3.20.22 \(\int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx\) [1922]

Optimal. Leaf size=81 \[ -\frac {39}{275} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac {13 (1-2 x)^{3/2}}{110 (3+5 x)}+\frac {39 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]

[Out]

-1/110*(1-2*x)^(5/2)/(3+5*x)^2-13/110*(1-2*x)^(3/2)/(3+5*x)+39/1375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1
/2)-39/275*(1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 43, 52, 65, 212} \begin {gather*} -\frac {(1-2 x)^{5/2}}{110 (5 x+3)^2}-\frac {13 (1-2 x)^{3/2}}{110 (5 x+3)}-\frac {39}{275} \sqrt {1-2 x}+\frac {39 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-39*Sqrt[1 - 2*x])/275 - (1 - 2*x)^(5/2)/(110*(3 + 5*x)^2) - (13*(1 - 2*x)^(3/2))/(110*(3 + 5*x)) + (39*ArcTa
nh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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} (2+3 x)}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{5/2}}{110 (3+5 x)^2}+\frac {13}{22} \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac {13 (1-2 x)^{3/2}}{110 (3+5 x)}-\frac {39}{110} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=-\frac {39}{275} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac {13 (1-2 x)^{3/2}}{110 (3+5 x)}-\frac {39}{50} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {39}{275} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac {13 (1-2 x)^{3/2}}{110 (3+5 x)}+\frac {39}{50} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {39}{275} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac {13 (1-2 x)^{3/2}}{110 (3+5 x)}+\frac {39 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 58, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {1-2 x} \left (82+205 x+120 x^2\right )}{50 (3+5 x)^2}+\frac {39 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/50*(Sqrt[1 - 2*x]*(82 + 205*x + 120*x^2))/(3 + 5*x)^2 + (39*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55]
)

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Maple [A]
time = 0.10, size = 57, normalized size = 0.70

method	result	size
risch	\(\frac {240 x^{3}+290 x^{2}-41 x -82}{50 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {39 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\)	\(51\)
derivativedivides	\(-\frac {12 \sqrt {1-2 x}}{125}-\frac {4 \left (-\frac {61 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {693 \sqrt {1-2 x}}{100}\right )}{5 \left (-6-10 x \right )^{2}}+\frac {39 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\)	\(57\)
default	\(-\frac {12 \sqrt {1-2 x}}{125}-\frac {4 \left (-\frac {61 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {693 \sqrt {1-2 x}}{100}\right )}{5 \left (-6-10 x \right )^{2}}+\frac {39 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\)	\(57\)
trager	\(-\frac {\left (120 x^{2}+205 x +82\right ) \sqrt {1-2 x}}{50 \left (3+5 x \right )^{2}}+\frac {39 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{2750}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12/125*(1-2*x)^(1/2)-4/5*(-61/20*(1-2*x)^(3/2)+693/100*(1-2*x)^(1/2))/(-6-10*x)^2+39/1375*arctanh(1/11*55^(1/
2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]
time = 0.51, size = 83, normalized size = 1.02 \begin {gather*} -\frac {39}{2750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {12}{125} \, \sqrt {-2 \, x + 1} + \frac {305 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 693 \, \sqrt {-2 \, x + 1}}{125 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-39/2750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 12/125*sqrt(-2*x + 1) +
1/125*(305*(-2*x + 1)^(3/2) - 693*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]
time = 0.61, size = 75, normalized size = 0.93 \begin {gather*} \frac {39 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (120 \, x^{2} + 205 \, x + 82\right )} \sqrt {-2 \, x + 1}}{2750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2750*(39*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(120*x^2 + 205
*x + 82)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [F(-1)] Timed out
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)*(2+3*x)/(3+5*x)**3,x)

[Out]

Timed out

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Giac [A]
time = 0.59, size = 77, normalized size = 0.95 \begin {gather*} -\frac {39}{2750} \, \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 {12}{125} \, \sqrt {-2 \, x + 1} + \frac {305 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 693 \, \sqrt {-2 \, x + 1}}{500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-39/2750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 12/125*sqrt(-2
*x + 1) + 1/500*(305*(-2*x + 1)^(3/2) - 693*sqrt(-2*x + 1))/(5*x + 3)^2

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Mupad [B]
time = 1.19, size = 63, normalized size = 0.78 \begin {gather*} \frac {39\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1375}-\frac {12\,\sqrt {1-2\,x}}{125}-\frac {\frac {693\,\sqrt {1-2\,x}}{3125}-\frac {61\,{\left (1-2\,x\right )}^{3/2}}{625}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(39*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/1375 - (12*(1 - 2*x)^(1/2))/125 - ((693*(1 - 2*x)^(1/2))/31
25 - (61*(1 - 2*x)^(3/2))/625)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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