Integrand size = 15, antiderivative size = 74 \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {x^6}{2 b \sqrt {a+b x^4}}+\frac {3 x^2 \sqrt {a+b x^4}}{4 b^2}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{5/2}} \] Output:
-1/2*x^6/b/(b*x^4+a)^(1/2)+3/4*x^2*(b*x^4+a)^(1/2)/b^2-3/4*a*arctanh(b^(1/ 2)*x^2/(b*x^4+a)^(1/2))/b^(5/2)
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {3 a x^2+b x^6}{4 b^2 \sqrt {a+b x^4}}-\frac {3 a \log \left (\sqrt {b} x^2+\sqrt {a+b x^4}\right )}{4 b^{5/2}} \] Input:
Integrate[x^9/(a + b*x^4)^(3/2),x]
Output:
(3*a*x^2 + b*x^6)/(4*b^2*Sqrt[a + b*x^4]) - (3*a*Log[Sqrt[b]*x^2 + Sqrt[a + b*x^4]])/(4*b^(5/2))
Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {807, 252, 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.
\(\displaystyle \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {x^8}{\left (b x^4+a\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \int \frac {x^4}{\sqrt {b x^4+a}}dx^2}{b}-\frac {x^6}{b \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \left (\frac {x^2 \sqrt {a+b x^4}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^4+a}}dx^2}{2 b}\right )}{b}-\frac {x^6}{b \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \left (\frac {x^2 \sqrt {a+b x^4}}{2 b}-\frac {a \int \frac {1}{1-b x^4}d\frac {x^2}{\sqrt {b x^4+a}}}{2 b}\right )}{b}-\frac {x^6}{b \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \left (\frac {x^2 \sqrt {a+b x^4}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}\right )}{b}-\frac {x^6}{b \sqrt {a+b x^4}}\right )\) |
Input:
Int[x^9/(a + b*x^4)^(3/2),x]
Output:
(-(x^6/(b*Sqrt[a + b*x^4])) + (3*((x^2*Sqrt[a + b*x^4])/(2*b) - (a*ArcTanh [(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*b^(3/2))))/b)/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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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 = 0.96 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {x^{6}}{4 b \sqrt {b \,x^{4}+a}}+\frac {3 a \,x^{2}}{4 b^{2} \sqrt {b \,x^{4}+a}}-\frac {3 a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {5}{2}}}\) | \(61\) |
risch | \(\frac {x^{2} \sqrt {b \,x^{4}+a}}{4 b^{2}}+\frac {a \,x^{2}}{2 b^{2} \sqrt {b \,x^{4}+a}}-\frac {3 a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {5}{2}}}\) | \(61\) |
elliptic | \(\frac {x^{6}}{4 b \sqrt {b \,x^{4}+a}}+\frac {3 a \,x^{2}}{4 b^{2} \sqrt {b \,x^{4}+a}}-\frac {3 a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {5}{2}}}\) | \(61\) |
pseudoelliptic | \(\frac {b^{\frac {3}{2}} x^{6}+3 a \,x^{2} \sqrt {b}-3 \sqrt {b \,x^{4}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{4}+a}}{\sqrt {b}\, x^{2}}\right ) a}{4 b^{\frac {5}{2}} \sqrt {b \,x^{4}+a}}\) | \(61\) |
Input:
int(x^9/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4*x^6/b/(b*x^4+a)^(1/2)+3/4*a/b^2*x^2/(b*x^4+a)^(1/2)-3/4*a/b^(5/2)*ln(b ^(1/2)*x^2+(b*x^4+a)^(1/2))
Time = 0.08 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.30 \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a b x^{4} + a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (b^{2} x^{6} + 3 \, a b x^{2}\right )} \sqrt {b x^{4} + a}}{8 \, {\left (b^{4} x^{4} + a b^{3}\right )}}, \frac {3 \, {\left (a b x^{4} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{4} + a} \sqrt {-b}}{b x^{2}}\right ) + {\left (b^{2} x^{6} + 3 \, a b x^{2}\right )} \sqrt {b x^{4} + a}}{4 \, {\left (b^{4} x^{4} + a b^{3}\right )}}\right ] \] Input:
integrate(x^9/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
