\(\int \frac {1}{(1-x)^{7/2} x^5} \, dx\) [216]

Optimal result

Integrand size = 13, antiderivative size = 118 \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=\frac {3003}{320 (1-x)^{5/2}}+\frac {1001}{64 (1-x)^{3/2}}+\frac {3003}{64 \sqrt {1-x}}-\frac {1}{4 (1-x)^{5/2} x^4}-\frac {13}{24 (1-x)^{5/2} x^3}-\frac {143}{96 (1-x)^{5/2} x^2}-\frac {429}{64 (1-x)^{5/2} x}-\frac {3003}{64} \text {arctanh}\left (\sqrt {1-x}\right ) \] Output:

3003/320/(1-x)^(5/2)+1001/64/(1-x)^(3/2)-1/4/(1-x)^(5/2)/x^4-13/24/(1-x)^( 
5/2)/x^3-143/96/(1-x)^(5/2)/x^2-429/64/(1-x)^(5/2)/x-3003/64*arctanh((1-x) 
^(1/2))+3003/64/(1-x)^(1/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.59 \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=-\frac {240+520 x+1430 x^2+6435 x^3-69069 x^4+105105 x^5-45045 x^6+45045 (1-x)^{5/2} x^4 \text {arctanh}\left (\sqrt {1-x}\right )}{960 (1-x)^{5/2} x^4} \] Input:

Integrate[1/((1 - x)^(7/2)*x^5),x]
 

Output:

-1/960*(240 + 520*x + 1430*x^2 + 6435*x^3 - 69069*x^4 + 105105*x^5 - 45045 
*x^6 + 45045*(1 - x)^(5/2)*x^4*ArcTanh[Sqrt[1 - x]])/((1 - x)^(5/2)*x^4)
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {52, 52, 52, 52, 61, 61, 61, 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.

\(\displaystyle \int \frac {1}{(1-x)^{7/2} x^5} \, dx\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {13}{8} \int \frac {1}{(1-x)^{7/2} x^4}dx-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \int \frac {1}{(1-x)^{7/2} x^3}dx-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \left (\frac {9}{4} \int \frac {1}{(1-x)^{7/2} x^2}dx-\frac {1}{2 (1-x)^{5/2} x^2}\right )-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \left (\frac {9}{4} \left (\frac {7}{2} \int \frac {1}{(1-x)^{7/2} x}dx-\frac {1}{(1-x)^{5/2} x}\right )-\frac {1}{2 (1-x)^{5/2} x^2}\right )-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \left (\frac {9}{4} \left (\frac {7}{2} \left (\int \frac {1}{(1-x)^{5/2} x}dx+\frac {2}{5 (1-x)^{5/2}}\right )-\frac {1}{(1-x)^{5/2} x}\right )-\frac {1}{2 (1-x)^{5/2} x^2}\right )-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \left (\frac {9}{4} \left (\frac {7}{2} \left (\int \frac {1}{(1-x)^{3/2} x}dx+\frac {2}{3 (1-x)^{3/2}}+\frac {2}{5 (1-x)^{5/2}}\right )-\frac {1}{(1-x)^{5/2} x}\right )-\frac {1}{2 (1-x)^{5/2} x^2}\right )-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \left (\frac {9}{4} \left (\frac {7}{2} \left (\int \frac {1}{\sqrt {1-x} x}dx+\frac {2}{\sqrt {1-x}}+\frac {2}{3 (1-x)^{3/2}}+\frac {2}{5 (1-x)^{5/2}}\right )-\frac {1}{(1-x)^{5/2} x}\right )-\frac {1}{2 (1-x)^{5/2} x^2}\right )-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \left (\frac {9}{4} \left (\frac {7}{2} \left (-2 \int \frac {1}{x}d\sqrt {1-x}+\frac {2}{\sqrt {1-x}}+\frac {2}{3 (1-x)^{3/2}}+\frac {2}{5 (1-x)^{5/2}}\right )-\frac {1}{(1-x)^{5/2} x}\right )-\frac {1}{2 (1-x)^{5/2} x^2}\right )-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {13}{8} \left (\frac {11}{6} \left (\frac {9}{4} \left (\frac {7}{2} \left (-2 \text {arctanh}\left (\sqrt {1-x}\right )+\frac {2}{\sqrt {1-x}}+\frac {2}{3 (1-x)^{3/2}}+\frac {2}{5 (1-x)^{5/2}}\right )-\frac {1}{(1-x)^{5/2} x}\right )-\frac {1}{2 (1-x)^{5/2} x^2}\right )-\frac {1}{3 (1-x)^{5/2} x^3}\right )-\frac {1}{4 (1-x)^{5/2} x^4}\)

