3.2823 \(\int \frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx\)

Optimal. Leaf size=160 \[ -\frac {4721 \sqrt {\frac {11}{3}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ),\frac {35}{33}\right )}{5250}-\frac {1}{7} \sqrt {1-2 x} (3 x+2)^{3/2} (5 x+3)^{3/2}-\frac {102}{175} \sqrt {1-2 x} \sqrt {3 x+2} (5 x+3)^{3/2}-\frac {4721 \sqrt {1-2 x} \sqrt {3 x+2} \sqrt {5 x+3}}{1050}-\frac {78472 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{2625} \]

[Out]

-78472/7875*EllipticE(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-4721/15750*EllipticF(1/7*21^(1/2)*(
1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-1/7*(2+3*x)^(3/2)*(3+5*x)^(3/2)*(1-2*x)^(1/2)-102/175*(3+5*x)^(3/2)*(1-
2*x)^(1/2)*(2+3*x)^(1/2)-4721/1050*(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 = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {101, 154, 158, 113, 119} \[ -\frac {1}{7} \sqrt {1-2 x} (3 x+2)^{3/2} (5 x+3)^{3/2}-\frac {102}{175} \sqrt {1-2 x} \sqrt {3 x+2} (5 x+3)^{3/2}-\frac {4721 \sqrt {1-2 x} \sqrt {3 x+2} \sqrt {5 x+3}}{1050}-\frac {4721 \sqrt {\frac {11}{3}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{5250}-\frac {78472 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{2625} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4721*Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*Sqrt[3 + 5*x])/1050 - (102*Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*(3 + 5*x)^(3/2))/175
- (Sqrt[1 - 2*x]*(2 + 3*x)^(3/2)*(3 + 5*x)^(3/2))/7 - (78472*Sqrt[11/3]*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[1 - 2
*x]], 35/33])/2625 - (4721*Sqrt[11/3]*EllipticF[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/5250

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && 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)^{3/2} (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx &=-\frac {1}{7} \sqrt {1-2 x} (2+3 x)^{3/2} (3+5 x)^{3/2}+\frac {1}{7} \int \frac {\sqrt {2+3 x} \sqrt {3+5 x} \left (\frac {129}{2}+102 x\right )}{\sqrt {1-2 x}} \, dx\\ &=-\frac {102}{175} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}-\frac {1}{7} \sqrt {1-2 x} (2+3 x)^{3/2} (3+5 x)^{3/2}-\frac {1}{175} \int \frac {\left (-4602-\frac {14163 x}{2}\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx\\ &=-\frac {4721 \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}}{1050}-\frac {102}{175} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}-\frac {1}{7} \sqrt {1-2 x} (2+3 x)^{3/2} (3+5 x)^{3/2}+\frac {\int \frac {\frac {596157}{4}+235416 x}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{1575}\\ &=-\frac {4721 \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}}{1050}-\frac {102}{175} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}-\frac {1}{7} \sqrt {1-2 x} (2+3 x)^{3/2} (3+5 x)^{3/2}+\frac {51931 \int \frac {1}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{10500}+\frac {78472 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx}{2625}\\ &=-\frac {4721 \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}}{1050}-\frac {102}{175} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}-\frac {1}{7} \sqrt {1-2 x} (2+3 x)^{3/2} (3+5 x)^{3/2}-\frac {78472 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{2625}-\frac {4721 \sqrt {\frac {11}{3}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{5250}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 100, normalized size = 0.62 \[ \frac {313888 E\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )|-\frac {33}{2}\right )-5 \left (31619 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ),-\frac {33}{2}\right )+3 \sqrt {2-4 x} \sqrt {3 x+2} \sqrt {5 x+3} \left (2250 x^2+5910 x+7457\right )\right )}{15750 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(313888*EllipticE[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2] - 5*(3*Sqrt[2 - 4*x]*Sqrt[2 + 3*x]*Sqrt[3 + 5*x]*(7
457 + 5910*x + 2250*x^2) + 31619*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2]))/(15750*Sqrt[2])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (15 \, x^{2} + 19 \, x + 6\right )} \sqrt {5 \, x + 3} \sqrt {3 \, x + 2} \sqrt {-2 \, x + 1}}{2 \, x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(15*x^2 + 19*x + 6)*sqrt(5*x + 3)*sqrt(3*x + 2)*sqrt(-2*x + 1)/(2*x - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 150, normalized size = 0.94 \[ \frac {\sqrt {3 x +2}\, \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (-2025000 x^{5}-6871500 x^{4}-10316700 x^{3}-3499230 x^{2}+2629770 x -313888 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+158095 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+1342260\right )}{945000 x^{3}+724500 x^{2}-220500 x -189000} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/31500*(3*x+2)^(1/2)*(5*x+3)^(1/2)*(-2*x+1)^(1/2)*(158095*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))-313888*2^(1/2)*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(-2*x+1)^(1/2)*Elli
pticE(1/11*(110*x+66)^(1/2),1/2*I*66^(1/2))-2025000*x^5-6871500*x^4-10316700*x^3-3499230*x^2+2629770*x+1342260
)/(30*x^3+23*x^2-7*x-6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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