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 \[ \frac{6 \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c e+d e x}}{\sqrt{e}}\right ),-1\right )}{5 d e^{7/2}}-\frac{6 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{5 d e^3 \sqrt{c e+d e x}}-\frac{2 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{5 d e (c e+d e x)^{5/2}}-\frac{6 E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d x e}}{\sqrt{e}}\right )\right |-1\right )}{5 d e^{7/2}} \]

[Out]

(-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))

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Rubi [A]  time = 0.118039, antiderivative size = 159, normalized size of antiderivative = 1., 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 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 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 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 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 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]

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]

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*}

Mathematica [C]  time = 0.0229086, 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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Maple [B]  time = 0.258, size = 768, normalized size = 4.8 \begin{align*}{\frac{1}{30\,{e}^{4} \left ( dx+c \right ) ^{3} \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2}-1 \right ) d} \left ( 5\,{\it EllipticF} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ){x}^{2}{d}^{2}\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}\sqrt{-2\,dx-2\,c+2}-33\,{\it EllipticE} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ){x}^{2}{d}^{2}\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}\sqrt{-2\,dx-2\,c+2}+5\,{\it EllipticF} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ){x}^{2}{d}^{2}\sqrt{2\,dx+2\,c+2}\sqrt{dx+c}\sqrt{-2\,dx-2\,c+2}+15\,{\it EllipticE} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ){x}^{2}{d}^{2}\sqrt{2\,dx+2\,c+2}\sqrt{dx+c}\sqrt{-2\,dx-2\,c+2}-36\,{d}^{4}{x}^{4}+10\,{\it EllipticF} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ) xcd\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}\sqrt{-2\,dx-2\,c+2}-66\,{\it EllipticE} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ) xcd\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}\sqrt{-2\,dx-2\,c+2}+10\,{\it EllipticF} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ) xcd\sqrt{2\,dx+2\,c+2}\sqrt{dx+c}\sqrt{-2\,dx-2\,c+2}+30\,{\it EllipticE} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ) xcd\sqrt{2\,dx+2\,c+2}\sqrt{dx+c}\sqrt{-2\,dx-2\,c+2}-144\,{x}^{3}c{d}^{3}+5\,{\it EllipticF} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ){c}^{2}\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}\sqrt{-2\,dx-2\,c+2}-33\,\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}\sqrt{-2\,dx-2\,c+2}{\it EllipticE} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ){c}^{2}+5\,{\it EllipticF} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ){c}^{2}\sqrt{2\,dx+2\,c+2}\sqrt{dx+c}\sqrt{-2\,dx-2\,c+2}+15\,\sqrt{2\,dx+2\,c+2}\sqrt{-2\,dx-2\,c+2}\sqrt{dx+c}{\it EllipticE} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ){c}^{2}-216\,{c}^{2}{d}^{2}{x}^{2}-144\,x{c}^{3}d+24\,{d}^{2}{x}^{2}-36\,{c}^{4}+48\,cdx+24\,{c}^{2}+12 \right ) \sqrt{-{d}^{2}{x}^{2}-2\,cdx-{c}^{2}+1}\sqrt{e \left ( dx+c \right ) }} \end{align*}

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]

1/30*(5*EllipticF(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)-33*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)+5*EllipticF(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)+15*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)-36*d^4*x^4+10*EllipticF(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)-66*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)+10*EllipticF(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)+30*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)-144*x^3*c*d^3+5*EllipticF(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)-33*(2*d*x+2*c+2)^(1/2)*(-d*x-c)^(1/2)*(-2*d*x-2*c+2)^(1/2)*EllipticE(1/2*(2*d*x+2*c+2)^(1/2),2
^(1/2))*c^2+5*EllipticF(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)+15*(2*d*x+2*c+2)^(1/2)*(-2*d*x-2*c+2)^(1/2)*(d*x+c)^(1/2)*EllipticE(1/2*(-2*d*x-2*c+2)^(1/2),2^(1/2))*c
^2-216*c^2*d^2*x^2-144*x*c^3*d+24*d^2*x^2-36*c^4+48*c*d*x+24*c^2+12)*(-d^2*x^2-2*c*d*x-c^2+1)^(1/2)*(e*(d*x+c)
)^(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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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

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)