3.1401 $$\int \frac{(c e+d e x)^{7/2}}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx$$

Optimal. Leaf size=124 $\frac{10 e^{7/2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c e+d e x}}{\sqrt{e}}\right ),-1\right )}{21 d}-\frac{10 e^3 \sqrt{-c^2-2 c d x-d^2 x^2+1} \sqrt{c e+d e x}}{21 d}-\frac{2 e \sqrt{-c^2-2 c d x-d^2 x^2+1} (c e+d e x)^{5/2}}{7 d}$

[Out]

(-10*e^3*Sqrt[c*e + d*e*x]*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(21*d) - (2*e*(c*e + d*e*x)^(5/2)*Sqrt[1 - c^2 -
2*c*d*x - d^2*x^2])/(7*d) + (10*e^(7/2)*EllipticF[ArcSin[Sqrt[c*e + d*e*x]/Sqrt[e]], -1])/(21*d)

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Rubi [A]  time = 0.0952692, antiderivative size = 124, 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 = {692, 689, 221} $-\frac{10 e^3 \sqrt{-c^2-2 c d x-d^2 x^2+1} \sqrt{c e+d e x}}{21 d}-\frac{2 e \sqrt{-c^2-2 c d x-d^2 x^2+1} (c e+d e x)^{5/2}}{7 d}+\frac{10 e^{7/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c e+d x e}}{\sqrt{e}}\right )\right |-1\right )}{21 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c*e + d*e*x)^(7/2)/Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2],x]

[Out]

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

Mathematica [C]  time = 0.063168, size = 86, normalized size = 0.69 $-\frac{2 e^3 \sqrt{e (c+d x)} \left (\sqrt{-c^2-2 c d x-d^2 x^2+1} \left (3 c^2+6 c d x+3 d^2 x^2+5\right )-5 \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};(c+d x)^2\right )\right )}{21 d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c*e + d*e*x)^(7/2)/Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2],x]

[Out]

(-2*e^3*Sqrt[e*(c + d*x)]*(Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]*(5 + 3*c^2 + 6*c*d*x + 3*d^2*x^2) - 5*Hypergeomet
ric2F1[1/4, 1/2, 5/4, (c + d*x)^2]))/(21*d)

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Maple [B]  time = 0.179, size = 463, normalized size = 3.7 \begin{align*}{\frac{{e}^{3}}{210\,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 ( -60\,{x}^{5}{d}^{5}-300\,{x}^{4}c{d}^{4}-600\,{x}^{3}{c}^{2}{d}^{3}-600\,{x}^{2}{c}^{3}{d}^{2}-40\,{x}^{3}{d}^{3}-300\,x{c}^{4}d+84\,\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-84\,\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-120\,{x}^{2}c{d}^{2}-60\,{c}^{5}-105\,\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 ) +105\,\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 ) -15\,\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 ) +35\,\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 ) -120\,x{c}^{2}d-40\,{c}^{3}+100\,dx+100\,c \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x)

[Out]

1/210*(e*(d*x+c))^(1/2)*(-d^2*x^2-2*c*d*x-c^2+1)^(1/2)*e^3*(-60*x^5*d^5-300*x^4*c*d^4-600*x^3*c^2*d^3-600*x^2*
c^3*d^2-40*x^3*d^3-300*x*c^4*d+84*(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-84*(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-120*x^2*c*d^2-60*c^5-105*(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))+105*(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))-15*(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))+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))-120*x*c^2*d-40*c^3+100*d*x+100*c)/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{7}{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)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^(7/2)/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^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}\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)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-(d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*x + c^3*e^3)*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(-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((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{{\left (d e x + c e\right )}^{\frac{7}{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)^(7/2)/(-d^2*x^2-2*c*d*x-c^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(7/2)/sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1), x)