Integrand size = 13, antiderivative size = 69 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} x \, dx=-\frac {3 b \sqrt {a+\frac {b}{x^4}}}{4 x^2}+\frac {1}{2} \left (a+\frac {b}{x^4}\right )^{3/2} x^2-\frac {3}{4} a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^4}} x^2}\right ) \]
1/2*(a+b/x^4)^(3/2)*x^2-3/4*a*arctanh(b^(1/2)/x^2/(a+b/x^4)^(1/2))*b^(1/2) -3/4*b*(a+b/x^4)^(1/2)/x^2
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} x \, dx=-\frac {\sqrt {a+\frac {b}{x^4}} \left (\left (b-2 a x^4\right ) \sqrt {b+a x^4}+3 a \sqrt {b} x^4 \text {arctanh}\left (\frac {\sqrt {b+a x^4}}{\sqrt {b}}\right )\right )}{4 x^2 \sqrt {b+a x^4}} \]
-1/4*(Sqrt[a + b/x^4]*((b - 2*a*x^4)*Sqrt[b + a*x^4] + 3*a*Sqrt[b]*x^4*Arc Tanh[Sqrt[b + a*x^4]/Sqrt[b]]))/(x^2*Sqrt[b + a*x^4])
Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {858, 807, 247, 211, 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.
\(\displaystyle \int x \left (a+\frac {b}{x^4}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \left (a+\frac {b}{x^4}\right )^{3/2} x^3d\frac {1}{x}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {1}{2} \int \left (a+\frac {b}{x^2}\right )^{3/2} x^2d\frac {1}{x^2}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{3/2}-3 b \int \sqrt {a+\frac {b}{x^2}}d\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{3/2}-3 b \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x^2}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{3/2}-3 b \left (\frac {1}{2} a \int \frac {1}{1-\frac {b}{\sqrt {a+\frac {b}{x^2}} x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x^2}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{3/2}-3 b \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^2}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )\right )\) |
((a + b/x^2)^(3/2)*x - 3*b*(Sqrt[a + b/x^2]/(2*x^2) + (a*ArcTanh[Sqrt[b]/( Sqrt[a + b/x^2]*x^2)])/(2*Sqrt[b])))/2
3.21.67.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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 + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x^{2} \left (3 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) \sqrt {b}\, a \,x^{4}-2 a \,x^{4} \sqrt {a \,x^{4}+b}+b \sqrt {a \,x^{4}+b}\right )}{4 \left (a \,x^{4}+b \right )^{\frac {3}{2}}}\) | \(85\) |
risch | \(-\frac {b \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{4 x^{2}}+\frac {\left (\frac {\sqrt {a \,x^{4}+b}\, a}{2}-\frac {3 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) \sqrt {b}\, a}{4}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{\sqrt {a \,x^{4}+b}}\) | \(89\) |
-1/4*((a*x^4+b)/x^4)^(3/2)*x^2*(3*ln(2*(b^(1/2)*(a*x^4+b)^(1/2)+b)/x^2)*b^ (1/2)*a*x^4-2*a*x^4*(a*x^4+b)^(1/2)+b*(a*x^4+b)^(1/2))/(a*x^4+b)^(3/2)
Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.07 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} x \, dx=\left [\frac {3 \, a \sqrt {b} x^{2} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) + 2 \, {\left (2 \, a x^{4} - b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{8 \, x^{2}}, \frac {3 \, a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right ) + {\left (2 \, a x^{4} - b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{4 \, x^{2}}\right ] \]
[1/8*(3*a*sqrt(b)*x^2*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^4) + 2 *b)/x^4) + 2*(2*a*x^4 - b)*sqrt((a*x^4 + b)/x^4))/x^2, 1/4*(3*a*sqrt(-b)*x ^2*arctan(sqrt(-b)*x^2*sqrt((a*x^4 + b)/x^4)/b) + (2*a*x^4 - b)*sqrt((a*x^ 4 + b)/x^4))/x^2]
Time = 1.52 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} x \, dx=\frac {a^{\frac {3}{2}} x^{2}}{2 \sqrt {1 + \frac {b}{a x^{4}}}} + \frac {\sqrt {a} b}{4 x^{2} \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {3 a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{4} - \frac {b^{2}}{4 \sqrt {a} x^{6} \sqrt {1 + \frac {b}{a x^{4}}}} \]
a**(3/2)*x**2/(2*sqrt(1 + b/(a*x**4))) + sqrt(a)*b/(4*x**2*sqrt(1 + b/(a*x **4))) - 3*a*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x**2))/4 - b**2/(4*sqrt(a)*x** 6*sqrt(1 + b/(a*x**4)))
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} x \, dx=\frac {1}{2} \, \sqrt {a + \frac {b}{x^{4}}} a x^{2} - \frac {\sqrt {a + \frac {b}{x^{4}}} a b x^{2}}{4 \, {\left ({\left (a + \frac {b}{x^{4}}\right )} x^{4} - b\right )}} + \frac {3}{8} \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right ) \]
1/2*sqrt(a + b/x^4)*a*x^2 - 1/4*sqrt(a + b/x^4)*a*b*x^2/((a + b/x^4)*x^4 - b) + 3/8*a*sqrt(b)*log((sqrt(a + b/x^4)*x^2 - sqrt(b))/(sqrt(a + b/x^4)*x ^2 + sqrt(b)))
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} x \, dx=\frac {\frac {3 \, a^{2} b \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, \sqrt {a x^{4} + b} a^{2} - \frac {\sqrt {a x^{4} + b} a b}{x^{4}}}{4 \, a} \]
1/4*(3*a^2*b*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) + 2*sqrt(a*x^4 + b) *a^2 - sqrt(a*x^4 + b)*a*b/x^4)/a
Timed out. \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} x \, dx=\int x\,{\left (a+\frac {b}{x^4}\right )}^{3/2} \,d x \]