3.20.1 \(\int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx\) [1901]

Optimal. Leaf size=56 \[ \frac {22}{25} \sqrt {1-2 x}+\frac {2}{15} (1-2 x)^{3/2}-\frac {22}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

2/15*(1-2*x)^(3/2)-22/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+22/25*(1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {52, 65, 212} \begin {gather*} \frac {2}{15} (1-2 x)^{3/2}+\frac {22}{25} \sqrt {1-2 x}-\frac {22}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(3/2)/(3 + 5*x),x]

[Out]

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

Rule 52

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 65

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 212

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

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx &=\frac {2}{15} (1-2 x)^{3/2}+\frac {11}{5} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {22}{25} \sqrt {1-2 x}+\frac {2}{15} (1-2 x)^{3/2}+\frac {121}{25} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {22}{25} \sqrt {1-2 x}+\frac {2}{15} (1-2 x)^{3/2}-\frac {121}{25} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {22}{25} \sqrt {1-2 x}+\frac {2}{15} (1-2 x)^{3/2}-\frac {22}{25} \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 = 0.06, size = 46, normalized size = 0.82 \begin {gather*} \frac {1}{375} \left (20 (19-5 x) \sqrt {1-2 x}-66 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(20*(19 - 5*x)*Sqrt[1 - 2*x] - 66*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/375

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Maple [A]
time = 0.10, size = 38, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{15}-\frac {22 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {22 \sqrt {1-2 x}}{25}\)	\(38\)
default	\(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{15}-\frac {22 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {22 \sqrt {1-2 x}}{25}\)	\(38\)
risch	\(\frac {4 \left (5 x -19\right ) \left (-1+2 x \right )}{75 \sqrt {1-2 x}}-\frac {22 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\)	\(39\)
trager	\(\left (-\frac {4 x}{15}+\frac {76}{75}\right ) \sqrt {1-2 x}-\frac {11 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{125}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(1-2*x)^(3/2)-22/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+22/25*(1-2*x)^(1/2)

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Maxima [A]
time = 0.49, size = 55, normalized size = 0.98 \begin {gather*} \frac {2}{15} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {22}{25} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(-2*x + 1)^(3/2) + 11/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 22
/25*sqrt(-2*x + 1)

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Fricas [A]
time = 0.92, size = 51, normalized size = 0.91 \begin {gather*} \frac {11}{125} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - \frac {4}{75} \, {\left (5 \, x - 19\right )} \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/125*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 4/75*(5*x - 19)*sqrt(-2*x
+ 1)

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Sympy [C] Result contains complex when optimal does not.
time = 0.96, size = 153, normalized size = 2.73 \begin {gather*} \begin {cases} - \frac {4 \sqrt {5} i \left (x + \frac {3}{5}\right ) \sqrt {10 x - 5}}{75} + \frac {88 \sqrt {5} i \sqrt {10 x - 5}}{375} + \frac {22 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {11}{10} \\- \frac {4 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )}{75} + \frac {88 \sqrt {5} \sqrt {5 - 10 x}}{375} + \frac {11 \sqrt {55} \log {\left (x + \frac {3}{5} \right )}}{125} - \frac {22 \sqrt {55} \log {\left (\sqrt {\frac {5}{11} - \frac {10 x}{11}} + 1 \right )}}{125} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-4*sqrt(5)*I*(x + 3/5)*sqrt(10*x - 5)/75 + 88*sqrt(5)*I*sqrt(10*x - 5)/375 + 22*sqrt(55)*I*asin(sqr
t(110)/(10*sqrt(x + 3/5)))/125, Abs(x + 3/5) > 11/10), (-4*sqrt(5)*sqrt(5 - 10*x)*(x + 3/5)/75 + 88*sqrt(5)*sq
rt(5 - 10*x)/375 + 11*sqrt(55)*log(x + 3/5)/125 - 22*sqrt(55)*log(sqrt(5/11 - 10*x/11) + 1)/125, True))

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Giac [A]
time = 0.64, size = 58, normalized size = 1.04 \begin {gather*} \frac {2}{15} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{125} \, \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 {22}{25} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(-2*x + 1)^(3/2) + 11/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x +
1))) + 22/25*sqrt(-2*x + 1)

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Mupad [B]
time = 1.17, size = 37, normalized size = 0.66 \begin {gather*} \frac {22\,\sqrt {1-2\,x}}{25}-\frac {22\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}+\frac {2\,{\left (1-2\,x\right )}^{3/2}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(3/2)/(5*x + 3),x)

[Out]

(22*(1 - 2*x)^(1/2))/25 - (22*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/125 + (2*(1 - 2*x)^(3/2))/15

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