\(\int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx\) [40]

Optimal result

Integrand size = 18, antiderivative size = 82 \[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=-\frac {3}{8} \sqrt {-1+x} \sqrt {1+x}+\frac {5}{24} (-1+x)^{3/2} \sqrt {1+x}-\frac {1}{12} (-1+x)^{5/2} \sqrt {1+x}+\frac {1}{4} (-1+x)^{3/2} x^2 \sqrt {1+x}+\frac {3 \text {arccosh}(x)}{8} \] Output:

-3/8*(-1+x)^(1/2)*(1+x)^(1/2)+5/24*(-1+x)^(3/2)*(1+x)^(1/2)-1/12*(-1+x)^(5 
/2)*(1+x)^(1/2)+1/4*(-1+x)^(3/2)*x^2*(1+x)^(1/2)+3/8*arccosh(x)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=\frac {\sqrt {\frac {-1+x}{1+x}} \left (\sqrt {-1+x} \left (-16-7 x+x^2-2 x^3+6 x^4\right )-18 \sqrt {1+x} \log \left (\sqrt {-1+x}-\sqrt {1+x}\right )\right )}{24 \sqrt {-1+x}} \] Input:

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

Output:

(Sqrt[(-1 + x)/(1 + x)]*(Sqrt[-1 + x]*(-16 - 7*x + x^2 - 2*x^3 + 6*x^4) - 
18*Sqrt[1 + x]*Log[Sqrt[-1 + x] - Sqrt[1 + x]]))/(24*Sqrt[-1 + x])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {111, 164, 60, 43}

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.

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

\(\Big \downarrow \) 111

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

\(\Big \downarrow \) 164

\(\displaystyle \frac {1}{4} \left (\frac {1}{6} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{2} \int \frac {\sqrt {x-1}}{\sqrt {x+1}}dx\right )+\frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{4} \left (\frac {1}{6} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{2} \left (\sqrt {x-1} \sqrt {x+1}-\int \frac {1}{\sqrt {x-1} \sqrt {x+1}}dx\right )\right )+\frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2\)

\(\Big \downarrow \) 43

\(\displaystyle \frac {1}{4} \left (\frac {1}{6} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{2} \left (\sqrt {x-1} \sqrt {x+1}-\text {arccosh}(x)\right )\right )+\frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2\)

Input:

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

Output:

((-1 + x)^(3/2)*x^2*Sqrt[1 + x])/4 + (((7 - 2*x)*(-1 + x)^(3/2)*Sqrt[1 + x 
])/6 - (3*(Sqrt[-1 + x]*Sqrt[1 + x] - ArcCosh[x]))/2)/4
 

Defintions of rubi rules used

rule 43

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

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 111

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(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 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> 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] + Simp[(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]
 

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {\left (6 x^{3}-8 x^{2}+9 x -16\right ) \sqrt {x -1}\, \sqrt {x +1}}{24}+\frac {3 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (x +1\right )}}{8 \sqrt {x -1}\, \sqrt {x +1}}\)	\(60\)
default	\(\frac {\sqrt {x -1}\, \sqrt {x +1}\, \left (6 x^{3} \sqrt {x^{2}-1}-8 x^{2} \sqrt {x^{2}-1}+9 x \sqrt {x^{2}-1}+9 \ln \left (x +\sqrt {x^{2}-1}\right )-16 \sqrt {x^{2}-1}\right )}{24 \sqrt {x^{2}-1}}\)	\(76\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=\frac {1}{24} \, {\left (6 \, x^{3} - 8 \, x^{2} + 9 \, x - 16\right )} \sqrt {x + 1} \sqrt {x - 1} - \frac {3}{8} \, \log \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) \] Input:

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

Output:

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)
 

Sympy [F]

\[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=\int \frac {x^{3} \sqrt {x - 1}}{\sqrt {x + 1}}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=\frac {1}{4} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} x - \frac {1}{3} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} + \frac {5}{8} \, \sqrt {x^{2} - 1} x - \sqrt {x^{2} - 1} + \frac {3}{8} \, \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \] Input:

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

Output:

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))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=\frac {1}{24} \, {\left ({\left (2 \, {\left (3 \, x - 10\right )} {\left (x + 1\right )} + 43\right )} {\left (x + 1\right )} - 39\right )} \sqrt {x + 1} \sqrt {x - 1} - \frac {3}{4} \, \log \left (\sqrt {x + 1} - \sqrt {x - 1}\right ) \] Input:

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

Output:

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))
 

Mupad [B] (verification not implemented)

Time = 18.83 (sec) , antiderivative size = 473, normalized size of antiderivative = 5.77 \[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=\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}} \] Input:

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

Output:

(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)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {-1+x} x^3}{\sqrt {1+x}} \, dx=\frac {\sqrt {x +1}\, \sqrt {x -1}\, x^{3}}{4}-\frac {\sqrt {x +1}\, \sqrt {x -1}\, x^{2}}{3}+\frac {3 \sqrt {x +1}\, \sqrt {x -1}\, x}{8}-\frac {2 \sqrt {x +1}\, \sqrt {x -1}}{3}+\frac {3 \,\mathrm {log}\left (\frac {\sqrt {x -1}+\sqrt {x +1}}{\sqrt {2}}\right )}{4} \] Input:

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

Output:

(6*sqrt(x + 1)*sqrt(x - 1)*x**3 - 8*sqrt(x + 1)*sqrt(x - 1)*x**2 + 9*sqrt( 
x + 1)*sqrt(x - 1)*x - 16*sqrt(x + 1)*sqrt(x - 1) + 18*log((sqrt(x - 1) + 
sqrt(x + 1))/sqrt(2)))/24