\(\int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx\) [1925]

Optimal result

Integrand size = 24, antiderivative size = 118 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=-\frac {103 \sqrt {1-2 x}}{6 (3+5 x)^2}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^2}+\frac {207 \sqrt {1-2 x}}{2 (3+5 x)}+204 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6933 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \]

[Out]

204*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-6933/55*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-103/6*(
1-2*x)^(1/2)/(3+5*x)^2+7/3*(1-2*x)^(1/2)/(2+3*x)/(3+5*x)^2+207/2*(1-2*x)^(1/2)/(3+5*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {100, 156, 162, 65, 212} \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=204 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6933 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}}+\frac {207 \sqrt {1-2 x}}{2 (5 x+3)}-\frac {103 \sqrt {1-2 x}}{6 (5 x+3)^2}+\frac {7 \sqrt {1-2 x}}{3 (3 x+2) (5 x+3)^2} \]

[In]

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

[Out]

(-103*Sqrt[1 - 2*x])/(6*(3 + 5*x)^2) + (7*Sqrt[1 - 2*x])/(3*(2 + 3*x)*(3 + 5*x)^2) + (207*Sqrt[1 - 2*x])/(2*(3
+ 5*x)) + 204*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - (6933*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] + Dist[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 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> 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] + Dist[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 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(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 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 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^2}+\frac {1}{3} \int \frac {124-171 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^3} \, dx \\ & = -\frac {103 \sqrt {1-2 x}}{6 (3+5 x)^2}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^2}-\frac {1}{66} \int \frac {8910-10197 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^2} \, dx \\ & = -\frac {103 \sqrt {1-2 x}}{6 (3+5 x)^2}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^2}+\frac {207 \sqrt {1-2 x}}{2 (3+5 x)}+\frac {1}{726} \int \frac {368082-225423 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx \\ & = -\frac {103 \sqrt {1-2 x}}{6 (3+5 x)^2}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^2}+\frac {207 \sqrt {1-2 x}}{2 (3+5 x)}-2142 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {6933}{2} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = -\frac {103 \sqrt {1-2 x}}{6 (3+5 x)^2}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^2}+\frac {207 \sqrt {1-2 x}}{2 (3+5 x)}+2142 \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {6933}{2} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = -\frac {103 \sqrt {1-2 x}}{6 (3+5 x)^2}+\frac {7 \sqrt {1-2 x}}{3 (2+3 x) (3+5 x)^2}+\frac {207 \sqrt {1-2 x}}{2 (3+5 x)}+204 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6933 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {\sqrt {1-2 x} \left (1178+3830 x+3105 x^2\right )}{2 (2+3 x) (3+5 x)^2}+204 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6933 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \]

[In]

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

[Out]

(Sqrt[1 - 2*x]*(1178 + 3830*x + 3105*x^2))/(2*(2 + 3*x)*(3 + 5*x)^2) + 204*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 -
2*x]] - (6933*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/Sqrt[55]

Maple [A] (verified)

Time = 3.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.64

method	result	size
risch	\(-\frac {6210 x^{3}+4555 x^{2}-1474 x -1178}{2 \left (3+5 x \right )^{2} \sqrt {1-2 x}\, \left (2+3 x \right )}+204 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}-\frac {6933 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{55}\)	\(76\)
derivativedivides	\(\frac {-685 \left (1-2 x \right )^{\frac {3}{2}}+1485 \sqrt {1-2 x}}{\left (-6-10 x \right )^{2}}-\frac {6933 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{55}-\frac {14 \sqrt {1-2 x}}{-\frac {4}{3}-2 x}+204 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}\)	\(82\)
default	\(\frac {-685 \left (1-2 x \right )^{\frac {3}{2}}+1485 \sqrt {1-2 x}}{\left (-6-10 x \right )^{2}}-\frac {6933 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{55}-\frac {14 \sqrt {1-2 x}}{-\frac {4}{3}-2 x}+204 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}\)	\(82\)
pseudoelliptic	\(\frac {22440 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \left (3+5 x \right )^{2} \sqrt {21}-13866 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right ) \left (3+5 x \right )^{2} \sqrt {55}+55 \sqrt {1-2 x}\, \left (3105 x^{2}+3830 x +1178\right )}{110 \left (2+3 x \right ) \left (3+5 x \right )^{2}}\)	\(97\)
trager	\(\frac {\left (3105 x^{2}+3830 x +1178\right ) \sqrt {1-2 x}}{2 \left (3+5 x \right )^{2} \left (2+3 x \right )}+\frac {6933 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{110}-102 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )\)	\(123\)

