3.1411 \(\int \frac {1}{(c e+d e x)^{7/2} \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx\)

Optimal. Leaf size=159 \[ -\frac {6 \sqrt {-c^2-2 c d x-d^2 x^2+1}}{5 d e^3 \sqrt {c e+d e x}}-\frac {2 \sqrt {-c^2-2 c d x-d^2 x^2+1}}{5 d e (c e+d e x)^{5/2}}+\frac {6 F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d x e}}{\sqrt {e}}\right )\right |-1\right )}{5 d e^{7/2}}-\frac {6 E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d x e}}{\sqrt {e}}\right )\right |-1\right )}{5 d e^{7/2}} \]

[Out]

-6/5*EllipticE((d*e*x+c*e)^(1/2)/e^(1/2),I)/d/e^(7/2)+6/5*EllipticF((d*e*x+c*e)^(1/2)/e^(1/2),I)/d/e^(7/2)-2/5
*(-d^2*x^2-2*c*d*x-c^2+1)^(1/2)/d/e/(d*e*x+c*e)^(5/2)-6/5*(-d^2*x^2-2*c*d*x-c^2+1)^(1/2)/d/e^3/(d*e*x+c*e)^(1/
2)

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Rubi [A]  time = 0.12, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {693, 690, 307, 221, 1199, 424} \[ -\frac {6 \sqrt {-c^2-2 c d x-d^2 x^2+1}}{5 d e^3 \sqrt {c e+d e x}}-\frac {2 \sqrt {-c^2-2 c d x-d^2 x^2+1}}{5 d e (c e+d e x)^{5/2}}+\frac {6 F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d x e}}{\sqrt {e}}\right )\right |-1\right )}{5 d e^{7/2}}-\frac {6 E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d x e}}{\sqrt {e}}\right )\right |-1\right )}{5 d e^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*e + d*e*x)^(7/2)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]),x]

[Out]

(-2*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(5*d*e*(c*e + d*e*x)^(5/2)) - (6*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(5*
d*e^3*Sqrt[c*e + d*e*x]) - (6*EllipticE[ArcSin[Sqrt[c*e + d*e*x]/Sqrt[e]], -1])/(5*d*e^(7/2)) + (6*EllipticF[A
rcSin[Sqrt[c*e + d*e*x]/Sqrt[e]], -1])/(5*d*e^(7/2))

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 690

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 - 4*a*
c))])/e, Subst[Int[x^2/Sqrt[Simp[1 - (b^2*x^4)/(d^2*(b^2 - 4*a*c)), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a
, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {1}{(c e+d e x)^{7/2} \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx &=-\frac {2 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e (c e+d e x)^{5/2}}+\frac {3 \int \frac {1}{(c e+d e x)^{3/2} \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx}{5 e^2}\\ &=-\frac {2 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e (c e+d e x)^{5/2}}-\frac {6 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e^3 \sqrt {c e+d e x}}-\frac {3 \int \frac {\sqrt {c e+d e x}}{\sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx}{5 e^4}\\ &=-\frac {2 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e (c e+d e x)^{5/2}}-\frac {6 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e^3 \sqrt {c e+d e x}}-\frac {6 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{e^2}}} \, dx,x,\sqrt {c e+d e x}\right )}{5 d e^5}\\ &=-\frac {2 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e (c e+d e x)^{5/2}}-\frac {6 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e^3 \sqrt {c e+d e x}}+\frac {6 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{e^2}}} \, dx,x,\sqrt {c e+d e x}\right )}{5 d e^4}-\frac {6 \operatorname {Subst}\left (\int \frac {1+\frac {x^2}{e}}{\sqrt {1-\frac {x^4}{e^2}}} \, dx,x,\sqrt {c e+d e x}\right )}{5 d e^4}\\ &=-\frac {2 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e (c e+d e x)^{5/2}}-\frac {6 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e^3 \sqrt {c e+d e x}}+\frac {6 F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d e x}}{\sqrt {e}}\right )\right |-1\right )}{5 d e^{7/2}}-\frac {6 \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {x^2}{e}}}{\sqrt {1-\frac {x^2}{e}}} \, dx,x,\sqrt {c e+d e x}\right )}{5 d e^4}\\ &=-\frac {2 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e (c e+d e x)^{5/2}}-\frac {6 \sqrt {1-c^2-2 c d x-d^2 x^2}}{5 d e^3 \sqrt {c e+d e x}}-\frac {6 E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d e x}}{\sqrt {e}}\right )\right |-1\right )}{5 d e^{7/2}}+\frac {6 F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d e x}}{\sqrt {e}}\right )\right |-1\right )}{5 d e^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 40, normalized size = 0.25 \[ -\frac {2 (c+d x) \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};(c+d x)^2\right )}{5 d (e (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*e + d*e*x)^(7/2)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]),x]

[Out]

(-2*(c + d*x)*Hypergeometric2F1[-5/4, 1/2, -1/4, (c + d*x)^2])/(5*d*(e*(c + d*x))^(7/2))

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} \sqrt {d e x + c e}}{d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + {\left (15 \, c^{2} - 1\right )} d^{4} e^{4} x^{4} + 4 \, {\left (5 \, c^{3} - c\right )} d^{3} e^{4} x^{3} + 3 \, {\left (5 \, c^{4} - 2 \, c^{2}\right )} d^{2} e^{4} x^{2} + 2 \, {\left (3 \, c^{5} - 2 \, c^{3}\right )} d e^{4} x + {\left (c^{6} - c^{4}\right )} e^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*sqrt(d*e*x + c*e)/(d^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + (15*c^2 - 1)*d
^4*e^4*x^4 + 4*(5*c^3 - c)*d^3*e^4*x^3 + 3*(5*c^4 - 2*c^2)*d^2*e^4*x^2 + 2*(3*c^5 - 2*c^3)*d*e^4*x + (c^6 - c^
4)*e^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (d e x + c e\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(d*e*x + c*e)^(7/2)), x)

