Integrand size = 15, antiderivative size = 103 \[ \int x^6 \sqrt [4]{a+b x^4} \, dx=\frac {a x^3 \sqrt [4]{a+b x^4}}{32 b}+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}+\frac {3 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{7/4}}-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{7/4}} \] Output:
1/32*a*x^3*(b*x^4+a)^(1/4)/b+1/8*x^7*(b*x^4+a)^(1/4)+3/64*a^2*arctan(b^(1/ 4)*x/(b*x^4+a)^(1/4))/b^(7/4)-3/64*a^2*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/ b^(7/4)
Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int x^6 \sqrt [4]{a+b x^4} \, dx=\frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4} \left (a+4 b x^4\right )+3 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-3 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{7/4}} \] Input:
Integrate[x^6*(a + b*x^4)^(1/4),x]
Output:
(2*b^(3/4)*x^3*(a + b*x^4)^(1/4)*(a + 4*b*x^4) + 3*a^2*ArcTan[(b^(1/4)*x)/ (a + b*x^4)^(1/4)] - 3*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^( 7/4))
Time = 0.39 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {811, 843, 854, 827, 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.
\(\displaystyle \int x^6 \sqrt [4]{a+b x^4} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{8} a \int \frac {x^6}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\) |
Input:
Int[x^6*(a + b*x^4)^(1/4),x]
Output:
(x^7*(a + b*x^4)^(1/4))/8 + (a*((x^3*(a + b*x^4)^(1/4))/(4*b) - (3*a*(-1/2 *ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/b^(3/4) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(3/4))))/(4*b)))/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[((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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]
Time = 0.81 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01
method | result | size |
pseudoelliptic | \(\frac {16 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{\frac {7}{4}} x^{7}+4 a \,x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{\frac {3}{4}}-3 \ln \left (\frac {b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{2}-6 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{2}}{128 b^{\frac {7}{4}}}\) | \(104\) |
Input:
int(x^6*(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/128*(16*(b*x^4+a)^(1/4)*b^(7/4)*x^7+4*a*x^3*(b*x^4+a)^(1/4)*b^(3/4)-3*ln ((b^(1/4)*x+(b*x^4+a)^(1/4))/(-b^(1/4)*x+(b*x^4+a)^(1/4)))*a^2-6*arctan(1/ b^(1/4)/x*(b*x^4+a)^(1/4))*a^2)/b^(7/4)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.12 \[ \int x^6 \sqrt [4]{a+b x^4} \, dx=-\frac {3 \, \left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (\frac {3 \, {\left (\left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) - 3 \, \left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (-\frac {3 \, {\left (\left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) - 3 i \, \left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (-\frac {3 \, {\left (i \, \left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) + 3 i \, \left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (-\frac {3 \, {\left (-i \, \left (\frac {a^{8}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) - 4 \, {\left (4 \, b x^{7} + a x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, b} \] Input:
integrate(x^6*(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
-1/128*(3*(a^8/b^7)^(1/4)*b*log(3*((a^8/b^7)^(1/4)*b^2*x + (b*x^4 + a)^(1/ 4)*a^2)/x) - 3*(a^8/b^7)^(1/4)*b*log(-3*((a^8/b^7)^(1/4)*b^2*x - (b*x^4 + a)^(1/4)*a^2)/x) - 3*I*(a^8/b^7)^(1/4)*b*log(-3*(I*(a^8/b^7)^(1/4)*b^2*x - (b*x^4 + a)^(1/4)*a^2)/x) + 3*I*(a^8/b^7)^(1/4)*b*log(-3*(-I*(a^8/b^7)^(1 /4)*b^2*x - (b*x^4 + a)^(1/4)*a^2)/x) - 4*(4*b*x^7 + a*x^3)*(b*x^4 + a)^(1 /4))/b
Result contains complex when optimal does not.
Time = 1.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.38 \[ \int x^6 \sqrt [4]{a+b x^4} \, dx=\frac {\sqrt [4]{a} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate(x**6*(b*x**4+a)**(1/4),x)
Output:
a**(1/4)*x**7*gamma(7/4)*hyper((-1/4, 7/4), (11/4,), b*x**4*exp_polar(I*pi )/a)/(4*gamma(11/4))
Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.48 \[ \int x^6 \sqrt [4]{a+b x^4} \, dx=\frac {\frac {3 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b}{x} + \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}}{x^{5}}}{32 \, {\left (b^{3} - \frac {2 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2} b}{x^{8}}\right )}} - \frac {3 \, {\left (\frac {2 \, a^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a^{2} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, b} \] Input:
integrate(x^6*(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
1/32*(3*(b*x^4 + a)^(1/4)*a^2*b/x + (b*x^4 + a)^(5/4)*a^2/x^5)/(b^3 - 2*(b *x^4 + a)*b^2/x^4 + (b*x^4 + a)^2*b/x^8) - 3/128*(2*a^2*arctan((b*x^4 + a) ^(1/4)/(b^(1/4)*x))/b^(3/4) - a^2*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^ (1/4) + (b*x^4 + a)^(1/4)/x))/b^(3/4))/b
\[ \int x^6 \sqrt [4]{a+b x^4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{6} \,d x } \] Input:
integrate(x^6*(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(1/4)*x^6, x)
Timed out. \[ \int x^6 \sqrt [4]{a+b x^4} \, dx=\int x^6\,{\left (b\,x^4+a\right )}^{1/4} \,d x \] Input:
int(x^6*(a + b*x^4)^(1/4),x)
Output:
int(x^6*(a + b*x^4)^(1/4), x)
\[ \int x^6 \sqrt [4]{a+b x^4} \, dx=\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{3}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{7}-3 \left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a^{2}}{32 b} \] Input:
int(x^6*(b*x^4+a)^(1/4),x)
Output:
((a + b*x**4)**(1/4)*a*x**3 + 4*(a + b*x**4)**(1/4)*b*x**7 - 3*int(((a + b *x**4)**(1/4)*x**2)/(a + b*x**4),x)*a**2)/(32*b)