3.8.52 \(\int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx\)
Optimal. Leaf size=124 \[ -\frac {b \left (12 a c+5 b^2\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{32 c^{7/2}}-\frac {\left (16 a c+15 b^2+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c} \]
________________________________________________________________________________________
Rubi [A] time = 0.11, antiderivative size = 124, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used =
{1114, 742, 779, 621, 204} \begin {gather*} -\frac {\left (16 a c+15 b^2+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}-\frac {b \left (12 a c+5 b^2\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{32 c^{7/2}}-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^7/Sqrt[a + b*x^2 - c*x^4],x]
[Out]
-(x^4*Sqrt[a + b*x^2 - c*x^4])/(6*c) - ((15*b^2 + 16*a*c + 10*b*c*x^2)*Sqrt[a + b*x^2 - c*x^4])/(48*c^3) - (b*
(5*b^2 + 12*a*c)*ArcTan[(b - 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4])])/(32*c^(7/2))
Rule 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 621
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 742
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]
Rule 779
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Rule 1114
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] && IntegerQ[(m - 1)/2]
Rubi steps
\begin {align*} \int \frac {x^7}{\sqrt {a+b x^2-c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\operatorname {Subst}\left (\int \frac {x \left (-2 a-\frac {5 b x}{2}\right )}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{6 c}\\ &=-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}+\frac {\left (b \left (5 b^2+12 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{32 c^3}\\ &=-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}+\frac {\left (b \left (5 b^2+12 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 c-x^2} \, dx,x,\frac {b-2 c x^2}{\sqrt {a+b x^2-c x^4}}\right )}{16 c^3}\\ &=-\frac {x^4 \sqrt {a+b x^2-c x^4}}{6 c}-\frac {\left (15 b^2+16 a c+10 b c x^2\right ) \sqrt {a+b x^2-c x^4}}{48 c^3}-\frac {b \left (5 b^2+12 a c\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{32 c^{7/2}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.08, size = 107, normalized size = 0.86 \begin {gather*} \frac {-2 \sqrt {c} \sqrt {a+b x^2-c x^4} \left (8 c \left (2 a+c x^4\right )+15 b^2+10 b c x^2\right )-3 b \left (12 a c+5 b^2\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{96 c^{7/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^7/Sqrt[a + b*x^2 - c*x^4],x]
[Out]
(-2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4]*(15*b^2 + 10*b*c*x^2 + 8*c*(2*a + c*x^4)) - 3*b*(5*b^2 + 12*a*c)*ArcTan[(b
