3.2837 \(\int \frac {(3+5 x)^{5/2}}{\sqrt {1-2 x} (2+3 x)^{7/2}} \, dx\)

Optimal. Leaf size=160 \[ -\frac {34154 \sqrt {\frac {11}{3}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ),\frac {35}{33}\right )}{46305}+\frac {2 \sqrt {1-2 x} (5 x+3)^{3/2}}{105 (3 x+2)^{5/2}}-\frac {53194 \sqrt {1-2 x} \sqrt {5 x+3}}{46305 \sqrt {3 x+2}}+\frac {544 \sqrt {1-2 x} \sqrt {5 x+3}}{6615 (3 x+2)^{3/2}}+\frac {53194 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{46305} \]

[Out]

53194/138915*EllipticE(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-34154/138915*EllipticF(1/7*21^(1/2
)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)+2/105*(3+5*x)^(3/2)*(1-2*x)^(1/2)/(2+3*x)^(5/2)+544/6615*(1-2*x)^(1/
2)*(3+5*x)^(1/2)/(2+3*x)^(3/2)-53194/46305*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 160, 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} \[ \frac {2 \sqrt {1-2 x} (5 x+3)^{3/2}}{105 (3 x+2)^{5/2}}-\frac {53194 \sqrt {1-2 x} \sqrt {5 x+3}}{46305 \sqrt {3 x+2}}+\frac {544 \sqrt {1-2 x} \sqrt {5 x+3}}{6615 (3 x+2)^{3/2}}-\frac {34154 \sqrt {\frac {11}{3}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{46305}+\frac {53194 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{46305} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(544*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(6615*(2 + 3*x)^(3/2)) - (53194*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(46305*Sqrt[2 +
3*x]) + (2*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(105*(2 + 3*x)^(5/2)) + (53194*Sqrt[11/3]*EllipticE[ArcSin[Sqrt[3/7
]*Sqrt[1 - 2*x]], 35/33])/46305 - (34154*Sqrt[11/3]*EllipticF[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/46305

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 {(3+5 x)^{5/2}}{\sqrt {1-2 x} (2+3 x)^{7/2}} \, dx &=\frac {2 \sqrt {1-2 x} (3+5 x)^{3/2}}{105 (2+3 x)^{5/2}}-\frac {2}{105} \int \frac {\left (-243-\frac {865 x}{2}\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^{5/2}} \, dx\\ &=\frac {544 \sqrt {1-2 x} \sqrt {3+5 x}}{6615 (2+3 x)^{3/2}}+\frac {2 \sqrt {1-2 x} (3+5 x)^{3/2}}{105 (2+3 x)^{5/2}}-\frac {4 \int \frac {-\frac {49871}{4}-\frac {88105 x}{4}}{\sqrt {1-2 x} (2+3 x)^{3/2} \sqrt {3+5 x}} \, dx}{6615}\\ &=\frac {544 \sqrt {1-2 x} \sqrt {3+5 x}}{6615 (2+3 x)^{3/2}}-\frac {53194 \sqrt {1-2 x} \sqrt {3+5 x}}{46305 \sqrt {2+3 x}}+\frac {2 \sqrt {1-2 x} (3+5 x)^{3/2}}{105 (2+3 x)^{5/2}}-\frac {8 \int \frac {-\frac {28265}{8}+\frac {132985 x}{4}}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{46305}\\ &=\frac {544 \sqrt {1-2 x} \sqrt {3+5 x}}{6615 (2+3 x)^{3/2}}-\frac {53194 \sqrt {1-2 x} \sqrt {3+5 x}}{46305 \sqrt {2+3 x}}+\frac {2 \sqrt {1-2 x} (3+5 x)^{3/2}}{105 (2+3 x)^{5/2}}-\frac {53194 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx}{46305}+\frac {187847 \int \frac {1}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{46305}\\ &=\frac {544 \sqrt {1-2 x} \sqrt {3+5 x}}{6615 (2+3 x)^{3/2}}-\frac {53194 \sqrt {1-2 x} \sqrt {3+5 x}}{46305 \sqrt {2+3 x}}+\frac {2 \sqrt {1-2 x} (3+5 x)^{3/2}}{105 (2+3 x)^{5/2}}+\frac {53194 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{46305}-\frac {34154 \sqrt {\frac {11}{3}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{46305}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 99, normalized size = 0.62 \[ \frac {\sqrt {2} \left (616735 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ),-\frac {33}{2}\right )-53194 E\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )|-\frac {33}{2}\right )\right )-\frac {6 \sqrt {1-2 x} \sqrt {5 x+3} \left (239373 x^2+311247 x+101257\right )}{(3 x+2)^{5/2}}}{138915} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(101257 + 311247*x + 239373*x^2))/(2 + 3*x)^(5/2) + Sqrt[2]*(-53194*EllipticE
[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2] + 616735*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2]))/138915

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (25 \, x^{2} + 30 \, x + 9\right )} \sqrt {5 \, x + 3} \sqrt {3 \, x + 2} \sqrt {-2 \, x + 1}}{162 \, x^{5} + 351 \, x^{4} + 216 \, x^{3} - 24 \, x^{2} - 64 \, x - 16}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(25*x^2 + 30*x + 9)*sqrt(5*x + 3)*sqrt(3*x + 2)*sqrt(-2*x + 1)/(162*x^5 + 351*x^4 + 216*x^3 - 24*x^2
- 64*x - 16), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 314, normalized size = 1.96 \[ -\frac {\left (14362380 x^{4}[phone] x^{3}-478746 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x^{2} \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+5550615 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x^{2} \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+3634188 x^{2}-638328 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+7400820 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, x \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-4994904 x -212776 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+2466940 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )-1822626\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{138915 \left (10 x^{2}+x -3\right ) \left (3 x +2\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/138915*(5550615*2^(1/2)*EllipticF(1/11*(110*x+66)^(1/2),1/2*I*66^(1/2))*x^2*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(-2
*x+1)^(1/2)-478746*2^(1/2)*EllipticE(1/11*(110*x+66)^(1/2),1/2*I*66^(1/2))*x^2*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(-2
*x+1)^(1/2)+7400820*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)-638328*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)+2466940*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))-212776*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)
)[phone]*x^4+20111058*x^3+3634188*x^2-4994904*x-1822626)*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(10*x^2+x-3)/(3*x+2)^(
5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (5\,x+3\right )}^{5/2}}{\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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