Integrand size = 11, antiderivative size = 52 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=\frac {15}{4 \sqrt {1+x}}-\frac {1}{2 x^2 \sqrt {1+x}}+\frac {5}{4 x \sqrt {1+x}}-\frac {15}{4} \text {arctanh}\left (\sqrt {1+x}\right ) \] Output:
-15/4*arctanh((1+x)^(1/2))+15/4/(1+x)^(1/2)-1/2/x^2/(1+x)^(1/2)+5/4/x/(1+x )^(1/2)
Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=\frac {1}{4} \left (\frac {-2+5 x+15 x^2}{x^2 \sqrt {1+x}}-15 \text {arctanh}\left (\sqrt {1+x}\right )\right ) \] Input:
Integrate[1/(x^3*(1 + x)^(3/2)),x]
Output:
((-2 + 5*x + 15*x^2)/(x^2*Sqrt[1 + x]) - 15*ArcTanh[Sqrt[1 + x]])/4
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {52, 52, 61, 73, 220}
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.
\(\displaystyle \int \frac {1}{x^3 (x+1)^{3/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5}{4} \int \frac {1}{x^2 (x+1)^{3/2}}dx-\frac {1}{2 x^2 \sqrt {x+1}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5}{4} \left (-\frac {3}{2} \int \frac {1}{x (x+1)^{3/2}}dx-\frac {1}{x \sqrt {x+1}}\right )-\frac {1}{2 x^2 \sqrt {x+1}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5}{4} \left (-\frac {3}{2} \left (\int \frac {1}{x \sqrt {x+1}}dx+\frac {2}{\sqrt {x+1}}\right )-\frac {1}{x \sqrt {x+1}}\right )-\frac {1}{2 x^2 \sqrt {x+1}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5}{4} \left (-\frac {3}{2} \left (2 \int \frac {1}{x}d\sqrt {x+1}+\frac {2}{\sqrt {x+1}}\right )-\frac {1}{x \sqrt {x+1}}\right )-\frac {1}{2 x^2 \sqrt {x+1}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {5}{4} \left (-\frac {3}{2} \left (\frac {2}{\sqrt {x+1}}-2 \text {arctanh}\left (\sqrt {x+1}\right )\right )-\frac {1}{x \sqrt {x+1}}\right )-\frac {1}{2 x^2 \sqrt {x+1}}\) |
Input:
Int[1/(x^3*(1 + x)^(3/2)),x]
Output:
-1/2*1/(x^2*Sqrt[1 + x]) - (5*(-(1/(x*Sqrt[1 + x])) - (3*(2/Sqrt[1 + x] - 2*ArcTanh[Sqrt[1 + x]]))/2))/4
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]
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]
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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {15 x^{2}+5 x -2}{4 x^{2} \sqrt {1+x}}-\frac {15 \,\operatorname {arctanh}\left (\sqrt {1+x}\right )}{4}\) | \(30\) |
trager | \(\frac {15 x^{2}+5 x -2}{4 x^{2} \sqrt {1+x}}+\frac {15 \ln \left (\frac {2 \sqrt {1+x}-2-x}{x}\right )}{8}\) | \(41\) |
derivativedivides | \(\frac {1}{8 \left (1+\sqrt {1+x}\right )^{2}}+\frac {7}{8 \left (1+\sqrt {1+x}\right )}-\frac {15 \ln \left (1+\sqrt {1+x}\right )}{8}+\frac {2}{\sqrt {1+x}}-\frac {1}{8 \left (-1+\sqrt {1+x}\right )^{2}}+\frac {7}{8 \left (-1+\sqrt {1+x}\right )}+\frac {15 \ln \left (-1+\sqrt {1+x}\right )}{8}\) | \(73\) |
default | \(\frac {1}{8 \left (1+\sqrt {1+x}\right )^{2}}+\frac {7}{8 \left (1+\sqrt {1+x}\right )}-\frac {15 \ln \left (1+\sqrt {1+x}\right )}{8}+\frac {2}{\sqrt {1+x}}-\frac {1}{8 \left (-1+\sqrt {1+x}\right )^{2}}+\frac {7}{8 \left (-1+\sqrt {1+x}\right )}+\frac {15 \ln \left (-1+\sqrt {1+x}\right )}{8}\) | \(73\) |
