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

Optimal. Leaf size=120 \[ -\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{252 (3 x+2)}-\frac {10}{27} \sqrt {10} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {4091 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{756 \sqrt {7}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 120, normalized size of antiderivative = 1.00, 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 = {97, 149, 157, 54, 216, 93, 204} \begin {gather*} -\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{6 (3 x+2)^2}-\frac {107 \sqrt {1-2 x} \sqrt {5 x+3}}{252 (3 x+2)}-\frac {10}{27} \sqrt {10} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {4091 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{756 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-107*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(252*(2 + 3*x)) - (Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(6*(2 + 3*x)^2) - (10*Sqr
t[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/27 - (4091*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(756*Sqrt[7]
)

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 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 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 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])

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 {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx &=-\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {1}{6} \int \frac {\left (\frac {9}{2}-20 x\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {3+5 x}}{252 (2+3 x)}-\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {1}{126} \int \frac {-\frac {503}{4}-700 x}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {3+5 x}}{252 (2+3 x)}-\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}-\frac {50}{27} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx+\frac {4091 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{1512}\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {3+5 x}}{252 (2+3 x)}-\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {4091}{756} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )-\frac {1}{27} \left (20 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=-\frac {107 \sqrt {1-2 x} \sqrt {3+5 x}}{252 (2+3 x)}-\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{6 (2+3 x)^2}-\frac {10}{27} \sqrt {10} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {4091 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{756 \sqrt {7}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 129, normalized size = 1.08 \begin {gather*} \frac {-21 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} (531 x+340)-4091 \sqrt {14 x-7} (3 x+2)^2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )+1960 \sqrt {10-20 x} (3 x+2)^2 \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{5292 \sqrt {2 x-1} (3 x+2)^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-21*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(340 + 531*x) + 1960*Sqrt[10 - 20*x]*(2 + 3*x)^2*ArcSinh[Sqrt[5/11]*Sqrt
[-1 + 2*x]] - 4091*(2 + 3*x)^2*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(5292*Sqrt[-1 +
2*x]*(2 + 3*x)^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.20, size = 126, normalized size = 1.05 \begin {gather*} -\frac {11 \sqrt {1-2 x} \left (\frac {107 (1-2 x)}{5 x+3}+1211\right )}{252 \sqrt {5 x+3} \left (\frac {1-2 x}{5 x+3}+7\right )^2}+\frac {10}{27} \sqrt {10} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {4091 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{756 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*Sqrt[1 - 2*x]*(1211 + (107*(1 - 2*x))/(3 + 5*x)))/(252*Sqrt[3 + 5*x]*(7 + (1 - 2*x)/(3 + 5*x))^2) + (10*S
qrt[10]*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/27 - (4091*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x
])])/(756*Sqrt[7])

________________________________________________________________________________________

fricas [A]  time = 1.57, size = 136, normalized size = 1.13 \begin {gather*} -\frac {4091 \, \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 ) - 1960 \, \sqrt {10} {\left (9 \, x^{2} + 12 \, x + 4\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 (531 \, x + 340\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{10584 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10584*(4091*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)) - 1960*sqrt(10)*(9*x^2 + 12*x + 4)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10
*x^2 + x - 3)) + 42*(531*x + 340)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

________________________________________________________________________________________

giac [B]  time = 2.16, size = 319, normalized size = 2.66 \begin {gather*} \frac {4091}{105840} \, \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 {5}{27} \, \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 \, \sqrt {10} {\left (107 \, {\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} + \frac {48440 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {193760 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{126 \, {\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4091/105840*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)))) - 5/27*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)))) - 11/126*sqrt(10)
*(107*((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 + 48440*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 193760*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x +
5) - sqrt(22)))/(((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)))^2 + 280)^2

________________________________________________________________________________________

maple [B]  time = 0.01, size = 191, normalized size = 1.59 \begin {gather*} -\frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (17640 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-36819 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+23520 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-49092 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+22302 \sqrt {-10 x^{2}-x +3}\, x +7840 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-16364 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+14280 \sqrt {-10 x^{2}-x +3}\right )}{10584 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10584*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(17640*10^(1/2)*arcsin(20/11*x+1/11)*x^2-36819*7^(1/2)*x^2*arctan(1/14*(
37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+23520*10^(1/2)*x*arcsin(20/11*x+1/11)-49092*7^(1/2)*x*arctan(1/14*(37*x+
20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+7840*10^(1/2)*arcsin(20/11*x+1/11)-16364*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2
)/(-10*x^2-x+3)^(1/2))+22302*(-10*x^2-x+3)^(1/2)*x+14280*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)^2

________________________________________________________________________________________

maxima [A]  time = 1.29, size = 101, normalized size = 0.84 \begin {gather*} -\frac {5}{27} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {4091}{10584} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {5}{63} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{14 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {103 \, \sqrt {-10 \, x^{2} - x + 3}}{252 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/27*sqrt(10)*arcsin(20/11*x + 1/11) + 4091/10584*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) -
5/63*sqrt(-10*x^2 - x + 3) - 1/14*(-10*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 103/252*sqrt(-10*x^2 - x + 3)/
(3*x + 2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{3/2}}{{\left (3\,x+2\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {3}{2}}}{\left (3 x + 2\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________