3.2564 \(\int \frac{(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=72 \[ -\frac{1229 \sqrt{1-2 x}}{1210 \sqrt{5 x+3}}+\frac{49}{22 \sqrt{1-2 x} \sqrt{5 x+3}}-\frac{9 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{5 \sqrt{10}} \]

[Out]

49/(22*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) - (1229*Sqrt[1 - 2*x])/(1210*Sqrt[3 + 5*x]) - (9*ArcSin[Sqrt[2/11]*Sqrt[3
+ 5*x]])/(5*Sqrt[10])

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Rubi [A]  time = 0.0156366, antiderivative size = 72, 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 = {89, 78, 54, 216} \[ -\frac{1229 \sqrt{1-2 x}}{1210 \sqrt{5 x+3}}+\frac{49}{22 \sqrt{1-2 x} \sqrt{5 x+3}}-\frac{9 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{5 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

49/(22*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) - (1229*Sqrt[1 - 2*x])/(1210*Sqrt[3 + 5*x]) - (9*ArcSin[Sqrt[2/11]*Sqrt[3
+ 5*x]])/(5*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 78

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

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}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx &=\frac{49}{22 \sqrt{1-2 x} \sqrt{3+5 x}}-\frac{1}{22} \int \frac{-\frac{127}{2}+99 x}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx\\ &=\frac{49}{22 \sqrt{1-2 x} \sqrt{3+5 x}}-\frac{1229 \sqrt{1-2 x}}{1210 \sqrt{3+5 x}}-\frac{9}{10} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=\frac{49}{22 \sqrt{1-2 x} \sqrt{3+5 x}}-\frac{1229 \sqrt{1-2 x}}{1210 \sqrt{3+5 x}}-\frac{9 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{5 \sqrt{5}}\\ &=\frac{49}{22 \sqrt{1-2 x} \sqrt{3+5 x}}-\frac{1229 \sqrt{1-2 x}}{1210 \sqrt{3+5 x}}-\frac{9 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{5 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0355093, size = 65, normalized size = 0.9 \[ \frac{12290 x+1089 \sqrt{10-20 x} \sqrt{5 x+3} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )+7330}{6050 \sqrt{1-2 x} \sqrt{5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7330 + 12290*x + 1089*Sqrt[10 - 20*x]*Sqrt[3 + 5*x]*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6050*Sqrt[1 - 2*x]*Sqr
t[3 + 5*x])

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Maple [A]  time = 0.012, size = 103, normalized size = 1.4 \begin{align*} -{\frac{1}{24200\,x-12100}\sqrt{1-2\,x} \left ( 10890\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}+1089\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-3267\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +24580\,x\sqrt{-10\,{x}^{2}-x+3}+14660\,\sqrt{-10\,{x}^{2}-x+3} \right ){\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)^2/(1-2*x)^(3/2)/(3+5*x)^(3/2),x)

[Out]

-1/12100*(1-2*x)^(1/2)*(10890*10^(1/2)*arcsin(20/11*x+1/11)*x^2+1089*10^(1/2)*arcsin(20/11*x+1/11)*x-3267*10^(
1/2)*arcsin(20/11*x+1/11)+24580*x*(-10*x^2-x+3)^(1/2)+14660*(-10*x^2-x+3)^(1/2))/(2*x-1)/(-10*x^2-x+3)^(1/2)/(
3+5*x)^(1/2)

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Maxima [A]  time = 3.46893, size = 55, normalized size = 0.76 \begin{align*} \frac{9}{100} \, \sqrt{10} \arcsin \left (-\frac{20}{11} \, x - \frac{1}{11}\right ) + \frac{1229 \, x}{605 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{733}{605 \, \sqrt{-10 \, x^{2} - x + 3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/100*sqrt(10)*arcsin(-20/11*x - 1/11) + 1229/605*x/sqrt(-10*x^2 - x + 3) + 733/605/sqrt(-10*x^2 - x + 3)

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Fricas [A]  time = 1.70284, size = 252, normalized size = 3.5 \begin{align*} \frac{1089 \, \sqrt{10}{\left (10 \, x^{2} + 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 (1229 \, x + 733\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{12100 \,{\left (10 \, x^{2} + x - 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12100*(1089*sqrt(10)*(10*x^2 + x - 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 +
x - 3)) - 20*(1229*x + 733)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(10*x^2 + x - 3)

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

[Out]

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

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Giac [B]  time = 2.7436, size = 142, normalized size = 1.97 \begin{align*} -\frac{9}{50} \, \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 )}}{6050 \, \sqrt{5 \, x + 3}} - \frac{49 \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{605 \,{\left (2 \, x - 1\right )}} + \frac{2 \, \sqrt{10} \sqrt{5 \, x + 3}}{3025 \,{\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)^2/(1-2*x)^(3/2)/(3+5*x)^(3/2),x, algorithm="giac")

[Out]

-9/50*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/6050*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt
(5*x + 3) - 49/605*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) + 2/3025*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sq
rt(-10*x + 5) - sqrt(22))