Integrand size = 17, antiderivative size = 98 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {\left (-b+a x^4\right )^{3/4} \left (-3 b x+4 a x^5\right )}{32 a}-\frac {3 b^2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}} \] Output:
1/32*(a*x^4-b)^(3/4)*(4*a*x^5-3*b*x)/a-3/64*b^2*arctan(a^(1/4)*x/(a*x^4-b) ^(1/4))/a^(5/4)-3/64*b^2*arctanh(a^(1/4)*x/(a*x^4-b)^(1/4))/a^(5/4)
Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.95 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {2 \sqrt [4]{a} x \left (-b+a x^4\right )^{3/4} \left (-3 b+4 a x^4\right )-3 b^2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )-3 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{64 a^{5/4}} \] Input:
Integrate[x^4*(-b + a*x^4)^(3/4),x]
Output:
(2*a^(1/4)*x*(-b + a*x^4)^(3/4)*(-3*b + 4*a*x^4) - 3*b^2*ArcTan[(a^(1/4)*x )/(-b + a*x^4)^(1/4)] - 3*b^2*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(64 *a^(5/4))
Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {811, 843, 770, 756, 216, 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.
\(\displaystyle \int x^4 \left (a x^4-b\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{8} x^5 \left (a x^4-b\right )^{3/4}-\frac {3}{8} b \int \frac {x^4}{\sqrt [4]{a x^4-b}}dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{8} x^5 \left (a x^4-b\right )^{3/4}-\frac {3}{8} b \left (\frac {b \int \frac {1}{\sqrt [4]{a x^4-b}}dx}{4 a}+\frac {x \left (a x^4-b\right )^{3/4}}{4 a}\right )\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {1}{8} x^5 \left (a x^4-b\right )^{3/4}-\frac {3}{8} b \left (\frac {b \int \frac {1}{1-\frac {a x^4}{a x^4-b}}d\frac {x}{\sqrt [4]{a x^4-b}}}{4 a}+\frac {x \left (a x^4-b\right )^{3/4}}{4 a}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{8} x^5 \left (a x^4-b\right )^{3/4}-\frac {3}{8} b \left (\frac {b \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}+1}d\frac {x}{\sqrt [4]{a x^4-b}}\right )}{4 a}+\frac {x \left (a x^4-b\right )^{3/4}}{4 a}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{8} x^5 \left (a x^4-b\right )^{3/4}-\frac {3}{8} b \left (\frac {b \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )}{4 a}+\frac {x \left (a x^4-b\right )^{3/4}}{4 a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{8} x^5 \left (a x^4-b\right )^{3/4}-\frac {3}{8} b \left (\frac {b \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )}{4 a}+\frac {x \left (a x^4-b\right )^{3/4}}{4 a}\right )\) |
Input:
Int[x^4*(-b + a*x^4)^(3/4),x]
Output:
(x^5*(-b + a*x^4)^(3/4))/8 - (3*b*((x*(-b + a*x^4)^(3/4))/(4*a) + (b*(ArcT an[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(1/4)) + ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(1/4))))/(4*a)))/8
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[((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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.18
method | result | size |
pseudoelliptic | \(\frac {16 a^{\frac {5}{4}} \left (a \,x^{4}-b \right )^{\frac {3}{4}} x^{5}-12 b x \,a^{\frac {1}{4}} \left (a \,x^{4}-b \right )^{\frac {3}{4}}+6 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b^{2}-3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) b^{2}}{128 a^{\frac {5}{4}}}\) | \(116\) |
Input:
int(x^4*(a*x^4-b)^(3/4),x,method=_RETURNVERBOSE)
Output:
1/128/a^(5/4)*(16*a^(5/4)*(a*x^4-b)^(3/4)*x^5-12*b*x*a^(1/4)*(a*x^4-b)^(3/ 4)+6*arctan(1/a^(1/4)/x*(a*x^4-b)^(1/4))*b^2-3*ln((-a^(1/4)*x-(a*x^4-b)^(1 /4))/(a^(1/4)*x-(a*x^4-b)^(1/4)))*b^2)
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.30 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=-\frac {3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} + \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 3 i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} + i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) + 3 i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} - i \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a \log \left (\frac {27 \, {\left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{6} - \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4} x\right )}}{x}\right ) - 4 \, {\left (4 \, a x^{5} - 3 \, b x\right )} {\left (a x^{4} - b\right )}^{\frac {3}{4}}}{128 \, a} \] Input:
integrate(x^4*(a*x^4-b)^(3/4),x, algorithm="fricas")
Output:
-1/128*(3*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*b^6 + (b^8/a^5)^(3/4 )*a^4*x)/x) - 3*I*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*b^6 + I*(b^8 /a^5)^(3/4)*a^4*x)/x) + 3*I*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^(1/4)*b^ 6 - I*(b^8/a^5)^(3/4)*a^4*x)/x) - 3*(b^8/a^5)^(1/4)*a*log(27*((a*x^4 - b)^ (1/4)*b^6 - (b^8/a^5)^(3/4)*a^4*x)/x) - 4*(4*a*x^5 - 3*b*x)*(a*x^4 - b)^(3 /4))/a
Result contains complex when optimal does not.