Input:

Int[1/((1 - x)^(7/2)*x^5),x]
 

Output:

-1/4*1/((1 - x)^(5/2)*x^4) + (13*(-1/3*1/((1 - x)^(5/2)*x^3) + (11*(-1/2*1 
/((1 - x)^(5/2)*x^2) + (9*(-(1/((1 - x)^(5/2)*x)) + (7*(2/(5*(1 - x)^(5/2) 
) + 2/(3*(1 - x)^(3/2)) + 2/Sqrt[1 - x] - 2*ArcTanh[Sqrt[1 - x]]))/2))/4)) 
/6))/8
 

Defintions of rubi rules used

rule 52

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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

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] - Simp[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] && 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 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 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])
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50

method	result	size
risch	\(\frac {45045 x^{6}-105105 x^{5}+69069 x^{4}-6435 x^{3}-1430 x^{2}-520 x -240}{960 \left (-1+x \right )^{2} \sqrt {1-x}\, x^{4}}-\frac {3003 \,\operatorname {arctanh}\left (\sqrt {1-x}\right )}{64}\)	\(59\)
trager	\(-\frac {\left (45045 x^{6}-105105 x^{5}+69069 x^{4}-6435 x^{3}-1430 x^{2}-520 x -240\right ) \sqrt {1-x}}{960 \left (-1+x \right )^{3} x^{4}}-\frac {3003 \ln \left (-\frac {2 \sqrt {1-x}+2-x}{x}\right )}{128}\)	\(71\)
pseudoelliptic	\(\frac {\frac {3003 x^{4} \sqrt {1-x}\, \left (-1+x \right )^{2} \ln \left (\sqrt {1-x}-1\right )}{128}-\frac {3003 x^{4} \sqrt {1-x}\, \left (-1+x \right )^{2} \ln \left (\sqrt {1-x}+1\right )}{128}+\frac {3003 x^{6}}{64}-\frac {7007 x^{5}}{64}+\frac {23023 x^{4}}{320}-\frac {429 x^{3}}{64}-\frac {143 x^{2}}{96}-\frac {13 x}{24}-\frac {1}{4}}{\left (\sqrt {1-x}-1\right )^{4} \left (\sqrt {1-x}+1\right )^{4} \left (1-x \right )^{\frac {5}{2}}}\)	\(115\)
meijerg	\(\frac {-\frac {\sqrt {\pi }}{4 x^{4}}-\frac {7 \sqrt {\pi }}{6 x^{3}}-\frac {63 \sqrt {\pi }}{16 x^{2}}-\frac {231 \sqrt {\pi }}{16 x}+\frac {3003 \left (\frac {329177}{180180}-2 \ln \left (2\right )+\ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{128}+\frac {\sqrt {\pi }\, \left (-329177 x^{4}+110880 x^{3}+30240 x^{2}+8960 x +1920\right )}{7680 x^{4}}-\frac {\sqrt {\pi }\, \left (-180180 x^{6}+420420 x^{5}-276276 x^{4}+25740 x^{3}+5720 x^{2}+2080 x +960\right )}{3840 x^{4} \left (1-x \right )^{\frac {5}{2}}}-\frac {3003 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-x}}{2}\right )}{64}}{\sqrt {\pi }}\)	\(146\)
derivativedivides	\(-\frac {1}{64 \left (\sqrt {1-x}-1\right )^{4}}+\frac {17}{96 \left (\sqrt {1-x}-1\right )^{3}}-\frac {159}{128 \left (\sqrt {1-x}-1\right )^{2}}+\frac {1083}{128 \left (\sqrt {1-x}-1\right )}+\frac {3003 \ln \left (\sqrt {1-x}-1\right )}{128}+\frac {1}{64 \left (\sqrt {1-x}+1\right )^{4}}+\frac {17}{96 \left (\sqrt {1-x}+1\right )^{3}}+\frac {159}{128 \left (\sqrt {1-x}+1\right )^{2}}+\frac {1083}{128 \left (\sqrt {1-x}+1\right )}-\frac {3003 \ln \left (\sqrt {1-x}+1\right )}{128}+\frac {2}{5 \left (1-x \right )^{\frac {5}{2}}}+\frac {10}{3 \left (1-x \right )^{\frac {3}{2}}}+\frac {30}{\sqrt {1-x}}\)	\(157\)
default	\(-\frac {1}{64 \left (\sqrt {1-x}-1\right )^{4}}+\frac {17}{96 \left (\sqrt {1-x}-1\right )^{3}}-\frac {159}{128 \left (\sqrt {1-x}-1\right )^{2}}+\frac {1083}{128 \left (\sqrt {1-x}-1\right )}+\frac {3003 \ln \left (\sqrt {1-x}-1\right )}{128}+\frac {1}{64 \left (\sqrt {1-x}+1\right )^{4}}+\frac {17}{96 \left (\sqrt {1-x}+1\right )^{3}}+\frac {159}{128 \left (\sqrt {1-x}+1\right )^{2}}+\frac {1083}{128 \left (\sqrt {1-x}+1\right )}-\frac {3003 \ln \left (\sqrt {1-x}+1\right )}{128}+\frac {2}{5 \left (1-x \right )^{\frac {5}{2}}}+\frac {10}{3 \left (1-x \right )^{\frac {3}{2}}}+\frac {30}{\sqrt {1-x}}\)	\(157\)

