3.2552 \(\int \frac{(2+3 x)^3}{(1-2 x)^{3/2} \sqrt{3+5 x}} \, dx\)

Optimal. Leaf size=84 \[ \frac{7 \sqrt{5 x+3} (3 x+2)^2}{11 \sqrt{1-2 x}}+\frac{3 \sqrt{1-2 x} \sqrt{5 x+3} (10380 x+25003)}{8800}-\frac{56421 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{800 \sqrt{10}} \]

[Out]

(7*(2 + 3*x)^2*Sqrt[3 + 5*x])/(11*Sqrt[1 - 2*x]) + (3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(25003 + 10380*x))/8800 - (5
6421*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(800*Sqrt[10])

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Rubi [A]  time = 0.0201763, 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} \[ \frac{7 \sqrt{5 x+3} (3 x+2)^2}{11 \sqrt{1-2 x}}+\frac{3 \sqrt{1-2 x} \sqrt{5 x+3} (10380 x+25003)}{8800}-\frac{56421 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{800 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2*Sqrt[3 + 5*x])/(11*Sqrt[1 - 2*x]) + (3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(25003 + 10380*x))/8800 - (5
6421*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(800*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}{(1-2 x)^{3/2} \sqrt{3+5 x}} \, dx &=\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{11 \sqrt{1-2 x}}-\frac{1}{11} \int \frac{(2+3 x) \left (159+\frac{519 x}{2}\right )}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{11 \sqrt{1-2 x}}+\frac{3 \sqrt{1-2 x} \sqrt{3+5 x} (25003+10380 x)}{8800}-\frac{56421 \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx}{1600}\\ &=\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{11 \sqrt{1-2 x}}+\frac{3 \sqrt{1-2 x} \sqrt{3+5 x} (25003+10380 x)}{8800}-\frac{56421 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{800 \sqrt{5}}\\ &=\frac{7 (2+3 x)^2 \sqrt{3+5 x}}{11 \sqrt{1-2 x}}+\frac{3 \sqrt{1-2 x} \sqrt{3+5 x} (25003+10380 x)}{8800}-\frac{56421 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{800 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0276325, size = 64, normalized size = 0.76 \[ \frac{620631 \sqrt{10-20 x} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-10 \sqrt{5 x+3} \left (11880 x^2+51678 x-97409\right )}{88000 \sqrt{1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[3 + 5*x]*(-97409 + 51678*x + 11880*x^2) + 620631*Sqrt[10 - 20*x]*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(
88000*Sqrt[1 - 2*x])

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Maple [A]  time = 0.012, size = 106, normalized size = 1.3 \begin{align*} -{\frac{1}{352000\,x-176000} \left ( 1241262\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-237600\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-620631\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -1033560\,x\sqrt{-10\,{x}^{2}-x+3}+1948180\,\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)^(3/2)/(3+5*x)^(1/2),x)

[Out]

-1/176000*(1241262*10^(1/2)*arcsin(20/11*x+1/11)*x-237600*x^2*(-10*x^2-x+3)^(1/2)-620631*10^(1/2)*arcsin(20/11
*x+1/11)-1033560*x*(-10*x^2-x+3)^(1/2)+1948180*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)*(1-2*x)^(1/2)/(2*x-1)/(-10*x
^2-x+3)^(1/2)

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Maxima [A]  time = 2.07783, size = 88, normalized size = 1.05 \begin{align*} -\frac{56421}{16000} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) + \frac{27}{40} \, \sqrt{-10 \, x^{2} - x + 3} x + \frac{2619}{800} \, \sqrt{-10 \, x^{2} - x + 3} - \frac{343 \, \sqrt{-10 \, x^{2} - x + 3}}{44 \,{\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)^(3/2)/(3+5*x)^(1/2),x, algorithm="maxima")

[Out]

-56421/16000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 27/40*sqrt(-10*x^2 - x + 3)*x + 2619/800*sqrt(-10*x^2 -
x + 3) - 343/44*sqrt(-10*x^2 - x + 3)/(2*x - 1)

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Fricas [A]  time = 1.76261, size = 258, normalized size = 3.07 \begin{align*} \frac{620631 \, \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 (11880 \, x^{2} + 51678 \, x - 97409\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{176000 \,{\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)^(3/2)/(3+5*x)^(1/2),x, algorithm="fricas")

[Out]

1/176000*(620631*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*(11880*x^2 + 51678*x - 97409)*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 )^{3}}{\left (1 - 2 x\right )^{\frac{3}{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)**(3/2)/(3+5*x)**(1/2),x)

[Out]

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

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Giac [A]  time = 2.20922, size = 96, normalized size = 1.14 \begin{align*} -\frac{56421}{8000} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) + \frac{{\left (594 \,{\left (4 \, \sqrt{5}{\left (5 \, x + 3\right )} + 63 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} - 620695 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{220000 \,{\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)^(3/2)/(3+5*x)^(1/2),x, algorithm="giac")

[Out]

-56421/8000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/220000*(594*(4*sqrt(5)*(5*x + 3) + 63*sqrt(5))*(5
*x + 3) - 620695*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)