Time = 1.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.43 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {b^{\frac {3}{4}} x^{5} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate(x**4*(a*x**4-b)**(3/4),x)
Output:
b**(3/4)*x**5*exp(3*I*pi/4)*gamma(5/4)*hyper((-3/4, 5/4), (9/4,), a*x**4/b )/(4*gamma(9/4))
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (78) = 156\).
Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.65 \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {3 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )}}{128 \, a} + \frac {\frac {{\left (a x^{4} - b\right )}^{\frac {3}{4}} a b^{2}}{x^{3}} + \frac {3 \, {\left (a x^{4} - b\right )}^{\frac {7}{4}} b^{2}}{x^{7}}}{32 \, {\left (a^{3} - \frac {2 \, {\left (a x^{4} - b\right )} a^{2}}{x^{4}} + \frac {{\left (a x^{4} - b\right )}^{2} a}{x^{8}}\right )}} \] Input:
integrate(x^4*(a*x^4-b)^(3/4),x, algorithm="maxima")
Output:
3/128*b^2*(2*arctan((a*x^4 - b)^(1/4)/(a^(1/4)*x))/a^(1/4) + log(-(a^(1/4) - (a*x^4 - b)^(1/4)/x)/(a^(1/4) + (a*x^4 - b)^(1/4)/x))/a^(1/4))/a + 1/32 *((a*x^4 - b)^(3/4)*a*b^2/x^3 + 3*(a*x^4 - b)^(7/4)*b^2/x^7)/(a^3 - 2*(a*x ^4 - b)*a^2/x^4 + (a*x^4 - b)^2*a/x^8)
\[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\int { {\left (a x^{4} - b\right )}^{\frac {3}{4}} x^{4} \,d x } \] Input:
integrate(x^4*(a*x^4-b)^(3/4),x, algorithm="giac")
Output:
integrate((a*x^4 - b)^(3/4)*x^4, x)
Timed out. \[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\int x^4\,{\left (a\,x^4-b\right )}^{3/4} \,d x \] Input:
int(x^4*(a*x^4 - b)^(3/4),x)
Output:
int(x^4*(a*x^4 - b)^(3/4), x)
\[ \int x^4 \left (-b+a x^4\right )^{3/4} \, dx=\frac {4 \left (a \,x^{4}-b \right )^{\frac {3}{4}} a \,x^{5}-3 \left (a \,x^{4}-b \right )^{\frac {3}{4}} b x -3 \left (\int \frac {1}{\left (a \,x^{4}-b \right )^{\frac {1}{4}}}d x \right ) b^{2}}{32 a} \] Input:
int(x^4*(a*x^4-b)^(3/4),x)
Output:
(4*(a*x**4 - b)**(3/4)*a*x**5 - 3*(a*x**4 - b)**(3/4)*b*x - 3*int((a*x**4 - b)**(3/4)/(a*x**4 - b),x)*b**2)/(32*a)