- 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4])])/(96*c^(7/2))
________________________________________________________________________________________
IntegrateAlgebraic [B] time = 23.35, size = 3247, normalized size = 26.19 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[x^7/Sqrt[a + b*x^2 - c*x^4],x]
[Out]
(Sqrt[a + b*x^2 - c*x^4]*(-196*a*b^12*c^(5/2) - 2352*a^2*b^10*c^(7/2) - 11136*a^3*b^8*c^(9/2) - 34304*a^4*b^6*
c^(11/2) - 95232*a^5*b^4*c^(13/2) - 159744*a^6*b^2*c^(15/2) - 65536*a^7*c^(17/2) + 15*b^14*Sqrt[-c]*Sqrt[-c^2]
+ 180*a*b^12*Sqrt[-c]*c*Sqrt[-c^2] + 1632*a^2*b^10*Sqrt[-c]*c^2*Sqrt[-c^2] + 11904*a^3*b^8*Sqrt[-c]*c^3*Sqrt[
-c^2] + 43776*a^4*b^6*Sqrt[-c]*c^4*Sqrt[-c^2] + 58368*a^5*b^4*Sqrt[-c]*c^5*Sqrt[-c^2] - 190*b^13*c^(5/2)*x^2 -
4944*a*b^11*c^(7/2)*x^2 - 36576*a^2*b^9*c^(9/2)*x^2 - 147968*a^3*b^7*c^(11/2)*x^2 - 477696*a^4*b^5*c^(13/2)*x
^2 - 921600*a^5*b^3*c^(15/2)*x^2 - 434176*a^6*b*c^(17/2)*x^2 + 180*b^13*Sqrt[-c]*c*Sqrt[-c^2]*x^2 + 2880*a*b^1
1*Sqrt[-c]*c^2*Sqrt[-c^2]*x^2 + 31104*a^2*b^9*Sqrt[-c]*c^3*Sqrt[-c^2]*x^2 + 156672*a^3*b^7*Sqrt[-c]*c^4*Sqrt[-
c^2]*x^2 + 267264*a^4*b^5*Sqrt[-c]*c^5*Sqrt[-c^2]*x^2 - 1880*b^12*c^(7/2)*x^4 - 38400*a*b^10*c^(9/2)*x^4 - 195
456*a^2*b^8*c^(11/2)*x^4 - 710656*a^3*b^6*c^(13/2)*x^4 - 1959936*a^4*b^4*c^(15/2)*x^4 - 1081344*a^5*b^2*c^(17/
2)*x^4 - 32768*a^6*c^(19/2)*x^4 + 1248*b^12*Sqrt[-c]*c^2*Sqrt[-c^2]*x^4 + 15744*a*b^10*Sqrt[-c]*c^3*Sqrt[-c^2]
*x^4 + 170496*a^2*b^8*Sqrt[-c]*c^4*Sqrt[-c^2]*x^4 + 485376*a^3*b^6*Sqrt[-c]*c^5*Sqrt[-c^2]*x^4 - 98304*a^4*b^4
*Sqrt[-c]*c^6*Sqrt[-c^2]*x^4 - 11712*b^11*c^(9/2)*x^6 - 88832*a*b^9*c^(11/2)*x^6 - 245760*a^2*b^7*c^(13/2)*x^6
- 1794048*a^3*b^5*c^(15/2)*x^6 - 1261568*a^4*b^3*c^(17/2)*x^6 - 196608*a^5*b*c^(19/2)*x^6 - 1920*b^11*Sqrt[-c
]*c^3*Sqrt[-c^2]*x^6 + 46080*a*b^9*Sqrt[-c]*c^4*Sqrt[-c^2]*x^6 + 436224*a^2*b^7*Sqrt[-c]*c^5*Sqrt[-c^2]*x^6 -
294912*a^3*b^5*Sqrt[-c]*c^6*Sqrt[-c^2]*x^6 - 9728*b^10*c^(11/2)*x^8 + 122880*a*b^8*c^(13/2)*x^8 - 688128*a^2*b
^6*c^(15/2)*x^8 - 720896*a^3*b^4*c^(17/2)*x^8 - 393216*a^4*b^2*c^(19/2)*x^8 - 3072*b^10*Sqrt[-c]*c^4*Sqrt[-c^2
]*x^8 + 208896*a*b^8*Sqrt[-c]*c^5*Sqrt[-c^2]*x^8 - 294912*a^2*b^6*Sqrt[-c]*c^6*Sqrt[-c^2]*x^8 + 45056*b^9*c^(1
3/2)*x^10 - 147456*a*b^7*c^(15/2)*x^10 - 196608*a^2*b^5*c^(17/2)*x^10 - 262144*a^3*b^3*c^(19/2)*x^10 + 49152*b
^9*Sqrt[-c]*c^5*Sqrt[-c^2]*x^10 - 98304*a*b^7*Sqrt[-c]*c^6*Sqrt[-c^2]*x^10))/(48*(b^12*c^(9/2) + 24*a*b^10*c^(
11/2) + 240*a^2*b^8*c^(13/2) + 1280*a^3*b^6*c^(15/2) + 3840*a^4*b^4*c^(17/2) + 6144*a^5*b^2*c^(19/2) + 4096*a^
