Integrand size = 14, antiderivative size = 74 \[ \int x \left (a-b x^4\right )^{3/2} \, dx=\frac {3}{16} a x^2 \sqrt {a-b x^4}+\frac {1}{8} x^2 \left (a-b x^4\right )^{3/2}+\frac {3 a^2 \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}\right )}{16 \sqrt {b}} \] Output:
3/16*a*x^2*(-b*x^4+a)^(1/2)+1/8*x^2*(-b*x^4+a)^(3/2)+3/16*a^2*arctan(b^(1/ 2)*x^2/(-b*x^4+a)^(1/2))/b^(1/2)
Result contains complex when optimal does not.
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int x \left (a-b x^4\right )^{3/2} \, dx=-\frac {1}{16} x^2 \sqrt {a-b x^4} \left (-5 a+2 b x^4\right )-\frac {3 i a^2 \log \left (i \sqrt {b} x^2+\sqrt {a-b x^4}\right )}{16 \sqrt {b}} \] Input:
Integrate[x*(a - b*x^4)^(3/2),x]
Output:
-1/16*(x^2*Sqrt[a - b*x^4]*(-5*a + 2*b*x^4)) - (((3*I)/16)*a^2*Log[I*Sqrt[ b]*x^2 + Sqrt[a - b*x^4]])/Sqrt[b]
Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {807, 211, 211, 224, 216}
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.
\(\displaystyle \int x \left (a-b x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \left (a-b x^4\right )^{3/2}dx^2\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} a \int \sqrt {a-b x^4}dx^2+\frac {1}{4} x^2 \left (a-b x^4\right )^{3/2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a-b x^4}}dx^2+\frac {1}{2} x^2 \sqrt {a-b x^4}\right )+\frac {1}{4} x^2 \left (a-b x^4\right )^{3/2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{b x^4+1}d\frac {x^2}{\sqrt {a-b x^4}}+\frac {1}{2} x^2 \sqrt {a-b x^4}\right )+\frac {1}{4} x^2 \left (a-b x^4\right )^{3/2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} a \left (\frac {a \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}\right )}{2 \sqrt {b}}+\frac {1}{2} x^2 \sqrt {a-b x^4}\right )+\frac {1}{4} x^2 \left (a-b x^4\right )^{3/2}\right )\) |
Input:
Int[x*(a - b*x^4)^(3/2),x]
Output:
((x^2*(a - b*x^4)^(3/2))/4 + (3*a*((x^2*Sqrt[a - b*x^4])/2 + (a*ArcTan[(Sq rt[b]*x^2)/Sqrt[a - b*x^4]])/(2*Sqrt[b])))/4)/2
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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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[(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 = 0.70 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {x^{2} \left (-2 b \,x^{4}+5 a \right ) \sqrt {-b \,x^{4}+a}}{16}+\frac {3 a^{2} \arctan \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {-b \,x^{4}+a}}\right )}{16 \sqrt {b}}\) | \(53\) |
default | \(\frac {3 a^{2} \arctan \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {-b \,x^{4}+a}}\right )}{16 \sqrt {b}}-\frac {x^{6} \sqrt {-b \,x^{4}+a}\, b}{8}+\frac {5 a \,x^{2} \sqrt {-b \,x^{4}+a}}{16}\) | \(60\) |
elliptic | \(\frac {3 a^{2} \arctan \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {-b \,x^{4}+a}}\right )}{16 \sqrt {b}}-\frac {x^{6} \sqrt {-b \,x^{4}+a}\, b}{8}+\frac {5 a \,x^{2} \sqrt {-b \,x^{4}+a}}{16}\) | \(60\) |
pseudoelliptic | \(-\frac {3 \arctan \left (\frac {\sqrt {-b \,x^{4}+a}}{\sqrt {b}\, x^{2}}\right ) a^{2}}{16 \sqrt {b}}+\frac {5 x^{2} \left (-\frac {2 b^{\frac {3}{2}} x^{4}}{5}+a \sqrt {b}\right ) \sqrt {-b \,x^{4}+a}}{16 \sqrt {b}}\) | \(60\) |
Input:
int(x*(-b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/16*x^2*(-2*b*x^4+5*a)*(-b*x^4+a)^(1/2)+3/16*a^2*arctan(b^(1/2)*x^2/(-b*x ^4+a)^(1/2))/b^(1/2)
Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.82 \[ \int x \left (a-b x^4\right )^{3/2} \, dx=\left [-\frac {3 \, a^{2} \sqrt {-b} \log \left (2 \, b x^{4} - 2 \, \sqrt {-b x^{4} + a} \sqrt {-b} x^{2} - a\right ) + 2 \, {\left (2 \, b^{2} x^{6} - 5 \, a b x^{2}\right )} \sqrt {-b x^{4} + a}}{32 \, b}, -\frac {3 \, a^{2} \sqrt {b} \arctan \left (\frac {\sqrt {-b x^{4} + a}}{\sqrt {b} x^{2}}\right ) + {\left (2 \, b^{2} x^{6} - 5 \, a b x^{2}\right )} \sqrt {-b x^{4} + a}}{16 \, b}\right ] \] Input:
integrate(x*(-b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
[-1/32*(3*a^2*sqrt(-b)*log(2*b*x^4 - 2*sqrt(-b*x^4 + a)*sqrt(-b)*x^2 - a) + 2*(2*b^2*x^6 - 5*a*b*x^2)*sqrt(-b*x^4 + a))/b, -1/16*(3*a^2*sqrt(b)*arct an(sqrt(-b*x^4 + a)/(sqrt(b)*x^2)) + (2*b^2*x^6 - 5*a*b*x^2)*sqrt(-b*x^4 + a))/b]
Result contains complex when optimal does not.
