3.2521 \(\int \frac{(2+3 x)^2 \sqrt{3+5 x}}{(1-2 x)^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ \frac{9}{40} \sqrt{1-2 x} (5 x+3)^{3/2}+\frac{49 (5 x+3)^{3/2}}{22 \sqrt{1-2 x}}+\frac{17951 \sqrt{1-2 x} \sqrt{5 x+3}}{1760}-\frac{17951 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{160 \sqrt{10}} \]

[Out]

(17951*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/1760 + (49*(3 + 5*x)^(3/2))/(22*Sqrt[1 - 2*x]) + (9*Sqrt[1 - 2*x]*(3 + 5*x
)^(3/2))/40 - (17951*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(160*Sqrt[10])

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Rubi [A]  time = 0.0241907, antiderivative size = 94, 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 = {89, 80, 50, 54, 216} \[ \frac{9}{40} \sqrt{1-2 x} (5 x+3)^{3/2}+\frac{49 (5 x+3)^{3/2}}{22 \sqrt{1-2 x}}+\frac{17951 \sqrt{1-2 x} \sqrt{5 x+3}}{1760}-\frac{17951 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{160 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(17951*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/1760 + (49*(3 + 5*x)^(3/2))/(22*Sqrt[1 - 2*x]) + (9*Sqrt[1 - 2*x]*(3 + 5*x
)^(3/2))/40 - (17951*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(160*Sqrt[10])

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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)^2 \sqrt{3+5 x}}{(1-2 x)^{3/2}} \, dx &=\frac{49 (3+5 x)^{3/2}}{22 \sqrt{1-2 x}}-\frac{1}{22} \int \frac{\sqrt{3+5 x} \left (\frac{853}{2}+99 x\right )}{\sqrt{1-2 x}} \, dx\\ &=\frac{49 (3+5 x)^{3/2}}{22 \sqrt{1-2 x}}+\frac{9}{40} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{17951}{880} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x}} \, dx\\ &=\frac{17951 \sqrt{1-2 x} \sqrt{3+5 x}}{1760}+\frac{49 (3+5 x)^{3/2}}{22 \sqrt{1-2 x}}+\frac{9}{40} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{17951}{320} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=\frac{17951 \sqrt{1-2 x} \sqrt{3+5 x}}{1760}+\frac{49 (3+5 x)^{3/2}}{22 \sqrt{1-2 x}}+\frac{9}{40} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{17951 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{160 \sqrt{5}}\\ &=\frac{17951 \sqrt{1-2 x} \sqrt{3+5 x}}{1760}+\frac{49 (3+5 x)^{3/2}}{22 \sqrt{1-2 x}}+\frac{9}{40} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{17951 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{160 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0294591, size = 64, normalized size = 0.68 \[ \frac{17951 \sqrt{10-20 x} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-10 \sqrt{5 x+3} \left (360 x^2+1518 x-2809\right )}{1600 \sqrt{1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[3 + 5*x]*(-2809 + 1518*x + 360*x^2) + 17951*Sqrt[10 - 20*x]*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(1600*
Sqrt[1 - 2*x])

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Maple [A]  time = 0.009, size = 106, normalized size = 1.1 \begin{align*} -{\frac{1}{6400\,x-3200} \left ( 35902\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-7200\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-17951\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -30360\,x\sqrt{-10\,{x}^{2}-x+3}+56180\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}\sqrt{3+5\,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)^2*(3+5*x)^(1/2)/(1-2*x)^(3/2),x)

[Out]

-1/3200*(35902*10^(1/2)*arcsin(20/11*x+1/11)*x-7200*x^2*(-10*x^2-x+3)^(1/2)-17951*10^(1/2)*arcsin(20/11*x+1/11
)-30360*x*(-10*x^2-x+3)^(1/2)+56180*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2*x-1)/(-10*x^2-x+3)^(1/
2)

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Maxima [A]  time = 2.57909, size = 88, normalized size = 0.94 \begin{align*} -\frac{17951}{3200} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) + \frac{9}{8} \, \sqrt{-10 \, x^{2} - x + 3} x + \frac{849}{160} \, \sqrt{-10 \, x^{2} - x + 3} - \frac{49 \, \sqrt{-10 \, x^{2} - x + 3}}{4 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17951/3200*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 9/8*sqrt(-10*x^2 - x + 3)*x + 849/160*sqrt(-10*x^2 - x +
3) - 49/4*sqrt(-10*x^2 - x + 3)/(2*x - 1)

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Fricas [A]  time = 1.85273, size = 248, normalized size = 2.64 \begin{align*} \frac{17951 \, \sqrt{10}{\left (2 \, 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 (360 \, x^{2} + 1518 \, x - 2809\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{3200 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3200*(17951*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)
) + 20*(360*x^2 + 1518*x - 2809)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.67173, size = 96, normalized size = 1.02 \begin{align*} -\frac{17951}{1600} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) + \frac{{\left (6 \,{\left (12 \, \sqrt{5}{\left (5 \, x + 3\right )} + 181 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} - 17951 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{4000 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17951/1600*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/4000*(6*(12*sqrt(5)*(5*x + 3) + 181*sqrt(5))*(5*x
+ 3) - 17951*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)