3.2811 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x)^{3/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac {64}{25} \sqrt {33} \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ),\frac {35}{33}\right )+\frac {14 (1-2 x)^{3/2}}{3 \sqrt {3 x+2} (5 x+3)^{3/2}}+\frac {6388 \sqrt {3 x+2} \sqrt {1-2 x}}{15 \sqrt {5 x+3}}-\frac {1012 \sqrt {3 x+2} \sqrt {1-2 x}}{15 (5 x+3)^{3/2}}-\frac {6388}{25} \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right ) \]

[Out]

-6388/75*EllipticE(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-64/25*EllipticF(1/7*21^(1/2)*(1-2*x)^(
1/2),1/33*1155^(1/2))*33^(1/2)+14/3*(1-2*x)^(3/2)/(3+5*x)^(3/2)/(2+3*x)^(1/2)-1012/15*(1-2*x)^(1/2)*(2+3*x)^(1
/2)/(3+5*x)^(3/2)+6388/15*(1-2*x)^(1/2)*(2+3*x)^(1/2)/(3+5*x)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {98, 150, 152, 158, 113, 119} \[ \frac {14 (1-2 x)^{3/2}}{3 \sqrt {3 x+2} (5 x+3)^{3/2}}+\frac {6388 \sqrt {3 x+2} \sqrt {1-2 x}}{15 \sqrt {5 x+3}}-\frac {1012 \sqrt {3 x+2} \sqrt {1-2 x}}{15 (5 x+3)^{3/2}}-\frac {64}{25} \sqrt {33} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )-\frac {6388}{25} \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14*(1 - 2*x)^(3/2))/(3*Sqrt[2 + 3*x]*(3 + 5*x)^(3/2)) - (1012*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/(15*(3 + 5*x)^(3/2
)) + (6388*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/(15*Sqrt[3 + 5*x]) - (6388*Sqrt[11/3]*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[
1 - 2*x]], 35/33])/25 - (64*Sqrt[33]*EllipticF[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/25

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 113

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

Rule 119

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

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 158

Int[((g_.) + (h_.)*(x_))/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol]
:> Dist[h/f, Int[Sqrt[e + f*x]/(Sqrt[a + b*x]*Sqrt[c + d*x]), x], x] + Dist[(f*g - e*h)/f, Int[1/(Sqrt[a + b*
x]*Sqrt[c + d*x]*Sqrt[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && SimplerQ[a + b*x, e + f*x] &&
SimplerQ[c + d*x, e + f*x]

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{5/2}}{(2+3 x)^{3/2} (3+5 x)^{5/2}} \, dx &=\frac {14 (1-2 x)^{3/2}}{3 \sqrt {2+3 x} (3+5 x)^{3/2}}+\frac {2}{3} \int \frac {(132-33 x) \sqrt {1-2 x}}{\sqrt {2+3 x} (3+5 x)^{5/2}} \, dx\\ &=\frac {14 (1-2 x)^{3/2}}{3 \sqrt {2+3 x} (3+5 x)^{3/2}}-\frac {1012 \sqrt {1-2 x} \sqrt {2+3 x}}{15 (3+5 x)^{3/2}}+\frac {4}{45} \int \frac {-\frac {7689}{2}+2376 x}{\sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}} \, dx\\ &=\frac {14 (1-2 x)^{3/2}}{3 \sqrt {2+3 x} (3+5 x)^{3/2}}-\frac {1012 \sqrt {1-2 x} \sqrt {2+3 x}}{15 (3+5 x)^{3/2}}+\frac {6388 \sqrt {1-2 x} \sqrt {2+3 x}}{15 \sqrt {3+5 x}}-\frac {8}{495} \int \frac {-\frac {100089}{2}-\frac {158103 x}{2}}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx\\ &=\frac {14 (1-2 x)^{3/2}}{3 \sqrt {2+3 x} (3+5 x)^{3/2}}-\frac {1012 \sqrt {1-2 x} \sqrt {2+3 x}}{15 (3+5 x)^{3/2}}+\frac {6388 \sqrt {1-2 x} \sqrt {2+3 x}}{15 \sqrt {3+5 x}}+\frac {1056}{25} \int \frac {1}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx+\frac {6388}{25} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx\\ &=\frac {14 (1-2 x)^{3/2}}{3 \sqrt {2+3 x} (3+5 x)^{3/2}}-\frac {1012 \sqrt {1-2 x} \sqrt {2+3 x}}{15 (3+5 x)^{3/2}}+\frac {6388 \sqrt {1-2 x} \sqrt {2+3 x}}{15 \sqrt {3+5 x}}-\frac {6388}{25} \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )-\frac {64}{25} \sqrt {33} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 100, normalized size = 0.63 \[ \frac {2}{75} \left (2 \sqrt {2} \left (1597 E\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )|-\frac {33}{2}\right )-805 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ),-\frac {33}{2}\right )\right )+\frac {5 \sqrt {1-2 x} \left (47910 x^2+59098 x+18187\right )}{\sqrt {3 x+2} (5 x+3)^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((5*Sqrt[1 - 2*x]*(18187 + 59098*x + 47910*x^2))/(Sqrt[2 + 3*x]*(3 + 5*x)^(3/2)) + 2*Sqrt[2]*(1597*Elliptic
E[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2] - 805*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2])))/75

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fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (4 \, x^{2} - 4 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {3 \, x + 2} \sqrt {-2 \, x + 1}}{1125 \, x^{5} + 3525 \, x^{4} + 4415 \, x^{3} + 2763 \, x^{2} + 864 \, x + 108}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((4*x^2 - 4*x + 1)*sqrt(5*x + 3)*sqrt(3*x + 2)*sqrt(-2*x + 1)/(1125*x^5 + 3525*x^4 + 4415*x^3 + 2763*x
^2 + 864*x + 108), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.03, size = 219, normalized size = 1.39 \[ \frac {2 \sqrt {-2 x +1}\, \sqrt {3 x +2}\, \left (479100 x^{3}+351430 x^{2}-15970 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+8050 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-113620 x -9582 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+4830 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-90935\right )}{75 \left (5 x +3\right )^{\frac {3}{2}} \left (6 x^{2}+x -2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/75*(-2*x+1)^(1/2)*(3*x+2)^(1/2)*(8050*2^(1/2)*EllipticF(1/11*(110*x+66)^(1/2),1/2*I*66^(1/2))*x*(5*x+3)^(1/2
)*(3*x+2)^(1/2)*(-2*x+1)^(1/2)-15970*2^(1/2)*EllipticE(1/11*(110*x+66)^(1/2),1/2*I*66^(1/2))*x*(5*x+3)^(1/2)*(
3*x+2)^(1/2)*(-2*x+1)^(1/2)+4830*2^(1/2)*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(-2*x+1)^(1/2)*EllipticF(1/11*(110*x+66)^
(1/2),1/2*I*66^(1/2))-9582*2^(1/2)*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(-2*x+1)^(1/2)*EllipticE(1/11*(110*x+66)^(1/2),
1/2*I*66^(1/2))+479100*x^3+351430*x^2-113620*x-90935)/(5*x+3)^(3/2)/(6*x^2+x-2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-2\,x\right )}^{5/2}}{{\left (3\,x+2\right )}^{3/2}\,{\left (5\,x+3\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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