3.2834 \(\int \frac{(3+5 x)^{5/2}}{\sqrt{1-2 x} \sqrt{2+3 x}} \, dx\)

Optimal. Leaf size=129 \[ -\frac{62}{135} \sqrt{\frac{11}{3}} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ),\frac{35}{33}\right )-\frac{1}{3} \sqrt{1-2 x} \sqrt{3 x+2} (5 x+3)^{3/2}-\frac{62}{27} \sqrt{1-2 x} \sqrt{3 x+2} \sqrt{5 x+3}-\frac{4141}{270} \sqrt{\frac{11}{3}} E\left (\sin ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )|\frac{35}{33}\right ) \]

[Out]

(-62*Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*Sqrt[3 + 5*x])/27 - (Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*(3 + 5*x)^(3/2))/3 - (4141*S
qrt[11/3]*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/270 - (62*Sqrt[11/3]*EllipticF[ArcSin[Sqrt[3/7]*S
qrt[1 - 2*x]], 35/33])/135

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Rubi [A]  time = 0.0397239, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {102, 154, 158, 113, 119} \[ -\frac{1}{3} \sqrt{1-2 x} \sqrt{3 x+2} (5 x+3)^{3/2}-\frac{62}{27} \sqrt{1-2 x} \sqrt{3 x+2} \sqrt{5 x+3}-\frac{62}{135} \sqrt{\frac{11}{3}} F\left (\sin ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )|\frac{35}{33}\right )-\frac{4141}{270} \sqrt{\frac{11}{3}} E\left (\sin ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )|\frac{35}{33}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-62*Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*Sqrt[3 + 5*x])/27 - (Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*(3 + 5*x)^(3/2))/3 - (4141*S
qrt[11/3]*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/270 - (62*Sqrt[11/3]*EllipticF[ArcSin[Sqrt[3/7]*S
qrt[1 - 2*x]], 35/33])/135

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

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]

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)]))

Rubi steps

\begin{align*} \int \frac{(3+5 x)^{5/2}}{\sqrt{1-2 x} \sqrt{2+3 x}} \, dx &=-\frac{1}{3} \sqrt{1-2 x} \sqrt{2+3 x} (3+5 x)^{3/2}-\frac{1}{15} \int \frac{\left (-\frac{405}{2}-310 x\right ) \sqrt{3+5 x}}{\sqrt{1-2 x} \sqrt{2+3 x}} \, dx\\ &=-\frac{62}{27} \sqrt{1-2 x} \sqrt{2+3 x} \sqrt{3+5 x}-\frac{1}{3} \sqrt{1-2 x} \sqrt{2+3 x} (3+5 x)^{3/2}+\frac{1}{135} \int \frac{\frac{13105}{2}+\frac{20705 x}{2}}{\sqrt{1-2 x} \sqrt{2+3 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{62}{27} \sqrt{1-2 x} \sqrt{2+3 x} \sqrt{3+5 x}-\frac{1}{3} \sqrt{1-2 x} \sqrt{2+3 x} (3+5 x)^{3/2}+\frac{341}{135} \int \frac{1}{\sqrt{1-2 x} \sqrt{2+3 x} \sqrt{3+5 x}} \, dx+\frac{4141}{270} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x} \sqrt{2+3 x}} \, dx\\ &=-\frac{62}{27} \sqrt{1-2 x} \sqrt{2+3 x} \sqrt{3+5 x}-\frac{1}{3} \sqrt{1-2 x} \sqrt{2+3 x} (3+5 x)^{3/2}-\frac{4141}{270} \sqrt{\frac{11}{3}} E\left (\sin ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )|\frac{35}{33}\right )-\frac{62}{135} \sqrt{\frac{11}{3}} F\left (\sin ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )|\frac{35}{33}\right )\\ \end{align*}

Mathematica [A]  time = 0.202548, size = 95, normalized size = 0.74 \[ \frac{4141 E\left (\sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )|-\frac{33}{2}\right )-5 \left (419 \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ),-\frac{33}{2}\right )+3 \sqrt{2-4 x} \sqrt{3 x+2} \sqrt{5 x+3} (45 x+89)\right )}{405 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4141*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]*(89
+ 45*x) + 419*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2]))/(405*Sqrt[2])

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Maple [C]  time = 0.013, size = 145, normalized size = 1.1 \begin{align*}{\frac{1}{24300\,{x}^{3}+18630\,{x}^{2}-5670\,x-4860}\sqrt{1-2\,x}\sqrt{2+3\,x}\sqrt{3+5\,x} \left ( 2095\,\sqrt{2}\sqrt{3+5\,x}\sqrt{2+3\,x}\sqrt{1-2\,x}{\it EllipticF} \left ( 1/11\,\sqrt{66+110\,x},i/2\sqrt{66} \right ) -4141\,\sqrt{2}\sqrt{3+5\,x}\sqrt{2+3\,x}\sqrt{1-2\,x}{\it EllipticE} \left ( 1/11\,\sqrt{66+110\,x},i/2\sqrt{66} \right ) -40500\,{x}^{4}-111150\,{x}^{3}-51960\,{x}^{2}+26790\,x+16020 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/810*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(2+3*x)^(1/2)*(2095*2^(1/2)*(3+5*x)^(1/2)*(2+3*x)^(1/2)*(1-2*x)^(1/2)*Ellipt
icF(1/11*(66+110*x)^(1/2),1/2*I*66^(1/2))-4141*2^(1/2)*(3+5*x)^(1/2)*(2+3*x)^(1/2)*(1-2*x)^(1/2)*EllipticE(1/1
1*(66+110*x)^(1/2),1/2*I*66^(1/2))-40500*x^4-111150*x^3-51960*x^2+26790*x+16020)/(30*x^3+23*x^2-7*x-6)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (5 \, x + 3\right )}^{\frac{5}{2}}}{\sqrt{3 \, x + 2} \sqrt{-2 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (25 \, x^{2} + 30 \, x + 9\right )} \sqrt{5 \, x + 3} \sqrt{3 \, x + 2} \sqrt{-2 \, x + 1}}{6 \, x^{2} + x - 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)^(5/2)/(1-2*x)^(1/2)/(2+3*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)/(6*x^2 + x - 2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (5 \, x + 3\right )}^{\frac{5}{2}}}{\sqrt{3 \, x + 2} \sqrt{-2 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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