3.5.74 \(\int \frac {x^9}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)
Optimal. Leaf size=196 \[ -\frac {3 a^2}{2 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^4}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a^3}{3 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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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} \begin {gather*} -\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}} \end {gather*}
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 \begin {gather*} \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}} \end {gather*}
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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IntegrateAlgebraic [B] time = 2.28, size = 3027, normalized size = 15.44 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[x^9/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
[Out]
((16*a^8)/(b^4*Sqrt[b^2]) + (224*a^7*x^2)/(3*b^3*Sqrt[b^2]) + (944*a^6*x^4)/(3*(b^2)^(3/2)) + (800*a^5*x^6)/(b
*Sqrt[b^2]) + (1200*a^4*x^8)/Sqrt[b^2] + (960*a^3*b*x^10)/Sqrt[b^2] + 320*a^2*Sqrt[b^2]*x^12 - (224*a^6*x^2*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*b^4) - (240*a^5*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^3 - (560*a^4*x^6*Sqrt
[a^2 + 2*a*b*x^2 + b^2*x^4])/b^2 - (640*a^3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b - 320*a^2*x^10*Sqrt[a^2 + 2
*a*b*x^2 + b^2*x^4] - (32*a^4*x^8*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] - (128*
a^3*b*x^10*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] - 192*a^2*Sqrt[b^2]*x^12*Log[-
a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] - (128*a*b^3*x^14*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a
*b*x^2 + b^2*x^4]])/Sqrt[b^2] - (32*b^4*x^16*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b
^2] + (32*a^3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/b
+ 96*a^2*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] + 96*
a*b*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] + 32*b^2*x^
14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] - (32*a^4*x^8*Log
[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] - (128*a^3*b*x^10*Log[a - Sqrt[b^2]*x^2 + Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] - 192*a^2*Sqrt[b^2]*x^12*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]] - (128*a*b^3*x^14*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] - (32*b^4*x
^16*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] + (32*a^3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/b + 96*a^2*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x
^4]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] + 96*a*b*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log
[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] + 32*b^2*x^14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sq
rt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/((-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a -
Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4) + ((-224*a^7*x^2)/(3*b^3*Sqrt[b^2]) - (944*a^6*x^4)/(3*(b
^2)^(3/2)) - (800*a^5*x^6)/(b*Sqrt[b^2]) - (4000*a^4*x^8)/(3*Sqrt[b^2]) - (4288*a^3*b*x^10)/(3*Sqrt[b^2]) - 89
6*a^2*Sqrt[b^2]*x^12 - (256*a*b^3*x^14)/Sqrt[b^2] + (16*a^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^5 + (176*a^6*x^
2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*b^4) + (256*a^5*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^3 + (544*a^4*x^6*
Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^2 + (2368*a^3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*b) + 640*a^2*x^10*Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4] + 256*a*b*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4] + (32*a^4*x^8*Log[a - Sqrt[b^2]*x^
2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/b + 128*a^3*x^10*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
] + 192*a^2*b*x^12*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] + 128*a*b^2*x^14*Log[a - Sqrt[b^2]
*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] + 32*b^3*x^16*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]]
- (32*a^3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[
b^2] - (96*a^2*b*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]]
)/Sqrt[b^2] - 96*a*Sqrt[b^2]*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]] - (32*b^3*x^14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4]])/Sqrt[b^2] - (32*a^4*x^8*Log[-(a*b^5) - b^5*Sqrt[b^2]*x^2 + b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/b - 1
28*a^3*x^10*Log[-(a*b^5) - b^5*Sqrt[b^2]*x^2 + b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] - 192*a^2*b*x^12*Log[-(a*b
^5) - b^5*Sqrt[b^2]*x^2 + b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] - 128*a*b^2*x^14*Log[-(a*b^5) - b^5*Sqrt[b^2]*x
^2 + b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] - 32*b^3*x^16*Log[-(a*b^5) - b^5*Sqrt[b^2]*x^2 + b^5*Sqrt[a^2 + 2*a*
b*x^2 + b^2*x^4]] + (32*a^3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-(a*b^5) - b^5*Sqrt[b^2]*x^2 + b^5*Sqrt[a^
2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] + (96*a^2*b*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-(a*b^5) - b^5*Sqrt[
b^2]*x^2 + b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] + 96*a*Sqrt[b^2]*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^
4]*Log[-(a*b^5) - b^5*Sqrt[b^2]*x^2 + b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]] + (32*b^3*x^14*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]*Log[-(a*b^5) - b^5*Sqrt[b^2]*x^2 + b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2])/((-a - Sqrt[b^
2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4)
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fricas [A] time = 2.77, size = 135, normalized size = 0.69 \begin {gather*} \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 {gather*}
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 \begin {gather*} \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 )} \end {gather*}
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 \begin {gather*} \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}} \end {gather*}
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 \begin {gather*} \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 {gather*}
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 \begin {gather*} \int \frac {x^9}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \end {gather*}
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 \begin {gather*} \int \frac {x^{9}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}
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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