\(\int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx\) [2109]

Optimal result

Integrand size = 22, antiderivative size = 41 \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {7}{11 \sqrt {1-2 x}}-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \]

[Out]

-2/605*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+7/11/(1-2*x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {79, 65, 212} \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {7}{11 \sqrt {1-2 x}}-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \]

[In]

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

[Out]

7/(11*Sqrt[1 - 2*x]) - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(11*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 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*} \text {integral}& = \frac {7}{11 \sqrt {1-2 x}}+\frac {1}{11} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = \frac {7}{11 \sqrt {1-2 x}}-\frac {1}{11} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = \frac {7}{11 \sqrt {1-2 x}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {7}{11 \sqrt {1-2 x}}-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \]

[In]

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

[Out]

7/(11*Sqrt[1 - 2*x]) - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(11*Sqrt[55])

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}+\frac {7}{11 \sqrt {1-2 x}}\)	\(29\)
default	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}+\frac {7}{11 \sqrt {1-2 x}}\)	\(29\)
risch	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}+\frac {7}{11 \sqrt {1-2 x}}\)	\(29\)
pseudoelliptic	\(-\frac {2 \left (\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}\, \sqrt {1-2 x}-\frac {385}{2}\right )}{605 \sqrt {1-2 x}}\)	\(36\)
trager	\(-\frac {7 \sqrt {1-2 x}}{11 \left (-1+2 x \right )}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{605}\)	\(62\)

[In]

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

[Out]

-2/605*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+7/11/(1-2*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.29 \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {\sqrt {55} {\left (2 \, x - 1\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 385 \, \sqrt {-2 \, x + 1}}{605 \, {\left (2 \, x - 1\right )}} \]

[In]

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

[Out]

1/605*(sqrt(55)*(2*x - 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 385*sqrt(-2*x + 1))/(2*x - 1)

Sympy [A] (verification not implemented)

Time = 1.82 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.24 \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {\sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{605} + \frac {7}{11 \sqrt {1 - 2 x}} \]

[In]

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

[Out]

sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/605 + 7/(11*sqrt(1 - 2*x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {1}{605} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {7}{11 \, \sqrt {-2 \, x + 1}} \]

[In]

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

[Out]

1/605*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 7/11/sqrt(-2*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.20 \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {1}{605} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {7}{11 \, \sqrt {-2 \, x + 1}} \]

[In]

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

[Out]

1/605*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 7/11/sqrt(-2*x +
1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int \frac {2+3 x}{(1-2 x)^{3/2} (3+5 x)} \, dx=\frac {7}{11\,\sqrt {1-2\,x}}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{605} \]

[In]

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

[Out]

7/(11*(1 - 2*x)^(1/2)) - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/605