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

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

[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])

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Rubi [A]  time = 0.162137, antiderivative size = 196, normalized size of antiderivative = 1., 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 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])

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 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])

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*}

Mathematica [A]  time = 0.0292026, size = 83, normalized size = 0.42 \[ \frac{a \left (88 a^2 b x^2+25 a^3+108 a b^2 x^4+48 b^3 x^6\right )+12 \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 b^5 \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

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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Maple [A]  time = 0.23, size = 141, normalized size = 0.7 \begin{align*}{\frac{ \left ( 12\,\ln \left ( b{x}^{2}+a \right ){x}^{8}{b}^{4}+48\,\ln \left ( b{x}^{2}+a \right ){x}^{6}a{b}^{3}+48\,a{b}^{3}{x}^{6}+72\,\ln \left ( b{x}^{2}+a \right ){x}^{4}{a}^{2}{b}^{2}+108\,{a}^{2}{b}^{2}{x}^{4}+48\,\ln \left ( b{x}^{2}+a \right ){x}^{2}{a}^{3}b+88\,{a}^{3}b{x}^{2}+12\,\ln \left ( b{x}^{2}+a \right ){a}^{4}+25\,{a}^{4} \right ) \left ( b{x}^{2}+a \right ) }{24\,{b}^{5}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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 = 1.02962, size = 134, normalized size = 0.68 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 1.26654, size = 282, normalized size = 1.44 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{9}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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]

sage0*x