### 3.1405 $$\int \frac{1}{(c e+d e x)^{9/2} \sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx$$

Optimal. Leaf size=126 $\frac{10 \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c e+d e x}}{\sqrt{e}}\right ),-1\right )}{21 d e^{9/2}}-\frac{10 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{21 d e^3 (c e+d e x)^{3/2}}-\frac{2 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{7 d e (c e+d e x)^{7/2}}$

[Out]

(-2*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(7*d*e*(c*e + d*e*x)^(7/2)) - (10*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(2
1*d*e^3*(c*e + d*e*x)^(3/2)) + (10*EllipticF[ArcSin[Sqrt[c*e + d*e*x]/Sqrt[e]], -1])/(21*d*e^(9/2))

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Rubi [A]  time = 0.0858568, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.081, Rules used = {693, 689, 221} $-\frac{10 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{21 d e^3 (c e+d e x)^{3/2}}-\frac{2 \sqrt{-c^2-2 c d x-d^2 x^2+1}}{7 d e (c e+d e x)^{7/2}}+\frac{10 F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d x e}}{\sqrt{e}}\right )\right |-1\right )}{21 d e^{9/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((c*e + d*e*x)^(9/2)*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])/(7*d*e*(c*e + d*e*x)^(7/2)) - (10*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(2
1*d*e^3*(c*e + d*e*x)^(3/2)) + (10*EllipticF[ArcSin[Sqrt[c*e + d*e*x]/Sqrt[e]], -1])/(21*d*e^(9/2))

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)^{9/2} \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}}{7 d e (c e+d e x)^{7/2}}+\frac{5 \int \frac{1}{(c e+d e x)^{5/2} \sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx}{7 e^2}\\ &=-\frac{2 \sqrt{1-c^2-2 c d x-d^2 x^2}}{7 d e (c e+d e x)^{7/2}}-\frac{10 \sqrt{1-c^2-2 c d x-d^2 x^2}}{21 d e^3 (c e+d e x)^{3/2}}+\frac{5 \int \frac{1}{\sqrt{c e+d e x} \sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx}{21 e^4}\\ &=-\frac{2 \sqrt{1-c^2-2 c d x-d^2 x^2}}{7 d e (c e+d e x)^{7/2}}-\frac{10 \sqrt{1-c^2-2 c d x-d^2 x^2}}{21 d e^3 (c e+d e x)^{3/2}}+\frac{10 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{e^2}}} \, dx,x,\sqrt{c e+d e x}\right )}{21 d e^5}\\ &=-\frac{2 \sqrt{1-c^2-2 c d x-d^2 x^2}}{7 d e (c e+d e x)^{7/2}}-\frac{10 \sqrt{1-c^2-2 c d x-d^2 x^2}}{21 d e^3 (c e+d e x)^{3/2}}+\frac{10 F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d e x}}{\sqrt{e}}\right )\right |-1\right )}{21 d e^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0233428, size = 40, normalized size = 0.32 $-\frac{2 (c+d x) \, _2F_1\left (-\frac{7}{4},\frac{1}{2};-\frac{3}{4};(c+d x)^2\right )}{7 d (e (c+d x))^{9/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((c*e + d*e*x)^(9/2)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]),x]

[Out]

(-2*(c + d*x)*Hypergeometric2F1[-7/4, 1/2, -3/4, (c + d*x)^2])/(7*d*(e*(c + d*x))^(9/2))

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Maple [B]  time = 0.254, size = 1002, normalized size = 8. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*e*x+c*e)^(9/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x)

[Out]

1/42*(12+8*c^2+21*(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+75*(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^2*c*d^2-63*(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^2*c*d^2+105*(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^2*c*d^2+63*(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^2*c*d^2+75*(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*c^2*d-63*(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*c^2*d+105*(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*c^2*d+63*(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*c^2*d-20*c^4+25*(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^3*d^3+8*d^2*x^2-80*x^3*c*d^3-80*x*c^3*d+35*(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+16*c*d*x+25*(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^3-120*c^2*d^2*x^2-21*(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^3-20*d^4*x^4+21*(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^3*d^3-21*(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^3*d^3+35*(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^3*d^3)/e^5*(-d^2*x^2-2*c*d*x-c
^2+1)^(1/2)*(e*(d*x+c))^(1/2)/(d*x+c)^4/(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{9}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(9/2)/(-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)^(9/2)), 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^{7} e^{5} x^{7} + 7 \, c d^{6} e^{5} x^{6} +{\left (21 \, c^{2} - 1\right )} d^{5} e^{5} x^{5} + 5 \,{\left (7 \, c^{3} - c\right )} d^{4} e^{5} x^{4} + 5 \,{\left (7 \, c^{4} - 2 \, c^{2}\right )} d^{3} e^{5} x^{3} +{\left (21 \, c^{5} - 10 \, c^{3}\right )} d^{2} e^{5} x^{2} +{\left (7 \, c^{6} - 5 \, c^{4}\right )} d e^{5} x +{\left (c^{7} - c^{5}\right )} e^{5}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(9/2)/(-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^7*e^5*x^7 + 7*c*d^6*e^5*x^6 + (21*c^2 - 1)*d
^5*e^5*x^5 + 5*(7*c^3 - c)*d^4*e^5*x^4 + 5*(7*c^4 - 2*c^2)*d^3*e^5*x^3 + (21*c^5 - 10*c^3)*d^2*e^5*x^2 + (7*c^
6 - 5*c^4)*d*e^5*x + (c^7 - c^5)*e^5), 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)**(9/2)/(-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{9}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^(9/2)/(-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)^(9/2)), x)