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

Optimal. Leaf size=130 \[ \frac {1}{9} \sqrt {1-2 x} (5 x+3)^{5/2}-\frac {5}{24} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {925}{864} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {6553 \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2592}+\frac {2}{81} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 130, 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 = {101, 154, 157, 54, 216, 93, 204} \begin {gather*} \frac {1}{9} \sqrt {1-2 x} (5 x+3)^{5/2}-\frac {5}{24} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {925}{864} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {6553 \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2592}+\frac {2}{81} \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[(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(2 + 3*x),x]

[Out]

(-925*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/864 - (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/24 + (Sqrt[1 - 2*x]*(3 + 5*x)^(5/2)
)/9 + (6553*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/2592 + (2*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3
 + 5*x])])/81

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 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[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 {\sqrt {1-2 x} (3+5 x)^{5/2}}{2+3 x} \, dx &=\frac {1}{9} \sqrt {1-2 x} (3+5 x)^{5/2}-\frac {1}{9} \int \frac {\left (-8-\frac {45 x}{2}\right ) (3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {5}{24} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1}{9} \sqrt {1-2 x} (3+5 x)^{5/2}+\frac {1}{108} \int \frac {\sqrt {3+5 x} \left (\frac {981}{2}+\frac {2775 x}{4}\right )}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {925}{864} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{24} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1}{9} \sqrt {1-2 x} (3+5 x)^{5/2}-\frac {1}{648} \int \frac {-\frac {32541}{4}-\frac {98295 x}{8}}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {925}{864} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{24} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1}{9} \sqrt {1-2 x} (3+5 x)^{5/2}-\frac {7}{81} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx+\frac {32765 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{5184}\\ &=-\frac {925}{864} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{24} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1}{9} \sqrt {1-2 x} (3+5 x)^{5/2}-\frac {14}{81} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {\left (6553 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{2592}\\ &=-\frac {925}{864} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{24} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1}{9} \sqrt {1-2 x} (3+5 x)^{5/2}+\frac {6553 \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2592}+\frac {2}{81} \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 = 0.13, size = 113, normalized size = 0.87 \begin {gather*} \frac {6 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} \left (2400 x^2+1980 x-601\right )+128 \sqrt {14 x-7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-6553 \sqrt {10-20 x} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{5184 \sqrt {2 x-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(6*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(-601 + 1980*x + 2400*x^2) - 6553*Sqrt[10 - 20*x]*ArcSinh[Sqrt[5/11]*Sqrt[
-1 + 2*x]] + 128*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(5184*Sqrt[-1 + 2*x])

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IntegrateAlgebraic [A]  time = 0.22, size = 145, normalized size = 1.12 \begin {gather*} -\frac {11 \sqrt {1-2 x} \left (\frac {23125 (1-2 x)^2}{(5 x+3)^2}+\frac {28400 (1-2 x)}{5 x+3}-3956\right )}{864 \sqrt {5 x+3} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^3}-\frac {6553 \sqrt {\frac {5}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{2592}+\frac {2}{81} \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[(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(2 + 3*x),x]

[Out]

(-11*Sqrt[1 - 2*x]*(-3956 + (23125*(1 - 2*x)^2)/(3 + 5*x)^2 + (28400*(1 - 2*x))/(3 + 5*x)))/(864*Sqrt[3 + 5*x]
*(2 + (5*(1 - 2*x))/(3 + 5*x))^3) - (6553*Sqrt[5/2]*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/2592 + (2
*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/81

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fricas [A]  time = 1.29, size = 113, normalized size = 0.87 \begin {gather*} \frac {1}{864} \, {\left (2400 \, x^{2} + 1980 \, x - 601\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {6553}{10368} \, \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}{81} \, \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 {gather*}

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),x, algorithm="fricas")

[Out]

1/864*(2400*x^2 + 1980*x - 601)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 6553/10368*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)) + 1/81*sqrt(7)*arctan(1/14*sqrt(7)*(37*x + 2
0)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))

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giac [B]  time = 1.37, size = 186, normalized size = 1.43 \begin {gather*} -\frac {1}{810} \, \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}{4320} \, {\left (12 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} - 15 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 925 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {6553}{10368} \, \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 {gather*}

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),x, algorithm="giac")

[Out]

-1/810*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)))) + 1/4320*(12*(8*sqrt(5)*(5*x + 3) - 15*sqrt(5))*(5*x + 3
) - 925*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) + 6553/10368*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))))

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maple [A]  time = 0.01, size = 115, normalized size = 0.88 \begin {gather*} \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (28800 \sqrt {-10 x^{2}-x +3}\, x^{2}+23760 \sqrt {-10 x^{2}-x +3}\, x +6553 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-128 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-7212 \sqrt {-10 x^{2}-x +3}\right )}{10368 \sqrt {-10 x^{2}-x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10368*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(28800*(-10*x^2-x+3)^(1/2)*x^2+6553*10^(1/2)*arcsin(20/11*x+1/11)-128*7^(
1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+23760*(-10*x^2-x+3)^(1/2)*x-7212*(-10*x^2-x+3)^(1/2))/
(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.12, size = 83, normalized size = 0.64 \begin {gather*} -\frac {5}{18} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {145}{72} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {6553}{10368} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {1}{81} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {119}{864} \, \sqrt {-10 \, x^{2} - x + 3} \end {gather*}

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),x, algorithm="maxima")

[Out]

-5/18*(-10*x^2 - x + 3)^(3/2) + 145/72*sqrt(-10*x^2 - x + 3)*x + 6553/10368*sqrt(10)*arcsin(20/11*x + 1/11) -
1/81*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 119/864*sqrt(-10*x^2 - x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 {5}{2}}}{3 x + 2}\, dx \end {gather*}

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),x)

[Out]

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

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