3.1972 \(\int \frac {(1-2 x)^{5/2}}{3+5 x} \, dx\)

Optimal. Leaf size=69 \[ \frac {2}{25} (1-2 x)^{5/2}+\frac {22}{75} (1-2 x)^{3/2}+\frac {242}{125} \sqrt {1-2 x}-\frac {242}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

22/75*(1-2*x)^(3/2)+2/25*(1-2*x)^(5/2)-242/625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+242/125*(1-2*x)^(
1/2)

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Rubi [A]  time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {50, 63, 206} \[ \frac {2}{25} (1-2 x)^{5/2}+\frac {22}{75} (1-2 x)^{3/2}+\frac {242}{125} \sqrt {1-2 x}-\frac {242}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(5/2)/(3 + 5*x),x]

[Out]

(242*Sqrt[1 - 2*x])/125 + (22*(1 - 2*x)^(3/2))/75 + (2*(1 - 2*x)^(5/2))/25 - (242*Sqrt[11/5]*ArcTanh[Sqrt[5/11
]*Sqrt[1 - 2*x]])/125

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{5/2}}{3+5 x} \, dx &=\frac {2}{25} (1-2 x)^{5/2}+\frac {11}{5} \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=\frac {22}{75} (1-2 x)^{3/2}+\frac {2}{25} (1-2 x)^{5/2}+\frac {121}{25} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {242}{125} \sqrt {1-2 x}+\frac {22}{75} (1-2 x)^{3/2}+\frac {2}{25} (1-2 x)^{5/2}+\frac {1331}{125} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {242}{125} \sqrt {1-2 x}+\frac {22}{75} (1-2 x)^{3/2}+\frac {2}{25} (1-2 x)^{5/2}-\frac {1331}{125} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {242}{125} \sqrt {1-2 x}+\frac {22}{75} (1-2 x)^{3/2}+\frac {2}{25} (1-2 x)^{5/2}-\frac {242}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 51, normalized size = 0.74 \[ \frac {2 \left (5 \sqrt {1-2 x} \left (60 x^2-170 x+433\right )-363 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )}{1875} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*Sqrt[1 - 2*x]*(433 - 170*x + 60*x^2) - 363*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]))/1875

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fricas [A]  time = 1.03, size = 56, normalized size = 0.81 \[ \frac {121}{625} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {2}{375} \, {\left (60 \, x^{2} - 170 \, x + 433\right )} \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/625*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 2/375*(60*x^2 - 170*x +
433)*sqrt(-2*x + 1)

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giac [A]  time = 0.97, size = 74, normalized size = 1.07 \[ \frac {2}{25} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {22}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{625} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {242}{125} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(2*x - 1)^2*sqrt(-2*x + 1) + 22/75*(-2*x + 1)^(3/2) + 121/625*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(
-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 242/125*sqrt(-2*x + 1)

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maple [A]  time = 0.00, size = 47, normalized size = 0.68 \[ -\frac {242 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{625}+\frac {22 \left (-2 x +1\right )^{\frac {3}{2}}}{75}+\frac {2 \left (-2 x +1\right )^{\frac {5}{2}}}{25}+\frac {242 \sqrt {-2 x +1}}{125} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

22/75*(-2*x+1)^(3/2)+2/25*(-2*x+1)^(5/2)-242/625*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)+242/125*(-2*x+
1)^(1/2)

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maxima [A]  time = 1.17, size = 64, normalized size = 0.93 \[ \frac {2}{25} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {22}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {242}{125} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(-2*x + 1)^(5/2) + 22/75*(-2*x + 1)^(3/2) + 121/625*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55)
+ 5*sqrt(-2*x + 1))) + 242/125*sqrt(-2*x + 1)

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mupad [B]  time = 0.07, size = 48, normalized size = 0.70 \[ \frac {242\,\sqrt {1-2\,x}}{125}+\frac {22\,{\left (1-2\,x\right )}^{3/2}}{75}+\frac {2\,{\left (1-2\,x\right )}^{5/2}}{25}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{625} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(5/2)/(5*x + 3),x)

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*242i)/625 + (242*(1 - 2*x)^(1/2))/125 + (22*(1 - 2*x)^(3/2))/
75 + (2*(1 - 2*x)^(5/2))/25

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sympy [A]  time = 2.77, size = 204, normalized size = 2.96 \[ \begin {cases} \frac {8 \sqrt {5} i \left (x + \frac {3}{5}\right )^{2} \sqrt {10 x - 5}}{125} - \frac {484 \sqrt {5} i \left (x + \frac {3}{5}\right ) \sqrt {10 x - 5}}{1875} + \frac {5566 \sqrt {5} i \sqrt {10 x - 5}}{9375} + \frac {242 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{625} & \text {for}\: \frac {10 \left |{x + \frac {3}{5}}\right |}{11} > 1 \\\frac {8 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )^{2}}{125} - \frac {484 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )}{1875} + \frac {5566 \sqrt {5} \sqrt {5 - 10 x}}{9375} + \frac {121 \sqrt {55} \log {\left (x + \frac {3}{5} \right )}}{625} - \frac {242 \sqrt {55} \log {\left (\sqrt {\frac {5}{11} - \frac {10 x}{11}} + 1 \right )}}{625} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*sqrt(5)*I*(x + 3/5)**2*sqrt(10*x - 5)/125 - 484*sqrt(5)*I*(x + 3/5)*sqrt(10*x - 5)/1875 + 5566*sq
rt(5)*I*sqrt(10*x - 5)/9375 + 242*sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/625, 10*Abs(x + 3/5)/11 > 1),
(8*sqrt(5)*sqrt(5 - 10*x)*(x + 3/5)**2/125 - 484*sqrt(5)*sqrt(5 - 10*x)*(x + 3/5)/1875 + 5566*sqrt(5)*sqrt(5 -
10*x)/9375 + 121*sqrt(55)*log(x + 3/5)/625 - 242*sqrt(55)*log(sqrt(5/11 - 10*x/11) + 1)/625, True))

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