Input:

int(1/(1-x)^(7/2)/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/960*(45045*x^6-105105*x^5+69069*x^4-6435*x^3-1430*x^2-520*x-240)/(-1+x)^ 
2/(1-x)^(1/2)/x^4-3003/64*arctanh((1-x)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=-\frac {45045 \, {\left (x^{7} - 3 \, x^{6} + 3 \, x^{5} - x^{4}\right )} \log \left (\sqrt {-x + 1} + 1\right ) - 45045 \, {\left (x^{7} - 3 \, x^{6} + 3 \, x^{5} - x^{4}\right )} \log \left (\sqrt {-x + 1} - 1\right ) + 2 \, {\left (45045 \, x^{6} - 105105 \, x^{5} + 69069 \, x^{4} - 6435 \, x^{3} - 1430 \, x^{2} - 520 \, x - 240\right )} \sqrt {-x + 1}}{1920 \, {\left (x^{7} - 3 \, x^{6} + 3 \, x^{5} - x^{4}\right )}} \] Input:

integrate(1/(1-x)^(7/2)/x^5,x, algorithm="fricas")
 

Output:

-1/1920*(45045*(x^7 - 3*x^6 + 3*x^5 - x^4)*log(sqrt(-x + 1) + 1) - 45045*( 
x^7 - 3*x^6 + 3*x^5 - x^4)*log(sqrt(-x + 1) - 1) + 2*(45045*x^6 - 105105*x 
^5 + 69069*x^4 - 6435*x^3 - 1430*x^2 - 520*x - 240)*sqrt(-x + 1))/(x^7 - 3 
*x^6 + 3*x^5 - x^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=\text {Timed out} \] Input:

integrate(1/(1-x)**(7/2)/x**5,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=\frac {45045 \, {\left (x - 1\right )}^{6} + 165165 \, {\left (x - 1\right )}^{5} + 219219 \, {\left (x - 1\right )}^{4} + 119691 \, {\left (x - 1\right )}^{3} + 18304 \, {\left (x - 1\right )}^{2} - 1664 \, x + 2048}{960 \, {\left ({\left (-x + 1\right )}^{\frac {13}{2}} - 4 \, {\left (-x + 1\right )}^{\frac {11}{2}} + 6 \, {\left (-x + 1\right )}^{\frac {9}{2}} - 4 \, {\left (-x + 1\right )}^{\frac {7}{2}} + {\left (-x + 1\right )}^{\frac {5}{2}}\right )}} - \frac {3003}{128} \, \log \left (\sqrt {-x + 1} + 1\right ) + \frac {3003}{128} \, \log \left (\sqrt {-x + 1} - 1\right ) \] Input:

integrate(1/(1-x)^(7/2)/x^5,x, algorithm="maxima")
 

Output:

