Integrand size = 15, antiderivative size = 74 \[ \int x^5 \sqrt {a+c x^4} \, dx=\frac {a x^2 \sqrt {a+c x^4}}{16 c}+\frac {1}{8} x^6 \sqrt {a+c x^4}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{16 c^{3/2}} \] Output:
1/16*a*x^2*(c*x^4+a)^(1/2)/c+1/8*x^6*(c*x^4+a)^(1/2)-1/16*a^2*arctanh(c^(1 /2)*x^2/(c*x^4+a)^(1/2))/c^(3/2)
Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int x^5 \sqrt {a+c x^4} \, dx=\frac {x^2 \sqrt {a+c x^4} \left (a+2 c x^4\right )}{16 c}-\frac {a^2 \log \left (\sqrt {c} x^2+\sqrt {a+c x^4}\right )}{16 c^{3/2}} \] Input:
Integrate[x^5*Sqrt[a + c*x^4],x]
Output:
(x^2*Sqrt[a + c*x^4]*(a + 2*c*x^4))/(16*c) - (a^2*Log[Sqrt[c]*x^2 + Sqrt[a + c*x^4]])/(16*c^(3/2))
Time = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {807, 248, 262, 224, 219}
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.
\(\displaystyle \int x^5 \sqrt {a+c x^4} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int x^4 \sqrt {c x^4+a}dx^2\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} a \int \frac {x^4}{\sqrt {c x^4+a}}dx^2+\frac {1}{4} x^6 \sqrt {a+c x^4}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} a \left (\frac {x^2 \sqrt {a+c x^4}}{2 c}-\frac {a \int \frac {1}{\sqrt {c x^4+a}}dx^2}{2 c}\right )+\frac {1}{4} x^6 \sqrt {a+c x^4}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} a \left (\frac {x^2 \sqrt {a+c x^4}}{2 c}-\frac {a \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+a}}}{2 c}\right )+\frac {1}{4} x^6 \sqrt {a+c x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} a \left (\frac {x^2 \sqrt {a+c x^4}}{2 c}-\frac {a \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 c^{3/2}}\right )+\frac {1}{4} x^6 \sqrt {a+c x^4}\right )\) |
Input:
Int[x^5*Sqrt[a + c*x^4],x]
Output:
((x^6*Sqrt[a + c*x^4])/4 + (a*((x^2*Sqrt[a + c*x^4])/(2*c) - (a*ArcTanh[(S qrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*c^(3/2))))/4)/2
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Time = 0.80 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {x^{2} \left (2 c \,x^{4}+a \right ) \sqrt {c \,x^{4}+a}}{16 c}-\frac {a^{2} \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{16 c^{\frac {3}{2}}}\) | \(53\) |
default | \(\frac {x^{2} \left (c \,x^{4}+a \right )^{\frac {3}{2}}}{8 c}-\frac {a \,x^{2} \sqrt {c \,x^{4}+a}}{16 c}-\frac {a^{2} \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{16 c^{\frac {3}{2}}}\) | \(63\) |
elliptic | \(\frac {x^{2} \left (c \,x^{4}+a \right )^{\frac {3}{2}}}{8 c}-\frac {a \,x^{2} \sqrt {c \,x^{4}+a}}{16 c}-\frac {a^{2} \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{16 c^{\frac {3}{2}}}\) | \(63\) |
pseudoelliptic | \(\frac {2 \sqrt {c \,x^{4}+a}\, c^{\frac {3}{2}} x^{6}+a \,x^{2} \sqrt {c \,x^{4}+a}\, \sqrt {c}-\operatorname {arctanh}\left (\frac {\sqrt {c \,x^{4}+a}}{x^{2} \sqrt {c}}\right ) a^{2}}{16 c^{\frac {3}{2}}}\) | \(63\) |
Input:
int(x^5*(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/16*x^2*(2*c*x^4+a)*(c*x^4+a)^(1/2)/c-1/16/c^(3/2)*a^2*ln(c^(1/2)*x^2+(c* x^4+a)^(1/2))
Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.76 \[ \int x^5 \sqrt {a+c x^4} \, dx=\left [\frac {a^{2} \sqrt {c} \log \left (-2 \, c x^{4} + 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) + 2 \, {\left (2 \, c^{2} x^{6} + a c x^{2}\right )} \sqrt {c x^{4} + a}}{32 \, c^{2}}, \frac {a^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + a} \sqrt {-c}}{c x^{2}}\right ) + {\left (2 \, c^{2} x^{6} + a c x^{2}\right )} \sqrt {c x^{4} + a}}{16 \, c^{2}}\right ] \] Input:
integrate(x^5*(c*x^4+a)^(1/2),x, algorithm="fricas")
Output:
[1/32*(a^2*sqrt(c)*log(-2*c*x^4 + 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a) + 2*( 2*c^2*x^6 + a*c*x^2)*sqrt(c*x^4 + a))/c^2, 1/16*(a^2*sqrt(-c)*arctan(sqrt( c*x^4 + a)*sqrt(-c)/(c*x^2)) + (2*c^2*x^6 + a*c*x^2)*sqrt(c*x^4 + a))/c^2]
Time = 2.01 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.28 \[ \int x^5 \sqrt {a+c x^4} \, dx=\frac {a^{\frac {3}{2}} x^{2}}{16 c \sqrt {1 + \frac {c x^{4}}{a}}} + \frac {3 \sqrt {a} x^{6}}{16 \sqrt {1 + \frac {c x^{4}}{a}}} - \frac {a^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{16 c^{\frac {3}{2}}} + \frac {c x^{10}}{8 \sqrt {a} \sqrt {1 + \frac {c x^{4}}{a}}} \] Input:
integrate(x**5*(c*x**4+a)**(1/2),x)
Output:
a**(3/2)*x**2/(16*c*sqrt(1 + c*x**4/a)) + 3*sqrt(a)*x**6/(16*sqrt(1 + c*x* *4/a)) - a**2*asinh(sqrt(c)*x**2/sqrt(a))/(16*c**(3/2)) + c*x**10/(8*sqrt( a)*sqrt(1 + c*x**4/a))
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (58) = 116\).
Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.62 \[ \int x^5 \sqrt {a+c x^4} \, dx=\frac {a^{2} \log \left (-\frac {\sqrt {c} - \frac {\sqrt {c x^{4} + a}}{x^{2}}}{\sqrt {c} + \frac {\sqrt {c x^{4} + a}}{x^{2}}}\right )}{32 \, c^{\frac {3}{2}}} + \frac {\frac {\sqrt {c x^{4} + a} a^{2} c}{x^{2}} + \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}} a^{2}}{x^{6}}}{16 \, {\left (c^{3} - \frac {2 \, {\left (c x^{4} + a\right )} c^{2}}{x^{4}} + \frac {{\left (c x^{4} + a\right )}^{2} c}{x^{8}}\right )}} \] Input:
integrate(x^5*(c*x^4+a)^(1/2),x, algorithm="maxima")
Output:
1/32*a^2*log(-(sqrt(c) - sqrt(c*x^4 + a)/x^2)/(sqrt(c) + sqrt(c*x^4 + a)/x ^2))/c^(3/2) + 1/16*(sqrt(c*x^4 + a)*a^2*c/x^2 + (c*x^4 + a)^(3/2)*a^2/x^6 )/(c^3 - 2*(c*x^4 + a)*c^2/x^4 + (c*x^4 + a)^2*c/x^8)
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.73 \[ \int x^5 \sqrt {a+c x^4} \, dx=\frac {1}{16} \, \sqrt {c x^{4} + a} {\left (2 \, x^{4} + \frac {a}{c}\right )} x^{2} + \frac {a^{2} \log \left ({\left | -\sqrt {c} x^{2} + \sqrt {c x^{4} + a} \right |}\right )}{16 \, c^{\frac {3}{2}}} \] Input:
integrate(x^5*(c*x^4+a)^(1/2),x, algorithm="giac")
Output:
1/16*sqrt(c*x^4 + a)*(2*x^4 + a/c)*x^2 + 1/16*a^2*log(abs(-sqrt(c)*x^2 + s qrt(c*x^4 + a)))/c^(3/2)
Timed out. \[ \int x^5 \sqrt {a+c x^4} \, dx=\int x^5\,\sqrt {c\,x^4+a} \,d x \] Input:
int(x^5*(a + c*x^4)^(1/2),x)
Output:
int(x^5*(a + c*x^4)^(1/2), x)
Time = 0.22 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.62 \[ \int x^5 \sqrt {a+c x^4} \, dx=\frac {-4 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) a^{3} x^{2}-8 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) a^{2} c \,x^{6}+\sqrt {c}\, \sqrt {c \,x^{4}+a}\, a^{3} x^{2}+10 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, a^{2} c \,x^{6}+24 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, a \,c^{2} x^{10}+16 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, c^{3} x^{14}-\mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) a^{4}-8 \,\mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) a^{3} c \,x^{4}-8 \,\mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) a^{2} c^{2} x^{8}+4 a^{3} c \,x^{4}+20 a^{2} c^{2} x^{8}+32 a \,c^{3} x^{12}+16 c^{4} x^{16}}{16 c \left (4 \sqrt {c \,x^{4}+a}\, a c \,x^{2}+8 \sqrt {c \,x^{4}+a}\, c^{2} x^{6}+\sqrt {c}\, a^{2}+8 \sqrt {c}\, a c \,x^{4}+8 \sqrt {c}\, c^{2} x^{8}\right )} \] Input:
int(x^5*(c*x^4+a)^(1/2),x)
Output:
( - 4*sqrt(c)*sqrt(a + c*x**4)*log((sqrt(a + c*x**4) + sqrt(c)*x**2)/sqrt( a))*a**3*x**2 - 8*sqrt(c)*sqrt(a + c*x**4)*log((sqrt(a + c*x**4) + sqrt(c) *x**2)/sqrt(a))*a**2*c*x**6 + sqrt(c)*sqrt(a + c*x**4)*a**3*x**2 + 10*sqrt (c)*sqrt(a + c*x**4)*a**2*c*x**6 + 24*sqrt(c)*sqrt(a + c*x**4)*a*c**2*x**1 0 + 16*sqrt(c)*sqrt(a + c*x**4)*c**3*x**14 - log((sqrt(a + c*x**4) + sqrt( c)*x**2)/sqrt(a))*a**4 - 8*log((sqrt(a + c*x**4) + sqrt(c)*x**2)/sqrt(a))* a**3*c*x**4 - 8*log((sqrt(a + c*x**4) + sqrt(c)*x**2)/sqrt(a))*a**2*c**2*x **8 + 4*a**3*c*x**4 + 20*a**2*c**2*x**8 + 32*a*c**3*x**12 + 16*c**4*x**16) /(16*c*(4*sqrt(a + c*x**4)*a*c*x**2 + 8*sqrt(a + c*x**4)*c**2*x**6 + sqrt( c)*a**2 + 8*sqrt(c)*a*c*x**4 + 8*sqrt(c)*c**2*x**8))