\(\int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx\) [1977]

Optimal result

Integrand size = 24, antiderivative size = 133 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=\frac {7 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {301 \sqrt {1-2 x}}{36 (2+3 x)^3}+\frac {4313 \sqrt {1-2 x}}{72 (2+3 x)^2}+\frac {100145 \sqrt {1-2 x}}{168 (2+3 x)}+\frac {3454265 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{84 \sqrt {21}}-1210 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

7/12*(1-2*x)^(3/2)/(2+3*x)^4+3454265/1764*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-1210*arctanh(1/11*55^(1
/2)*(1-2*x)^(1/2))*55^(1/2)+301/36*(1-2*x)^(1/2)/(2+3*x)^3+4313/72*(1-2*x)^(1/2)/(2+3*x)^2+100145/168*(1-2*x)^
(1/2)/(2+3*x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {100, 154, 156, 162, 65, 212} \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=\frac {3454265 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{84 \sqrt {21}}-1210 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )+\frac {7 (1-2 x)^{3/2}}{12 (3 x+2)^4}+\frac {100145 \sqrt {1-2 x}}{168 (3 x+2)}+\frac {4313 \sqrt {1-2 x}}{72 (3 x+2)^2}+\frac {301 \sqrt {1-2 x}}{36 (3 x+2)^3} \]

[In]

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

[Out]

(7*(1 - 2*x)^(3/2))/(12*(2 + 3*x)^4) + (301*Sqrt[1 - 2*x])/(36*(2 + 3*x)^3) + (4313*Sqrt[1 - 2*x])/(72*(2 + 3*
x)^2) + (100145*Sqrt[1 - 2*x])/(168*(2 + 3*x)) + (3454265*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(84*Sqrt[21]) - 12
10*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*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 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 154

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*((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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]

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 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {1}{12} \int \frac {(195-159 x) \sqrt {1-2 x}}{(2+3 x)^4 (3+5 x)} \, dx \\ & = \frac {7 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {301 \sqrt {1-2 x}}{36 (2+3 x)^3}-\frac {1}{108} \int \frac {-15777+21621 x}{\sqrt {1-2 x} (2+3 x)^3 (3+5 x)} \, dx \\ & = \frac {7 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {301 \sqrt {1-2 x}}{36 (2+3 x)^3}+\frac {4313 \sqrt {1-2 x}}{72 (2+3 x)^2}-\frac {\int \frac {-1197315+1358595 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)} \, dx}{1512} \\ & = \frac {7 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {301 \sqrt {1-2 x}}{36 (2+3 x)^3}+\frac {4313 \sqrt {1-2 x}}{72 (2+3 x)^2}+\frac {100145 \sqrt {1-2 x}}{168 (2+3 x)}-\frac {\int \frac {-51509115+31545675 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx}{10584} \\ & = \frac {7 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {301 \sqrt {1-2 x}}{36 (2+3 x)^3}+\frac {4313 \sqrt {1-2 x}}{72 (2+3 x)^2}+\frac {100145 \sqrt {1-2 x}}{168 (2+3 x)}-\frac {3454265}{168} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+33275 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = \frac {7 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {301 \sqrt {1-2 x}}{36 (2+3 x)^3}+\frac {4313 \sqrt {1-2 x}}{72 (2+3 x)^2}+\frac {100145 \sqrt {1-2 x}}{168 (2+3 x)}+\frac {3454265}{168} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-33275 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = \frac {7 (1-2 x)^{3/2}}{12 (2+3 x)^4}+\frac {301 \sqrt {1-2 x}}{36 (2+3 x)^3}+\frac {4313 \sqrt {1-2 x}}{72 (2+3 x)^2}+\frac {100145 \sqrt {1-2 x}}{168 (2+3 x)}+\frac {3454265 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{84 \sqrt {21}}-1210 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=\frac {\sqrt {1-2 x} \left (844322+3730002 x+5498403 x^2+2703915 x^3\right )}{168 (2+3 x)^4}+\frac {3454265 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{84 \sqrt {21}}-1210 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[In]

