\(\int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\) [2129]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}-\frac {2589 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}} \]

[Out]

-2589/366025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/22/(3+5*x)^2/(1-2*x)^(1/2)-613/605*(1-2*x)^(1/2)
/(3+5*x)^2-2589/13310*(1-2*x)^(1/2)/(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {91, 79, 44, 65, 212} \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=-\frac {2589 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}}-\frac {2589 \sqrt {1-2 x}}{13310 (5 x+3)}-\frac {613 \sqrt {1-2 x}}{605 (5 x+3)^2}+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^2} \]

[In]

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

[Out]

49/(22*Sqrt[1 - 2*x]*(3 + 5*x)^2) - (613*Sqrt[1 - 2*x])/(605*(3 + 5*x)^2) - (2589*Sqrt[1 - 2*x])/(13310*(3 + 5
*x)) - (2589*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6655*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 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}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {1}{22} \int \frac {-431+99 x}{\sqrt {1-2 x} (3+5 x)^3} \, dx \\ & = \frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}+\frac {2589 \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx}{1210} \\ & = \frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}+\frac {2589 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{13310} \\ & = \frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}-\frac {2589 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{13310} \\ & = \frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}-\frac {2589 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {\frac {55 \left (8392+29561 x+25890 x^2\right )}{\sqrt {1-2 x} (3+5 x)^2}-5178 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{732050} \]

[In]

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

[Out]

((55*(8392 + 29561*x + 25890*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)^2) - 5178*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/732050

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.52

method	result	size
risch	\(\frac {25890 x^{2}+29561 x +8392}{13310 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {2589 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}\)	\(46\)
pseudoelliptic	\(-\frac {2589 \left (\sqrt {55}\, \left (x +\frac {3}{5}\right )^{2} \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )-11 x^{2}-\frac {325171 x}{25890}-\frac {46156}{12945}\right )}{14641 \sqrt {1-2 x}\, \left (3+5 x \right )^{2}}\)	\(56\)
derivativedivides	\(\frac {\frac {139 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {141 \sqrt {1-2 x}}{605}}{\left (-6-10 x \right )^{2}}-\frac {2589 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}+\frac {98}{1331 \sqrt {1-2 x}}\)	\(57\)
default	\(\frac {\frac {139 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {141 \sqrt {1-2 x}}{605}}{\left (-6-10 x \right )^{2}}-\frac {2589 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}+\frac {98}{1331 \sqrt {1-2 x}}\)	\(57\)
trager	\(-\frac {\left (25890 x^{2}+29561 x +8392\right ) \sqrt {1-2 x}}{13310 \left (3+5 x \right )^{2} \left (-1+2 x \right )}-\frac {2589 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{732050}\)	\(79\)

[In]

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

[Out]

1/13310*(25890*x^2+29561*x+8392)/(3+5*x)^2/(1-2*x)^(1/2)-2589/366025*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 = 84, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {2589 \, \sqrt {55} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (25890 \, x^{2} + 29561 \, x + 8392\right )} \sqrt {-2 \, x + 1}}{732050 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]

[In]

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

[Out]

1/732050*(2589*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(
25890*x^2 + 29561*x + 8392)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)

Sympy [A] (verification not implemented)

Time = 107.70 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.89 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {49 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{14641} - \frac {272 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{605} + \frac {8 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{55} + \frac {98}{1331 \sqrt {1 - 2 x}} \]

[In]

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

[Out]

49*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/14641 - 272*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)))/605 + 8*Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/1
1 + 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(5
5)*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)))/55 + 98/(1331*sqrt(1 - 2*x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {2589}{732050} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {12945 \, {\left (2 \, x - 1\right )}^{2} + 110902 \, x + 3839}{6655 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \]

[In]

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

[Out]

2589/732050*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/6655*(12945*(2*x -
1)^2 + 110902*x + 3839)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {2589}{732050} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {98}{1331 \, \sqrt {-2 \, x + 1}} + \frac {695 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1551 \, \sqrt {-2 \, x + 1}}{26620 \, {\left (5 \, x + 3\right )}^{2}} \]

[In]

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

[Out]

2589/732050*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 98/1331/sqr
t(-2*x + 1) + 1/26620*(695*(-2*x + 1)^(3/2) - 1551*sqrt(-2*x + 1))/(5*x + 3)^2

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {\frac {10082\,x}{15125}+\frac {2589\,{\left (2\,x-1\right )}^2}{33275}+\frac {349}{15125}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}}-\frac {2589\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{366025} \]

[In]

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

[Out]

((10082*x)/15125 + (2589*(2*x - 1)^2)/33275 + 349/15125)/((121*(1 - 2*x)^(1/2))/25 - (22*(1 - 2*x)^(3/2))/5 +
(1 - 2*x)^(5/2)) - (2589*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/366025