3.2870 \(\int \frac{\sqrt{2+3 x}}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=81 \[ \frac{2 \sqrt{\frac{7}{5}} \sqrt{-5 x-3} E\left (\sin ^{-1}\left (\sqrt{5} \sqrt{3 x+2}\right )|\frac{2}{35}\right )}{11 \sqrt{5 x+3}}-\frac{2 \sqrt{1-2 x} \sqrt{3 x+2}}{11 \sqrt{5 x+3}} \]

[Out]

(-2*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/(11*Sqrt[3 + 5*x]) + (2*Sqrt[7/5]*Sqrt[-3 - 5*x]*EllipticE[ArcSin[Sqrt[5]*Sqr
t[2 + 3*x]], 2/35])/(11*Sqrt[3 + 5*x])

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Rubi [A]  time = 0.0247858, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {99, 21, 114, 113} \[ \frac{2 \sqrt{\frac{7}{5}} \sqrt{-5 x-3} E\left (\sin ^{-1}\left (\sqrt{5} \sqrt{3 x+2}\right )|\frac{2}{35}\right )}{11 \sqrt{5 x+3}}-\frac{2 \sqrt{1-2 x} \sqrt{3 x+2}}{11 \sqrt{5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/(11*Sqrt[3 + 5*x]) + (2*Sqrt[7/5]*Sqrt[-3 - 5*x]*EllipticE[ArcSin[Sqrt[5]*Sqr
t[2 + 3*x]], 2/35])/(11*Sqrt[3 + 5*x])

Rule 99

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 114

Int[Sqrt[(e_.) + (f_.)*(x_)]/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*
x]*Sqrt[(b*(c + d*x))/(b*c - a*d)])/(Sqrt[c + d*x]*Sqrt[(b*(e + f*x))/(b*e - a*f)]), Int[Sqrt[(b*e)/(b*e - a*f
) + (b*f*x)/(b*e - a*f)]/(Sqrt[a + b*x]*Sqrt[(b*c)/(b*c - a*d) + (b*d*x)/(b*c - a*d)]), x], 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]

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])

Rubi steps

\begin{align*} \int \frac{\sqrt{2+3 x}}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx &=-\frac{2 \sqrt{1-2 x} \sqrt{2+3 x}}{11 \sqrt{3+5 x}}+\frac{2}{11} \int \frac{\frac{3}{2}-3 x}{\sqrt{1-2 x} \sqrt{2+3 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{2 \sqrt{1-2 x} \sqrt{2+3 x}}{11 \sqrt{3+5 x}}+\frac{3}{11} \int \frac{\sqrt{1-2 x}}{\sqrt{2+3 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{2 \sqrt{1-2 x} \sqrt{2+3 x}}{11 \sqrt{3+5 x}}+\frac{\left (3 \sqrt{7} \sqrt{-3-5 x}\right ) \int \frac{\sqrt{\frac{3}{7}-\frac{6 x}{7}}}{\sqrt{-9-15 x} \sqrt{2+3 x}} \, dx}{11 \sqrt{3+5 x}}\\ &=-\frac{2 \sqrt{1-2 x} \sqrt{2+3 x}}{11 \sqrt{3+5 x}}+\frac{2 \sqrt{\frac{7}{5}} \sqrt{-3-5 x} E\left (\sin ^{-1}\left (\sqrt{5} \sqrt{2+3 x}\right )|\frac{2}{35}\right )}{11 \sqrt{3+5 x}}\\ \end{align*}

Mathematica [C]  time = 0.0848123, size = 61, normalized size = 0.75 \[ \frac{2}{55} \left (-\frac{5 \sqrt{1-2 x} \sqrt{3 x+2}}{\sqrt{5 x+3}}-i \sqrt{33} E\left (i \sinh ^{-1}\left (\sqrt{15 x+9}\right )|-\frac{2}{33}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-5*Sqrt[1 - 2*x]*Sqrt[2 + 3*x])/Sqrt[3 + 5*x] - I*Sqrt[33]*EllipticE[I*ArcSinh[Sqrt[9 + 15*x]], -2/33]))/
55

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Maple [C]  time = 0.017, size = 135, normalized size = 1.7 \begin{align*} -{\frac{1}{1650\,{x}^{3}+1265\,{x}^{2}-385\,x-330}\sqrt{1-2\,x}\sqrt{2+3\,x}\sqrt{3+5\,x} \left ( 35\,\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 ) -2\,\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 ) +60\,{x}^{2}+10\,x-20 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/55*(2+3*x)^(1/2)*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(35*2^(1/2)*(3+5*x)^(1/2)*(2+3*x)^(1/2)*(1-2*x)^(1/2)*Elliptic
F(1/11*(66+110*x)^(1/2),1/2*I*66^(1/2))-2*2^(1/2)*(3+5*x)^(1/2)*(2+3*x)^(1/2)*(1-2*x)^(1/2)*EllipticE(1/11*(66
+110*x)^(1/2),1/2*I*66^(1/2))+60*x^2+10*x-20)/(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{\sqrt{3 \, x + 2}}{{\left (5 \, x + 3\right )}^{\frac{3}{2}} \sqrt{-2 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{5 \, x + 3} \sqrt{3 \, x + 2} \sqrt{-2 \, x + 1}}{50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(5*x + 3)*sqrt(3*x + 2)*sqrt(-2*x + 1)/(50*x^3 + 35*x^2 - 12*x - 9), 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((2+3*x)**(1/2)/(3+5*x)**(3/2)/(1-2*x)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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