6*c^(21/2) + 24*b^11*c^(11/2)*x^2 + 480*a*b^9*c^(13/2)*x^2 + 3840*a^2*b^7*c^(15/2)*x^2 + 15360*a^3*b^5*c^(17/2
)*x^2 + 30720*a^4*b^3*c^(19/2)*x^2 + 24576*a^5*b*c^(21/2)*x^2 + 192*b^10*c^(13/2)*x^4 + 3072*a*b^8*c^(15/2)*x^
4 + 18432*a^2*b^6*c^(17/2)*x^4 + 49152*a^3*b^4*c^(19/2)*x^4 + 49152*a^4*b^2*c^(21/2)*x^4 + 512*b^9*c^(15/2)*x^
6 + 6144*a*b^7*c^(17/2)*x^6 + 24576*a^2*b^5*c^(19/2)*x^6 + 32768*a^3*b^3*c^(21/2)*x^6)) + (9280*a^4*b^12*c^3*S
qrt[-c^2] + 222720*a^5*b^10*c^4*Sqrt[-c^2] + 2227200*a^6*b^8*c^5*Sqrt[-c^2] + 11878400*a^7*b^6*c^6*Sqrt[-c^2]
+ 35635200*a^8*b^4*c^7*Sqrt[-c^2] + 57016320*a^9*b^2*c^8*Sqrt[-c^2] + 38010880*a^10*c^9*Sqrt[-c^2] - 5*b^19*Sq
rt[-c]*Sqrt[c]*x^2 + 60*a*b^17*Sqrt[-c]*c^(3/2)*x^2 - 544*a^2*b^15*Sqrt[-c]*c^(5/2)*x^2 - 6272*a^3*b^13*Sqrt[-
c]*c^(7/2)*x^2 + 220416*a^4*b^11*Sqrt[-c]*c^(9/2)*x^2 + 3021824*a^5*b^9*Sqrt[-c]*c^(11/2)*x^2 + 15040512*a^6*b
^7*Sqrt[-c]*c^(13/2)*x^2 + 34013184*a^7*b^5*Sqrt[-c]*c^(15/2)*x^2 + 5*b^19*Sqrt[-c^2]*x^2 - 60*a*b^17*c*Sqrt[-
c^2]*x^2 + 544*a^2*b^15*c^2*Sqrt[-c^2]*x^2 + 6272*a^3*b^13*c^3*Sqrt[-c^2]*x^2 + 2304*a^4*b^11*c^4*Sqrt[-c^2]*x
^2 + 1432576*a^5*b^9*c^5*Sqrt[-c^2]*x^2 + 20594688*a^6*b^7*c^6*Sqrt[-c^2]*x^2 + 108527616*a^7*b^5*c^7*Sqrt[-c^
2]*x^2 + 285081600*a^8*b^3*c^8*Sqrt[-c^2]*x^2 + 228065280*a^9*b*c^9*Sqrt[-c^2]*x^2 + 60*b^18*Sqrt[-c]*c^(3/2)*
x^4 - 704*a*b^16*Sqrt[-c]*c^(5/2)*x^4 - 17280*a^2*b^14*Sqrt[-c]*c^(7/2)*x^4 - 15360*a^3*b^12*Sqrt[-c]*c^(9/2)*
x^4 + 1936384*a^4*b^10*Sqrt[-c]*c^(11/2)*x^4 - 60*b^18*c*Sqrt[-c^2]*x^4 + 704*a*b^16*c^2*Sqrt[-c^2]*x^4 + 1728
0*a^2*b^14*c^3*Sqrt[-c^2]*x^4 + 15360*a^3*b^12*c^4*Sqrt[-c^2]*x^4 - 154624*a^4*b^10*c^5*Sqrt[-c^2]*x^4 + 28508
160*a^5*b^8*c^6*Sqrt[-c^2]*x^4 + 171048960*a^6*b^6*c^7*Sqrt[-c^2]*x^4 + 456130560*a^7*b^4*c^8*Sqrt[-c^2]*x^4 +
456130560*a^8*b^2*c^9*Sqrt[-c^2]*x^4 - 160*b^17*Sqrt[-c]*c^(5/2)*x^6 - 12672*a*b^15*Sqrt[-c]*c^(7/2)*x^6 - 30
208*a^2*b^13*Sqrt[-c]*c^(9/2)*x^6 + 260096*a^3*b^11*Sqrt[-c]*c^(11/2)*x^6 + 4718592*a^4*b^9*Sqrt[-c]*c^(13/2)*
x^6 + 160*b^17*c^2*Sqrt[-c^2]*x^6 + 12672*a*b^15*c^3*Sqrt[-c^2]*x^6 + 30208*a^2*b^13*c^4*Sqrt[-c^2]*x^6 - 2600
96*a^3*b^11*c^5*Sqrt[-c^2]*x^6 + 32768*a^4*b^9*c^6*Sqrt[-c^2]*x^6 + 57016320*a^5*b^7*c^7*Sqrt[-c^2]*x^6 + 2280
65280*a^6*b^5*c^8*Sqrt[-c^2]*x^6 + 304087040*a^7*b^3*c^9*Sqrt[-c^2]*x^6 - 2176*b^16*Sqrt[-c]*c^(7/2)*x^8 - 235
52*a*b^14*Sqrt[-c]*c^(9/2)*x^8 + 210944*a^2*b^12*Sqrt[-c]*c^(11/2)*x^8 - 98304*a^3*b^10*Sqrt[-c]*c^(13/2)*x^8
+ 2176*b^16*c^3*Sqrt[-c^2]*x^8 + 23552*a*b^14*c^4*Sqrt[-c^2]*x^8 - 210944*a^2*b^12*c^5*Sqrt[-c^2]*x^8 + 98304*
a^3*b^10*c^6*Sqrt[-c^2]*x^8 - 9216*b^15*Sqrt[-c]*c^(9/2)*x^10 + 86016*a*b^13*Sqrt[-c]*c^(11/2)*x^10 - 98304*a^
