### 3.1408 $$\int \frac{(c e+d e x)^{5/2}}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx$$

Optimal. Leaf size=111 $-\frac{6 e^{5/2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c e+d e x}}{\sqrt{e}}\right ),-1\right )}{5 d}-\frac{2 e \sqrt{-c^2-2 c d x-d^2 x^2+1} (c e+d e x)^{3/2}}{5 d}+\frac{6 e^{5/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d x e}}{\sqrt{e}}\right )\right |-1\right )}{5 d}$

[Out]

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

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Rubi [A]  time = 0.090528, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.162, Rules used = {692, 690, 307, 221, 1199, 424} $-\frac{2 e \sqrt{-c^2-2 c d x-d^2 x^2+1} (c e+d e x)^{3/2}}{5 d}-\frac{6 e^{5/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d x e}}{\sqrt{e}}\right )\right |-1\right )}{5 d}+\frac{6 e^{5/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d x e}}{\sqrt{e}}\right )\right |-1\right )}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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

Mathematica [C]  time = 0.0293227, size = 54, normalized size = 0.49 $-\frac{2 e (e (c+d x))^{3/2} \left (\sqrt{1-(c+d x)^2}-\, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};(c+d x)^2\right )\right )}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*e*(e*(c + d*x))^(3/2)*(Sqrt[1 - (c + d*x)^2] - Hypergeometric2F1[1/2, 3/4, 7/4, (c + d*x)^2]))/(5*d)

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Maple [B]  time = 0.176, size = 333, normalized size = 3. \begin{align*}{\frac{{e}^{2}}{30\,d \left ({x}^{3}{d}^{3}+3\,{x}^{2}c{d}^{2}+3\,x{c}^{2}d+{c}^{3}-dx-c \right ) }\sqrt{e \left ( dx+c \right ) }\sqrt{-{d}^{2}{x}^{2}-2\,cdx-{c}^{2}+1} \left ( -12\,{d}^{4}{x}^{4}-48\,{x}^{3}c{d}^{3}-72\,{c}^{2}{d}^{2}{x}^{2}-48\,x{c}^{3}d+5\,\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 ) +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 ) +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 ) +15\,\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 ) +12\,{d}^{2}{x}^{2}-12\,{c}^{4}+24\,cdx+12\,{c}^{2} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(e*(d*x+c))^(1/2)*(-d^2*x^2-2*c*d*x-c^2+1)^(1/2)*e^2*(-12*d^4*x^4-48*x^3*c*d^3-72*c^2*d^2*x^2-48*x*c^3*d+
5*(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))+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))+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))+15*(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))+12*d^2*x^2-12*c^4+24*c*d*x+12*c^2)/
d/(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3-d*x-c)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Integral((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{{\left (d e x + c e\right )}^{\frac{5}{2}}}{\sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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