3.2294 \(\int \frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{(2+3 x)^4} \, dx\)

Optimal. Leaf size=149 \[ -\frac{\sqrt{1-2 x} (5 x+3)^{5/2}}{9 (3 x+2)^3}-\frac{59 \sqrt{1-2 x} (5 x+3)^{3/2}}{252 (3 x+2)^2}-\frac{6401 \sqrt{1-2 x} \sqrt{5 x+3}}{10584 (3 x+2)}-\frac{50}{81} \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )-\frac{250433 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{31752 \sqrt{7}} \]

[Out]

(-6401*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(10584*(2 + 3*x)) - (59*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(252*(2 + 3*x)^2) -
(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(9*(2 + 3*x)^3) - (50*Sqrt[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/81 - (250433
*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(31752*Sqrt[7])

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Rubi [A]  time = 0.053743, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {97, 149, 157, 54, 216, 93, 204} \[ -\frac{\sqrt{1-2 x} (5 x+3)^{5/2}}{9 (3 x+2)^3}-\frac{59 \sqrt{1-2 x} (5 x+3)^{3/2}}{252 (3 x+2)^2}-\frac{6401 \sqrt{1-2 x} \sqrt{5 x+3}}{10584 (3 x+2)}-\frac{50}{81} \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )-\frac{250433 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{31752 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(2 + 3*x)^4,x]

[Out]

(-6401*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(10584*(2 + 3*x)) - (59*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(252*(2 + 3*x)^2) -
(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(9*(2 + 3*x)^3) - (50*Sqrt[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/81 - (250433
*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(31752*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 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)^4} \, dx &=-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{9 (2+3 x)^3}+\frac{1}{9} \int \frac{\left (\frac{19}{2}-30 x\right ) (3+5 x)^{3/2}}{\sqrt{1-2 x} (2+3 x)^3} \, dx\\ &=-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{252 (2+3 x)^2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{9 (2+3 x)^3}+\frac{1}{378} \int \frac{\left (\frac{801}{4}-2100 x\right ) \sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)^2} \, dx\\ &=-\frac{6401 \sqrt{1-2 x} \sqrt{3+5 x}}{10584 (2+3 x)}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{252 (2+3 x)^2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{9 (2+3 x)^3}+\frac{\int \frac{-\frac{141567}{8}-73500 x}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{7938}\\ &=-\frac{6401 \sqrt{1-2 x} \sqrt{3+5 x}}{10584 (2+3 x)}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{252 (2+3 x)^2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{9 (2+3 x)^3}-\frac{250}{81} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx+\frac{250433 \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{63504}\\ &=-\frac{6401 \sqrt{1-2 x} \sqrt{3+5 x}}{10584 (2+3 x)}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{252 (2+3 x)^2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{9 (2+3 x)^3}+\frac{250433 \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )}{31752}-\frac{1}{81} \left (100 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )\\ &=-\frac{6401 \sqrt{1-2 x} \sqrt{3+5 x}}{10584 (2+3 x)}-\frac{59 \sqrt{1-2 x} (3+5 x)^{3/2}}{252 (2+3 x)^2}-\frac{\sqrt{1-2 x} (3+5 x)^{5/2}}{9 (2+3 x)^3}-\frac{50}{81} \sqrt{10} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )-\frac{250433 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{31752 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.150884, size = 126, normalized size = 0.85 \[ \frac{21 \sqrt{5 x+3} \left (248358 x^3+194169 x^2-57062 x-51056\right )+137200 \sqrt{10-20 x} (3 x+2)^3 \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-250433 \sqrt{7-14 x} (3 x+2)^3 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{222264 \sqrt{1-2 x} (3 x+2)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(2 + 3*x)^4,x]

[Out]

(21*Sqrt[3 + 5*x]*(-51056 - 57062*x + 194169*x^2 + 248358*x^3) + 137200*Sqrt[10 - 20*x]*(2 + 3*x)^3*ArcSin[Sqr
t[5/11]*Sqrt[1 - 2*x]] - 250433*Sqrt[7 - 14*x]*(2 + 3*x)^3*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(222
264*Sqrt[1 - 2*x]*(2 + 3*x)^3)

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Maple [B]  time = 0.011, size = 253, normalized size = 1.7 \begin{align*} -{\frac{1}{444528\, \left ( 2+3\,x \right ) ^{3}}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( 3704400\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{3}-6761691\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{3}+7408800\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}-13523382\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}+4939200\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-9015588\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+5215518\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+1097600\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -2003464\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +6685308\,x\sqrt{-10\,{x}^{2}-x+3}+2144352\,\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)^4,x)

[Out]

-1/444528*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(3704400*10^(1/2)*arcsin(20/11*x+1/11)*x^3-6761691*7^(1/2)*arctan(1/14*(
37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3+7408800*10^(1/2)*arcsin(20/11*x+1/11)*x^2-13523382*7^(1/2)*arctan(1/
14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+4939200*10^(1/2)*arcsin(20/11*x+1/11)*x-9015588*7^(1/2)*arctan(1
/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+5215518*x^2*(-10*x^2-x+3)^(1/2)+1097600*10^(1/2)*arcsin(20/11*x+1
/11)-2003464*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+6685308*x*(-10*x^2-x+3)^(1/2)+2144352*
(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(2+3*x)^3

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Maxima [A]  time = 1.97487, size = 178, normalized size = 1.19 \begin{align*} -\frac{25}{81} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) + \frac{250433}{444528} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) - \frac{515}{2646} \, \sqrt{-10 \, x^{2} - x + 3} + \frac{{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{63 \,{\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} - \frac{103 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{588 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac{5989 \, \sqrt{-10 \, x^{2} - x + 3}}{10584 \,{\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)^4,x, algorithm="maxima")

[Out]

-25/81*sqrt(10)*arcsin(20/11*x + 1/11) + 250433/444528*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2
)) - 515/2646*sqrt(-10*x^2 - x + 3) + 1/63*(-10*x^2 - x + 3)^(3/2)/(27*x^3 + 54*x^2 + 36*x + 8) - 103/588*(-10
*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 5989/10584*sqrt(-10*x^2 - x + 3)/(3*x + 2)

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Fricas [A]  time = 1.60169, size = 482, normalized size = 3.23 \begin{align*} -\frac{250433 \, \sqrt{7}{\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\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 ) - 137200 \, \sqrt{10}{\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\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 ) + 42 \,{\left (124179 \, x^{2} + 159174 \, x + 51056\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{444528 \,{\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\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)^4,x, algorithm="fricas")

[Out]

-1/444528*(250433*sqrt(7)*(27*x^3 + 54*x^2 + 36*x + 8)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x
+ 1)/(10*x^2 + x - 3)) - 137200*sqrt(10)*(27*x^3 + 54*x^2 + 36*x + 8)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*
x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 42*(124179*x^2 + 159174*x + 51056)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(27
*x^3 + 54*x^2 + 36*x + 8)

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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)**4,x)

[Out]

Timed out

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Giac [B]  time = 3.28873, size = 520, normalized size = 3.49 \begin{align*} \frac{250433}{4445280} \, \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}{81} \, \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{11 \,{\left (6401 \, \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 )}^{5} + 4674880 \, \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} + 1034801600 \, \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 )}}{5292 \,{\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 )}^{3}} \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)^4,x, algorithm="giac")

[Out]

250433/4445280*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)))) - 25/81*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)))) - 11/5292*(64
01*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - s
qrt(22)))^5 + 4674880*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*
sqrt(-10*x + 5) - sqrt(22)))^3 + 1034801600*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*s
qrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*s
qrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^3