3.1994 \(\int \frac{(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=96 \[ -\frac{(1-2 x)^{7/2}}{110 (5 x+3)^2}-\frac{63 (1-2 x)^{5/2}}{550 (5 x+3)}-\frac{21}{275} (1-2 x)^{3/2}-\frac{63}{125} \sqrt{1-2 x}+\frac{63}{125} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(-63*Sqrt[1 - 2*x])/125 - (21*(1 - 2*x)^(3/2))/275 - (1 - 2*x)^(7/2)/(110*(3 + 5*x)^2) - (63*(1 - 2*x)^(5/2))/
(550*(3 + 5*x)) + (63*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

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Rubi [A]  time = 0.021367, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {78, 47, 50, 63, 206} \[ -\frac{(1-2 x)^{7/2}}{110 (5 x+3)^2}-\frac{63 (1-2 x)^{5/2}}{550 (5 x+3)}-\frac{21}{275} (1-2 x)^{3/2}-\frac{63}{125} \sqrt{1-2 x}+\frac{63}{125} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-63*Sqrt[1 - 2*x])/125 - (21*(1 - 2*x)^(3/2))/275 - (1 - 2*x)^(7/2)/(110*(3 + 5*x)^2) - (63*(1 - 2*x)^(5/2))/
(550*(3 + 5*x)) + (63*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

Rule 78

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] || LtQ
[p, n]))))

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 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} (2+3 x)}{(3+5 x)^3} \, dx &=-\frac{(1-2 x)^{7/2}}{110 (3+5 x)^2}+\frac{63}{110} \int \frac{(1-2 x)^{5/2}}{(3+5 x)^2} \, dx\\ &=-\frac{(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac{63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac{63}{110} \int \frac{(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=-\frac{21}{275} (1-2 x)^{3/2}-\frac{(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac{63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac{63}{50} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=-\frac{63}{125} \sqrt{1-2 x}-\frac{21}{275} (1-2 x)^{3/2}-\frac{(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac{63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac{693}{250} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{63}{125} \sqrt{1-2 x}-\frac{21}{275} (1-2 x)^{3/2}-\frac{(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac{63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac{693}{250} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{63}{125} \sqrt{1-2 x}-\frac{21}{275} (1-2 x)^{3/2}-\frac{(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac{63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac{63}{125} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [C]  time = 0.0150294, size = 48, normalized size = 0.5 \[ -\frac{(1-2 x)^{7/2} \left (36 (5 x+3)^2 \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{5}{11} (1-2 x)\right )+121\right )}{13310 (5 x+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - 2*x)^(7/2)*(121 + 36*(3 + 5*x)^2*Hypergeometric2F1[2, 7/2, 9/2, (5*(1 - 2*x))/11]))/(13310*(3 + 5*x)^2)

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Maple [A]  time = 0.007, size = 66, normalized size = 0.7 \begin{align*} -{\frac{4}{125} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{256}{625}\sqrt{1-2\,x}}-{\frac{44}{25\, \left ( -10\,x-6 \right ) ^{2}} \left ( -{\frac{57}{20} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{649}{100}\sqrt{1-2\,x}} \right ) }+{\frac{63\,\sqrt{55}}{625}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(1-2*x)^(3/2)-256/625*(1-2*x)^(1/2)-44/25*(-57/20*(1-2*x)^(3/2)+649/100*(1-2*x)^(1/2))/(-10*x-6)^2+63/6
25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.91194, size = 124, normalized size = 1.29 \begin{align*} -\frac{4}{125} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - \frac{63}{1250} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{256}{625} \, \sqrt{-2 \, x + 1} + \frac{11 \,{\left (285 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 649 \, \sqrt{-2 \, x + 1}\right )}}{625 \,{\left (25 \,{\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) -
256/625*sqrt(-2*x + 1) + 11/625*(285*(-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]  time = 1.38081, size = 251, normalized size = 2.61 \begin{align*} \frac{63 \, \sqrt{11} \sqrt{5}{\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (-\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 5 \,{\left (400 \, x^{3} - 2280 \, x^{2} - 3795 \, x - 1394\right )} \sqrt{-2 \, x + 1}}{1250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1250*(63*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) +
5*(400*x^3 - 2280*x^2 - 3795*x - 1394)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 2.61956, size = 116, normalized size = 1.21 \begin{align*} -\frac{4}{125} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - \frac{63}{1250} \, \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{256}{625} \, \sqrt{-2 \, x + 1} + \frac{11 \,{\left (285 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 649 \, \sqrt{-2 \, x + 1}\right )}}{2500 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*
x + 1))) - 256/625*sqrt(-2*x + 1) + 11/2500*(285*(-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(5*x + 3)^2