3.2482 \(\int \frac{(3+5 x)^{5/2}}{\sqrt{1-2 x} (2+3 x)} \, dx\)

Optimal. Leaf size=108 \[ -\frac{5}{12} \sqrt{1-2 x} (5 x+3)^{3/2}-\frac{455}{144} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{3035}{432} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )+\frac{2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{27 \sqrt{7}} \]

[Out]

(-455*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/144 - (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/12 + (3035*Sqrt[5/2]*ArcSin[Sqrt[2/
11]*Sqrt[3 + 5*x]])/432 + (2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(27*Sqrt[7])

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Rubi [A]  time = 0.0414162, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {102, 154, 157, 54, 216, 93, 204} \[ -\frac{5}{12} \sqrt{1-2 x} (5 x+3)^{3/2}-\frac{455}{144} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{3035}{432} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )+\frac{2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{27 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-455*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/144 - (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/12 + (3035*Sqrt[5/2]*ArcSin[Sqrt[2/
11]*Sqrt[3 + 5*x]])/432 + (2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(27*Sqrt[7])

Rule 102

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

Rule 154

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

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(3+5 x)^{5/2}}{\sqrt{1-2 x} (2+3 x)} \, dx &=-\frac{5}{12} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{1}{12} \int \frac{\left (-153-\frac{455 x}{2}\right ) \sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=-\frac{455}{144} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{5}{12} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{1}{72} \int \frac{\frac{5053}{2}+\frac{15175 x}{4}}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{455}{144} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{5}{12} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{1}{27} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx+\frac{15175}{864} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{455}{144} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{5}{12} \sqrt{1-2 x} (3+5 x)^{3/2}-\frac{2}{27} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )+\frac{1}{432} \left (3035 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )\\ &=-\frac{455}{144} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{5}{12} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{3035}{432} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )+\frac{2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{27 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0405747, size = 87, normalized size = 0.81 \[ \frac{-210 \sqrt{1-2 x} \sqrt{5 x+3} (60 x+127)-21245 \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )+64 \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{6048} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-210*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(127 + 60*x) - 21245*Sqrt[10]*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]] + 64*Sqrt[7]*
ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/6048

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Maple [A]  time = 0.012, size = 98, normalized size = 0.9 \begin{align*}{\frac{1}{12096}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( 21245\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -64\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) -25200\,x\sqrt{-10\,{x}^{2}-x+3}-53340\,\sqrt{-10\,{x}^{2}-x+3} \right ){\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12096*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(21245*10^(1/2)*arcsin(20/11*x+1/11)-64*7^(1/2)*arctan(1/14*(37*x+20)*7^(1
/2)/(-10*x^2-x+3)^(1/2))-25200*x*(-10*x^2-x+3)^(1/2)-53340*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)

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Maxima [A]  time = 3.96018, size = 93, normalized size = 0.86 \begin{align*} -\frac{25}{12} \, \sqrt{-10 \, x^{2} - x + 3} x + \frac{3035}{1728} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{1}{189} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) - \frac{635}{144} \, \sqrt{-10 \, x^{2} - x + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/12*sqrt(-10*x^2 - x + 3)*x + 3035/1728*sqrt(10)*arcsin(20/11*x + 1/11) - 1/189*sqrt(7)*arcsin(37/11*x/abs(
3*x + 2) + 20/11/abs(3*x + 2)) - 635/144*sqrt(-10*x^2 - x + 3)

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Fricas [A]  time = 1.78748, size = 350, normalized size = 3.24 \begin{align*} -\frac{5}{144} \,{\left (60 \, x + 127\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1} - \frac{3035}{1728} \, \sqrt{5} \sqrt{2} \arctan \left (\frac{\sqrt{5} \sqrt{2}{\left (20 \, x + 1\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{20 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) + \frac{1}{189} \, \sqrt{7} \arctan \left (\frac{\sqrt{7}{\left (37 \, x + 20\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{14 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/144*(60*x + 127)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 3035/1728*sqrt(5)*sqrt(2)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x
+ 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 1/189*sqrt(7)*arctan(1/14*sqrt(7)*(37*x + 20)*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 (5 x + 3\right )^{\frac{5}{2}}}{\sqrt{1 - 2 x} \left (3 x + 2\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.69298, size = 234, normalized size = 2.17 \begin{align*} -\frac{1}{1890} \, \sqrt{70} \sqrt{10}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{70} \sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} - \frac{1}{144} \,{\left (12 \, \sqrt{5}{\left (5 \, x + 3\right )} + 91 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} + \frac{3035}{1728} \, \sqrt{10}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1890*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2
/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 1/144*(12*sqrt(5)*(5*x + 3) + 91*sqrt(5))*sqrt(5*x +
3)*sqrt(-10*x + 5) + 3035/1728*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22)
)^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))))