3.2610 \(\int \frac{(2+3 x)^3}{(1-2 x)^{5/2} \sqrt{3+5 x}} \, dx\)

Optimal. Leaf size=84 \[ \frac{7 \sqrt{5 x+3} (3 x+2)^2}{33 (1-2 x)^{3/2}}-\frac{(95621-33462 x) \sqrt{5 x+3}}{14520 \sqrt{1-2 x}}+\frac{1593 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{40 \sqrt{10}} \]

[Out]

-((95621 - 33462*x)*Sqrt[3 + 5*x])/(14520*Sqrt[1 - 2*x]) + (7*(2 + 3*x)^2*Sqrt[3 + 5*x])/(33*(1 - 2*x)^(3/2))
+ (1593*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(40*Sqrt[10])

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Rubi [A]  time = 0.0204362, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {98, 143, 54, 216} \[ \frac{7 \sqrt{5 x+3} (3 x+2)^2}{33 (1-2 x)^{3/2}}-\frac{(95621-33462 x) \sqrt{5 x+3}}{14520 \sqrt{1-2 x}}+\frac{1593 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{40 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((95621 - 33462*x)*Sqrt[3 + 5*x])/(14520*Sqrt[1 - 2*x]) + (7*(2 + 3*x)^2*Sqrt[3 + 5*x])/(33*(1 - 2*x)^(3/2))
+ (1593*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(40*Sqrt[10])

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 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{(2+3 x)^3}{(1-2 x)^{5/2} \sqrt{3+5 x}} \, dx &=\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}-\frac{1}{33} \int \frac{(2+3 x) \left (155+\frac{507 x}{2}\right )}{(1-2 x)^{3/2} \sqrt{3+5 x}} \, dx\\ &=-\frac{(95621-33462 x) \sqrt{3+5 x}}{14520 \sqrt{1-2 x}}+\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}+\frac{1593}{80} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{(95621-33462 x) \sqrt{3+5 x}}{14520 \sqrt{1-2 x}}+\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}+\frac{1593 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{40 \sqrt{5}}\\ &=-\frac{(95621-33462 x) \sqrt{3+5 x}}{14520 \sqrt{1-2 x}}+\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}+\frac{1593 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{40 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0559079, size = 69, normalized size = 0.82 \[ \frac{578259 \sqrt{10-20 x} (2 x-1) \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-10 \sqrt{5 x+3} \left (39204 x^2-261664 x+83301\right )}{145200 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[3 + 5*x]*(83301 - 261664*x + 39204*x^2) + 578259*Sqrt[10 - 20*x]*(-1 + 2*x)*ArcSin[Sqrt[5/11]*Sqrt[1
- 2*x]])/(145200*(1 - 2*x)^(3/2))

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Maple [A]  time = 0.013, size = 120, normalized size = 1.4 \begin{align*}{\frac{1}{290400\, \left ( 2\,x-1 \right ) ^{2}} \left ( 2313036\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}-2313036\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-784080\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+578259\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +5233280\,x\sqrt{-10\,{x}^{2}-x+3}-1666020\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{3+5\,x}\sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/290400*(2313036*10^(1/2)*arcsin(20/11*x+1/11)*x^2-2313036*10^(1/2)*arcsin(20/11*x+1/11)*x-784080*x^2*(-10*x^
2-x+3)^(1/2)+578259*10^(1/2)*arcsin(20/11*x+1/11)+5233280*x*(-10*x^2-x+3)^(1/2)-1666020*(-10*x^2-x+3)^(1/2))*(
3+5*x)^(1/2)*(1-2*x)^(1/2)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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Maxima [A]  time = 3.13333, size = 103, normalized size = 1.23 \begin{align*} \frac{1593}{800} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{27}{40} \, \sqrt{-10 \, x^{2} - x + 3} + \frac{343 \, \sqrt{-10 \, x^{2} - x + 3}}{132 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac{11123 \, \sqrt{-10 \, x^{2} - x + 3}}{1452 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1593/800*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 27/40*sqrt(-10*x^2 - x + 3) + 343/132*sqrt(-10*x^2 - x + 3)/
(4*x^2 - 4*x + 1) + 11123/1452*sqrt(-10*x^2 - x + 3)/(2*x - 1)

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Fricas [A]  time = 1.53956, size = 282, normalized size = 3.36 \begin{align*} -\frac{578259 \, \sqrt{10}{\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac{\sqrt{10}{\left (20 \, x + 1\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{20 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \,{\left (39204 \, x^{2} - 261664 \, x + 83301\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{290400 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/290400*(578259*sqrt(10)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*
x^2 + x - 3)) + 20*(39204*x^2 - 261664*x + 83301)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.90274, size = 96, normalized size = 1.14 \begin{align*} \frac{1593}{400} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{{\left (4 \,{\left (9801 \, \sqrt{5}{\left (5 \, x + 3\right )} - 385886 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} + 6360321 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{1815000 \,{\left (2 \, x - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1593/400*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/1815000*(4*(9801*sqrt(5)*(5*x + 3) - 385886*sqrt(5))
*(5*x + 3) + 6360321*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2