meijerg | \(\frac {-\frac {\sqrt {\pi }}{2 x^{2}}+\frac {3 \sqrt {\pi }}{2 x}+\frac {15 \left (\frac {47}{30}-2 \ln \left (2\right )+\ln \left (x \right )\right ) \sqrt {\pi }}{8}+\frac {\sqrt {\pi }\, \left (-47 x^{2}-24 x +8\right )}{16 x^{2}}-\frac {\sqrt {\pi }\, \left (-60 x^{2}-20 x +8\right )}{16 x^{2} \sqrt {1+x}}-\frac {15 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+x}}{2}\right )}{4}}{\sqrt {\pi }}\) | \(92\) |
Input:
int(1/x^3/(1+x)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4*(15*x^2+5*x-2)/x^2/(1+x)^(1/2)-15/4*arctanh((1+x)^(1/2))
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=-\frac {15 \, {\left (x^{3} + x^{2}\right )} \log \left (\sqrt {x + 1} + 1\right ) - 15 \, {\left (x^{3} + x^{2}\right )} \log \left (\sqrt {x + 1} - 1\right ) - 2 \, {\left (15 \, x^{2} + 5 \, x - 2\right )} \sqrt {x + 1}}{8 \, {\left (x^{3} + x^{2}\right )}} \] Input:
integrate(1/x^3/(1+x)^(3/2),x, algorithm="fricas")
Output:
-1/8*(15*(x^3 + x^2)*log(sqrt(x + 1) + 1) - 15*(x^3 + x^2)*log(sqrt(x + 1) - 1) - 2*(15*x^2 + 5*x - 2)*sqrt(x + 1))/(x^3 + x^2)
Result contains complex when optimal does not.
Time = 1.48 (sec) , antiderivative size = 3966, normalized size of antiderivative = 76.27 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/x**3/(1+x)**(3/2),x)
Output:
Piecewise((-30*(x + 1)**(17/2)*acoth(sqrt(x + 1))/(8*(x + 1)**(17/2) - 64* (x + 1)**(15/2) + 224*(x + 1)**(13/2) - 448*(x + 1)**(11/2) + 560*(x + 1)* *(9/2) - 448*(x + 1)**(7/2) + 224*(x + 1)**(5/2) - 64*(x + 1)**(3/2) + 8*s qrt(x + 1)) - 15*I*pi*(x + 1)**(17/2)/(8*(x + 1)**(17/2) - 64*(x + 1)**(15 /2) + 224*(x + 1)**(13/2) - 448*(x + 1)**(11/2) + 560*(x + 1)**(9/2) - 448 *(x + 1)**(7/2) + 224*(x + 1)**(5/2) - 64*(x + 1)**(3/2) + 8*sqrt(x + 1)) + 240*(x + 1)**(15/2)*acoth(sqrt(x + 1))/(8*(x + 1)**(17/2) - 64*(x + 1)** (15/2) + 224*(x + 1)**(13/2) - 448*(x + 1)**(11/2) + 560*(x + 1)**(9/2) - 448*(x + 1)**(7/2) + 224*(x + 1)**(5/2) - 64*(x + 1)**(3/2) + 8*sqrt(x + 1 )) + 120*I*pi*(x + 1)**(15/2)/(8*(x + 1)**(17/2) - 64*(x + 1)**(15/2) + 22 4*(x + 1)**(13/2) - 448*(x + 1)**(11/2) + 560*(x + 1)**(9/2) - 448*(x + 1) **(7/2) + 224*(x + 1)**(5/2) - 64*(x + 1)**(3/2) + 8*sqrt(x + 1)) - 840*(x + 1)**(13/2)*acoth(sqrt(x + 1))/(8*(x + 1)**(17/2) - 64*(x + 1)**(15/2) + 224*(x + 1)**(13/2) - 448*(x + 1)**(11/2) + 560*(x + 1)**(9/2) - 448*(x + 1)**(7/2) + 224*(x + 1)**(5/2) - 64*(x + 1)**(3/2) + 8*sqrt(x + 1)) - 420 *I*pi*(x + 1)**(13/2)/(8*(x + 1)**(17/2) - 64*(x + 1)**(15/2) + 224*(x + 1 )**(13/2) - 448*(x + 1)**(11/2) + 560*(x + 1)**(9/2) - 448*(x + 1)**(7/2) + 224*(x + 1)**(5/2) - 64*(x + 1)**(3/2) + 8*sqrt(x + 1)) + 1680*(x + 1)** (11/2)*acoth(sqrt(x + 1))/(8*(x + 1)**(17/2) - 64*(x + 1)**(15/2) + 224*(x + 1)**(13/2) - 448*(x + 1)**(11/2) + 560*(x + 1)**(9/2) - 448*(x + 1)*...
