3.15.4 \(\int \frac {1}{(c e+d e x)^{5/2} \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx\) [1404]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 = {707, 703, 227} \begin {gather*} \frac {2 F\left (\left .\text {ArcSin}\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}} \end {gather*}

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 227

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 703

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(4/e)*Sqrt[-c/(b^2
- 4*a*c)], 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 707

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])

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 \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 40, normalized size = 0.50 \begin {gather*} -\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}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(182\) vs. \(2(66)=132\).
time = 0.76, size = 183, normalized size = 2.29

method	result	size
default	\(\frac {\left (\sqrt {-2 d x -2 c +2}\, \sqrt {d x +c}\, \sqrt {2 d x +2 c +2}\, \EllipticF \left (\frac {\sqrt {-2 d x -2 c +2}}{2}, \sqrt {2}\right ) d x +\sqrt {-2 d x -2 c +2}\, \sqrt {d x +c}\, \sqrt {2 d x +2 c +2}\, \EllipticF \left (\frac {\sqrt {-2 d x -2 c +2}}{2}, \sqrt {2}\right ) c -2 d^{2} x^{2}-4 c d x -2 c^{2}+2\right ) \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, \sqrt {e \left (d x +c \right )}}{3 e^{3} \left (d x +c \right )^{2} \left (d^{2} x^{2}+2 c d x +c^{2}-1\right ) d}\)	\(183\)
elliptic	\(\frac {\sqrt {-e \left (d x +c \right ) \left (d^{2} x^{2}+2 c d x +c^{2}-1\right )}\, \left (-\frac {2 \sqrt {-d^{3} e \,x^{3}-3 c \,d^{2} e \,x^{2}-3 c^{2} d e x -c^{3} e +d x e +c e}}{3 d^{3} e^{3} \left (x +\frac {c}{d}\right )^{2}}+\frac {2 \left (-\frac {c +1}{d}+\frac {c -1}{d}\right ) \sqrt {\frac {x +\frac {c -1}{d}}{-\frac {c +1}{d}+\frac {c -1}{d}}}\, \sqrt {\frac {x +\frac {c}{d}}{-\frac {c -1}{d}+\frac {c}{d}}}\, \sqrt {\frac {x +\frac {c +1}{d}}{-\frac {c -1}{d}+\frac {c +1}{d}}}\, \EllipticF \left (\sqrt {\frac {x +\frac {c -1}{d}}{-\frac {c +1}{d}+\frac {c -1}{d}}}, \sqrt {\frac {-\frac {c -1}{d}+\frac {c +1}{d}}{-\frac {c -1}{d}+\frac {c}{d}}}\right )}{3 e^{2} \sqrt {-d^{3} e \,x^{3}-3 c \,d^{2} e \,x^{2}-3 c^{2} d e x -c^{3} e +d x e +c e}}\right )}{\sqrt {e \left (d x +c \right )}\, \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}}\)	\(337\)

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,method=_RETURNVERBOSE)

[Out]

1/3*((-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))*d*x+(-
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-2*d^2*x^2-4
*c*d*x-2*c^2+2)/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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*x*e + c*e)^(5/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.62, size = 106, normalized size = 1.32 \begin {gather*} -\frac {2 \, {\left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} \sqrt {d x + c} d^{2} e^{\frac {1}{2}} + {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \sqrt {-d^{3} e} {\rm weierstrassPInverse}\left (\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right )} e^{\left (-3\right )}}{3 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end {gather*}

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]

-2/3*(sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*sqrt(d*x + c)*d^2*e^(1/2) + (d^2*x^2 + 2*c*d*x + c^2)*sqrt(-d^3*e)*we
ierstrassPInverse(4/d^2, 0, (d*x + c)/d))*e^(-3)/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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*x*e + c*e)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,e+d\,e\,x\right )}^{5/2}\,\sqrt {-c^2-2\,c\,d\,x-d^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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