3.1404 \(\int \frac{1}{(c e+d e x)^{5/2} \sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx\)

Optimal. Leaf size=80 \[ \frac{2 \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c e+d e x}}{\sqrt{e}}\right ),-1\right )}{3 d e^{5/2}}-\frac{2 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{3 d e (c e+d e x)^{3/2}} \]

[Out]

(-2*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(3*d*e*(c*e + d*e*x)^(3/2)) + (2*EllipticF[ArcSin[Sqrt[c*e + d*e*x]/Sqr
t[e]], -1])/(3*d*e^(5/2))

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Rubi [A]  time = 0.0579532, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.081, Rules used = {693, 689, 221} \[ \frac{2 F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d x e}}{\sqrt{e}}\right )\right |-1\right )}{3 d e^{5/2}}-\frac{2 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{3 d e (c e+d e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*e + d*e*x)^(5/2)*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])/(3*d*e*(c*e + d*e*x)^(3/2)) + (2*EllipticF[ArcSin[Sqrt[c*e + d*e*x]/Sqr
t[e]], -1])/(3*d*e^(5/2))

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 689

Int[1/(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[1/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 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]

Rubi steps

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

Mathematica [C]  time = 0.0201096, size = 40, normalized size = 0.5 \[ -\frac{2 (c+d x) \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};(c+d x)^2\right )}{3 d (e (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*e + d*e*x)^(5/2)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]),x]

[Out]

(-2*(c + d*x)*Hypergeometric2F1[-3/4, 1/2, 1/4, (c + d*x)^2])/(3*d*(e*(c + d*x))^(5/2))

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Maple [B]  time = 0.286, size = 494, normalized size = 6.2 \begin{align*}{\frac{1}{6\,{e}^{3} \left ( dx+c \right ) ^{2} \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2}-1 \right ) d} \left ( 5\,\sqrt{-2\,dx-2\,c+2}\sqrt{dx+c}\sqrt{2\,dx+2\,c+2}{\it EllipticF} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ) xd+3\,\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 ) xd+3\,\sqrt{-2\,dx-2\,c+2}\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}{\it EllipticF} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ) xd-3\,\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 ) xd+5\,\sqrt{-2\,dx-2\,c+2}\sqrt{dx+c}\sqrt{2\,dx+2\,c+2}{\it EllipticF} \left ( 1/2\,\sqrt{-2\,dx-2\,c+2},\sqrt{2} \right ) c+3\,\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+3\,\sqrt{-2\,dx-2\,c+2}\sqrt{2\,dx+2\,c+2}\sqrt{-dx-c}{\it EllipticF} \left ( 1/2\,\sqrt{2\,dx+2\,c+2},\sqrt{2} \right ) c-3\,\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-4\,{d}^{2}{x}^{2}-8\,cdx-4\,{c}^{2}+4 \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)^(5/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x)

[Out]

1/6*(5*(-2*d*x-2*c+2)^(1/2)*(d*x+c)^(1/2)*(2*d*x+2*c+2)^(1/2)*EllipticF(1/2*(-2*d*x-2*c+2)^(1/2),2^(1/2))*x*d+
3*(-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))*x*d+3*(-2
*d*x-2*c+2)^(1/2)*(2*d*x+2*c+2)^(1/2)*(-d*x-c)^(1/2)*EllipticF(1/2*(2*d*x+2*c+2)^(1/2),2^(1/2))*x*d-3*(-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))*x*d+5*(-2*d*x-2*c+2
)^(1/2)*(d*x+c)^(1/2)*(2*d*x+2*c+2)^(1/2)*EllipticF(1/2*(-2*d*x-2*c+2)^(1/2),2^(1/2))*c+3*(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+3*(-2*d*x-2*c+2)^(1/2)*(2*d*x
+2*c+2)^(1/2)*(-d*x-c)^(1/2)*EllipticF(1/2*(2*d*x+2*c+2)^(1/2),2^(1/2))*c-3*(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-4*d^2*x^2-8*c*d*x-4*c^2+4)/e^3*(-d^2*x^2-2*
c*d*x-c^2+1)^(1/2)*(e*(d*x+c))^(1/2)/(d*x+c)^2/(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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(5/2)/(-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)^(5/2)), 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^{5} e^{3} x^{5} + 5 \, c d^{4} e^{3} x^{4} +{\left (10 \, c^{2} - 1\right )} d^{3} e^{3} x^{3} +{\left (10 \, c^{3} - 3 \, c\right )} d^{2} e^{3} x^{2} +{\left (5 \, c^{4} - 3 \, c^{2}\right )} d e^{3} x +{\left (c^{5} - c^{3}\right )} e^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(5/2)/(-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^5*e^3*x^5 + 5*c*d^4*e^3*x^4 + (10*c^2 - 1)*d
^3*e^3*x^3 + (10*c^3 - 3*c)*d^2*e^3*x^2 + (5*c^4 - 3*c^2)*d*e^3*x + (c^5 - c^3)*e^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \left (c + d x\right )\right )^{\frac{5}{2}} \sqrt{- \left (c + d x - 1\right ) \left (c + d x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)**(5/2)/(-d**2*x**2-2*c*d*x-c**2+1)**(1/2),x)

[Out]

Integral(1/((e*(c + d*x))**(5/2)*sqrt(-(c + d*x - 1)*(c + d*x + 1))), x)

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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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(5/2)/(-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)^(5/2)), x)