\(\int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx\) [2039]

Optimal result

Integrand size = 24, antiderivative size = 54 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {111}{50} \sqrt {1-2 x}+\frac {3}{10} (1-2 x)^{3/2}-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]

[Out]

3/10*(1-2*x)^(3/2)-2/1375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-111/50*(1-2*x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {90, 65, 212} \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}}+\frac {3}{10} (1-2 x)^{3/2}-\frac {111}{50} \sqrt {1-2 x} \]

[In]

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

[Out]

(-111*Sqrt[1 - 2*x])/50 + (3*(1 - 2*x)^(3/2))/10 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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 90

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

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*} \text {integral}& = \int \left (\frac {111}{50 \sqrt {1-2 x}}-\frac {9}{10} \sqrt {1-2 x}+\frac {1}{25 \sqrt {1-2 x} (3+5 x)}\right ) \, dx \\ & = -\frac {111}{50} \sqrt {1-2 x}+\frac {3}{10} (1-2 x)^{3/2}+\frac {1}{25} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = -\frac {111}{50} \sqrt {1-2 x}+\frac {3}{10} (1-2 x)^{3/2}-\frac {1}{25} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = -\frac {111}{50} \sqrt {1-2 x}+\frac {3}{10} (1-2 x)^{3/2}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {3}{25} \sqrt {1-2 x} (16+5 x)-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]

[In]

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

[Out]

(-3*Sqrt[1 - 2*x]*(16 + 5*x))/25 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.63

method	result	size
pseudoelliptic	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}+\frac {3 \left (-16-5 x \right ) \sqrt {1-2 x}}{25}\)	\(34\)
derivativedivides	\(\frac {3 \left (1-2 x \right )^{\frac {3}{2}}}{10}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}-\frac {111 \sqrt {1-2 x}}{50}\)	\(38\)
default	\(\frac {3 \left (1-2 x \right )^{\frac {3}{2}}}{10}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}-\frac {111 \sqrt {1-2 x}}{50}\)	\(38\)
risch	\(\frac {3 \left (5 x +16\right ) \left (-1+2 x \right )}{25 \sqrt {1-2 x}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\)	\(39\)
trager	\(\left (-\frac {3 x}{5}-\frac {48}{25}\right ) \sqrt {1-2 x}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1375}\)	\(59\)

[In]

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

[Out]

-2/1375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+3/25*(-16-5*x)*(1-2*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {3}{25} \, {\left (5 \, x + 16\right )} \sqrt {-2 \, x + 1} + \frac {1}{1375} \, \sqrt {55} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) \]

[In]

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

[Out]

-3/25*(5*x + 16)*sqrt(-2*x + 1) + 1/1375*sqrt(55)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3))

Sympy [A] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=\frac {3 \left (1 - 2 x\right )^{\frac {3}{2}}}{10} - \frac {111 \sqrt {1 - 2 x}}{50} + \frac {\sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{1375} \]

[In]

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

[Out]

3*(1 - 2*x)**(3/2)/10 - 111*sqrt(1 - 2*x)/50 + sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) +
sqrt(55)/5))/1375

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=\frac {3}{10} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{1375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {111}{50} \, \sqrt {-2 \, x + 1} \]

[In]

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

[Out]

3/10*(-2*x + 1)^(3/2) + 1/1375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 11
1/50*sqrt(-2*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.07 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=\frac {3}{10} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{1375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {111}{50} \, \sqrt {-2 \, x + 1} \]

[In]

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

[Out]

3/10*(-2*x + 1)^(3/2) + 1/1375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x +
1))) - 111/50*sqrt(-2*x + 1)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx=\frac {3\,{\left (1-2\,x\right )}^{3/2}}{10}-\frac {111\,\sqrt {1-2\,x}}{50}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1375} \]

[In]

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

[Out]

(3*(1 - 2*x)^(3/2))/10 - (111*(1 - 2*x)^(1/2))/50 - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/1375