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

Optimal. Leaf size=88 \[ -\frac {\sqrt {1-x} (x+1)^{5/2}}{3 x^3}-\frac {\sqrt {1-x} (x+1)^{3/2}}{3 x^2}-\frac {\sqrt {1-x} \sqrt {x+1}}{x}-\tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \]

[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

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Rubi [A]  time = 0.02, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {96, 94, 92, 206} \[ -\frac {\sqrt {1-x} (x+1)^{5/2}}{3 x^3}-\frac {\sqrt {1-x} (x+1)^{3/2}}{3 x^2}-\frac {\sqrt {1-x} \sqrt {x+1}}{x}-\tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \]

Antiderivative was successfully verified.

[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 92

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 94

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[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx &=-\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}-\operatorname {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*}

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Mathematica [A]  time = 0.02, size = 66, normalized size = 0.75 \[ -\frac {-5 x^4-3 x^3+4 x^2+3 \sqrt {1-x^2} x^3 \tanh ^{-1}\left (\sqrt {1-x^2}\right )+3 x+1}{3 x^3 \sqrt {1-x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 55, normalized size = 0.62 \[ \frac {3 \, x^{3} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - {\left (5 \, x^{2} + 3 \, x + 1\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,-4,0,%%%{
4,[2]%%%}] at parameters values [-93.616423693]Warning, choosing root of [1,0,-4,0,%%%{4,[2]%%%}] at parameter
s values [-17.8804557086]1/3*(12*(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1
))^5-128*(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))^3+576*(2*sqrt(x+1)/(-
2*sqrt(-x+1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1)))/((2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-1/2*
(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))^2-4)^3-ln(abs(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))+2-1/2*(-2*sqrt(-x+1)+
2*sqrt(2))/sqrt(x+1)))+ln(abs(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-2-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))
)

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maple [A]  time = 0.02, size = 78, normalized size = 0.89 \[ -\frac {\sqrt {x +1}\, \sqrt {-x +1}\, \left (3 x^{3} \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )+5 \sqrt {-x^{2}+1}\, x^{2}+3 \sqrt {-x^{2}+1}\, x +\sqrt {-x^{2}+1}\right )}{3 \sqrt {-x^{2}+1}\, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.98, size = 68, normalized size = 0.77 \[ -\frac {5 \, \sqrt {-x^{2} + 1}}{3 \, x} - \frac {\sqrt {-x^{2} + 1}}{x^{2}} - \frac {\sqrt {-x^{2} + 1}}{3 \, x^{3}} - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (x+1\right )}^{3/2}}{x^4\,\sqrt {1-x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x + 1\right )^{\frac {3}{2}}}{x^{4} \sqrt {1 - x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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