3.735 \(\int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx\)

Optimal. Leaf size=69 \[ \frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2+\frac {1}{24} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{8} \sqrt {x-1} \sqrt {x+1}+\frac {3}{8} \cosh ^{-1}(x) \]

[Out]

3/8*arccosh(x)+1/24*(7-2*x)*(-1+x)^(3/2)*(1+x)^(1/2)+1/4*(-1+x)^(3/2)*x^2*(1+x)^(1/2)-3/8*(-1+x)^(1/2)*(1+x)^(
1/2)

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Rubi [A]  time = 0.01, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {100, 147, 50, 52} \[ \frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2+\frac {1}{24} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{8} \sqrt {x-1} \sqrt {x+1}+\frac {3}{8} \cosh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x]*x^3)/Sqrt[1 + x],x]

[Out]

(-3*Sqrt[-1 + x]*Sqrt[1 + x])/8 + ((7 - 2*x)*(-1 + x)^(3/2)*Sqrt[1 + x])/24 + ((-1 + x)^(3/2)*x^2*Sqrt[1 + x])
/4 + (3*ArcCosh[x])/8

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 100

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx &=\frac {1}{4} (-1+x)^{3/2} x^2 \sqrt {1+x}+\frac {1}{4} \int \frac {(2-x) \sqrt {-1+x} x}{\sqrt {1+x}} \, dx\\ &=\frac {1}{24} (7-2 x) (-1+x)^{3/2} \sqrt {1+x}+\frac {1}{4} (-1+x)^{3/2} x^2 \sqrt {1+x}-\frac {3}{8} \int \frac {\sqrt {-1+x}}{\sqrt {1+x}} \, dx\\ &=-\frac {3}{8} \sqrt {-1+x} \sqrt {1+x}+\frac {1}{24} (7-2 x) (-1+x)^{3/2} \sqrt {1+x}+\frac {1}{4} (-1+x)^{3/2} x^2 \sqrt {1+x}+\frac {3}{8} \int \frac {1}{\sqrt {-1+x} \sqrt {1+x}} \, dx\\ &=-\frac {3}{8} \sqrt {-1+x} \sqrt {1+x}+\frac {1}{24} (7-2 x) (-1+x)^{3/2} \sqrt {1+x}+\frac {1}{4} (-1+x)^{3/2} x^2 \sqrt {1+x}+\frac {3}{8} \cosh ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 76, normalized size = 1.10 \[ \frac {\sqrt {\frac {x-1}{x+1}} \left (6 x^5-8 x^4+3 x^3-8 x^2-18 \sqrt {1-x^2} \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )-9 x+16\right )}{24 (x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(-1 + x)/(1 + x)]*(16 - 9*x - 8*x^2 + 3*x^3 - 8*x^4 + 6*x^5 - 18*Sqrt[1 - x^2]*ArcSin[Sqrt[1 - x]/Sqrt[2
]]))/(24*(-1 + x))

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fricas [A]  time = 0.42, size = 46, normalized size = 0.67 \[ \frac {1}{24} \, {\left (6 \, x^{3} - 8 \, x^{2} + 9 \, x - 16\right )} \sqrt {x + 1} \sqrt {x - 1} - \frac {3}{8} \, \log \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*x^3 - 8*x^2 + 9*x - 16)*sqrt(x + 1)*sqrt(x - 1) - 3/8*log(sqrt(x + 1)*sqrt(x - 1) - x)

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giac [A]  time = 0.33, size = 47, normalized size = 0.68 \[ \frac {1}{24} \, {\left ({\left (2 \, {\left (3 \, x - 10\right )} {\left (x + 1\right )} + 43\right )} {\left (x + 1\right )} - 39\right )} \sqrt {x + 1} \sqrt {x - 1} - \frac {3}{4} \, \log \left (\sqrt {x + 1} - \sqrt {x - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((2*(3*x - 10)*(x + 1) + 43)*(x + 1) - 39)*sqrt(x + 1)*sqrt(x - 1) - 3/4*log(sqrt(x + 1) - sqrt(x - 1))

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maple [A]  time = 0.01, size = 76, normalized size = 1.10 \[ \frac {\sqrt {x -1}\, \sqrt {x +1}\, \left (6 \sqrt {x^{2}-1}\, x^{3}-8 \sqrt {x^{2}-1}\, x^{2}+9 \sqrt {x^{2}-1}\, x +9 \ln \left (x +\sqrt {x^{2}-1}\right )-16 \sqrt {x^{2}-1}\right )}{24 \sqrt {x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(x-1)^(1/2)*(x+1)^(1/2)*(6*x^3*(x^2-1)^(1/2)-8*(x^2-1)^(1/2)*x^2+9*x*(x^2-1)^(1/2)+9*ln(x+(x^2-1)^(1/2))-
16*(x^2-1)^(1/2))/(x^2-1)^(1/2)

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maxima [A]  time = 0.88, size = 55, normalized size = 0.80 \[ \frac {1}{4} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} x - \frac {1}{3} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} + \frac {5}{8} \, \sqrt {x^{2} - 1} x - \sqrt {x^{2} - 1} + \frac {3}{8} \, \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^2 - 1)^(3/2)*x - 1/3*(x^2 - 1)^(3/2) + 5/8*sqrt(x^2 - 1)*x - sqrt(x^2 - 1) + 3/8*log(2*x + 2*sqrt(x^2 -
1))

