3.20.15 \(\int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx\) [1915]

3.20.15.1 Optimal result

Integrand size = 24, antiderivative size = 102 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {68 \sqrt {1-2 x}}{3 (3+5 x)}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)}-134 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+138 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

output

138/5*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-134/3*arctanh(1/7*21^( 
1/2)*(1-2*x)^(1/2))*21^(1/2)-68/3*(1-2*x)^(1/2)/(3+5*x)+7/3*(1-2*x)^(1/2)/ 
(2+3*x)/(3+5*x)
 

3.20.15.2 Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x} (43+68 x)}{6+19 x+15 x^2}-134 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+138 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

input

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

output

-((Sqrt[1 - 2*x]*(43 + 68*x))/(6 + 19*x + 15*x^2)) - 134*Sqrt[7/3]*ArcTanh 
[Sqrt[3/7]*Sqrt[1 - 2*x]] + 138*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x 
]]
 

3.20.15.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {109, 168, 27, 174, 73, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {1}{3} \int \frac {89-101 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^2}dx+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (-\frac {1}{11} \int \frac {33 (111-68 x)}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {68 \sqrt {1-2 x}}{5 x+3}\right )+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-3 \int \frac {111-68 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {68 \sqrt {1-2 x}}{5 x+3}\right )+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (-3 \left (759 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-469 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )-\frac {68 \sqrt {1-2 x}}{5 x+3}\right )+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (-3 \left (469 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-759 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {68 \sqrt {1-2 x}}{5 x+3}\right )+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (-3 \left (134 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-138 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {68 \sqrt {1-2 x}}{5 x+3}\right )+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)}\)

input

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

output

(7*Sqrt[1 - 2*x])/(3*(2 + 3*x)*(3 + 5*x)) + ((-68*Sqrt[1 - 2*x])/(3 + 5*x) 
 - 3*(134*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 138*Sqrt[11/5]*ArcT 
anh[Sqrt[5/11]*Sqrt[1 - 2*x]]))/3
 

3.20.15.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

3.20.15.4 Maple [A] (verified)

Time = 3.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {\left (43+68 x \right ) \left (-1+2 x \right )}{\left (15 x^{2}+19 x +6\right ) \sqrt {1-2 x}}-\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}+\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{5}\)	\(68\)
derivativedivides	\(\frac {22 \sqrt {1-2 x}}{5 \left (-\frac {6}{5}-2 x \right )}+\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{5}+\frac {14 \sqrt {1-2 x}}{3 \left (-\frac {4}{3}-2 x \right )}-\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\)	\(70\)
default	\(\frac {22 \sqrt {1-2 x}}{5 \left (-\frac {6}{5}-2 x \right )}+\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{5}+\frac {14 \sqrt {1-2 x}}{3 \left (-\frac {4}{3}-2 x \right )}-\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\)	\(70\)
pseudoelliptic	\(\frac {-670 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (15 x^{2}+19 x +6\right ) \sqrt {21}+414 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (15 x^{2}+19 x +6\right ) \sqrt {55}-15 \sqrt {1-2 x}\, \left (43+68 x \right )}{225 x^{2}+285 x +90}\)	\(85\)
trager	\(-\frac {\left (43+68 x \right ) \sqrt {1-2 x}}{15 x^{2}+19 x +6}+\frac {67 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{3}+\frac {69 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{5}\)	\(116\)

input

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

output

(43+68*x)*(-1+2*x)/(15*x^2+19*x+6)/(1-2*x)^(1/2)-134/3*arctanh(1/7*21^(1/2 
)*(1-2*x)^(1/2))*21^(1/2)+138/5*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1 
/2)
 

3.20.15.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {207 \, \sqrt {11} \sqrt {5} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 335 \, \sqrt {7} \sqrt {3} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 15 \, {\left (68 \, x + 43\right )} \sqrt {-2 \, x + 1}}{15 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \]

input

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

output

1/15*(207*sqrt(11)*sqrt(5)*(15*x^2 + 19*x + 6)*log(-(sqrt(11)*sqrt(5)*sqrt 
(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 335*sqrt(7)*sqrt(3)*(15*x^2 + 19*x + 6) 
*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 15*(68*x + 43 
)*sqrt(-2*x + 1))/(15*x^2 + 19*x + 6)
 

3.20.15.6 Sympy [A] (verification not implemented)

Time = 31.54 (sec) , antiderivative size = 318, normalized size of antiderivative = 3.12 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=22 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right ) - 14 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right ) - 196 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right ) - 484 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right ) \]

input

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

output

22*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21 
)/3)) - 14*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + 
 sqrt(55)/5)) - 196*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x)/7 - 1 
)/4 + log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/4 - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 
 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (sqrt(1 - 2*x) > -sqrt 
(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3))) - 484*Piecewise((sqrt(55)*(-log(s 
qrt(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)))
 

3.20.15.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {69}{5} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {67}{3} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {4 \, {\left (34 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \]

input

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

output

-69/5*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x 
+ 1))) + 67/3*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sq 
rt(-2*x + 1))) + 4*(34*(-2*x + 1)^(3/2) - 77*sqrt(-2*x + 1))/(15*(2*x - 1) 
^2 + 136*x + 9)
 

3.20.15.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.14 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {69}{5} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {67}{3} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {4 \, {\left (34 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \]

input

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

output

-69/5*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5* 
sqrt(-2*x + 1))) + 67/3*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1 
))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4*(34*(-2*x + 1)^(3/2) - 77*sqrt(-2*x 
+ 1))/(15*(2*x - 1)^2 + 136*x + 9)
 

3.20.15.9 Mupad [B] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {138\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{5}-\frac {134\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {\frac {308\,\sqrt {1-2\,x}}{15}-\frac {136\,{\left (1-2\,x\right )}^{3/2}}{15}}{\frac {136\,x}{15}+{\left (2\,x-1\right )}^2+\frac {3}{5}} \]

input

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

output

(138*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/5 - (134*21^(1/2)*atan 
h((21^(1/2)*(1 - 2*x)^(1/2))/7))/3 - ((308*(1 - 2*x)^(1/2))/15 - (136*(1 - 
 2*x)^(3/2))/15)/((136*x)/15 + (2*x - 1)^2 + 3/5)