3.3.15 \(\int \frac {1}{x^3 (1+x)^{3/2}} \, dx\) [215]

Optimal. Leaf size=52 \[ \frac {15}{4 \sqrt {1+x}}-\frac {1}{2 x^2 \sqrt {1+x}}+\frac {5}{4 x \sqrt {1+x}}-\frac {15}{4} \tanh ^{-1}\left (\sqrt {1+x}\right ) \]

[Out]

-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)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {44, 53, 65, 213} \begin {gather*} -\frac {1}{2 x^2 \sqrt {x+1}}+\frac {5}{4 x \sqrt {x+1}}+\frac {15}{4 \sqrt {x+1}}-\frac {15}{4} \tanh ^{-1}\left (\sqrt {x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + x)^(3/2)),x]

[Out]

15/(4*Sqrt[1 + x]) - 1/(2*x^2*Sqrt[1 + x]) + 5/(4*x*Sqrt[1 + x]) - (15*ArcTanh[Sqrt[1 + x]])/4

Rule 44

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && 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 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 213

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])

Rubi steps

\begin {align*} \int \frac {1}{x^3 (1+x)^{3/2}} \, dx &=\frac {2}{x^2 \sqrt {1+x}}+5 \int \frac {1}{x^3 \sqrt {1+x}} \, dx\\ &=\frac {2}{x^2 \sqrt {1+x}}-\frac {5 \sqrt {1+x}}{2 x^2}-\frac {15}{4} \int \frac {1}{x^2 \sqrt {1+x}} \, dx\\ &=\frac {2}{x^2 \sqrt {1+x}}-\frac {5 \sqrt {1+x}}{2 x^2}+\frac {15 \sqrt {1+x}}{4 x}+\frac {15}{8} \int \frac {1}{x \sqrt {1+x}} \, dx\\ &=\frac {2}{x^2 \sqrt {1+x}}-\frac {5 \sqrt {1+x}}{2 x^2}+\frac {15 \sqrt {1+x}}{4 x}+\frac {15}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {2}{x^2 \sqrt {1+x}}-\frac {5 \sqrt {1+x}}{2 x^2}+\frac {15 \sqrt {1+x}}{4 x}-\frac {15}{4} \tanh ^{-1}\left (\sqrt {1+x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 36, normalized size = 0.69 \begin {gather*} \frac {1}{4} \left (\frac {-2+5 x+15 x^2}{x^2 \sqrt {1+x}}-15 \tanh ^{-1}\left (\sqrt {1+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 + x)^(3/2)),x]

[Out]

((-2 + 5*x + 15*x^2)/(x^2*Sqrt[1 + x]) - 15*ArcTanh[Sqrt[1 + x]])/4

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 73, normalized size = 1.40

method	result	size
risch	\(\frac {15 x^{2}+5 x -2}{4 \sqrt {1+x}\, x^{2}}-\frac {15 \arctanh \left (\sqrt {1+x}\right )}{4}\)	\(30\)
trager	\(\frac {15 x^{2}+5 x -2}{4 \sqrt {1+x}\, x^{2}}+\frac {15 \ln \left (\frac {2 \sqrt {1+x}-2-x}{x}\right )}{8}\)	\(41\)
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	\(\frac {-\frac {\sqrt {\pi }}{2 x^{2}}+\frac {3 \sqrt {\pi }}{2 x}+\frac {15 \left (\frac {47}{30}-2 \ln \left (2\right )+\ln \left (x \right )\right ) \sqrt {\pi }}{8}+\frac {\sqrt {\pi }\, \left (-47 x^{2}-24 x +8\right )}{16 x^{2}}-\frac {\sqrt {\pi }\, \left (-60 x^{2}-20 x +8\right )}{16 x^{2} \sqrt {1+x}}-\frac {15 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+x}}{2}\right )}{4}}{\sqrt {\pi }}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/(-1+(1+x)^(1/2))^2+7/8/(-1+(1+x)^(1/2))+15/8*ln(-1+(1+x)^(1/2))+2/(1+x)^(1/2)+1/8/(1+(1+x)^(1/2))^2+7/8/(
1+(1+x)^(1/2))-15/8*ln(1+(1+x)^(1/2))

________________________________________________________________________________________

Maxima [A]
time = 1.04, size = 55, normalized size = 1.06 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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) + 1
5/8*log(sqrt(x + 1) - 1)

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 63, normalized size = 1.21 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 1.53, size = 3966, normalized size = 76.27 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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
*sqrt(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) - 4
48*(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) + 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))
- 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)) + 16
80*(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)**(7/2) + 224*(x + 1)**(5/2) - 64*(x + 1)**(3/2) + 8*sqrt(x + 1)
) + 840*I*pi*(x + 1)**(11/2)/(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)) - 2100*
(x + 1)**(9/2)*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)) -
1050*I*pi*(x + 1)**(9/2)/(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)**(7/2)*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)) + 840*
I*pi*(x + 1)**(7/2)/(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)) - 840*(x + 1)**(
5/2)*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)**(5/2)/(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)**(3/2)*a
coth(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)
**(3/2)/(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)) - 30*sqrt(x + 1)*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)) - 15*I*pi*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)) + 30*(x + 1)**8/(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)) - 230*(x + 1)**7/(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)) + 766*(x + 1)**6/(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*sqr
t(x + 1)) - 1446*(x + 1)**5/(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)) + 1690*(
x + 1)**4/(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)) - 1250*(x + 1)**3/(8*(x +
1)**(17/2) - 64*(x + 1)**(15/2) + 224*(x + 1)**...

________________________________________________________________________________________

Giac [A]
time = 0.52, size = 49, normalized size = 0.94 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

________________________________________________________________________________________

Mupad [B]
time = 0.05, size = 43, normalized size = 0.83 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(x + 1)^(3/2)),x)

[Out]

- (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))

________________________________________________________________________________________