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mupad [B]  time = 12.86, size = 473, normalized size = 6.86 \[ \frac {3\,\mathrm {atanh}\left (\frac {\sqrt {x-1}-\mathrm {i}}{\sqrt {x+1}-1}\right )}{2}+\frac {\frac {23\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^3}{2\,{\left (\sqrt {x+1}-1\right )}^3}-\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^4\,64{}\mathrm {i}}{{\left (\sqrt {x+1}-1\right )}^4}+\frac {333\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^5}{2\,{\left (\sqrt {x+1}-1\right )}^5}+\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^6\,256{}\mathrm {i}}{3\,{\left (\sqrt {x+1}-1\right )}^6}+\frac {671\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^7}{2\,{\left (\sqrt {x+1}-1\right )}^7}-\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^8\,128{}\mathrm {i}}{3\,{\left (\sqrt {x+1}-1\right )}^8}+\frac {671\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^9}{2\,{\left (\sqrt {x+1}-1\right )}^9}+\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^{10}\,256{}\mathrm {i}}{3\,{\left (\sqrt {x+1}-1\right )}^{10}}+\frac {333\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^{11}}{2\,{\left (\sqrt {x+1}-1\right )}^{11}}-\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^{12}\,64{}\mathrm {i}}{{\left (\sqrt {x+1}-1\right )}^{12}}+\frac {23\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^{13}}{2\,{\left (\sqrt {x+1}-1\right )}^{13}}-\frac {3\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^{15}}{2\,{\left (\sqrt {x+1}-1\right )}^{15}}-\frac {3\,\left (\sqrt {x-1}-\mathrm {i}\right )}{2\,\left (\sqrt {x+1}-1\right )}}{1+\frac {28\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {x+1}-1\right )}^4}-\frac {56\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {x+1}-1\right )}^6}+\frac {70\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {x+1}-1\right )}^8}-\frac {56\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {x+1}-1\right )}^{10}}+\frac {28\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {x+1}-1\right )}^{12}}-\frac {8\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^{14}}{{\left (\sqrt {x+1}-1\right )}^{14}}+\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^{16}}{{\left (\sqrt {x+1}-1\right )}^{16}}-\frac {8\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*atanh(((x - 1)^(1/2) - 1i)/((x + 1)^(1/2) - 1)))/2 + ((23*((x - 1)^(1/2) - 1i)^3)/(2*((x + 1)^(1/2) - 1)^3)
- (((x - 1)^(1/2) - 1i)^4*64i)/((x + 1)^(1/2) - 1)^4 + (333*((x - 1)^(1/2) - 1i)^5)/(2*((x + 1)^(1/2) - 1)^5)
+ (((x - 1)^(1/2) - 1i)^6*256i)/(3*((x + 1)^(1/2) - 1)^6) + (671*((x - 1)^(1/2) - 1i)^7)/(2*((x + 1)^(1/2) -
1)^7) - (((x - 1)^(1/2) - 1i)^8*128i)/(3*((x + 1)^(1/2) - 1)^8) + (671*((x - 1)^(1/2) - 1i)^9)/(2*((x + 1)^(1/
2) - 1)^9) + (((x - 1)^(1/2) - 1i)^10*256i)/(3*((x + 1)^(1/2) - 1)^10) + (333*((x - 1)^(1/2) - 1i)^11)/(2*((x
+ 1)^(1/2) - 1)^11) - (((x - 1)^(1/2) - 1i)^12*64i)/((x + 1)^(1/2) - 1)^12 + (23*((x - 1)^(1/2) - 1i)^13)/(2*(
(x + 1)^(1/2) - 1)^13) - (3*((x - 1)^(1/2) - 1i)^15)/(2*((x + 1)^(1/2) - 1)^15) - (3*((x - 1)^(1/2) - 1i))/(2*
((x + 1)^(1/2) - 1)))/((28*((x - 1)^(1/2) - 1i)^4)/((x + 1)^(1/2) - 1)^4 - (8*((x - 1)^(1/2) - 1i)^2)/((x + 1)
^(1/2) - 1)^2 - (56*((x - 1)^(1/2) - 1i)^6)/((x + 1)^(1/2) - 1)^6 + (70*((x - 1)^(1/2) - 1i)^8)/((x + 1)^(1/2)
- 1)^8 - (56*((x - 1)^(1/2) - 1i)^10)/((x + 1)^(1/2) - 1)^10 + (28*((x - 1)^(1/2) - 1i)^12)/((x + 1)^(1/2) -
1)^12 - (8*((x - 1)^(1/2) - 1i)^14)/((x + 1)^(1/2) - 1)^14 + ((x - 1)^(1/2) - 1i)^16/((x + 1)^(1/2) - 1)^16 +
1)

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sympy [A]  time = 13.85, size = 83, normalized size = 1.20 \[ \frac {\left (x - 1\right )^{\frac {7}{2}} \sqrt {x + 1}}{4} + \frac {5 \left (x - 1\right )^{\frac {5}{2}} \sqrt {x + 1}}{12} + \frac {11 \left (x - 1\right )^{\frac {3}{2}} \sqrt {x + 1}}{24} - \frac {3 \sqrt {x - 1} \sqrt {x + 1}}{8} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {x - 1}}{2} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - 1)**(7/2)*sqrt(x + 1)/4 + 5*(x - 1)**(5/2)*sqrt(x + 1)/12 + 11*(x - 1)**(3/2)*sqrt(x + 1)/24 - 3*sqrt(x -
1)*sqrt(x + 1)/8 + 3*asinh(sqrt(2)*sqrt(x - 1)/2)/4

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