3.2293 \(\int \frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\sqrt{1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}-\frac{59 \sqrt{1-2 x} (5 x+3)^{3/2}}{84 (3 x+2)}+\frac{215}{84} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{25}{9} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )+\frac{2119 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{252 \sqrt{7}} \]

[Out]

(215*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/84 - (59*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(84*(2 + 3*x)) - (Sqrt[1 - 2*x]*(3 +
5*x)^(5/2))/(6*(2 + 3*x)^2) + (25*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/9 + (2119*ArcTan[Sqrt[1 - 2*x]/
(Sqrt[7]*Sqrt[3 + 5*x])])/(252*Sqrt[7])

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Rubi [A]  time = 0.0536854, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {97, 149, 154, 157, 54, 216, 93, 204} \[ -\frac{\sqrt{1-2 x} (5 x+3)^{5/2}}{6 (3 x+2)^2}-\frac{59 \sqrt{1-2 x} (5 x+3)^{3/2}}{84 (3 x+2)}+\frac{215}{84} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{25}{9} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )+\frac{2119 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{252 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(215*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/84 - (59*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(84*(2 + 3*x)) - (Sqrt[1 - 2*x]*(3 +
5*x)^(5/2))/(6*(2 + 3*x)^2) + (25*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/9 + (2119*ArcTan[Sqrt[1 - 2*x]/
(Sqrt[7]*Sqrt[3 + 5*x])])/(252*Sqrt[7])

Rule 97

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

Rule 149

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

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{\sqrt{1-2 x} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx &=-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac{1}{6} \int \frac{\left (\frac{19}{2}-30 x\right ) (3+5 x)^{3/2}}{\sqrt{1-2 x} (2+3 x)^2} \, dx\\ &=-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{84 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac{1}{126} \int \frac{\left (\frac{1197}{4}-1935 x\right ) \sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=\frac{215}{84} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{84 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{6 (2+3 x)^2}-\frac{1}{756} \int \frac{-\frac{14643}{2}-15750 x}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=\frac{215}{84} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{84 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{6 (2+3 x)^2}-\frac{2119}{504} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx+\frac{125}{18} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=\frac{215}{84} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{84 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{6 (2+3 x)^2}-\frac{2119}{252} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )+\frac{1}{9} \left (25 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )\\ &=\frac{215}{84} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{84 (2+3 x)}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac{25}{9} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )+\frac{2119 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{252 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.13012, size = 126, normalized size = 0.88 \[ \frac{21 \sqrt{5 x+3} \left (-1400 x^3-1378 x^2+279 x+380\right )-2450 \sqrt{10-20 x} (3 x+2)^2 \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )+2119 \sqrt{7-14 x} (3 x+2)^2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{1764 \sqrt{1-2 x} (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*Sqrt[3 + 5*x]*(380 + 279*x - 1378*x^2 - 1400*x^3) - 2450*Sqrt[10 - 20*x]*(2 + 3*x)^2*ArcSin[Sqrt[5/11]*Sqr
t[1 - 2*x]] + 2119*Sqrt[7 - 14*x]*(2 + 3*x)^2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1764*Sqrt[1 - 2*
x]*(2 + 3*x)^2)

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Maple [A]  time = 0.01, size = 208, normalized size = 1.4 \begin{align*} -{\frac{1}{3528\, \left ( 2+3\,x \right ) ^{2}}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( 19071\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-22050\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}+25428\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x-29400\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-29400\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+8476\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) -9800\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -43638\,x\sqrt{-10\,{x}^{2}-x+3}-15960\,\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)*(1-2*x)^(1/2)/(2+3*x)^3,x)

[Out]

-1/3528*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(19071*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-2205
0*10^(1/2)*arcsin(20/11*x+1/11)*x^2+25428*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x-29400*1
0^(1/2)*arcsin(20/11*x+1/11)*x-29400*x^2*(-10*x^2-x+3)^(1/2)+8476*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x
^2-x+3)^(1/2))-9800*10^(1/2)*arcsin(20/11*x+1/11)-43638*x*(-10*x^2-x+3)^(1/2)-15960*(-10*x^2-x+3)^(1/2))/(-10*
x^2-x+3)^(1/2)/(2+3*x)^2

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Maxima [A]  time = 2.07426, size = 136, normalized size = 0.94 \begin{align*} \frac{25}{36} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{2119}{3528} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) + \frac{20}{21} \, \sqrt{-10 \, x^{2} - x + 3} + \frac{{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{42 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac{9 \, \sqrt{-10 \, x^{2} - x + 3}}{28 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/36*sqrt(10)*arcsin(20/11*x + 1/11) - 2119/3528*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) +
20/21*sqrt(-10*x^2 - x + 3) + 1/42*(-10*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) + 9/28*sqrt(-10*x^2 - x + 3)/(3*
x + 2)

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Fricas [A]  time = 1.62548, size = 443, normalized size = 3.08 \begin{align*} -\frac{2450 \, \sqrt{5} \sqrt{2}{\left (9 \, x^{2} + 12 \, x + 4\right )} \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 ) - 2119 \, \sqrt{7}{\left (9 \, x^{2} + 12 \, x + 4\right )} \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 ) - 42 \,{\left (700 \, x^{2} + 1039 \, x + 380\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{3528 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3528*(2450*sqrt(5)*sqrt(2)*(9*x^2 + 12*x + 4)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*
x + 1)/(10*x^2 + x - 3)) - 2119*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(
-2*x + 1)/(10*x^2 + x - 3)) - 42*(700*x^2 + 1039*x + 380)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 2.39748, size = 463, normalized size = 3.22 \begin{align*} -\frac{2119}{35280} \, \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{25}{36} \, \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 )} + \frac{5}{27} \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} + \frac{11 \,{\left (247 \, \sqrt{10}{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}^{3} + 87640 \, \sqrt{10}{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}\right )}}{378 \,{\left ({\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}^{2} + 280\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2119/35280*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)))) + 25/36*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)))) + 5/27*sqrt(5)*s
qrt(5*x + 3)*sqrt(-10*x + 5) + 11/378*(247*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sq
rt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 87640*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqr
t(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqr
t(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2