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maple [B]  time = 0.13, size = 447, normalized size = 2.81 \[ \frac {\left (-6 d^{4} x^{4}-24 c \,d^{3} x^{3}-36 c^{2} d^{2} x^{2}-24 c^{3} d x -10 \sqrt {d x +c}\, \sqrt {2 d x +2 c +2}\, \sqrt {-2 d x -2 c +2}\, d^{2} x^{2} \EllipticE \left (\frac {\sqrt {-2 d x -2 c +2}}{2}, \sqrt {2}\right )+7 \sqrt {2 d x +2 c +2}\, \sqrt {-d x -c}\, \sqrt {-2 d x -2 c +2}\, d^{2} x^{2} \EllipticE \left (\frac {\sqrt {2 d x +2 c +2}}{2}, \sqrt {2}\right )-6 c^{4}-20 \sqrt {d x +c}\, \sqrt {2 d x +2 c +2}\, \sqrt {-2 d x -2 c +2}\, c d x \EllipticE \left (\frac {\sqrt {-2 d x -2 c +2}}{2}, \sqrt {2}\right )+14 \sqrt {2 d x +2 c +2}\, \sqrt {-d x -c}\, \sqrt {-2 d x -2 c +2}\, c d x \EllipticE \left (\frac {\sqrt {2 d x +2 c +2}}{2}, \sqrt {2}\right )+4 d^{2} x^{2}-10 \sqrt {d x +c}\, \sqrt {2 d x +2 c +2}\, \sqrt {-2 d x -2 c +2}\, c^{2} \EllipticE \left (\frac {\sqrt {-2 d x -2 c +2}}{2}, \sqrt {2}\right )+7 \sqrt {2 d x +2 c +2}\, \sqrt {-d x -c}\, \sqrt {-2 d x -2 c +2}\, c^{2} \EllipticE \left (\frac {\sqrt {2 d x +2 c +2}}{2}, \sqrt {2}\right )+8 c d x +4 c^{2}+2\right ) \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, \sqrt {\left (d x +c \right ) e}}{5 \left (d x +c \right )^{3} \left (d^{2} x^{2}+2 c d x +c^{2}-1\right ) d \,e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*e*x+c*e)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x)

[Out]

1/5*(7*EllipticE(1/2*(2*d*x+2*c+2)^(1/2),2^(1/2))*x^2*d^2*(2*d*x+2*c+2)^(1/2)*(-d*x-c)^(1/2)*(-2*d*x-2*c+2)^(1
/2)-10*EllipticE(1/2*(-2*d*x-2*c+2)^(1/2),2^(1/2))*x^2*d^2*(d*x+c)^(1/2)*(2*d*x+2*c+2)^(1/2)*(-2*d*x-2*c+2)^(1
/2)-6*d^4*x^4+14*EllipticE(1/2*(2*d*x+2*c+2)^(1/2),2^(1/2))*x*c*d*(2*d*x+2*c+2)^(1/2)*(-d*x-c)^(1/2)*(-2*d*x-2
*c+2)^(1/2)-20*EllipticE(1/2*(-2*d*x-2*c+2)^(1/2),2^(1/2))*x*c*d*(d*x+c)^(1/2)*(2*d*x+2*c+2)^(1/2)*(-2*d*x-2*c
+2)^(1/2)-24*c*d^3*x^3+7*EllipticE(1/2*(2*d*x+2*c+2)^(1/2),2^(1/2))*c^2*(2*d*x+2*c+2)^(1/2)*(-d*x-c)^(1/2)*(-2
*d*x-2*c+2)^(1/2)-10*EllipticE(1/2*(-2*d*x-2*c+2)^(1/2),2^(1/2))*c^2*(d*x+c)^(1/2)*(2*d*x+2*c+2)^(1/2)*(-2*d*x
-2*c+2)^(1/2)-36*c^2*d^2*x^2-24*c^3*d*x+4*d^2*x^2-6*c^4+8*c*d*x+4*c^2+2)*(-d^2*x^2-2*c*d*x-c^2+1)^(1/2)*((d*x+
c)*e)^(1/2)/e^4/(d*x+c)^3/(d^2*x^2+2*c*d*x+c^2-1)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (d e x + c e\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(d*e*x + c*e)^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (c\,e+d\,e\,x\right )}^{7/2}\,\sqrt {-c^2-2\,c\,d\,x-d^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*e + d*e*x)^(7/2)*(1 - d^2*x^2 - 2*c*d*x - c^2)^(1/2)),x)

[Out]

int(1/((c*e + d*e*x)^(7/2)*(1 - d^2*x^2 - 2*c*d*x - c^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \left (c + d x\right )\right )^{\frac {7}{2}} \sqrt {- \left (c + d x - 1\right ) \left (c + d x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)**(7/2)/(-d**2*x**2-2*c*d*x-c**2+1)**(1/2),x)

[Out]

Integral(1/((e*(c + d*x))**(7/2)*sqrt(-(c + d*x - 1)*(c + d*x + 1))), x)

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