\(\int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx\) [2062]

Optimal result

Integrand size = 24, antiderivative size = 68 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}-\frac {27 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {2313 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

[Out]

-2313/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/550*(1-2*x)^(1/2)/(3+5*x)^2-27/1210*(1-2*x)^(1/2)
/(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {91, 79, 65, 212} \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {2313 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}}-\frac {27 \sqrt {1-2 x}}{1210 (5 x+3)}-\frac {\sqrt {1-2 x}}{550 (5 x+3)^2} \]

[In]

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

[Out]

-1/550*Sqrt[1 - 2*x]/(3 + 5*x)^2 - (27*Sqrt[1 - 2*x])/(1210*(3 + 5*x)) - (2313*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/(3025*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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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}& = -\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}+\frac {1}{550} \int \frac {729+990 x}{\sqrt {1-2 x} (3+5 x)^2} \, dx \\ & = -\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}-\frac {27 \sqrt {1-2 x}}{1210 (3+5 x)}+\frac {2313 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{6050} \\ & = -\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}-\frac {27 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {2313 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{6050} \\ & = -\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}-\frac {27 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {2313 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {-\frac {55 \sqrt {1-2 x} (416+675 x)}{(3+5 x)^2}-4626 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{332750} \]

[In]

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

[Out]

((-55*Sqrt[1 - 2*x]*(416 + 675*x))/(3 + 5*x)^2 - 4626*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/332750

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {1350 x^{2}+157 x -416}{6050 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {2313 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}\)	\(46\)
derivativedivides	\(\frac {\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{121}-\frac {137 \sqrt {1-2 x}}{275}}{\left (-6-10 x \right )^{2}}-\frac {2313 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}\)	\(48\)
default	\(\frac {\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{121}-\frac {137 \sqrt {1-2 x}}{275}}{\left (-6-10 x \right )^{2}}-\frac {2313 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}\)	\(48\)
pseudoelliptic	\(\frac {-4626 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}-55 \sqrt {1-2 x}\, \left (675 x +416\right )}{332750 \left (3+5 x \right )^{2}}\)	\(50\)
trager	\(-\frac {\left (675 x +416\right ) \sqrt {1-2 x}}{6050 \left (3+5 x \right )^{2}}-\frac {2313 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{332750}\)	\(67\)

[In]

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

[Out]

1/6050*(1350*x^2+157*x-416)/(3+5*x)^2/(1-2*x)^(1/2)-2313/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {2313 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (675 \, x + 416\right )} \sqrt {-2 \, x + 1}}{332750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/332750*(2313*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(675*x + 4
16)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (58) = 116\).

Time = 95.58 (sec) , antiderivative size = 330, normalized size of antiderivative = 4.85 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {9 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{1375} - \frac {24 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} + \frac {8 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} \]

[In]

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

[Out]

9*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/1375 - 24*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, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)
/5)))/25 + 8*Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/11 +
1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55)*s
qrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt
(1 - 2*x) < sqrt(55)/5)))/25

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {2313}{332750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {675 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1507 \, \sqrt {-2 \, x + 1}}{3025 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

[In]

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

[Out]

2313/332750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/3025*(675*(-2*x + 1
)^(3/2) - 1507*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {2313}{332750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {675 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1507 \, \sqrt {-2 \, x + 1}}{12100 \, {\left (5 \, x + 3\right )}^{2}} \]

[In]

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

[Out]

2313/332750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/12100*(67
5*(-2*x + 1)^(3/2) - 1507*sqrt(-2*x + 1))/(5*x + 3)^2

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.79 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {2313\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{166375}-\frac {\frac {137\,\sqrt {1-2\,x}}{6875}-\frac {27\,{\left (1-2\,x\right )}^{3/2}}{3025}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]

[In]

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

[Out]

- (2313*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/166375 - ((137*(1 - 2*x)^(1/2))/6875 - (27*(1 - 2*x)^(3
/2))/3025)/((44*x)/5 + (2*x - 1)^2 + 11/25)