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

[Out]

(Sqrt[1 - 2*x]*(844322 + 3730002*x + 5498403*x^2 + 2703915*x^3))/(168*(2 + 3*x)^4) + (3454265*ArcTanh[Sqrt[3/7
]*Sqrt[1 - 2*x]])/(84*Sqrt[21]) - 1210*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.56

method	result	size
risch	\(-\frac {5407830 x^{4}+8292891 x^{3}+1961601 x^{2}-2041358 x -844322}{168 \left (2+3 x \right )^{4} \sqrt {1-2 x}}+\frac {3454265 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1764}-1210 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}\)	\(74\)
derivativedivides	\(-1210 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}-\frac {162 \left (\frac {100145 \left (1-2 x \right )^{\frac {7}{2}}}{504}-\frac {909931 \left (1-2 x \right )^{\frac {5}{2}}}{648}+\frac {2144065 \left (1-2 x \right )^{\frac {3}{2}}}{648}-\frac {5053615 \sqrt {1-2 x}}{1944}\right )}{\left (-4-6 x \right )^{4}}+\frac {3454265 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1764}\)	\(84\)
default	\(-1210 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}-\frac {162 \left (\frac {100145 \left (1-2 x \right )^{\frac {7}{2}}}{504}-\frac {909931 \left (1-2 x \right )^{\frac {5}{2}}}{648}+\frac {2144065 \left (1-2 x \right )^{\frac {3}{2}}}{648}-\frac {5053615 \sqrt {1-2 x}}{1944}\right )}{\left (-4-6 x \right )^{4}}+\frac {3454265 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1764}\)	\(84\)
pseudoelliptic	\(\frac {6908530 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right )^{4} \sqrt {21}-4268880 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right )^{4} \sqrt {55}+21 \sqrt {1-2 x}\, \left (2703915 x^{3}+5498403 x^{2}+3730002 x +844322\right )}{3528 \left (2+3 x \right )^{4}}\)	\(85\)
trager	\(\frac {\left (2703915 x^{3}+5498403 x^{2}+3730002 x +844322\right ) \sqrt {1-2 x}}{168 \left (2+3 x \right )^{4}}+605 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )+\frac {3454265 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{3528}\)	\(121\)

[In]

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

[Out]

-1/168*(5407830*x^4+8292891*x^3+1961601*x^2-2041358*x-844322)/(2+3*x)^4/(1-2*x)^(1/2)+3454265/1764*arctanh(1/7
*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-1210*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 = 150, normalized size of antiderivative = 1.13 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=\frac {2134440 \, \sqrt {55} {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 3454265 \, \sqrt {21} {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (2703915 \, x^{3} + 5498403 \, x^{2} + 3730002 \, x + 844322\right )} \sqrt {-2 \, x + 1}}{3528 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

1/3528*(2134440*sqrt(55)*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x
+ 3)) + 3454265*sqrt(21)*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*
x + 2)) + 21*(2703915*x^3 + 5498403*x^2 + 3730002*x + 844322)*sqrt(-2*x + 1))/(81*x^4 + 216*x^3 + 216*x^2 + 96
*x + 16)

Sympy [A] (verification not implemented)

