3.1840 \(\int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=88 \[ -\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}+\frac {21}{125} \sqrt {1-2 x} (3 x+2)^2-\frac {294}{625} \sqrt {1-2 x}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]

[Out]

-196/34375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-294/625*(1-2*x)^(1/2)+21/125*(2+3*x)^2*(1-2*x)^(1/2)-
1/5*(2+3*x)^3*(1-2*x)^(1/2)/(3+5*x)

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Rubi [A]  time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {97, 153, 12, 80, 63, 206} \[ -\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}+\frac {21}{125} \sqrt {1-2 x} (3 x+2)^2-\frac {294}{625} \sqrt {1-2 x}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-294*Sqrt[1 - 2*x])/625 + (21*Sqrt[1 - 2*x]*(2 + 3*x)^2)/125 - (Sqrt[1 - 2*x]*(2 + 3*x)^3)/(5*(3 + 5*x)) - (1
96*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(625*Sqrt[55])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 80

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

Rule 97

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

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

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 {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}+\frac {1}{5} \int \frac {(7-21 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {1}{125} \int -\frac {98 (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}+\frac {98}{125} \int \frac {2+3 x}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}+\frac {98}{625} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {98}{625} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 63, normalized size = 0.72 \[ \frac {\sqrt {1-2 x} \left (1350 x^3+2385 x^2-90 x-622\right )}{625 (5 x+3)}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(-622 - 90*x + 2385*x^2 + 1350*x^3))/(625*(3 + 5*x)) - (196*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/
(625*Sqrt[55])

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fricas [A]  time = 1.11, size = 69, normalized size = 0.78 \[ \frac {98 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (1350 \, x^{3} + 2385 \, x^{2} - 90 \, x - 622\right )} \sqrt {-2 \, x + 1}}{34375 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/34375*(98*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(1350*x^3 + 2385*x^2 -
90*x - 622)*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.27, size = 90, normalized size = 1.02 \[ \frac {27}{250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(2*x - 1)^2*sqrt(-2*x + 1) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*
sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/625*sqrt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 63, normalized size = 0.72 \[ -\frac {196 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{34375}+\frac {27 \left (-2 x +1\right )^{\frac {5}{2}}}{250}-\frac {117 \left (-2 x +1\right )^{\frac {3}{2}}}{250}+\frac {18 \sqrt {-2 x +1}}{625}+\frac {2 \sqrt {-2 x +1}}{3125 \left (-2 x -\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(-2*x+1)^(5/2)-117/250*(-2*x+1)^(3/2)+18/625*(-2*x+1)^(1/2)+2/3125*(-2*x+1)^(1/2)/(-2*x-6/5)-196/34375*
arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.30, size = 80, normalized size = 0.91 \[ \frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(-2*x + 1)^(5/2) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqr
t(55) + 5*sqrt(-2*x + 1))) + 18/625*sqrt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 0.05, size = 64, normalized size = 0.73 \[ \frac {18\,\sqrt {1-2\,x}}{625}-\frac {2\,\sqrt {1-2\,x}}{3125\,\left (2\,x+\frac {6}{5}\right )}-\frac {117\,{\left (1-2\,x\right )}^{3/2}}{250}+\frac {27\,{\left (1-2\,x\right )}^{5/2}}{250}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,196{}\mathrm {i}}{34375} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*196i)/34375 - (2*(1 - 2*x)^(1/2))/(3125*(2*x + 6/5)) + (18*(1
- 2*x)^(1/2))/625 - (117*(1 - 2*x)^(3/2))/250 + (27*(1 - 2*x)^(5/2))/250

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sympy [A]  time = 133.29, size = 202, normalized size = 2.30 \[ \frac {27 \left (1 - 2 x\right )^{\frac {5}{2}}}{250} - \frac {117 \left (1 - 2 x\right )^{\frac {3}{2}}}{250} + \frac {18 \sqrt {1 - 2 x}}{625} - \frac {44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: x \leq \frac {1}{2} \wedge x > - \frac {3}{5} \end {cases}\right )}{625} + \frac {194 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right )}{625} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27*(1 - 2*x)**(5/2)/250 - 117*(1 - 2*x)**(3/2)/250 + 18*sqrt(1 - 2*x)/625 - 44*Piecewise((sqrt(55)*(-log(sqrt(
55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1
/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (x <= 1/2) & (x > -3/5)))/625 + 194*Piecewise((-sqrt(55)*acoth(sqrt
(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))
/625

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