1/960*(45045*(x - 1)^6 + 165165*(x - 1)^5 + 219219*(x - 1)^4 + 119691*(x - 
 1)^3 + 18304*(x - 1)^2 - 1664*x + 2048)/((-x + 1)^(13/2) - 4*(-x + 1)^(11 
/2) + 6*(-x + 1)^(9/2) - 4*(-x + 1)^(7/2) + (-x + 1)^(5/2)) - 3003/128*log 
(sqrt(-x + 1) + 1) + 3003/128*log(sqrt(-x + 1) - 1)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=\frac {2 \, {\left (225 \, {\left (x - 1\right )}^{2} - 25 \, x + 28\right )}}{15 \, {\left (x - 1\right )}^{2} \sqrt {-x + 1}} - \frac {3249 \, {\left (x - 1\right )}^{3} \sqrt {-x + 1} + 10633 \, {\left (x - 1\right )}^{2} \sqrt {-x + 1} - 11767 \, {\left (-x + 1\right )}^{\frac {3}{2}} + 4431 \, \sqrt {-x + 1}}{192 \, x^{4}} - \frac {3003}{128} \, \log \left (\sqrt {-x + 1} + 1\right ) + \frac {3003}{128} \, \log \left ({\left | \sqrt {-x + 1} - 1 \right |}\right ) \] Input:

integrate(1/(1-x)^(7/2)/x^5,x, algorithm="giac")
 

Output:

2/15*(225*(x - 1)^2 - 25*x + 28)/((x - 1)^2*sqrt(-x + 1)) - 1/192*(3249*(x 
 - 1)^3*sqrt(-x + 1) + 10633*(x - 1)^2*sqrt(-x + 1) - 11767*(-x + 1)^(3/2) 
 + 4431*sqrt(-x + 1))/x^4 - 3003/128*log(sqrt(-x + 1) + 1) + 3003/128*log( 
abs(sqrt(-x + 1) - 1))
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=\frac {\frac {286\,{\left (x-1\right )}^2}{15}-\frac {26\,x}{15}+\frac {39897\,{\left (x-1\right )}^3}{320}+\frac {73073\,{\left (x-1\right )}^4}{320}+\frac {11011\,{\left (x-1\right )}^5}{64}+\frac {3003\,{\left (x-1\right )}^6}{64}+\frac {32}{15}}{{\left (1-x\right )}^{5/2}-4\,{\left (1-x\right )}^{7/2}+6\,{\left (1-x\right )}^{9/2}-4\,{\left (1-x\right )}^{11/2}+{\left (1-x\right )}^{13/2}}-\frac {3003\,\mathrm {atanh}\left (\sqrt {1-x}\right )}{64} \] Input:

int(1/(x^5*(1 - x)^(7/2)),x)
 

Output:

((286*(x - 1)^2)/15 - (26*x)/15 + (39897*(x - 1)^3)/320 + (73073*(x - 1)^4 
)/320 + (11011*(x - 1)^5)/64 + (3003*(x - 1)^6)/64 + 32/15)/((1 - x)^(5/2) 
 - 4*(1 - x)^(7/2) + 6*(1 - x)^(9/2) - 4*(1 - x)^(11/2) + (1 - x)^(13/2)) 
- (3003*atanh((1 - x)^(1/2)))/64
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(1-x)^{7/2} x^5} \, dx=\frac {45045 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {1-x}-1\right ) x^{6}-90090 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {1-x}-1\right ) x^{5}+45045 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {1-x}-1\right ) x^{4}-45045 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {1-x}+1\right ) x^{6}+90090 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {1-x}+1\right ) x^{5}-45045 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {1-x}+1\right ) x^{4}+90090 x^{6}-210210 x^{5}+138138 x^{4}-12870 x^{3}-2860 x^{2}-1040 x -480}{1920 \sqrt {1-x}\, x^{4} \left (x^{2}-2 x +1\right )} \] Input:

int(1/(1-x)^(7/2)/x^5,x)
 

Output:

(45045*sqrt( - x + 1)*log(sqrt( - x + 1) - 1)*x**6 - 90090*sqrt( - x + 1)* 
log(sqrt( - x + 1) - 1)*x**5 + 45045*sqrt( - x + 1)*log(sqrt( - x + 1) - 1 
)*x**4 - 45045*sqrt( - x + 1)*log(sqrt( - x + 1) + 1)*x**6 + 90090*sqrt( - 
 x + 1)*log(sqrt( - x + 1) + 1)*x**5 - 45045*sqrt( - x + 1)*log(sqrt( - x 
+ 1) + 1)*x**4 + 90090*x**6 - 210210*x**5 + 138138*x**4 - 12870*x**3 - 286 
0*x**2 - 1040*x - 480)/(1920*sqrt( - x + 1)*x**4*(x**2 - 2*x + 1))