3.5.45 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{12}} \, dx\)

Optimal. Leaf size=251 \[ -\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {2 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5 \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1112, 270} \begin {gather*} -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {2 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^12,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*x^11*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*x^
9*(a + b*x^2)) - (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*x^7*(a + b*x^2)) - (2*a^2*b^3*Sqrt[a^2 + 2*a*
b*x^2 + b^2*x^4])/(x^5*(a + b*x^2)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*x^3*(a + b*x^2)) - (b^5*Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4])/(x*(a + b*x^2))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1112

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

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{12}} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^5}{x^{12}} \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5 b^5}{x^{12}}+\frac {5 a^4 b^6}{x^{10}}+\frac {10 a^3 b^7}{x^8}+\frac {10 a^2 b^8}{x^6}+\frac {5 a b^9}{x^4}+\frac {b^{10}}{x^2}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {2 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (63 a^5+385 a^4 b x^2+990 a^3 b^2 x^4+1386 a^2 b^3 x^6+1155 a b^4 x^8+693 b^5 x^{10}\right )}{693 x^{11} \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^12,x]

[Out]

-1/693*(Sqrt[(a + b*x^2)^2]*(63*a^5 + 385*a^4*b*x^2 + 990*a^3*b^2*x^4 + 1386*a^2*b^3*x^6 + 1155*a*b^4*x^8 + 69
3*b^5*x^10))/(x^11*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 18.96, size = 83, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (-63 a^5-385 a^4 b x^2-990 a^3 b^2 x^4-1386 a^2 b^3 x^6-1155 a b^4 x^8-693 b^5 x^{10}\right )}{693 x^{11} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^12,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(-63*a^5 - 385*a^4*b*x^2 - 990*a^3*b^2*x^4 - 1386*a^2*b^3*x^6 - 1155*a*b^4*x^8 - 693*b^5*
x^10))/(693*x^11*(a + b*x^2))

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fricas [A]  time = 0.94, size = 59, normalized size = 0.24 \begin {gather*} -\frac {693 \, b^{5} x^{10} + 1155 \, a b^{4} x^{8} + 1386 \, a^{2} b^{3} x^{6} + 990 \, a^{3} b^{2} x^{4} + 385 \, a^{4} b x^{2} + 63 \, a^{5}}{693 \, x^{11}} \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^12,x, algorithm="fricas")

[Out]

-1/693*(693*b^5*x^10 + 1155*a*b^4*x^8 + 1386*a^2*b^3*x^6 + 990*a^3*b^2*x^4 + 385*a^4*b*x^2 + 63*a^5)/x^11

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giac [A]  time = 0.16, size = 107, normalized size = 0.43 \begin {gather*} -\frac {693 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 1155 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 1386 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 990 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 385 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 63 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{693 \, x^{11}} \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^12,x, algorithm="giac")

[Out]

-1/693*(693*b^5*x^10*sgn(b*x^2 + a) + 1155*a*b^4*x^8*sgn(b*x^2 + a) + 1386*a^2*b^3*x^6*sgn(b*x^2 + a) + 990*a^
3*b^2*x^4*sgn(b*x^2 + a) + 385*a^4*b*x^2*sgn(b*x^2 + a) + 63*a^5*sgn(b*x^2 + a))/x^11

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maple [A]  time = 0.01, size = 80, normalized size = 0.32 \begin {gather*} -\frac {\left (693 b^{5} x^{10}+1155 a \,b^{4} x^{8}+1386 a^{2} b^{3} x^{6}+990 a^{3} b^{2} x^{4}+385 a^{4} b \,x^{2}+63 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{693 \left (b \,x^{2}+a \right )^{5} x^{11}} \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^12,x)

[Out]

-1/693*(693*b^5*x^10+1155*a*b^4*x^8+1386*a^2*b^3*x^6+990*a^3*b^2*x^4+385*a^4*b*x^2+63*a^5)*((b*x^2+a)^2)^(5/2)
/x^11/(b*x^2+a)^5

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maxima [A]  time = 1.33, size = 57, normalized size = 0.23 \begin {gather*} -\frac {b^{5}}{x} - \frac {5 \, a b^{4}}{3 \, x^{3}} - \frac {2 \, a^{2} b^{3}}{x^{5}} - \frac {10 \, a^{3} b^{2}}{7 \, x^{7}} - \frac {5 \, a^{4} b}{9 \, x^{9}} - \frac {a^{5}}{11 \, x^{11}} \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^12,x, algorithm="maxima")

[Out]

-b^5/x - 5/3*a*b^4/x^3 - 2*a^2*b^3/x^5 - 10/7*a^3*b^2/x^7 - 5/9*a^4*b/x^9 - 1/11*a^5/x^11

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mupad [B]  time = 4.22, size = 231, normalized size = 0.92 \begin {gather*} -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^3\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^9\,\left (b\,x^2+a\right )}-\frac {2\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^5\,\left (b\,x^2+a\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^7\,\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^12,x)

[Out]

- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(11*x^11*(a + b*x^2)) - (b^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(x*(
a + b*x^2)) - (5*a*b^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(3*x^3*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 + 2*a*
b*x^2)^(1/2))/(9*x^9*(a + b*x^2)) - (2*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(x^5*(a + b*x^2)) - (10*a^3*
b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(7*x^7*(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^{12}}\, 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**12,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**12, x)

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