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 \[ -\frac {10 e^3 \sqrt {-c^2-2 c d x-d^2 x^2+1} \sqrt {c e+d e x}}{21 d}-\frac {2 e \sqrt {-c^2-2 c d x-d^2 x^2+1} (c e+d e x)^{5/2}}{7 d}+\frac {10 e^{7/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c e+d x e}}{\sqrt {e}}\right )\right |-1\right )}{21 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 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 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 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])

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*}

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Mathematica [C]  time = 0.06, 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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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} \sqrt {d e x + c e}}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{\frac {7}{2}}}{\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}\,{d x} \]

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)

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maple [B]  time = 0.07, size = 263, normalized size = 2.12 \[ -\frac {\sqrt {\left (d x +c \right ) e}\, \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, \left (6 d^{5} x^{5}+30 c \,d^{4} x^{4}+60 c^{2} d^{3} x^{3}+60 c^{3} d^{2} x^{2}+30 c^{4} d x +4 d^{3} x^{3}+6 c^{5}+12 c \,d^{2} x^{2}+12 c^{2} d x +4 c^{3}-10 d x -10 c +2 \sqrt {-2 d x -2 c +2}\, \sqrt {d x +c}\, \sqrt {2 d x +2 c +2}\, \EllipticF \left (\frac {\sqrt {-2 d x -2 c +2}}{2}, \sqrt {2}\right )+7 \sqrt {-2 d x -2 c +2}\, \sqrt {2 d x +2 c +2}\, \sqrt {-d x -c}\, \EllipticF \left (\frac {\sqrt {2 d x +2 c +2}}{2}, \sqrt {2}\right )\right ) e^{3}}{21 \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}-d x -c \right ) d} \]

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]

-1/21*((d*x+c)*e)^(1/2)*(-d^2*x^2-2*c*d*x-c^2+1)^(1/2)*e^3*(6*d^5*x^5+30*c*d^4*x^4+60*c^2*d^3*x^3+60*c^3*d^2*x
^2+4*d^3*x^3+30*c^4*d*x+12*c*d^2*x^2+6*c^5+2*(-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))+7*(-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))+12*c^2*d*x+4*c^3-10*d*x-10*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{\frac {7}{2}}}{\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^{7/2}}{\sqrt {-c^2-2\,c\,d\,x-d^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \left (c + d x\right )\right )^{\frac {7}{2}}}{\sqrt {- \left (c + d x - 1\right ) \left (c + d x + 1\right )}}\, dx \]

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]

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

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