3.2983 \(\int \frac {(2+3 x)^{5/2}}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac {41 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ),\frac {35}{33}\right )}{605 \sqrt {33}}+\frac {7 (3 x+2)^{3/2}}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}+\frac {974 \sqrt {1-2 x} \sqrt {3 x+2}}{3993 \sqrt {5 x+3}}-\frac {203 \sqrt {3 x+2}}{363 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {974 E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{605 \sqrt {33}} \]

[Out]

-974/19965*EllipticE(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-41/19965*EllipticF(1/7*21^(1/2)*(1-2
*x)^(1/2),1/33*1155^(1/2))*33^(1/2)+7/33*(2+3*x)^(3/2)/(1-2*x)^(3/2)/(3+5*x)^(1/2)-203/363*(2+3*x)^(1/2)/(1-2*
x)^(1/2)/(3+5*x)^(1/2)+974/3993*(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 = 156, 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} \[ \frac {7 (3 x+2)^{3/2}}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}+\frac {974 \sqrt {1-2 x} \sqrt {3 x+2}}{3993 \sqrt {5 x+3}}-\frac {203 \sqrt {3 x+2}}{363 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {41 F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{605 \sqrt {33}}-\frac {974 E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{605 \sqrt {33}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-203*Sqrt[2 + 3*x])/(363*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (974*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/(3993*Sqrt[3 + 5*x]
) + (7*(2 + 3*x)^(3/2))/(33*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) - (974*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 3
5/33])/(605*Sqrt[33]) - (41*EllipticF[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/(605*Sqrt[33])

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 {(2+3 x)^{5/2}}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx &=\frac {7 (2+3 x)^{3/2}}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {1}{33} \int \frac {\sqrt {2+3 x} \left (\frac {107}{2}+96 x\right )}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx\\ &=-\frac {203 \sqrt {2+3 x}}{363 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {7 (2+3 x)^{3/2}}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {1}{363} \int \frac {\frac {121}{2}-\frac {123 x}{2}}{\sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}} \, dx\\ &=-\frac {203 \sqrt {2+3 x}}{363 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {974 \sqrt {1-2 x} \sqrt {2+3 x}}{3993 \sqrt {3+5 x}}+\frac {7 (2+3 x)^{3/2}}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {2 \int \frac {\frac {3777}{4}+1461 x}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{3993}\\ &=-\frac {203 \sqrt {2+3 x}}{363 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {974 \sqrt {1-2 x} \sqrt {2+3 x}}{3993 \sqrt {3+5 x}}+\frac {7 (2+3 x)^{3/2}}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {41 \int \frac {1}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{1210}+\frac {974 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx}{6655}\\ &=-\frac {203 \sqrt {2+3 x}}{363 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {974 \sqrt {1-2 x} \sqrt {2+3 x}}{3993 \sqrt {3+5 x}}+\frac {7 (2+3 x)^{3/2}}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {974 E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{605 \sqrt {33}}-\frac {41 F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{605 \sqrt {33}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 102, normalized size = 0.65 \[ \frac {-595 \sqrt {2} \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ),-\frac {33}{2}\right )+\frac {10 \sqrt {3 x+2} \left (3896 x^2+3111 x+435\right )}{(1-2 x)^{3/2} \sqrt {5 x+3}}+1948 \sqrt {2} E\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )|-\frac {33}{2}\right )}{39930} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((10*Sqrt[2 + 3*x]*(435 + 3111*x + 3896*x^2))/((1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) + 1948*Sqrt[2]*EllipticE[ArcSin[
Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2] - 595*Sqrt[2]*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2])/39930

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (9 \, x^{2} + 12 \, x + 4\right )} \sqrt {5 \, x + 3} \sqrt {3 \, x + 2} \sqrt {-2 \, x + 1}}{200 \, x^{5} - 60 \, x^{4} - 138 \, x^{3} + 47 \, x^{2} + 24 \, x - 9}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(9*x^2 + 12*x + 4)*sqrt(5*x + 3)*sqrt(3*x + 2)*sqrt(-2*x + 1)/(200*x^5 - 60*x^4 - 138*x^3 + 47*x^2 +
24*x - 9), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.03, size = 228, normalized size = 1.46 \[ \frac {\left (116880 x^{3}+171250 x^{2}-3896 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+1190 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+75270 x +1948 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-595 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+8700\right ) \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \sqrt {3 x +2}}{39930 \left (15 x^{2}+19 x +6\right ) \left (2 x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/39930*(1190*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)-3896*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)
-595*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))+1948*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))+116880*x^3+1
71250*x^2+75270*x+8700)*(5*x+3)^(1/2)*(-2*x+1)^(1/2)*(3*x+2)^(1/2)/(15*x^2+19*x+6)/(2*x-1)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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