3.2451 \(\int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ -\frac {2 (1-2 x)^{5/2}}{15 (5 x+3)^{3/2}}+\frac {4 (1-2 x)^{3/2}}{15 \sqrt {5 x+3}}+\frac {4}{25} \sqrt {5 x+3} \sqrt {1-2 x}+\frac {22}{25} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ) \]

[Out]

-2/15*(1-2*x)^(5/2)/(3+5*x)^(3/2)+22/125*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+4/15*(1-2*x)^(3/2)/(3+5*
x)^(1/2)+4/25*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {47, 50, 54, 216} \[ -\frac {2 (1-2 x)^{5/2}}{15 (5 x+3)^{3/2}}+\frac {4 (1-2 x)^{3/2}}{15 \sqrt {5 x+3}}+\frac {4}{25} \sqrt {5 x+3} \sqrt {1-2 x}+\frac {22}{25} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - 2*x)^(5/2))/(15*(3 + 5*x)^(3/2)) + (4*(1 - 2*x)^(3/2))/(15*Sqrt[3 + 5*x]) + (4*Sqrt[1 - 2*x]*Sqrt[3 +
5*x])/25 + (22*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/25

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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]

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx &=-\frac {2 (1-2 x)^{5/2}}{15 (3+5 x)^{3/2}}-\frac {2}{3} \int \frac {(1-2 x)^{3/2}}{(3+5 x)^{3/2}} \, dx\\ &=-\frac {2 (1-2 x)^{5/2}}{15 (3+5 x)^{3/2}}+\frac {4 (1-2 x)^{3/2}}{15 \sqrt {3+5 x}}+\frac {4}{5} \int \frac {\sqrt {1-2 x}}{\sqrt {3+5 x}} \, dx\\ &=-\frac {2 (1-2 x)^{5/2}}{15 (3+5 x)^{3/2}}+\frac {4 (1-2 x)^{3/2}}{15 \sqrt {3+5 x}}+\frac {4}{25} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {22}{25} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {2 (1-2 x)^{5/2}}{15 (3+5 x)^{3/2}}+\frac {4 (1-2 x)^{3/2}}{15 \sqrt {3+5 x}}+\frac {4}{25} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {44 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{25 \sqrt {5}}\\ &=-\frac {2 (1-2 x)^{5/2}}{15 (3+5 x)^{3/2}}+\frac {4 (1-2 x)^{3/2}}{15 \sqrt {3+5 x}}+\frac {4}{25} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {22}{25} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.41 \[ -\frac {4}{847} \sqrt {\frac {2}{11}} (1-2 x)^{7/2} \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};-\frac {5}{11} (2 x-1)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[2/11]*(1 - 2*x)^(7/2)*Hypergeometric2F1[5/2, 7/2, 9/2, (-5*(-1 + 2*x))/11])/847

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fricas [A]  time = 0.75, size = 97, normalized size = 1.01 \[ -\frac {33 \, \sqrt {5} \sqrt {2} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 10 \, {\left (30 \, x^{2} + 190 \, x + 79\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{375 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/375*(33*sqrt(5)*sqrt(2)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x
+ 1)/(10*x^2 + x - 3)) - 10*(30*x^2 + 190*x + 79)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [B]  time = 1.70, size = 158, normalized size = 1.65 \[ -\frac {11}{30000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {108 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} + \frac {4}{625} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {22}{125} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {11 \, \sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {27 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{1875 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/30000*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 108*(sqrt(2)*sqrt(-10*x + 5) - sq
rt(22))/sqrt(5*x + 3)) + 4/625*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 22/125*sqrt(10)*arcsin(1/11*sqrt(22)*sq
rt(5*x + 3)) - 11/1875*sqrt(10)*(5*x + 3)^(3/2)*(27*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqr
t(2)*sqrt(-10*x + 5) - sqrt(22))^3

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {\left (-2 x +1\right )^{\frac {5}{2}}}{\left (5 x +3\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.44, size = 129, normalized size = 1.34 \[ \frac {11}{125} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{5 \, {\left (625 \, x^{4} + 1500 \, x^{3} + 1350 \, x^{2} + 540 \, x + 81\right )}} - \frac {11 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{30 \, {\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} - \frac {121 \, \sqrt {-10 \, x^{2} - x + 3}}{150 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {77 \, \sqrt {-10 \, x^{2} - x + 3}}{75 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/125*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 1/5*(-10*x^2 - x + 3)^(5/2)/(625*x^4 + 1500*x^3 + 1350*x^2 + 5
40*x + 81) - 11/30*(-10*x^2 - x + 3)^(3/2)/(125*x^3 + 225*x^2 + 135*x + 27) - 121/150*sqrt(-10*x^2 - x + 3)/(2
5*x^2 + 30*x + 9) + 77/75*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.38, size = 257, normalized size = 2.68 \[ \begin {cases} \frac {4 \sqrt {10} \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )}{125} + \frac {308 \sqrt {10} \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{1875} - \frac {242 \sqrt {10} \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{9375 \left (x + \frac {3}{5}\right )} + \frac {11 \sqrt {10} i \log {\left (\frac {1}{x + \frac {3}{5}} \right )}}{125} + \frac {11 \sqrt {10} i \log {\left (x + \frac {3}{5} \right )}}{125} + \frac {22 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{125} & \text {for}\: \frac {11}{10 \left |{x + \frac {3}{5}}\right |} > 1 \\\frac {4 \sqrt {10} i \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )}{125} + \frac {308 \sqrt {10} i \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{1875} - \frac {242 \sqrt {10} i \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{9375 \left (x + \frac {3}{5}\right )} + \frac {11 \sqrt {10} i \log {\left (\frac {1}{x + \frac {3}{5}} \right )}}{125} - \frac {22 \sqrt {10} i \log {\left (\sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} + 1 \right )}}{125} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((4*sqrt(10)*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)/125 + 308*sqrt(10)*sqrt(-1 + 11/(10*(x + 3/5)))/1
875 - 242*sqrt(10)*sqrt(-1 + 11/(10*(x + 3/5)))/(9375*(x + 3/5)) + 11*sqrt(10)*I*log(1/(x + 3/5))/125 + 11*sqr
t(10)*I*log(x + 3/5)/125 + 22*sqrt(10)*asin(sqrt(110)*sqrt(x + 3/5)/11)/125, 11/(10*Abs(x + 3/5)) > 1), (4*sqr
t(10)*I*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)/125 + 308*sqrt(10)*I*sqrt(1 - 11/(10*(x + 3/5)))/1875 - 242*sqrt
(10)*I*sqrt(1 - 11/(10*(x + 3/5)))/(9375*(x + 3/5)) + 11*sqrt(10)*I*log(1/(x + 3/5))/125 - 22*sqrt(10)*I*log(s
qrt(1 - 11/(10*(x + 3/5))) + 1)/125, True))

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