3.2609 \(\int \frac{(2+3 x)^4}{(1-2 x)^{5/2} \sqrt{3+5 x}} \, dx\)

Optimal. Leaf size=113 \[ \frac{7 \sqrt{5 x+3} (3 x+2)^3}{33 (1-2 x)^{3/2}}-\frac{1589 \sqrt{5 x+3} (3 x+2)^2}{726 \sqrt{1-2 x}}-\frac{\sqrt{1-2 x} \sqrt{5 x+3} (2380020 x+5735477)}{193600}+\frac{392283 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{1600 \sqrt{10}} \]

[Out]

(-1589*(2 + 3*x)^2*Sqrt[3 + 5*x])/(726*Sqrt[1 - 2*x]) + (7*(2 + 3*x)^3*Sqrt[3 + 5*x])/(33*(1 - 2*x)^(3/2)) - (
Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(5735477 + 2380020*x))/193600 + (392283*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(1600*Sq
rt[10])

________________________________________________________________________________________

Rubi [A]  time = 0.0290464, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {98, 150, 147, 54, 216} \[ \frac{7 \sqrt{5 x+3} (3 x+2)^3}{33 (1-2 x)^{3/2}}-\frac{1589 \sqrt{5 x+3} (3 x+2)^2}{726 \sqrt{1-2 x}}-\frac{\sqrt{1-2 x} \sqrt{5 x+3} (2380020 x+5735477)}{193600}+\frac{392283 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{1600 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^4/((1 - 2*x)^(5/2)*Sqrt[3 + 5*x]),x]

[Out]

(-1589*(2 + 3*x)^2*Sqrt[3 + 5*x])/(726*Sqrt[1 - 2*x]) + (7*(2 + 3*x)^3*Sqrt[3 + 5*x])/(33*(1 - 2*x)^(3/2)) - (
Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(5735477 + 2380020*x))/193600 + (392283*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(1600*Sq
rt[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 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 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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)^4}{(1-2 x)^{5/2} \sqrt{3+5 x}} \, dx &=\frac{7 (2+3 x)^3 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}-\frac{1}{33} \int \frac{(2+3 x)^2 \left (218+\frac{717 x}{2}\right )}{(1-2 x)^{3/2} \sqrt{3+5 x}} \, dx\\ &=-\frac{1589 (2+3 x)^2 \sqrt{3+5 x}}{726 \sqrt{1-2 x}}+\frac{7 (2+3 x)^3 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}-\frac{1}{363} \int \frac{\left (-\frac{36489}{2}-\frac{119001 x}{4}\right ) (2+3 x)}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{1589 (2+3 x)^2 \sqrt{3+5 x}}{726 \sqrt{1-2 x}}+\frac{7 (2+3 x)^3 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}-\frac{\sqrt{1-2 x} \sqrt{3+5 x} (5735477+2380020 x)}{193600}+\frac{392283 \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx}{3200}\\ &=-\frac{1589 (2+3 x)^2 \sqrt{3+5 x}}{726 \sqrt{1-2 x}}+\frac{7 (2+3 x)^3 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}-\frac{\sqrt{1-2 x} \sqrt{3+5 x} (5735477+2380020 x)}{193600}+\frac{392283 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{1600 \sqrt{5}}\\ &=-\frac{1589 (2+3 x)^2 \sqrt{3+5 x}}{726 \sqrt{1-2 x}}+\frac{7 (2+3 x)^3 \sqrt{3+5 x}}{33 (1-2 x)^{3/2}}-\frac{\sqrt{1-2 x} \sqrt{3+5 x} (5735477+2380020 x)}{193600}+\frac{392283 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{1600 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.108697, size = 74, normalized size = 0.65 \[ \frac{142398729 \sqrt{10-20 x} (2 x-1) \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-10 \sqrt{5 x+3} \left (2352240 x^3+14544684 x^2-61036064 x+21305631\right )}{5808000 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^4/((1 - 2*x)^(5/2)*Sqrt[3 + 5*x]),x]

[Out]

(-10*Sqrt[3 + 5*x]*(21305631 - 61036064*x + 14544684*x^2 + 2352240*x^3) + 142398729*Sqrt[10 - 20*x]*(-1 + 2*x)
*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5808000*(1 - 2*x)^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 137, normalized size = 1.2 \begin{align*}{\frac{1}{11616000\, \left ( 2\,x-1 \right ) ^{2}} \left ( 569594916\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}-47044800\,{x}^{3}\sqrt{-10\,{x}^{2}-x+3}-569594916\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-290893680\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}[phone]\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) [phone]\,x\sqrt{-10\,{x}^{2}-x+3}-426112620\,\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)^4/(1-2*x)^(5/2)/(3+5*x)^(1/2),x)

[Out]

1/11616000*(569594916*10^(1/2)*arcsin(20/11*x+1/11)*x^2-47044800*x^3*(-10*x^2-x+3)^(1/2)-569594916*10^(1/2)*ar
csin(20/11*x+1/11)*x-290893680*x^2*(-10*x^2-x+3)^(1/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)[phone]*x*(-
10*x^2-x+3)^(1/2)-426112620*(-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)

________________________________________________________________________________________

Maxima [A]  time = 1.6482, size = 123, normalized size = 1.09 \begin{align*} \frac{392283}{32000} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{81}{80} \, \sqrt{-10 \, x^{2} - x + 3} x - \frac{11637}{1600} \, \sqrt{-10 \, x^{2} - x + 3} + \frac{2401 \, \sqrt{-10 \, x^{2} - x + 3}}{264 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac{55909 \, \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)^4/(1-2*x)^(5/2)/(3+5*x)^(1/2),x, algorithm="maxima")

[Out]

392283/32000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 81/80*sqrt(-10*x^2 - x + 3)*x - 11637/1600*sqrt(-10*x^2
- x + 3) + 2401/264*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1) + 55909/1452*sqrt(-10*x^2 - x + 3)/(2*x - 1)

________________________________________________________________________________________

Fricas [A]  time = 1.54287, size = 319, normalized size = 2.82 \begin{align*} -\frac{142398729 \, \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 (2352240 \, x^{3} + 14544684 \, x^{2} - 61036064 \, x + 21305631\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{11616000 \,{\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)^4/(1-2*x)^(5/2)/(3+5*x)^(1/2),x, algorithm="fricas")

[Out]

-1/11616000*(142398729*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*(2352240*x^3 + 14544684*x^2 - 61036064*x + 21305631)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*
x^2 - 4*x + 1)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (3 x + 2\right )^{4}}{\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)**4/(1-2*x)**(5/2)/(3+5*x)**(1/2),x)

[Out]

Integral((3*x + 2)**4/((1 - 2*x)**(5/2)*sqrt(5*x + 3)), x)

________________________________________________________________________________________

Giac [A]  time = 2.76373, size = 113, normalized size = 1. \begin{align*} \frac{392283}{16000} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{{\left (4 \,{\left (9801 \,{\left (12 \, \sqrt{5}{\left (5 \, x + 3\right )} + 263 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} - 94936582 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} + 1566381795 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{72600000 \,{\left (2 \, x - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x)^(1/2),x, algorithm="giac")

[Out]

392283/16000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/72600000*(4*(9801*(12*sqrt(5)*(5*x + 3) + 263*sq
rt(5))*(5*x + 3) - 94936582*sqrt(5))*(5*x + 3) + 1566381795*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2