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=\frac {15 \, {\left (x + 1\right )}^{2} - 25 \, x - 17}{4 \, {\left ({\left (x + 1\right )}^{\frac {5}{2}} - 2 \, {\left (x + 1\right )}^{\frac {3}{2}} + \sqrt {x + 1}\right )}} - \frac {15}{8} \, \log \left (\sqrt {x + 1} + 1\right ) + \frac {15}{8} \, \log \left (\sqrt {x + 1} - 1\right ) \] Input:
integrate(1/x^3/(1+x)^(3/2),x, algorithm="maxima")
Output:
1/4*(15*(x + 1)^2 - 25*x - 17)/((x + 1)^(5/2) - 2*(x + 1)^(3/2) + sqrt(x + 1)) - 15/8*log(sqrt(x + 1) + 1) + 15/8*log(sqrt(x + 1) - 1)
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=\frac {2}{\sqrt {x + 1}} + \frac {7 \, {\left (x + 1\right )}^{\frac {3}{2}} - 9 \, \sqrt {x + 1}}{4 \, x^{2}} - \frac {15}{8} \, \log \left (\sqrt {x + 1} + 1\right ) + \frac {15}{8} \, \log \left ({\left | \sqrt {x + 1} - 1 \right |}\right ) \] Input:
integrate(1/x^3/(1+x)^(3/2),x, algorithm="giac")
Output:
2/sqrt(x + 1) + 1/4*(7*(x + 1)^(3/2) - 9*sqrt(x + 1))/x^2 - 15/8*log(sqrt( x + 1) + 1) + 15/8*log(abs(sqrt(x + 1) - 1))
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=-\frac {15\,\mathrm {atanh}\left (\sqrt {x+1}\right )}{4}-\frac {\frac {25\,x}{4}-\frac {15\,{\left (x+1\right )}^2}{4}+\frac {17}{4}}{\sqrt {x+1}-2\,{\left (x+1\right )}^{3/2}+{\left (x+1\right )}^{5/2}} \] Input:
int(1/(x^3*(x + 1)^(3/2)),x)
Output:
- (15*atanh((x + 1)^(1/2)))/4 - ((25*x)/4 - (15*(x + 1)^2)/4 + 17/4)/((x + 1)^(1/2) - 2*(x + 1)^(3/2) + (x + 1)^(5/2))
Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^3 (1+x)^{3/2}} \, dx=\frac {15 \sqrt {x +1}\, \mathrm {log}\left (\sqrt {x +1}-1\right ) x^{2}-15 \sqrt {x +1}\, \mathrm {log}\left (\sqrt {x +1}+1\right ) x^{2}+30 x^{2}+10 x -4}{8 \sqrt {x +1}\, x^{2}} \] Input:
int(1/x^3/(1+x)^(3/2),x)
Output:
(15*sqrt(x + 1)*log(sqrt(x + 1) - 1)*x**2 - 15*sqrt(x + 1)*log(sqrt(x + 1) + 1)*x**2 + 30*x**2 + 10*x - 4)/(8*sqrt(x + 1)*x**2)