[In]

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

[Out]

-1/2*(6210*x^3+4555*x^2-1474*x-1178)/(3+5*x)^2/(1-2*x)^(1/2)/(2+3*x)+204*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*2
1^(1/2)-6933/55*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {6933 \, \sqrt {55} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 11220 \, \sqrt {21} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 55 \, {\left (3105 \, x^{2} + 3830 \, x + 1178\right )} \sqrt {-2 \, x + 1}}{110 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \]

[In]

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

[Out]

1/110*(6933*sqrt(55)*(75*x^3 + 140*x^2 + 87*x + 18)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 11220
*sqrt(21)*(75*x^3 + 140*x^2 + 87*x + 18)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 55*(3105*x^2 + 3
830*x + 1178)*sqrt(-2*x + 1))/(75*x^3 + 140*x^2 + 87*x + 18)

Sympy [A] (verification not implemented)

Time = 63.41 (sec) , antiderivative size = 486, normalized size of antiderivative = 4.12 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=- 101 \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 ) + \frac {707 \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 )}{11} + 588 \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 ) + 3080 \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 ) + 968 \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 ) \]

[In]

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

[Out]

-101*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21)/3)) + 707*sqrt(55)*(log(sqrt(1 -
2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/11 + 588*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))) + 3080*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))) + 968*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)))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.08 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {6933}{110} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - 102 \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {3105 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 13870 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 15477 \, \sqrt {-2 \, x + 1}}{75 \, {\left (2 \, x - 1\right )}^{3} + 505 \, {\left (2 \, x - 1\right )}^{2} + 2266 \, x - 286} \]

[In]

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

[Out]

6933/110*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 102*sqrt(21)*log(-(sqrt(
21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + (3105*(-2*x + 1)^(5/2) - 13870*(-2*x + 1)^(3/2) + 154
77*sqrt(-2*x + 1))/(75*(2*x - 1)^3 + 505*(2*x - 1)^2 + 2266*x - 286)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.04 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {6933}{110} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - 102 \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {21 \, \sqrt {-2 \, x + 1}}{3 \, x + 2} - \frac {5 \, {\left (137 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 297 \, \sqrt {-2 \, x + 1}\right )}}{4 \, {\left (5 \, x + 3\right )}^{2}} \]

[In]

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

[Out]

6933/110*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 102*sqrt(21)*l
og(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 21*sqrt(-2*x + 1)/(3*x + 2) - 5/4*
(137*(-2*x + 1)^(3/2) - 297*sqrt(-2*x + 1))/(5*x + 3)^2

Mupad [B] (verification not implemented)

Time = 1.60 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^3} \, dx=204\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )-\frac {6933\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{55}+\frac {\frac {5159\,\sqrt {1-2\,x}}{25}-\frac {2774\,{\left (1-2\,x\right )}^{3/2}}{15}+\frac {207\,{\left (1-2\,x\right )}^{5/2}}{5}}{\frac {2266\,x}{75}+\frac {101\,{\left (2\,x-1\right )}^2}{15}+{\left (2\,x-1\right )}^3-\frac {286}{75}} \]

[In]

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

[Out]

204*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7) - (6933*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/55 + (
(5159*(1 - 2*x)^(1/2))/25 - (2774*(1 - 2*x)^(3/2))/15 + (207*(1 - 2*x)^(5/2))/5)/((2266*x)/75 + (101*(2*x - 1)
^2)/15 + (2*x - 1)^3 - 286/75)