\(\int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx\) [2194]

Optimal result

Integrand size = 24, antiderivative size = 101 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {2873}{73205 \sqrt {1-2 x}}+\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}-\frac {2873}{39930 \sqrt {1-2 x} (3+5 x)}-\frac {2873 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641 \sqrt {55}} \]

[Out]

49/66/(1-2*x)^(3/2)/(3+5*x)^2-2873/805255*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+2873/73205/(1-2*x)^(1/
2)-614/1815/(3+5*x)^2/(1-2*x)^(1/2)-2873/39930/(3+5*x)/(1-2*x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {91, 79, 44, 53, 65, 212} \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=-\frac {2873 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641 \sqrt {55}}+\frac {2873}{73205 \sqrt {1-2 x}}-\frac {2873}{39930 \sqrt {1-2 x} (5 x+3)}-\frac {614}{1815 \sqrt {1-2 x} (5 x+3)^2}+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)^2} \]

[In]

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

[Out]

2873/(73205*Sqrt[1 - 2*x]) + 49/(66*(1 - 2*x)^(3/2)*(3 + 5*x)^2) - 614/(1815*Sqrt[1 - 2*x]*(3 + 5*x)^2) - 2873
/(39930*Sqrt[1 - 2*x]*(3 + 5*x)) - (2873*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(14641*Sqrt[55])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {1}{66} \int \frac {-313+297 x}{(1-2 x)^{3/2} (3+5 x)^3} \, dx \\ & = \frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}+\frac {2873 \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^2} \, dx}{3630} \\ & = \frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}-\frac {2873}{39930 \sqrt {1-2 x} (3+5 x)}+\frac {2873 \int \frac {1}{(1-2 x)^{3/2} (3+5 x)} \, dx}{13310} \\ & = \frac {2873}{73205 \sqrt {1-2 x}}+\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}-\frac {2873}{39930 \sqrt {1-2 x} (3+5 x)}+\frac {2873 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{29282} \\ & = \frac {2873}{73205 \sqrt {1-2 x}}+\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}-\frac {2873}{39930 \sqrt {1-2 x} (3+5 x)}-\frac {2873 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{29282} \\ & = \frac {2873}{73205 \sqrt {1-2 x}}+\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}-\frac {2873}{39930 \sqrt {1-2 x} (3+5 x)}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641 \sqrt {55}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {-\frac {55 \left (-47568-107127 x+57460 x^2+172380 x^3\right )}{2 (1-2 x)^{3/2} (3+5 x)^2}-8619 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{2415765} \]

[In]

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

[Out]

((-55*(-47568 - 107127*x + 57460*x^2 + 172380*x^3))/(2*(1 - 2*x)^(3/2)*(3 + 5*x)^2) - 8619*Sqrt[55]*ArcTanh[Sq
rt[5/11]*Sqrt[1 - 2*x]])/2415765

Maple [A] (verified)

Time = 3.54 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.57

method	result	size
risch	\(\frac {172380 x^{3}+57460 x^{2}-107127 x -47568}{87846 \left (3+5 x \right )^{2} \sqrt {1-2 x}\, \left (-1+2 x \right )}-\frac {2873 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{805255}\)	\(58\)
derivativedivides	\(\frac {\frac {65 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {145 \sqrt {1-2 x}}{1331}}{\left (-6-10 x \right )^{2}}-\frac {2873 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{805255}+\frac {98}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {546}{14641 \sqrt {1-2 x}}\)	\(66\)
default	\(\frac {\frac {65 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {145 \sqrt {1-2 x}}{1331}}{\left (-6-10 x \right )^{2}}-\frac {2873 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{805255}+\frac {98}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {546}{14641 \sqrt {1-2 x}}\)	\(66\)
pseudoelliptic	\(\frac {\frac {2873 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (-1+2 x \right ) \left (3+5 x \right )^{2} \sqrt {55}}{805255}-\frac {28730 x^{3}}{14641}-\frac {28730 x^{2}}{43923}+\frac {35709 x}{29282}+\frac {7928}{14641}}{\left (1-2 x \right )^{\frac {3}{2}} \left (3+5 x \right )^{2}}\)	\(69\)
trager	\(-\frac {\left (172380 x^{3}+57460 x^{2}-107127 x -47568\right ) \sqrt {1-2 x}}{87846 \left (10 x^{2}+x -3\right )^{2}}+\frac {2873 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1610510}\)	\(80\)

[In]

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

[Out]

1/87846*(172380*x^3+57460*x^2-107127*x-47568)/(3+5*x)^2/(1-2*x)^(1/2)/(-1+2*x)-2873/805255*arctanh(1/11*55^(1/
2)*(1-2*x)^(1/2))*55^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {8619 \, \sqrt {55} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (172380 \, x^{3} + 57460 \, x^{2} - 107127 \, x - 47568\right )} \sqrt {-2 \, x + 1}}{4831530 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \]

[In]

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

[Out]

1/4831530*(8619*sqrt(55)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x +
3)) - 55*(172380*x^3 + 57460*x^2 - 107127*x - 47568)*sqrt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)

Sympy [A] (verification not implemented)

Time = 109.03 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.50 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {273 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{161051} - \frac {280 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{1331} + \frac {8 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{121} + \frac {546}{14641 \sqrt {1 - 2 x}} + \frac {98}{3993 \left (1 - 2 x\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

273*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/161051 - 280*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) < sqr
t(55)/5)))/1331 + 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*(sqr
t(55)*sqrt(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)))/121 + 546/(14641*sqrt(1 - 2*x)) + 98/(3993*(1 - 2*x)**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.91 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {2873}{1610510} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {43095 \, {\left (2 \, x - 1\right )}^{3} + 158015 \, {\left (2 \, x - 1\right )}^{2} + 159236 \, x - 210056}{43923 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 121 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]

[In]

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

[Out]

2873/1610510*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/43923*(43095*(2*x
- 1)^3 + 158015*(2*x - 1)^2 + 159236*x - 210056)/(25*(-2*x + 1)^(7/2) - 110*(-2*x + 1)^(5/2) + 121*(-2*x + 1)^
(3/2))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {2873}{1610510} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {28 \, {\left (117 \, x - 97\right )}}{43923 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {5 \, {\left (13 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 29 \, \sqrt {-2 \, x + 1}\right )}}{5324 \, {\left (5 \, x + 3\right )}^{2}} \]

[In]

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

[Out]

2873/1610510*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 28/43923*(
117*x - 97)/((2*x - 1)*sqrt(-2*x + 1)) + 5/5324*(13*(-2*x + 1)^(3/2) - 29*sqrt(-2*x + 1))/(5*x + 3)^2

Mupad [B] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=-\frac {2873\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{805255}-\frac {\frac {1316\,x}{9075}+\frac {2873\,{\left (2\,x-1\right )}^2}{19965}+\frac {2873\,{\left (2\,x-1\right )}^3}{73205}-\frac {1736}{9075}}{\frac {121\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {22\,{\left (1-2\,x\right )}^{5/2}}{5}+{\left (1-2\,x\right )}^{7/2}} \]

[In]

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

[Out]

- (2873*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/805255 - ((1316*x)/9075 + (2873*(2*x - 1)^2)/19965 + (2
873*(2*x - 1)^3)/73205 - 1736/9075)/((121*(1 - 2*x)^(3/2))/25 - (22*(1 - 2*x)^(5/2))/5 + (1 - 2*x)^(7/2))