Time = 1.74 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.55 \[ \int x \left (a-b x^4\right )^{3/2} \, dx=\begin {cases} - \frac {5 i a^{\frac {3}{2}} x^{2}}{16 \sqrt {-1 + \frac {b x^{4}}{a}}} + \frac {7 i \sqrt {a} b x^{6}}{16 \sqrt {-1 + \frac {b x^{4}}{a}}} - \frac {3 i a^{2} \operatorname {acosh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 \sqrt {b}} - \frac {i b^{2} x^{10}}{8 \sqrt {a} \sqrt {-1 + \frac {b x^{4}}{a}}} & \text {for}\: \left |{\frac {b x^{4}}{a}}\right | > 1 \\\frac {5 a^{\frac {3}{2}} x^{2} \sqrt {1 - \frac {b x^{4}}{a}}}{16} - \frac {\sqrt {a} b x^{6} \sqrt {1 - \frac {b x^{4}}{a}}}{8} + \frac {3 a^{2} \operatorname {asin}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 \sqrt {b}} & \text {otherwise} \end {cases} \] Input:
integrate(x*(-b*x**4+a)**(3/2),x)
Output:
Piecewise((-5*I*a**(3/2)*x**2/(16*sqrt(-1 + b*x**4/a)) + 7*I*sqrt(a)*b*x** 6/(16*sqrt(-1 + b*x**4/a)) - 3*I*a**2*acosh(sqrt(b)*x**2/sqrt(a))/(16*sqrt (b)) - I*b**2*x**10/(8*sqrt(a)*sqrt(-1 + b*x**4/a)), Abs(b*x**4/a) > 1), ( 5*a**(3/2)*x**2*sqrt(1 - b*x**4/a)/16 - sqrt(a)*b*x**6*sqrt(1 - b*x**4/a)/ 8 + 3*a**2*asin(sqrt(b)*x**2/sqrt(a))/(16*sqrt(b)), True))
Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.39 \[ \int x \left (a-b x^4\right )^{3/2} \, dx=-\frac {3 \, a^{2} \arctan \left (\frac {\sqrt {-b x^{4} + a}}{\sqrt {b} x^{2}}\right )}{16 \, \sqrt {b}} + \frac {\frac {3 \, \sqrt {-b x^{4} + a} a^{2} b}{x^{2}} + \frac {5 \, {\left (-b x^{4} + a\right )}^{\frac {3}{2}} a^{2}}{x^{6}}}{16 \, {\left (b^{2} - \frac {2 \, {\left (b x^{4} - a\right )} b}{x^{4}} + \frac {{\left (b x^{4} - a\right )}^{2}}{x^{8}}\right )}} \] Input:
integrate(x*(-b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
-3/16*a^2*arctan(sqrt(-b*x^4 + a)/(sqrt(b)*x^2))/sqrt(b) + 1/16*(3*sqrt(-b *x^4 + a)*a^2*b/x^2 + 5*(-b*x^4 + a)^(3/2)*a^2/x^6)/(b^2 - 2*(b*x^4 - a)*b /x^4 + (b*x^4 - a)^2/x^8)
Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int x \left (a-b x^4\right )^{3/2} \, dx=\frac {1}{4} \, {\left (\sqrt {-b x^{4} + a} x^{2} - \frac {a \log \left ({\left | -\sqrt {-b} x^{2} + \sqrt {-b x^{4} + a} \right |}\right )}{\sqrt {-b}}\right )} a - \frac {1}{16} \, {\left (\sqrt {-b x^{4} + a} {\left (2 \, x^{4} - \frac {a}{b}\right )} x^{2} - \frac {a^{2} \log \left ({\left | -\sqrt {-b} x^{2} + \sqrt {-b x^{4} + a} \right |}\right )}{\sqrt {-b} b}\right )} b \] Input:
integrate(x*(-b*x^4+a)^(3/2),x, algorithm="giac")
Output:
1/4*(sqrt(-b*x^4 + a)*x^2 - a*log(abs(-sqrt(-b)*x^2 + sqrt(-b*x^4 + a)))/s qrt(-b))*a - 1/16*(sqrt(-b*x^4 + a)*(2*x^4 - a/b)*x^2 - a^2*log(abs(-sqrt( -b)*x^2 + sqrt(-b*x^4 + a)))/(sqrt(-b)*b))*b
Timed out. \[ \int x \left (a-b x^4\right )^{3/2} \, dx=\int x\,{\left (a-b\,x^4\right )}^{3/2} \,d x \] Input:
int(x*(a - b*x^4)^(3/2),x)
Output:
int(x*(a - b*x^4)^(3/2), x)
Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.77 \[ \int x \left (a-b x^4\right )^{3/2} \, dx=\frac {3 \mathit {asin} \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a^{2}+5 \sqrt {b}\, \sqrt {-b \,x^{4}+a}\, a \,x^{2}-2 \sqrt {b}\, \sqrt {-b \,x^{4}+a}\, b \,x^{6}}{16 \sqrt {b}} \] Input:
int(x*(-b*x^4+a)^(3/2),x)
Output:
(3*asin((sqrt(b)*x**2)/sqrt(a))*a**2 + 5*sqrt(b)*sqrt(a - b*x**4)*a*x**2 - 2*sqrt(b)*sqrt(a - b*x**4)*b*x**6)/(16*sqrt(b))