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

Optimal. Leaf size=135 \[ -\frac {1}{3} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}+\frac {107}{36} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {1649 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{108 \sqrt {10}}+\frac {37}{27} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 135, normalized size of antiderivative = 1.00, 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, 154, 157, 54, 216, 93, 204} \begin {gather*} -\frac {1}{3} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {(1-2 x)^{3/2} (5 x+3)^{3/2}}{3 (3 x+2)}+\frac {107}{36} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {1649 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{108 \sqrt {10}}+\frac {37}{27} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(107*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/36 - (Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/3 - ((1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))/(
3*(2 + 3*x)) + (1649*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(108*Sqrt[10]) + (37*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt
[7]*Sqrt[3 + 5*x])])/27

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 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 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 {(1-2 x)^{3/2} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx &=-\frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{3 (2+3 x)}+\frac {1}{3} \int \frac {\left (-\frac {3}{2}-30 x\right ) \sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x} \, dx\\ &=-\frac {1}{3} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{3 (2+3 x)}+\frac {1}{90} \int \frac {(225-1605 x) \sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {107}{36} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{3} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{3 (2+3 x)}-\frac {1}{540} \int \frac {-5655-\frac {24735 x}{2}}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {107}{36} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{3} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{3 (2+3 x)}-\frac {259}{54} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx+\frac {1649}{216} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {107}{36} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{3} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{3 (2+3 x)}-\frac {259}{27} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {1649 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{108 \sqrt {5}}\\ &=\frac {107}{36} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{3} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{3 (2+3 x)}+\frac {1649 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{108 \sqrt {10}}+\frac {37}{27} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 130, normalized size = 0.96 \begin {gather*} \frac {-30 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} \left (60 x^2-105 x-106\right )+1480 (3 x+2) \sqrt {14 x-7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-1649 \sqrt {10-20 x} (3 x+2) \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{1080 \sqrt {2 x-1} (3 x+2)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-30*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(-106 - 105*x + 60*x^2) - 1649*Sqrt[10 - 20*x]*(2 + 3*x)*ArcSinh[Sqrt[5/
11]*Sqrt[-1 + 2*x]] + 1480*(2 + 3*x)*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1080*Sqrt
[-1 + 2*x]*(2 + 3*x))

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IntegrateAlgebraic [A]  time = 0.24, size = 160, normalized size = 1.19 \begin {gather*} \frac {11 \sqrt {1-2 x} \left (\frac {535 (1-2 x)^2}{(5 x+3)^2}+\frac {2375 (1-2 x)}{5 x+3}+574\right )}{36 \sqrt {5 x+3} \left (\frac {1-2 x}{5 x+3}+7\right ) \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^2}-\frac {1649 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{108 \sqrt {10}}+\frac {37}{27} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*Sqrt[1 - 2*x]*(574 + (535*(1 - 2*x)^2)/(3 + 5*x)^2 + (2375*(1 - 2*x))/(3 + 5*x)))/(36*Sqrt[3 + 5*x]*(7 + (
1 - 2*x)/(3 + 5*x))*(2 + (5*(1 - 2*x))/(3 + 5*x))^2) - (1649*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/
(108*Sqrt[10]) + (37*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/27

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fricas [A]  time = 1.78, size = 126, normalized size = 0.93 \begin {gather*} \frac {1480 \, \sqrt {7} {\left (3 \, x + 2\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 ) - 1649 \, \sqrt {10} {\left (3 \, x + 2\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 ) - 60 \, {\left (60 \, x^{2} - 105 \, x - 106\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{2160 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2160*(1480*sqrt(7)*(3*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
- 1649*sqrt(10)*(3*x + 2)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 60*
(60*x^2 - 105*x - 106)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)

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giac [B]  time = 1.95, size = 292, normalized size = 2.16 \begin {gather*} -\frac {37}{540} \, \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}{540} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} - 181 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {1649}{2160} \, \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 {154 \, \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 )}}{27 \, {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-37/540*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)))) - 1/540*(12*sqrt(5)*(5*x + 3) - 181*sqrt(5))*sqrt(5*x +
3)*sqrt(-10*x + 5) + 1649/2160*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)))) + 154/27*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)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(2
2))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)

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maple [A]  time = 0.01, size = 163, normalized size = 1.21 \begin {gather*} \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (-3600 \sqrt {-10 x^{2}-x +3}\, x^{2}+4947 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-4440 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+6300 \sqrt {-10 x^{2}-x +3}\, x +3298 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-2960 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+6360 \sqrt {-10 x^{2}-x +3}\right )}{2160 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2160*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(4947*10^(1/2)*x*arcsin(20/11*x+1/11)-4440*7^(1/2)*x*arctan(1/14*(37*x+20)
*7^(1/2)/(-10*x^2-x+3)^(1/2))-3600*(-10*x^2-x+3)^(1/2)*x^2+3298*10^(1/2)*arcsin(20/11*x+1/11)-2960*7^(1/2)*arc
tan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+6300*(-10*x^2-x+3)^(1/2)*x+6360*(-10*x^2-x+3)^(1/2))/(-10*x^2-
x+3)^(1/2)/(3*x+2)

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maxima [A]  time = 1.33, size = 90, normalized size = 0.67 \begin {gather*} -\frac {5}{3} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {1649}{2160} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {37}{54} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {71}{36} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{3 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/3*sqrt(-10*x^2 - x + 3)*x + 1649/2160*sqrt(10)*arcsin(20/11*x + 1/11) - 37/54*sqrt(7)*arcsin(37/11*x/abs(3*
x + 2) + 20/11/abs(3*x + 2)) + 71/36*sqrt(-10*x^2 - x + 3) - 1/3*(-10*x^2 - x + 3)^(3/2)/(3*x + 2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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