3.2938 \(\int \frac {(2+3 x)^{5/2}}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac {412 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ),\frac {35}{33}\right )}{3025 \sqrt {33}}+\frac {7 (3 x+2)^{3/2}}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {4157 \sqrt {1-2 x} \sqrt {3 x+2}}{19965 \sqrt {5 x+3}}-\frac {107 \sqrt {1-2 x} \sqrt {3 x+2}}{1815 (5 x+3)^{3/2}}+\frac {4157 E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{3025 \sqrt {33}} \]

[Out]

4157/99825*EllipticE(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-412/99825*EllipticF(1/7*21^(1/2)*(1-
2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)+7/11*(2+3*x)^(3/2)/(3+5*x)^(3/2)/(1-2*x)^(1/2)-107/1815*(1-2*x)^(1/2)*(2+
3*x)^(1/2)/(3+5*x)^(3/2)-4157/19965*(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} \[ \frac {7 (3 x+2)^{3/2}}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {4157 \sqrt {1-2 x} \sqrt {3 x+2}}{19965 \sqrt {5 x+3}}-\frac {107 \sqrt {1-2 x} \sqrt {3 x+2}}{1815 (5 x+3)^{3/2}}-\frac {412 F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{3025 \sqrt {33}}+\frac {4157 E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{3025 \sqrt {33}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-107*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/(1815*(3 + 5*x)^(3/2)) + (7*(2 + 3*x)^(3/2))/(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3
/2)) - (4157*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/(19965*Sqrt[3 + 5*x]) + (4157*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*
x]], 35/33])/(3025*Sqrt[33]) - (412*EllipticF[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/(3025*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)^{3/2} (3+5 x)^{5/2}} \, dx &=\frac {7 (2+3 x)^{3/2}}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {1}{11} \int \frac {\left (-\frac {25}{2}-3 x\right ) \sqrt {2+3 x}}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {2+3 x}}{1815 (3+5 x)^{3/2}}+\frac {7 (2+3 x)^{3/2}}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {2 \int \frac {-\frac {1573}{4}-309 x}{\sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}} \, dx}{1815}\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {2+3 x}}{1815 (3+5 x)^{3/2}}+\frac {7 (2+3 x)^{3/2}}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {4157 \sqrt {1-2 x} \sqrt {2+3 x}}{19965 \sqrt {3+5 x}}+\frac {4 \int \frac {-\frac {6123}{4}-\frac {12471 x}{4}}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{19965}\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {2+3 x}}{1815 (3+5 x)^{3/2}}+\frac {7 (2+3 x)^{3/2}}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {4157 \sqrt {1-2 x} \sqrt {2+3 x}}{19965 \sqrt {3+5 x}}+\frac {206 \int \frac {1}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{3025}-\frac {4157 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx}{33275}\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {2+3 x}}{1815 (3+5 x)^{3/2}}+\frac {7 (2+3 x)^{3/2}}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {4157 \sqrt {1-2 x} \sqrt {2+3 x}}{19965 \sqrt {3+5 x}}+\frac {4157 E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{3025 \sqrt {33}}-\frac {412 F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{3025 \sqrt {33}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 97, normalized size = 0.62 \[ \frac {\sqrt {2} \left (10955 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ),-\frac {33}{2}\right )+\frac {5 \sqrt {6 x+4} \left (20785 x^2+22313 x+5881\right )}{\sqrt {1-2 x} (5 x+3)^{3/2}}-4157 E\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )|-\frac {33}{2}\right )\right )}{99825} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*((5*Sqrt[4 + 6*x]*(5881 + 22313*x + 20785*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - 4157*EllipticE[ArcS
in[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2] + 10955*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2]))/99825

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (9 \, x^{2} + 12 \, x + 4\right )} \sqrt {5 \, x + 3} \sqrt {3 \, x + 2} \sqrt {-2 \, x + 1}}{500 \, x^{5} + 400 \, x^{4} - 235 \, x^{3} - 207 \, x^{2} + 27 \, x + 27}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((9*x^2 + 12*x + 4)*sqrt(5*x + 3)*sqrt(3*x + 2)*sqrt(-2*x + 1)/(500*x^5 + 400*x^4 - 235*x^3 - 207*x^2
+ 27*x + 27), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.03, size = 219, normalized size = 1.40 \[ \frac {\sqrt {3 x +2}\, \sqrt {-2 x +1}\, \left (-623550 x^{3}-1085090 x^{2}+20785 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-54775 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-622690 x +12471 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-32865 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-117620\right )}{99825 \left (5 x +3\right )^{\frac {3}{2}} \left (6 x^{2}+x -2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/99825*(3*x+2)^(1/2)*(-2*x+1)^(1/2)*(20785*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)-54775*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)+12471*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))-32865*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))-623550*x^3-1085090*x^2-622690*x-117620)/(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 \[ \int \frac {{\left (3 \, x + 2\right )}^{\frac {5}{2}}}{{\left (5 \, x + 3\right )}^{\frac {5}{2}} {\left (-2 \, x + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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