3.2499 \(\int \frac{(2+3 x)^3}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{2 \sqrt{1-2 x} (3 x+2)^2}{55 \sqrt{5 x+3}}-\frac{3 \sqrt{1-2 x} \sqrt{5 x+3} (300 x+979)}{4400}+\frac{2493 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{400 \sqrt{10}} \]

[Out]

(-2*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(55*Sqrt[3 + 5*x]) - (3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(979 + 300*x))/4400 + (2493
*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(400*Sqrt[10])

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Rubi [A]  time = 0.019888, 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, 147, 54, 216} \[ -\frac{2 \sqrt{1-2 x} (3 x+2)^2}{55 \sqrt{5 x+3}}-\frac{3 \sqrt{1-2 x} \sqrt{5 x+3} (300 x+979)}{4400}+\frac{2493 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{400 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(55*Sqrt[3 + 5*x]) - (3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(979 + 300*x))/4400 + (2493
*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(400*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 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)^3}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{55 \sqrt{3+5 x}}-\frac{2}{55} \int \frac{\left (-39-\frac{75 x}{2}\right ) (2+3 x)}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{55 \sqrt{3+5 x}}-\frac{3 \sqrt{1-2 x} \sqrt{3+5 x} (979+300 x)}{4400}+\frac{2493}{800} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{55 \sqrt{3+5 x}}-\frac{3 \sqrt{1-2 x} \sqrt{3+5 x} (979+300 x)}{4400}+\frac{2493 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{400 \sqrt{5}}\\ &=-\frac{2 \sqrt{1-2 x} (2+3 x)^2}{55 \sqrt{3+5 x}}-\frac{3 \sqrt{1-2 x} \sqrt{3+5 x} (979+300 x)}{4400}+\frac{2493 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{400 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0294162, size = 64, normalized size = 0.76 \[ \frac{-10 \sqrt{1-2 x} \left (5940 x^2+19305 x+9451\right )-27423 \sqrt{50 x+30} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{44000 \sqrt{5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x]*(9451 + 19305*x + 5940*x^2) - 27423*Sqrt[30 + 50*x]*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(4400
0*Sqrt[3 + 5*x])

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Maple [A]  time = 0.012, size = 99, normalized size = 1.2 \begin{align*}{\frac{1}{88000} \left ( 137115\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-118800\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+82269\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -386100\,x\sqrt{-10\,{x}^{2}-x+3}-189020\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}{\frac{1}{\sqrt{3+5\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/88000*(137115*10^(1/2)*arcsin(20/11*x+1/11)*x-118800*x^2*(-10*x^2-x+3)^(1/2)+82269*10^(1/2)*arcsin(20/11*x+1
/11)-386100*x*(-10*x^2-x+3)^(1/2)-189020*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)

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Maxima [A]  time = 2.97037, size = 88, normalized size = 1.05 \begin{align*} \frac{2493}{8000} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{27}{100} \, \sqrt{-10 \, x^{2} - x + 3} x - \frac{1431}{2000} \, \sqrt{-10 \, x^{2} - x + 3} - \frac{2 \, \sqrt{-10 \, x^{2} - x + 3}}{1375 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2493/8000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 27/100*sqrt(-10*x^2 - x + 3)*x - 1431/2000*sqrt(-10*x^2 - x
+ 3) - 2/1375*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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Fricas [A]  time = 1.82662, size = 254, normalized size = 3.02 \begin{align*} -\frac{27423 \, \sqrt{10}{\left (5 \, x + 3\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 (5940 \, x^{2} + 19305 \, x + 9451\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{88000 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/88000*(27423*sqrt(10)*(5*x + 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x -
3)) + 20*(5940*x^2 + 19305*x + 9451)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.91453, size = 150, normalized size = 1.79 \begin{align*} -\frac{27}{10000} \,{\left (4 \, \sqrt{5}{\left (5 \, x + 3\right )} + 41 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} + \frac{2493}{4000} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{13750 \, \sqrt{5 \, x + 3}} + \frac{2 \, \sqrt{10} \sqrt{5 \, x + 3}}{6875 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/10000*(4*sqrt(5)*(5*x + 3) + 41*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) + 2493/4000*sqrt(10)*arcsin(1/11*sq
rt(22)*sqrt(5*x + 3)) - 1/13750*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 2/6875*sqrt(10)*
sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))