Time = 125.25 (sec) , antiderivative size = 838, normalized size of antiderivative = 6.30 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-6655*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21)/3))/7 + 605*sqrt(55)*(log(sqrt(
1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5)) + 15972*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)*sq
rt(1 - 2*x)/7 - 1)))/147, (sqrt(1 - 2*x) > -sqrt(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3))) - 57512*Piecewise((sq
rt(21)*(3*log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/16 - 3*log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/16 + 3/(16*(sqrt(21)*sqrt
(1 - 2*x)/7 + 1)) + 1/(16*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)**2) + 3/(16*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)) - 1/(16*(
sqrt(21)*sqrt(1 - 2*x)/7 - 1)**2))/1029, (sqrt(1 - 2*x) > -sqrt(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3)))/9 + 22
736*Piecewise((sqrt(21)*(-5*log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/32 + 5*log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/32 - 5/
(32*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(16*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)**2) - 1/(48*(sqrt(21)*sqrt(1 - 2*x)
/7 + 1)**3) - 5/(32*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)) + 1/(16*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)**2) - 1/(48*(sqrt(2
1)*sqrt(1 - 2*x)/7 - 1)**3))/7203, (sqrt(1 - 2*x) > -sqrt(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3)))/9 - 10976*Pi
ecewise((sqrt(21)*(35*log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/256 - 35*log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/256 + 35/(2
56*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)) + 15/(256*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)**2) + 5/(192*(sqrt(21)*sqrt(1 - 2*
x)/7 + 1)**3) + 1/(128*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)**4) + 35/(256*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)) - 15/(256*
(sqrt(21)*sqrt(1 - 2*x)/7 - 1)**2) + 5/(192*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)**3) - 1/(128*(sqrt(21)*sqrt(1 - 2*x
)/7 - 1)**4))/50421, (sqrt(1 - 2*x) > -sqrt(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3)))/9

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.10 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=605 \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {3454265}{3528} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2703915 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 19108551 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 45025365 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 35375305 \, \sqrt {-2 \, x + 1}}{84 \, {\left (81 \, {\left (2 \, x - 1\right )}^{4} + 756 \, {\left (2 \, x - 1\right )}^{3} + 2646 \, {\left (2 \, x - 1\right )}^{2} + 8232 \, x - 1715\right )}} \]

[In]

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

[Out]

605*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 3454265/3528*sqrt(21)*log(-(s
qrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/84*(2703915*(-2*x + 1)^(7/2) - 19108551*(-2*x +
1)^(5/2) + 45025365*(-2*x + 1)^(3/2) - 35375305*sqrt(-2*x + 1))/(81*(2*x - 1)^4 + 756*(2*x - 1)^3 + 2646*(2*x
- 1)^2 + 8232*x - 1715)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=605 \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {3454265}{3528} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {2703915 \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + 19108551 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 45025365 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 35375305 \, \sqrt {-2 \, x + 1}}{1344 \, {\left (3 \, x + 2\right )}^{4}} \]

[In]

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

[Out]

605*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 3454265/3528*sqrt(2
1)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/1344*(2703915*(2*x - 1)^3*sq
rt(-2*x + 1) + 19108551*(2*x - 1)^2*sqrt(-2*x + 1) - 45025365*(-2*x + 1)^(3/2) + 35375305*sqrt(-2*x + 1))/(3*x
+ 2)^4

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^5 (3+5 x)} \, dx=\frac {3454265\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{1764}-1210\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )+\frac {\frac {5053615\,\sqrt {1-2\,x}}{972}-\frac {2144065\,{\left (1-2\,x\right )}^{3/2}}{324}+\frac {909931\,{\left (1-2\,x\right )}^{5/2}}{324}-\frac {100145\,{\left (1-2\,x\right )}^{7/2}}{252}}{\frac {2744\,x}{27}+\frac {98\,{\left (2\,x-1\right )}^2}{3}+\frac {28\,{\left (2\,x-1\right )}^3}{3}+{\left (2\,x-1\right )}^4-\frac {1715}{81}} \]

[In]

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

[Out]

(3454265*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/1764 - 1210*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/1
1) + ((5053615*(1 - 2*x)^(1/2))/972 - (2144065*(1 - 2*x)^(3/2))/324 + (909931*(1 - 2*x)^(5/2))/324 - (100145*(
1 - 2*x)^(7/2))/252)/((2744*x)/27 + (98*(2*x - 1)^2)/3 + (28*(2*x - 1)^3)/3 + (2*x - 1)^4 - 1715/81)