3.65 \(\int \frac {1}{\sqrt {2-3 x} \sqrt {-5+2 x} \sqrt {1+4 x} (7+5 x)} \, dx\)

Optimal. Leaf size=51 \[ -\frac {3 \sqrt {5-2 x} \Pi \left (\frac {55}{124};\sin ^{-1}\left (\frac {2 \sqrt {2-3 x}}{\sqrt {11}}\right )|-\frac {1}{2}\right )}{31 \sqrt {11} \sqrt {2 x-5}} \]

[Out]

-3/341*EllipticPi(2/11*(2-3*x)^(1/2)*11^(1/2),55/124,1/2*I*2^(1/2))*(5-2*x)^(1/2)*11^(1/2)/(-5+2*x)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {168, 538, 537} \[ -\frac {3 \sqrt {5-2 x} \Pi \left (\frac {55}{124};\sin ^{-1}\left (\frac {2 \sqrt {2-3 x}}{\sqrt {11}}\right )|-\frac {1}{2}\right )}{31 \sqrt {11} \sqrt {2 x-5}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 - 3*x]*Sqrt[-5 + 2*x]*Sqrt[1 + 4*x]*(7 + 5*x)),x]

[Out]

(-3*Sqrt[5 - 2*x]*EllipticPi[55/124, ArcSin[(2*Sqrt[2 - 3*x])/Sqrt[11]], -1/2])/(31*Sqrt[11]*Sqrt[-5 + 2*x])

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 538

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d*x^2)/c]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[c, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {2-3 x} \sqrt {-5+2 x} \sqrt {1+4 x} (7+5 x)} \, dx &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{\left (31-5 x^2\right ) \sqrt {\frac {11}{3}-\frac {4 x^2}{3}} \sqrt {-\frac {11}{3}-\frac {2 x^2}{3}}} \, dx,x,\sqrt {2-3 x}\right )\right )\\ &=-\frac {\left (2 \sqrt {\frac {3}{11}} \sqrt {5-2 x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (31-5 x^2\right ) \sqrt {\frac {11}{3}-\frac {4 x^2}{3}} \sqrt {1+\frac {2 x^2}{11}}} \, dx,x,\sqrt {2-3 x}\right )}{\sqrt {-5+2 x}}\\ &=-\frac {3 \sqrt {5-2 x} \Pi \left (\frac {55}{124};\sin ^{-1}\left (\frac {2 \sqrt {2-3 x}}{\sqrt {11}}\right )|-\frac {1}{2}\right )}{31 \sqrt {11} \sqrt {-5+2 x}}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 99, normalized size = 1.94 \[ -\frac {3 (3 x-2) \sqrt {\frac {8 x^2-18 x-5}{(2-3 x)^2}} \left (\operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt {11}}{2 \sqrt {2-3 x}}\right ),-2\right )-\Pi \left (\frac {124}{55};\left .\sin ^{-1}\left (\frac {\sqrt {11}}{2 \sqrt {2-3 x}}\right )\right |-2\right )\right )}{31 \sqrt {4 x+1} \sqrt {11 x-\frac {55}{2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[2 - 3*x]*Sqrt[-5 + 2*x]*Sqrt[1 + 4*x]*(7 + 5*x)),x]

[Out]

(-3*(-2 + 3*x)*Sqrt[(-5 - 18*x + 8*x^2)/(2 - 3*x)^2]*(EllipticF[ArcSin[Sqrt[11]/(2*Sqrt[2 - 3*x])], -2] - Elli
pticPi[124/55, ArcSin[Sqrt[11]/(2*Sqrt[2 - 3*x])], -2]))/(31*Sqrt[1 + 4*x]*Sqrt[-55/2 + 11*x])

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {4 \, x + 1} \sqrt {2 \, x - 5} \sqrt {-3 \, x + 2}}{120 \, x^{4} - 182 \, x^{3} - 385 \, x^{2} + 197 \, x + 70}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(7+5*x)/(2-3*x)^(1/2)/(-5+2*x)^(1/2)/(1+4*x)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(4*x + 1)*sqrt(2*x - 5)*sqrt(-3*x + 2)/(120*x^4 - 182*x^3 - 385*x^2 + 197*x + 70), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (5 \, x + 7\right )} \sqrt {4 \, x + 1} \sqrt {2 \, x - 5} \sqrt {-3 \, x + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(7+5*x)/(2-3*x)^(1/2)/(-5+2*x)^(1/2)/(1+4*x)^(1/2),x, algorithm="giac")

[Out]

integrate(1/((5*x + 7)*sqrt(4*x + 1)*sqrt(2*x - 5)*sqrt(-3*x + 2)), x)

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maple [A]  time = 0.02, size = 37, normalized size = 0.73 \[ -\frac {3 \sqrt {-2 x +5}\, \sqrt {11}\, \EllipticPi \left (\frac {2 \sqrt {-33 x +22}}{11}, \frac {55}{124}, \frac {i \sqrt {2}}{2}\right )}{341 \sqrt {2 x -5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x+7)/(-3*x+2)^(1/2)/(2*x-5)^(1/2)/(4*x+1)^(1/2),x)

[Out]

-3/341*EllipticPi(2/11*(-33*x+22)^(1/2),55/124,1/2*I*2^(1/2))*(-2*x+5)^(1/2)*11^(1/2)/(2*x-5)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (5 \, x + 7\right )} \sqrt {4 \, x + 1} \sqrt {2 \, x - 5} \sqrt {-3 \, x + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(7+5*x)/(2-3*x)^(1/2)/(-5+2*x)^(1/2)/(1+4*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((5*x + 7)*sqrt(4*x + 1)*sqrt(2*x - 5)*sqrt(-3*x + 2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {2-3\,x}\,\sqrt {4\,x+1}\,\sqrt {2\,x-5}\,\left (5\,x+7\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2 - 3*x)^(1/2)*(4*x + 1)^(1/2)*(2*x - 5)^(1/2)*(5*x + 7)),x)

[Out]

int(1/((2 - 3*x)^(1/2)*(4*x + 1)^(1/2)*(2*x - 5)^(1/2)*(5*x + 7)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 - 3 x} \sqrt {2 x - 5} \sqrt {4 x + 1} \left (5 x + 7\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(7+5*x)/(2-3*x)**(1/2)/(-5+2*x)**(1/2)/(1+4*x)**(1/2),x)

[Out]

Integral(1/(sqrt(2 - 3*x)*sqrt(2*x - 5)*sqrt(4*x + 1)*(5*x + 7)), x)

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