[1/8*(3*(a*b*x^4 + a^2)*sqrt(b)*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x ^2 - a) + 2*(b^2*x^6 + 3*a*b*x^2)*sqrt(b*x^4 + a))/(b^4*x^4 + a*b^3), 1/4* (3*(a*b*x^4 + a^2)*sqrt(-b)*arctan(sqrt(b*x^4 + a)*sqrt(-b)/(b*x^2)) + (b^ 2*x^6 + 3*a*b*x^2)*sqrt(b*x^4 + a))/(b^4*x^4 + a*b^3)]
Time = 1.81 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {3 \sqrt {a} x^{2}}{4 b^{2} \sqrt {1 + \frac {b x^{4}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {x^{6}}{4 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}} \] Input:
integrate(x**9/(b*x**4+a)**(3/2),x)
Output:
3*sqrt(a)*x**2/(4*b**2*sqrt(1 + b*x**4/a)) - 3*a*asinh(sqrt(b)*x**2/sqrt(a ))/(4*b**(5/2)) + x**6/(4*sqrt(a)*b*sqrt(1 + b*x**4/a))
Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.39 \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {2 \, a b - \frac {3 \, {\left (b x^{4} + a\right )} a}{x^{4}}}{4 \, {\left (\frac {\sqrt {b x^{4} + a} b^{3}}{x^{2}} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} b^{2}}{x^{6}}\right )}} + \frac {3 \, a \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{4} + a}}{x^{2}}}{\sqrt {b} + \frac {\sqrt {b x^{4} + a}}{x^{2}}}\right )}{8 \, b^{\frac {5}{2}}} \] Input:
integrate(x^9/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
1/4*(2*a*b - 3*(b*x^4 + a)*a/x^4)/(sqrt(b*x^4 + a)*b^3/x^2 - (b*x^4 + a)^( 3/2)*b^2/x^6) + 3/8*a*log(-(sqrt(b) - sqrt(b*x^4 + a)/x^2)/(sqrt(b) + sqrt (b*x^4 + a)/x^2))/b^(5/2)
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.74 \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {{\left (\frac {x^{4}}{b} + \frac {3 \, a}{b^{2}}\right )} x^{2}}{4 \, \sqrt {b x^{4} + a}} + \frac {3 \, a \log \left ({\left | -\sqrt {b} x^{2} + \sqrt {b x^{4} + a} \right |}\right )}{4 \, b^{\frac {5}{2}}} \] Input:
integrate(x^9/(b*x^4+a)^(3/2),x, algorithm="giac")
Output:
1/4*(x^4/b + 3*a/b^2)*x^2/sqrt(b*x^4 + a) + 3/4*a*log(abs(-sqrt(b)*x^2 + s qrt(b*x^4 + a)))/b^(5/2)
Timed out. \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^9}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:
int(x^9/(a + b*x^4)^(3/2),x)
Output:
int(x^9/(a + b*x^4)^(3/2), x)
Time = 0.24 (sec) , antiderivative size = 312, normalized size of antiderivative = 4.22 \[ \int \frac {x^9}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-36 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a^{2} x^{2}-48 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a b \,x^{6}+39 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{2} x^{2}+88 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a b \,x^{6}+16 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, b^{2} x^{10}-12 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a^{3}-60 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a^{2} b \,x^{4}-48 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a \,b^{2} x^{8}+9 a^{3}+81 a^{2} b \,x^{4}+96 a \,b^{2} x^{8}+16 b^{3} x^{12}}{16 b^{2} \left (3 \sqrt {b \,x^{4}+a}\, a b \,x^{2}+4 \sqrt {b \,x^{4}+a}\, b^{2} x^{6}+\sqrt {b}\, a^{2}+5 \sqrt {b}\, a b \,x^{4}+4 \sqrt {b}\, b^{2} x^{8}\right )} \] Input:
int(x^9/(b*x^4+a)^(3/2),x)
Output:
( - 36*sqrt(b)*sqrt(a + b*x**4)*log((sqrt(a + b*x**4) + sqrt(b)*x**2)/sqrt (a))*a**2*x**2 - 48*sqrt(b)*sqrt(a + b*x**4)*log((sqrt(a + b*x**4) + sqrt( b)*x**2)/sqrt(a))*a*b*x**6 + 39*sqrt(b)*sqrt(a + b*x**4)*a**2*x**2 + 88*sq rt(b)*sqrt(a + b*x**4)*a*b*x**6 + 16*sqrt(b)*sqrt(a + b*x**4)*b**2*x**10 - 12*log((sqrt(a + b*x**4) + sqrt(b)*x**2)/sqrt(a))*a**3 - 60*log((sqrt(a + b*x**4) + sqrt(b)*x**2)/sqrt(a))*a**2*b*x**4 - 48*log((sqrt(a + b*x**4) + sqrt(b)*x**2)/sqrt(a))*a*b**2*x**8 + 9*a**3 + 81*a**2*b*x**4 + 96*a*b**2* x**8 + 16*b**3*x**12)/(16*b**2*(3*sqrt(a + b*x**4)*a*b*x**2 + 4*sqrt(a + b *x**4)*b**2*x**6 + sqrt(b)*a**2 + 5*sqrt(b)*a*b*x**4 + 4*sqrt(b)*b**2*x**8 ))