2*b^11*Sqrt[-c]*c^(13/2)*x^10 + 9216*b^15*c^4*Sqrt[-c^2]*x^10 - 86016*a*b^13*c^5*Sqrt[-c^2]*x^10 + 98304*a^2*b
^11*c^6*Sqrt[-c^2]*x^10 + 16384*b^14*Sqrt[-c]*c^(11/2)*x^12 - 32768*a*b^12*Sqrt[-c]*c^(13/2)*x^12 - 16384*b^14
*c^5*Sqrt[-c^2]*x^12 + 32768*a*b^12*c^6*Sqrt[-c^2]*x^12)/(16*b^5*c^(7/2)*(b^2 + 4*a*c)^3*(b^2 + 4*a*c + 8*b*c*
x^2)^3) + ((5*b^3 + 12*a*b*c)*ArcTan[(-2*Sqrt[-c]*Sqrt[c]*x^2 + 2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4])/b])/(32*c^(
7/2)) + (Sqrt[-c]*(5*b^3 + 12*a*b*c)*Log[b^2 + 4*a*c + 4*b*c*x^2 - 8*c^2*x^4 - 8*Sqrt[-c]*c*x^2*Sqrt[a + b*x^2
- c*x^4]])/(64*c^4)
________________________________________________________________________________________
fricas [A] time = 1.21, size = 249, normalized size = 2.01 \begin {gather*} \left [-\frac {3 \, {\left (5 \, b^{3} + 12 \, a b c\right )} \sqrt {-c} \log \left (8 \, c^{2} x^{4} - 8 \, b c x^{2} + b^{2} - 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {-c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} x^{4} + 10 \, b c^{2} x^{2} + 15 \, b^{2} c + 16 \, a c^{2}\right )} \sqrt {-c x^{4} + b x^{2} + a}}{192 \, c^{4}}, -\frac {3 \, {\left (5 \, b^{3} + 12 \, a b c\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {c}}{2 \, {\left (c^{2} x^{4} - b c x^{2} - a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 10 \, b c^{2} x^{2} + 15 \, b^{2} c + 16 \, a c^{2}\right )} \sqrt {-c x^{4} + b x^{2} + a}}{96 \, c^{4}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
[Out]
[-1/192*(3*(5*b^3 + 12*a*b*c)*sqrt(-c)*log(8*c^2*x^4 - 8*b*c*x^2 + b^2 - 4*sqrt(-c*x^4 + b*x^2 + a)*(2*c*x^2 -
b)*sqrt(-c) - 4*a*c) + 4*(8*c^3*x^4 + 10*b*c^2*x^2 + 15*b^2*c + 16*a*c^2)*sqrt(-c*x^4 + b*x^2 + a))/c^4, -1/9
6*(3*(5*b^3 + 12*a*b*c)*sqrt(c)*arctan(1/2*sqrt(-c*x^4 + b*x^2 + a)*(2*c*x^2 - b)*sqrt(c)/(c^2*x^4 - b*c*x^2 -
a*c)) + 2*(8*c^3*x^4 + 10*b*c^2*x^2 + 15*b^2*c + 16*a*c^2)*sqrt(-c*x^4 + b*x^2 + a))/c^4]
________________________________________________________________________________________
giac [A] time = 0.26, size = 112, normalized size = 0.90 \begin {gather*} -\frac {1}{48} \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{c} + \frac {5 \, b}{c^{2}}\right )} + \frac {15 \, b^{2} + 16 \, a c}{c^{3}}\right )} - \frac {{\left (5 \, b^{3} + 12 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {-c} x^{2} - \sqrt {-c x^{4} + b x^{2} + a}\right )} \sqrt {-c} + b \right |}\right )}{32 \, \sqrt {-c} c^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
[Out]
-1/48*sqrt(-c*x^4 + b*x^2 + a)*(2*x^2*(4*x^2/c + 5*b/c^2) + (15*b^2 + 16*a*c)/c^3) - 1/32*(5*b^3 + 12*a*b*c)*l
og(abs(2*(sqrt(-c)*x^2 - sqrt(-c*x^4 + b*x^2 + a))*sqrt(-c) + b))/(sqrt(-c)*c^3)
