3.646 \(\int \frac {x^9}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=196 \[ -\frac {3 a^2}{2 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^4}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a^3}{3 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

2*a/b^5/((b*x^2+a)^2)^(1/2)-1/8*a^4/b^5/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)+2/3*a^3/b^5/(b*x^2+a)^2/((b*x^2+a)^2)^
(1/2)-3/2*a^2/b^5/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+1/2*(b*x^2+a)*ln(b*x^2+a)/b^5/((b*x^2+a)^2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 196, 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} \[ -\frac {a^4}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a^3}{3 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a^2}{2 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(2*a)/(b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - a^4/(8*b^5*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (2*a
^3)/(3*b^5*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (3*a^2)/(2*b^5*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) + ((a + b*x^2)*Log[a + b*x^2])/(2*b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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

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Mathematica [A]  time = 0.03, size = 83, normalized size = 0.42 \[ \frac {a \left (25 a^3+88 a^2 b x^2+108 a b^2 x^4+48 b^3 x^6\right )+12 \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 b^5 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(a*(25*a^3 + 88*a^2*b*x^2 + 108*a*b^2*x^4 + 48*b^3*x^6) + 12*(a + b*x^2)^4*Log[a + b*x^2])/(24*b^5*(a + b*x^2)
^3*Sqrt[(a + b*x^2)^2])

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fricas [A]  time = 1.07, size = 135, normalized size = 0.69 \[ \frac {48 \, a b^{3} x^{6} + 108 \, a^{2} b^{2} x^{4} + 88 \, a^{3} b x^{2} + 25 \, a^{4} + 12 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/24*(48*a*b^3*x^6 + 108*a^2*b^2*x^4 + 88*a^3*b*x^2 + 25*a^4 + 12*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a
^3*b*x^2 + a^4)*log(b*x^2 + a))/(b^9*x^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + a^4*b^5)

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giac [A]  time = 0.21, size = 84, normalized size = 0.43 \[ \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {48 \, a b^{2} x^{6} + 108 \, a^{2} b x^{4} + 88 \, a^{3} x^{2} + \frac {25 \, a^{4}}{b}}{24 \, {\left (b x^{2} + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/2*log(abs(b*x^2 + a))/(b^5*sgn(b*x^2 + a)) + 1/24*(48*a*b^2*x^6 + 108*a^2*b*x^4 + 88*a^3*x^2 + 25*a^4/b)/((b
*x^2 + a)^4*b^4*sgn(b*x^2 + a))

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maple [A]  time = 0.02, size = 141, normalized size = 0.72 \[ \frac {\left (12 b^{4} x^{8} \ln \left (b \,x^{2}+a \right )+48 a \,b^{3} x^{6} \ln \left (b \,x^{2}+a \right )+48 a \,b^{3} x^{6}+72 a^{2} b^{2} x^{4} \ln \left (b \,x^{2}+a \right )+108 a^{2} b^{2} x^{4}+48 a^{3} b \,x^{2} \ln \left (b \,x^{2}+a \right )+88 a^{3} b \,x^{2}+12 a^{4} \ln \left (b \,x^{2}+a \right )+25 a^{4}\right ) \left (b \,x^{2}+a \right )}{24 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/24*(12*ln(b*x^2+a)*x^8*b^4+48*ln(b*x^2+a)*x^6*a*b^3+48*a*b^3*x^6+72*ln(b*x^2+a)*x^4*a^2*b^2+108*a^2*b^2*x^4+
48*ln(b*x^2+a)*x^2*a^3*b+88*a^3*b*x^2+12*ln(b*x^2+a)*a^4+25*a^4)*(b*x^2+a)/b^5/((b*x^2+a)^2)^(5/2)

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maxima [A]  time = 1.46, size = 99, normalized size = 0.51 \[ \frac {48 \, a b^{3} x^{6} + 108 \, a^{2} b^{2} x^{4} + 88 \, a^{3} b x^{2} + 25 \, a^{4}}{24 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/24*(48*a*b^3*x^6 + 108*a^2*b^2*x^4 + 88*a^3*b*x^2 + 25*a^4)/(b^9*x^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b
^6*x^2 + a^4*b^5) + 1/2*log(b*x^2 + a)/b^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^9}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^9/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{9}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**9/((a + b*x**2)**2)**(5/2), x)

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