^3*x^6*sgn(b*x^2 + a) + 20*a*b^2*x^4*sgn(b*x^2 + a) + 15*a^2*b*x^2*sgn(b*x^2 + a) + 4*a^3*sgn(b*x^2
+ a))/x^10

________________________________________________________________________________________

maple [A]  time = 0.01, size = 58, normalized size = 0.81 \begin {gather*} -\frac {\left (10 b^{3} x^{6}+20 a \,b^{2} x^{4}+15 a^{2} b \,x^{2}+4 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{40 \left (b \,x^{2}+a \right )^{3} x^{10}} \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)^(3/2)/x^11,x)

[Out]

-1/40*(10*b^3*x^6+20*a*b^2*x^4+15*a^2*b*x^2+4*a^3)*((b*x^2+a)^2)^(3/2)/x^10/(b*x^2+a)^3

________________________________________________________________________________________

maxima [A]  time = 1.43, size = 35, normalized size = 0.49 \begin {gather*} -\frac {b^{3}}{4 \, x^{4}} - \frac {a b^{2}}{2 \, x^{6}} - \frac {3 \, a^{2} b}{8 \, x^{8}} - \frac {a^{3}}{10 \, x^{10}} \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)^(3/2)/x^11,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.20, size = 151, normalized size = 2.10 \begin {gather*} -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^4\,\left (b\,x^2+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^6\,\left (b\,x^2+a\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,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)^(3/2)/x^11,x)

[Out]

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

________________________________________________________________________________________

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 {3}{2}}}{x^{11}}\, 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)**(3/2)/x**11,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**11, x)

________________________________________________________________________________________