3.1912 \(\int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=76 \[ -\frac {(1-2 x)^{5/2}}{55 (5 x+3)}+\frac {4}{55} (1-2 x)^{3/2}+\frac {12}{25} \sqrt {1-2 x}-\frac {12}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

4/55*(1-2*x)^(3/2)-1/55*(1-2*x)^(5/2)/(3+5*x)-12/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+12/25*(1-2*
x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 50, 63, 206} \[ -\frac {(1-2 x)^{5/2}}{55 (5 x+3)}+\frac {4}{55} (1-2 x)^{3/2}+\frac {12}{25} \sqrt {1-2 x}-\frac {12}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*Sqrt[1 - 2*x])/25 + (4*(1 - 2*x)^(3/2))/55 - (1 - 2*x)^(5/2)/(55*(3 + 5*x)) - (12*Sqrt[11/5]*ArcTanh[Sqrt[
5/11]*Sqrt[1 - 2*x]])/25

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 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 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)^{3/2} (2+3 x)}{(3+5 x)^2} \, dx &=-\frac {(1-2 x)^{5/2}}{55 (3+5 x)}+\frac {6}{11} \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=\frac {4}{55} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{55 (3+5 x)}+\frac {6}{5} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {12}{25} \sqrt {1-2 x}+\frac {4}{55} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{55 (3+5 x)}+\frac {66}{25} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {12}{25} \sqrt {1-2 x}+\frac {4}{55} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{55 (3+5 x)}-\frac {66}{25} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {12}{25} \sqrt {1-2 x}+\frac {4}{55} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{55 (3+5 x)}-\frac {12}{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 = 0.04, size = 58, normalized size = 0.76 \[ \frac {1}{125} \left (\frac {5 \sqrt {1-2 x} \left (-20 x^2+60 x+41\right )}{5 x+3}-12 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(41 + 60*x - 20*x^2))/(3 + 5*x) - 12*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

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fricas [A]  time = 0.97, size = 70, normalized size = 0.92 \[ \frac {6 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 5 \, {\left (20 \, x^{2} - 60 \, x - 41\right )} \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/125*(6*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 5*(20*x^2 - 6
0*x - 41)*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 0.97, size = 74, normalized size = 0.97 \[ \frac {2}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {6}{125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {62}{125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(-2*x + 1)^(3/2) + 6/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x +
1))) + 62/125*sqrt(-2*x + 1) - 11/125*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 54, normalized size = 0.71 \[ -\frac {12 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{125}+\frac {2 \left (-2 x +1\right )^{\frac {3}{2}}}{25}+\frac {62 \sqrt {-2 x +1}}{125}+\frac {22 \sqrt {-2 x +1}}{625 \left (-2 x -\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(-2*x+1)^(3/2)+62/125*(-2*x+1)^(1/2)+22/625*(-2*x+1)^(1/2)/(-2*x-6/5)-12/125*arctanh(1/11*55^(1/2)*(-2*x+
1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.22, size = 71, normalized size = 0.93 \[ \frac {2}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {6}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {62}{125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(-2*x + 1)^(3/2) + 6/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 62/
125*sqrt(-2*x + 1) - 11/125*sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 0.07, size = 55, normalized size = 0.72 \[ \frac {62\,\sqrt {1-2\,x}}{125}-\frac {22\,\sqrt {1-2\,x}}{625\,\left (2\,x+\frac {6}{5}\right )}+\frac {2\,{\left (1-2\,x\right )}^{3/2}}{25}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,12{}\mathrm {i}}{125} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*12i)/125 - (22*(1 - 2*x)^(1/2))/(625*(2*x + 6/5)) + (62*(1 -
2*x)^(1/2))/125 + (2*(1 - 2*x)^(3/2))/25

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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