Integrand size = 16, antiderivative size = 99 \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=-\frac {a^2 x^2 \sqrt {a-b x^4}}{32 b}+\frac {1}{16} a x^6 \sqrt {a-b x^4}+\frac {1}{12} x^6 \left (a-b x^4\right )^{3/2}+\frac {a^3 \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}\right )}{32 b^{3/2}} \] Output:
-1/32*a^2*x^2*(-b*x^4+a)^(1/2)/b+1/16*a*x^6*(-b*x^4+a)^(1/2)+1/12*x^6*(-b* x^4+a)^(3/2)+1/32*a^3*arctan(b^(1/2)*x^2/(-b*x^4+a)^(1/2))/b^(3/2)
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=-\frac {x^2 \sqrt {a-b x^4} \left (3 a^2-14 a b x^4+8 b^2 x^8\right )}{96 b}-\frac {i a^3 \log \left (i \sqrt {b} x^2+\sqrt {a-b x^4}\right )}{32 b^{3/2}} \] Input:
Integrate[x^5*(a - b*x^4)^(3/2),x]
Output:
-1/96*(x^2*Sqrt[a - b*x^4]*(3*a^2 - 14*a*b*x^4 + 8*b^2*x^8))/b - ((I/32)*a ^3*Log[I*Sqrt[b]*x^2 + Sqrt[a - b*x^4]])/b^(3/2)
Time = 0.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {807, 248, 248, 262, 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.
\(\displaystyle \int x^5 \left (a-b x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int x^4 \left (a-b x^4\right )^{3/2}dx^2\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} a \int x^4 \sqrt {a-b x^4}dx^2+\frac {1}{6} x^6 \left (a-b x^4\right )^{3/2}\right )\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} a \left (\frac {1}{4} a \int \frac {x^4}{\sqrt {a-b x^4}}dx^2+\frac {1}{4} x^6 \sqrt {a-b x^4}\right )+\frac {1}{6} x^6 \left (a-b x^4\right )^{3/2}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {a \int \frac {1}{\sqrt {a-b x^4}}dx^2}{2 b}-\frac {x^2 \sqrt {a-b x^4}}{2 b}\right )+\frac {1}{4} x^6 \sqrt {a-b x^4}\right )+\frac {1}{6} x^6 \left (a-b x^4\right )^{3/2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {a \int \frac {1}{b x^4+1}d\frac {x^2}{\sqrt {a-b x^4}}}{2 b}-\frac {x^2 \sqrt {a-b x^4}}{2 b}\right )+\frac {1}{4} x^6 \sqrt {a-b x^4}\right )+\frac {1}{6} x^6 \left (a-b x^4\right )^{3/2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {a \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}\right )}{2 b^{3/2}}-\frac {x^2 \sqrt {a-b x^4}}{2 b}\right )+\frac {1}{4} x^6 \sqrt {a-b x^4}\right )+\frac {1}{6} x^6 \left (a-b x^4\right )^{3/2}\right )\) |
Input:
Int[x^5*(a - b*x^4)^(3/2),x]
Output:
((x^6*(a - b*x^4)^(3/2))/6 + (a*((x^6*Sqrt[a - b*x^4])/4 + (a*(-1/2*(x^2*S qrt[a - b*x^4])/b + (a*ArcTan[(Sqrt[b]*x^2)/Sqrt[a - b*x^4]])/(2*b^(3/2))) )/4))/2)/2
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[((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 = 0.78 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68
method | result | size |
risch | \(-\frac {x^{2} \left (8 b^{2} x^{8}-14 a b \,x^{4}+3 a^{2}\right ) \sqrt {-b \,x^{4}+a}}{96 b}+\frac {a^{3} \arctan \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {-b \,x^{4}+a}}\right )}{32 b^{\frac {3}{2}}}\) | \(67\) |
default | \(-\frac {x^{10} \sqrt {-b \,x^{4}+a}\, b}{12}+\frac {7 a \,x^{6} \sqrt {-b \,x^{4}+a}}{48}-\frac {a^{2} x^{2} \sqrt {-b \,x^{4}+a}}{32 b}+\frac {a^{3} \arctan \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {-b \,x^{4}+a}}\right )}{32 b^{\frac {3}{2}}}\) | \(81\) |