________________________________________________________________________________________
maple [A] time = 0.02, size = 168, normalized size = 1.35 \begin {gather*} -\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}\, x^{4}}{6 c}-\frac {5 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, b \,x^{2}}{24 c^{2}}+\frac {3 a b \arctan \left (\frac {\left (x^{2}-\frac {b}{2 c}\right ) \sqrt {c}}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{8 c^{\frac {5}{2}}}+\frac {5 b^{3} \arctan \left (\frac {\left (x^{2}-\frac {b}{2 c}\right ) \sqrt {c}}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{32 c^{\frac {7}{2}}}-\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}\, a}{3 c^{2}}-\frac {5 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, b^{2}}{16 c^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^7/(-c*x^4+b*x^2+a)^(1/2),x)
[Out]
-1/6*x^4*(-c*x^4+b*x^2+a)^(1/2)/c-5/24*b/c^2*x^2*(-c*x^4+b*x^2+a)^(1/2)-5/16*b^2/c^3*(-c*x^4+b*x^2+a)^(1/2)+5/
32*b^3/c^(7/2)*arctan(c^(1/2)*(x^2-1/2*b/c)/(-c*x^4+b*x^2+a)^(1/2))+3/8*b/c^(5/2)*a*arctan(c^(1/2)*(x^2-1/2*b/
c)/(-c*x^4+b*x^2+a)^(1/2))-1/3/c^2*a*(-c*x^4+b*x^2+a)^(1/2)
________________________________________________________________________________________
maxima [A] time = 2.43, size = 153, normalized size = 1.23 \begin {gather*} -\frac {\sqrt {-c x^{4} + b x^{2} + a} x^{4}}{6 \, c} - \frac {5 \, \sqrt {-c x^{4} + b x^{2} + a} b x^{2}}{24 \, c^{2}} - \frac {5 \, b^{3} \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{32 \, c^{\frac {7}{2}}} - \frac {3 \, a b \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{8 \, c^{\frac {5}{2}}} - \frac {5 \, \sqrt {-c x^{4} + b x^{2} + a} b^{2}}{16 \, c^{3}} - \frac {\sqrt {-c x^{4} + b x^{2} + a} a}{3 \, c^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7/(-c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")
[Out]
-1/6*sqrt(-c*x^4 + b*x^2 + a)*x^4/c - 5/24*sqrt(-c*x^4 + b*x^2 + a)*b*x^2/c^2 - 5/32*b^3*arcsin(-(2*c*x^2 - b)
/sqrt(b^2 + 4*a*c))/c^(7/2) - 3/8*a*b*arcsin(-(2*c*x^2 - b)/sqrt(b^2 + 4*a*c))/c^(5/2) - 5/16*sqrt(-c*x^4 + b*
x^2 + a)*b^2/c^3 - 1/3*sqrt(-c*x^4 + b*x^2 + a)*a/c^2
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^7}{\sqrt {-c\,x^4+b\,x^2+a}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^7/(a + b*x^2 - c*x^4)^(1/2),x)
[Out]
int(x^7/(a + b*x^2 - c*x^4)^(1/2), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\sqrt {a + b x^{2} - c x^{4}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**7/(-c*x**4+b*x**2+a)**(1/2),x)
[Out]
Integral(x**7/sqrt(a + b*x**2 - c*x**4), x)
________________________________________________________________________________________