\(\int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx\) [731]

Optimal result

Integrand size = 20, antiderivative size = 88 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=-\frac {\sqrt {1-x} \sqrt {1+x}}{x}-\frac {\sqrt {1-x} (1+x)^{3/2}}{3 x^2}-\frac {\sqrt {1-x} (1+x)^{5/2}}{3 x^3}-\text {arctanh}\left (\sqrt {1-x} \sqrt {1+x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {98, 96, 94, 212} \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=-\text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )-\frac {\sqrt {1-x} (x+1)^{5/2}}{3 x^3}-\frac {\sqrt {1-x} (x+1)^{3/2}}{3 x^2}-\frac {\sqrt {1-x} \sqrt {x+1}}{x} \]

[In]

Int[(1 + x)^(3/2)/(Sqrt[1 - x]*x^4),x]

[Out]

-((Sqrt[1 - x]*Sqrt[1 + x])/x) - (Sqrt[1 - x]*(1 + x)^(3/2))/(3*x^2) - (Sqrt[1 - x]*(1 + x)^(5/2))/(3*x^3) - A
rcTanh[Sqrt[1 - x]*Sqrt[1 + x]]

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.67 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=-\frac {\sqrt {1-x} \left (1+4 x+8 x^2+5 x^3\right )}{3 x^3 \sqrt {1+x}}-2 \text {arctanh}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \]

[In]

Integrate[(1 + x)^(3/2)/(Sqrt[1 - x]*x^4),x]

[Out]

-1/3*(Sqrt[1 - x]*(1 + 4*x + 8*x^2 + 5*x^3))/(x^3*Sqrt[1 + x]) - 2*ArcTanh[Sqrt[1 - x]/Sqrt[1 + x]]

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{3}+5 x^{2} \sqrt {-x^{2}+1}+3 x \sqrt {-x^{2}+1}+\sqrt {-x^{2}+1}\right )}{3 x^{3} \sqrt {-x^{2}+1}}\)	\(78\)
risch	\(\frac {\left (-1+x \right ) \sqrt {1+x}\, \left (5 x^{2}+3 x +1\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{3 x^{3} \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\)	\(88\)

[In]

int((1+x)^(3/2)/x^4/(1-x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.62 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=\frac {3 \, x^{3} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - {\left (5 \, x^{2} + 3 \, x + 1\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, x^{3}} \]

[In]

integrate((1+x)^(3/2)/x^4/(1-x)^(1/2),x, algorithm="fricas")

[Out]

1/3*(3*x^3*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - (5*x^2 + 3*x + 1)*sqrt(x + 1)*sqrt(-x + 1))/x^3

Sympy [F]

\[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=\int \frac {\left (x + 1\right )^{\frac {3}{2}}}{x^{4} \sqrt {1 - x}}\, dx \]

[In]

integrate((1+x)**(3/2)/x**4/(1-x)**(1/2),x)

[Out]

Integral((x + 1)**(3/2)/(x**4*sqrt(1 - x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=-\frac {5 \, \sqrt {-x^{2} + 1}}{3 \, x} - \frac {\sqrt {-x^{2} + 1}}{x^{2}} - \frac {\sqrt {-x^{2} + 1}}{3 \, x^{3}} - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

[In]

integrate((1+x)^(3/2)/x^4/(1-x)^(1/2),x, algorithm="maxima")

[Out]

-5/3*sqrt(-x^2 + 1)/x - sqrt(-x^2 + 1)/x^2 - 1/3*sqrt(-x^2 + 1)/x^3 - log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (68) = 136\).

Time = 0.35 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.16 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=-\frac {4 \, {\left (3 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{5} - 32 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{3} + \frac {144 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}{\sqrt {x + 1}} - \frac {144 \, \sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}}{3 \, {\left ({\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{2} - 4\right )}^{3}} - \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} + 2 \right |}\right ) + \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \right |}\right ) \]

[In]

integrate((1+x)^(3/2)/x^4/(1-x)^(1/2),x, algorithm="giac")

[Out]

-4/3*(3*((sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))^5 - 32*((sqrt(2) - sqrt(
-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))^3 + 144*(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - 14
4*sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))/(((sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-
x + 1)))^2 - 4)^3 - log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)) + 2))
+ log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)) - 2))

Mupad [F(-1)]

Timed out. \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx=\int \frac {{\left (x+1\right )}^{3/2}}{x^4\,\sqrt {1-x}} \,d x \]

[In]

int((x + 1)^(3/2)/(x^4*(1 - x)^(1/2)),x)

[Out]

int((x + 1)^(3/2)/(x^4*(1 - x)^(1/2)), x)