elliptic | \(-\frac {x^{10} \sqrt {-b \,x^{4}+a}\, b}{12}+\frac {7 a \,x^{6} \sqrt {-b \,x^{4}+a}}{48}-\frac {a^{2} x^{2} \sqrt {-b \,x^{4}+a}}{32 b}+\frac {a^{3} \arctan \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {-b \,x^{4}+a}}\right )}{32 b^{\frac {3}{2}}}\) | \(81\) |
pseudoelliptic | \(\frac {-8 \sqrt {-b \,x^{4}+a}\, b^{\frac {5}{2}} x^{10}+14 a \,b^{\frac {3}{2}} x^{6} \sqrt {-b \,x^{4}+a}-3 a^{2} x^{2} \sqrt {-b \,x^{4}+a}\, \sqrt {b}-3 \arctan \left (\frac {\sqrt {-b \,x^{4}+a}}{\sqrt {b}\, x^{2}}\right ) a^{3}}{96 b^{\frac {3}{2}}}\) | \(88\) |
Input:
int(x^5*(-b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/96*x^2*(8*b^2*x^8-14*a*b*x^4+3*a^2)/b*(-b*x^4+a)^(1/2)+1/32*a^3*arctan( b^(1/2)*x^2/(-b*x^4+a)^(1/2))/b^(3/2)
Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.59 \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=\left [-\frac {3 \, a^{3} \sqrt {-b} \log \left (2 \, b x^{4} - 2 \, \sqrt {-b x^{4} + a} \sqrt {-b} x^{2} - a\right ) + 2 \, {\left (8 \, b^{3} x^{10} - 14 \, a b^{2} x^{6} + 3 \, a^{2} b x^{2}\right )} \sqrt {-b x^{4} + a}}{192 \, b^{2}}, -\frac {3 \, a^{3} \sqrt {b} \arctan \left (\frac {\sqrt {-b x^{4} + a}}{\sqrt {b} x^{2}}\right ) + {\left (8 \, b^{3} x^{10} - 14 \, a b^{2} x^{6} + 3 \, a^{2} b x^{2}\right )} \sqrt {-b x^{4} + a}}{96 \, b^{2}}\right ] \] Input:
integrate(x^5*(-b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
[-1/192*(3*a^3*sqrt(-b)*log(2*b*x^4 - 2*sqrt(-b*x^4 + a)*sqrt(-b)*x^2 - a) + 2*(8*b^3*x^10 - 14*a*b^2*x^6 + 3*a^2*b*x^2)*sqrt(-b*x^4 + a))/b^2, -1/9 6*(3*a^3*sqrt(b)*arctan(sqrt(-b*x^4 + a)/(sqrt(b)*x^2)) + (8*b^3*x^10 - 14 *a*b^2*x^6 + 3*a^2*b*x^2)*sqrt(-b*x^4 + a))/b^2]
Result contains complex when optimal does not.
Time = 3.68 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.65 \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=\begin {cases} \frac {i a^{\frac {5}{2}} x^{2}}{32 b \sqrt {-1 + \frac {b x^{4}}{a}}} - \frac {17 i a^{\frac {3}{2}} x^{6}}{96 \sqrt {-1 + \frac {b x^{4}}{a}}} + \frac {11 i \sqrt {a} b x^{10}}{48 \sqrt {-1 + \frac {b x^{4}}{a}}} - \frac {i a^{3} \operatorname {acosh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{32 b^{\frac {3}{2}}} - \frac {i b^{2} x^{14}}{12 \sqrt {a} \sqrt {-1 + \frac {b x^{4}}{a}}} & \text {for}\: \left |{\frac {b x^{4}}{a}}\right | > 1 \\- \frac {a^{\frac {5}{2}} x^{2}}{32 b \sqrt {1 - \frac {b x^{4}}{a}}} + \frac {17 a^{\frac {3}{2}} x^{6}}{96 \sqrt {1 - \frac {b x^{4}}{a}}} - \frac {11 \sqrt {a} b x^{10}}{48 \sqrt {1 - \frac {b x^{4}}{a}}} + \frac {a^{3} \operatorname {asin}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{32 b^{\frac {3}{2}}} + \frac {b^{2} x^{14}}{12 \sqrt {a} \sqrt {1 - \frac {b x^{4}}{a}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(-b*x**4+a)**(3/2),x)
Output:
