3.1375 \(\int \frac{1}{(3-2 x)^{5/2} \sqrt{1-3 x+x^2}} \, dx\)

Optimal. Leaf size=79 \[ -\frac{2 \sqrt{-x^2+3 x-1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{3-2 x}}{\sqrt [4]{5}}\right ),-1\right )}{15 \sqrt [4]{5} \sqrt{x^2-3 x+1}}-\frac{4 \sqrt{x^2-3 x+1}}{15 (3-2 x)^{3/2}} \]

[Out]

(-4*Sqrt[1 - 3*x + x^2])/(15*(3 - 2*x)^(3/2)) - (2*Sqrt[-1 + 3*x - x^2]*EllipticF[ArcSin[Sqrt[3 - 2*x]/5^(1/4)
], -1])/(15*5^(1/4)*Sqrt[1 - 3*x + x^2])

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Rubi [A]  time = 0.0344653, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {693, 691, 689, 221} \[ -\frac{4 \sqrt{x^2-3 x+1}}{15 (3-2 x)^{3/2}}-\frac{2 \sqrt{-x^2+3 x-1} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{15 \sqrt [4]{5} \sqrt{x^2-3 x+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/((3 - 2*x)^(5/2)*Sqrt[1 - 3*x + x^2]),x]

[Out]

(-4*Sqrt[1 - 3*x + x^2])/(15*(3 - 2*x)^(3/2)) - (2*Sqrt[-1 + 3*x - x^2]*EllipticF[ArcSin[Sqrt[3 - 2*x]/5^(1/4)
], -1])/(15*5^(1/4)*Sqrt[1 - 3*x + x^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 691

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[-((c*(a + b*x + c
*x^2))/(b^2 - 4*a*c))]/Sqrt[a + b*x + c*x^2], Int[(d + e*x)^m/Sqrt[-((a*c)/(b^2 - 4*a*c)) - (b*c*x)/(b^2 - 4*a
*c) - (c^2*x^2)/(b^2 - 4*a*c)], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
 0] && EqQ[m^2, 1/4]

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}{(3-2 x)^{5/2} \sqrt{1-3 x+x^2}} \, dx &=-\frac{4 \sqrt{1-3 x+x^2}}{15 (3-2 x)^{3/2}}+\frac{1}{15} \int \frac{1}{\sqrt{3-2 x} \sqrt{1-3 x+x^2}} \, dx\\ &=-\frac{4 \sqrt{1-3 x+x^2}}{15 (3-2 x)^{3/2}}+\frac{\sqrt{-1+3 x-x^2} \int \frac{1}{\sqrt{3-2 x} \sqrt{-\frac{1}{5}+\frac{3 x}{5}-\frac{x^2}{5}}} \, dx}{15 \sqrt{5} \sqrt{1-3 x+x^2}}\\ &=-\frac{4 \sqrt{1-3 x+x^2}}{15 (3-2 x)^{3/2}}-\frac{\left (2 \sqrt{-1+3 x-x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{5}}} \, dx,x,\sqrt{3-2 x}\right )}{15 \sqrt{5} \sqrt{1-3 x+x^2}}\\ &=-\frac{4 \sqrt{1-3 x+x^2}}{15 (3-2 x)^{3/2}}-\frac{2 \sqrt{-1+3 x-x^2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{3-2 x}}{\sqrt [4]{5}}\right )\right |-1\right )}{15 \sqrt [4]{5} \sqrt{1-3 x+x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0147454, size = 65, normalized size = 0.82 \[ \frac{2 \sqrt{-x^2+3 x-1} \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};\frac{1}{5} (3-2 x)^2\right )}{3 \sqrt{5} (3-2 x)^{3/2} \sqrt{x^2-3 x+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((3 - 2*x)^(5/2)*Sqrt[1 - 3*x + x^2]),x]

[Out]

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

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Maple [B]  time = 0.206, size = 172, normalized size = 2.2 \begin{align*}{\frac{1}{75\, \left ( -3+2\,x \right ) ^{2}} \left ( 2\,\sqrt{ \left ( -2\,x+3+\sqrt{5} \right ) \sqrt{5}}\sqrt{ \left ( -3+2\,x \right ) \sqrt{5}}\sqrt{ \left ( 2\,x-3+\sqrt{5} \right ) \sqrt{5}}{\it EllipticF} \left ( 1/10\,\sqrt{2}\sqrt{5}\sqrt{ \left ( -2\,x+3+\sqrt{5} \right ) \sqrt{5}},\sqrt{2} \right ) x-3\,\sqrt{ \left ( -2\,x+3+\sqrt{5} \right ) \sqrt{5}}\sqrt{ \left ( -3+2\,x \right ) \sqrt{5}}\sqrt{ \left ( 2\,x-3+\sqrt{5} \right ) \sqrt{5}}{\it EllipticF} \left ( 1/10\,\sqrt{2}\sqrt{5}\sqrt{ \left ( -2\,x+3+\sqrt{5} \right ) \sqrt{5}},\sqrt{2} \right ) -20\,{x}^{2}+60\,x-20 \right ) \sqrt{3-2\,x}{\frac{1}{\sqrt{{x}^{2}-3\,x+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x)

[Out]

1/75*(2*((-2*x+3+5^(1/2))*5^(1/2))^(1/2)*((-3+2*x)*5^(1/2))^(1/2)*((2*x-3+5^(1/2))*5^(1/2))^(1/2)*EllipticF(1/
10*2^(1/2)*5^(1/2)*((-2*x+3+5^(1/2))*5^(1/2))^(1/2),2^(1/2))*x-3*((-2*x+3+5^(1/2))*5^(1/2))^(1/2)*((-3+2*x)*5^
(1/2))^(1/2)*((2*x-3+5^(1/2))*5^(1/2))^(1/2)*EllipticF(1/10*2^(1/2)*5^(1/2)*((-2*x+3+5^(1/2))*5^(1/2))^(1/2),2
^(1/2))-20*x^2+60*x-20)/(x^2-3*x+1)^(1/2)*(3-2*x)^(1/2)/(-3+2*x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{2} - 3 \, x + 1}{\left (-2 \, x + 3\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^2 - 3*x + 1)*(-2*x + 3)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{x^{2} - 3 \, x + 1} \sqrt{-2 \, x + 3}}{8 \, x^{5} - 60 \, x^{4} + 170 \, x^{3} - 225 \, x^{2} + 135 \, x - 27}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(x^2 - 3*x + 1)*sqrt(-2*x + 3)/(8*x^5 - 60*x^4 + 170*x^3 - 225*x^2 + 135*x - 27), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (3 - 2 x\right )^{\frac{5}{2}} \sqrt{x^{2} - 3 x + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)**(5/2)/(x**2-3*x+1)**(1/2),x)

[Out]

Integral(1/((3 - 2*x)**(5/2)*sqrt(x**2 - 3*x + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{2} - 3 \, x + 1}{\left (-2 \, x + 3\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)^(5/2)/(x^2-3*x+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^2 - 3*x + 1)*(-2*x + 3)^(5/2)), x)