Piecewise((I*a**(5/2)*x**2/(32*b*sqrt(-1 + b*x**4/a)) - 17*I*a**(3/2)*x**6 /(96*sqrt(-1 + b*x**4/a)) + 11*I*sqrt(a)*b*x**10/(48*sqrt(-1 + b*x**4/a)) - I*a**3*acosh(sqrt(b)*x**2/sqrt(a))/(32*b**(3/2)) - I*b**2*x**14/(12*sqrt (a)*sqrt(-1 + b*x**4/a)), Abs(b*x**4/a) > 1), (-a**(5/2)*x**2/(32*b*sqrt(1 - b*x**4/a)) + 17*a**(3/2)*x**6/(96*sqrt(1 - b*x**4/a)) - 11*sqrt(a)*b*x* *10/(48*sqrt(1 - b*x**4/a)) + a**3*asin(sqrt(b)*x**2/sqrt(a))/(32*b**(3/2) ) + b**2*x**14/(12*sqrt(a)*sqrt(1 - b*x**4/a)), True))
Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.48 \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=-\frac {a^{3} \arctan \left (\frac {\sqrt {-b x^{4} + a}}{\sqrt {b} x^{2}}\right )}{32 \, b^{\frac {3}{2}}} + \frac {\frac {3 \, \sqrt {-b x^{4} + a} a^{3} b^{2}}{x^{2}} + \frac {8 \, {\left (-b x^{4} + a\right )}^{\frac {3}{2}} a^{3} b}{x^{6}} - \frac {3 \, {\left (-b x^{4} + a\right )}^{\frac {5}{2}} a^{3}}{x^{10}}}{96 \, {\left (b^{4} - \frac {3 \, {\left (b x^{4} - a\right )} b^{3}}{x^{4}} + \frac {3 \, {\left (b x^{4} - a\right )}^{2} b^{2}}{x^{8}} - \frac {{\left (b x^{4} - a\right )}^{3} b}{x^{12}}\right )}} \] Input:
integrate(x^5*(-b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
-1/32*a^3*arctan(sqrt(-b*x^4 + a)/(sqrt(b)*x^2))/b^(3/2) + 1/96*(3*sqrt(-b *x^4 + a)*a^3*b^2/x^2 + 8*(-b*x^4 + a)^(3/2)*a^3*b/x^6 - 3*(-b*x^4 + a)^(5 /2)*a^3/x^10)/(b^4 - 3*(b*x^4 - a)*b^3/x^4 + 3*(b*x^4 - a)^2*b^2/x^8 - (b* x^4 - a)^3*b/x^12)
Time = 0.14 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.48 \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=\frac {1}{16} \, {\left (\sqrt {-b x^{4} + a} {\left (2 \, x^{4} - \frac {a}{b}\right )} x^{2} - \frac {a^{2} \log \left ({\left | -\sqrt {-b} x^{2} + \sqrt {-b x^{4} + a} \right |}\right )}{\sqrt {-b} b}\right )} a - \frac {1}{96} \, {\left ({\left (2 \, {\left (4 \, x^{4} - \frac {a}{b}\right )} x^{4} - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {-b x^{4} + a} x^{2} - \frac {3 \, a^{3} \log \left ({\left | -\sqrt {-b} x^{2} + \sqrt {-b x^{4} + a} \right |}\right )}{\sqrt {-b} b^{2}}\right )} b \] Input:
integrate(x^5*(-b*x^4+a)^(3/2),x, algorithm="giac")
Output:
1/16*(sqrt(-b*x^4 + a)*(2*x^4 - a/b)*x^2 - a^2*log(abs(-sqrt(-b)*x^2 + sqr t(-b*x^4 + a)))/(sqrt(-b)*b))*a - 1/96*((2*(4*x^4 - a/b)*x^4 - 3*a^2/b^2)* sqrt(-b*x^4 + a)*x^2 - 3*a^3*log(abs(-sqrt(-b)*x^2 + sqrt(-b*x^4 + a)))/(s qrt(-b)*b^2))*b
Timed out. \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=\int x^5\,{\left (a-b\,x^4\right )}^{3/2} \,d x \] Input:
int(x^5*(a - b*x^4)^(3/2),x)
Output:
int(x^5*(a - b*x^4)^(3/2), x)
Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83 \[ \int x^5 \left (a-b x^4\right )^{3/2} \, dx=\frac {3 \mathit {asin} \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a^{3}-3 \sqrt {b}\, \sqrt {-b \,x^{4}+a}\, a^{2} x^{2}+14 \sqrt {b}\, \sqrt {-b \,x^{4}+a}\, a b \,x^{6}-8 \sqrt {b}\, \sqrt {-b \,x^{4}+a}\, b^{2} x^{10}}{96 \sqrt {b}\, b} \] Input:
int(x^5*(-b*x^4+a)^(3/2),x)
Output:
(3*asin((sqrt(b)*x**2)/sqrt(a))*a**3 - 3*sqrt(b)*sqrt(a - b*x**4)*a**2*x** 2 + 14*sqrt(b)*sqrt(a - b*x**4)*a*b*x**6 - 8*sqrt(b)*sqrt(a - b*x**4)*b**2 *x**10)/(96*sqrt(b)*b)