Integrand size = 28, antiderivative size = 162 \[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\frac {\arctan \left (1-\frac {\sqrt {c}}{\sqrt [4]{d} x \sqrt [4]{c+d x^4}}\right )}{8 a \sqrt {c} d^{3/4}}-\frac {\arctan \left (1+\frac {\sqrt {c}}{\sqrt [4]{d} x \sqrt [4]{c+d x^4}}\right )}{8 a \sqrt {c} d^{3/4}}-\frac {\text {arctanh}\left (\frac {2 \sqrt {c} \sqrt [4]{d}}{x \sqrt [4]{c+d x^4} \left (2 \sqrt {d}+\frac {c}{x^2 \sqrt {c+d x^4}}\right )}\right )}{8 a \sqrt {c} d^{3/4}} \] Output:
1/8*arctan(1-c^(1/2)/d^(1/4)/x/(d*x^4+c)^(1/4))/a/c^(1/2)/d^(3/4)-1/8*arct an(1+c^(1/2)/d^(1/4)/x/(d*x^4+c)^(1/4))/a/c^(1/2)/d^(3/4)-1/8*arctanh(2*c^ (1/2)*d^(1/4)/x/(d*x^4+c)^(1/4)/(2*d^(1/2)+c/x^2/(d*x^4+c)^(1/2)))/a/c^(1/ 2)/d^(3/4)
Time = 8.34 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\frac {\arctan \left (\frac {-\frac {\sqrt {c}}{2 \sqrt [4]{d}}+\frac {\sqrt [4]{d} x^2 \sqrt {c+d x^4}}{\sqrt {c}}}{x \sqrt [4]{c+d x^4}}\right )-\text {arctanh}\left (\frac {2 \sqrt {c} \sqrt [4]{d} x \sqrt [4]{c+d x^4}}{c+2 \sqrt {d} x^2 \sqrt {c+d x^4}}\right )}{8 a \sqrt {c} d^{3/4}} \] Input:
Integrate[x^2/((c + d*x^4)^(1/4)*(a*c + 2*a*d*x^4)),x]
Output:
(ArcTan[(-1/2*Sqrt[c]/d^(1/4) + (d^(1/4)*x^2*Sqrt[c + d*x^4])/Sqrt[c])/(x* (c + d*x^4)^(1/4))] - ArcTanh[(2*Sqrt[c]*d^(1/4)*x*(c + d*x^4)^(1/4))/(c + 2*Sqrt[d]*x^2*Sqrt[c + d*x^4])])/(8*a*Sqrt[c]*d^(3/4))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.41, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1013, 27, 1012}
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.
\(\displaystyle \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt [4]{\frac {d x^4}{c}+1} \int \frac {x^2}{a \left (2 d x^4+c\right ) \sqrt [4]{\frac {d x^4}{c}+1}}dx}{\sqrt [4]{c+d x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt [4]{\frac {d x^4}{c}+1} \int \frac {x^2}{\left (2 d x^4+c\right ) \sqrt [4]{\frac {d x^4}{c}+1}}dx}{a \sqrt [4]{c+d x^4}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^3 \sqrt [4]{\frac {d x^4}{c}+1} \operatorname {AppellF1}\left (\frac {3}{4},1,\frac {1}{4},\frac {7}{4},-\frac {2 d x^4}{c},-\frac {d x^4}{c}\right )}{3 a c \sqrt [4]{c+d x^4}}\) |
Input:
Int[x^2/((c + d*x^4)^(1/4)*(a*c + 2*a*d*x^4)),x]
Output:
(x^3*(1 + (d*x^4)/c)^(1/4)*AppellF1[3/4, 1, 1/4, 7/4, (-2*d*x^4)/c, -((d*x ^4)/c)])/(3*a*c*(c + d*x^4)^(1/4))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {x^{2}}{\left (d \,x^{4}+c \right )^{\frac {1}{4}} \left (2 a d \,x^{4}+a c \right )}d x\]
Input:
int(x^2/(d*x^4+c)^(1/4)/(2*a*d*x^4+a*c),x)
Output:
int(x^2/(d*x^4+c)^(1/4)/(2*a*d*x^4+a*c),x)
Timed out. \[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate(x^2/(d*x^4+c)^(1/4)/(2*a*d*x^4+a*c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\frac {\int \frac {x^{2}}{c \sqrt [4]{c + d x^{4}} + 2 d x^{4} \sqrt [4]{c + d x^{4}}}\, dx}{a} \] Input:
integrate(x**2/(d*x**4+c)**(1/4)/(2*a*d*x**4+a*c),x)
Output:
Integral(x**2/(c*(c + d*x**4)**(1/4) + 2*d*x**4*(c + d*x**4)**(1/4)), x)/a
\[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (2 \, a d x^{4} + a c\right )} {\left (d x^{4} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x^2/(d*x^4+c)^(1/4)/(2*a*d*x^4+a*c),x, algorithm="maxima")
Output:
integrate(x^2/((2*a*d*x^4 + a*c)*(d*x^4 + c)^(1/4)), x)
\[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (2 \, a d x^{4} + a c\right )} {\left (d x^{4} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x^2/(d*x^4+c)^(1/4)/(2*a*d*x^4+a*c),x, algorithm="giac")
Output:
integrate(x^2/((2*a*d*x^4 + a*c)*(d*x^4 + c)^(1/4)), x)
Timed out. \[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\int \frac {x^2}{{\left (d\,x^4+c\right )}^{1/4}\,\left (2\,a\,d\,x^4+a\,c\right )} \,d x \] Input:
int(x^2/((c + d*x^4)^(1/4)*(a*c + 2*a*d*x^4)),x)
Output:
int(x^2/((c + d*x^4)^(1/4)*(a*c + 2*a*d*x^4)), x)
\[ \int \frac {x^2}{\sqrt [4]{c+d x^4} \left (a c+2 a d x^4\right )} \, dx=\frac {\int \frac {x^{2}}{\left (d \,x^{4}+c \right )^{\frac {1}{4}} c +2 \left (d \,x^{4}+c \right )^{\frac {1}{4}} d \,x^{4}}d x}{a} \] Input:
int(x^2/(d*x^4+c)^(1/4)/(2*a*d*x^4+a*c),x)
Output:
int(x**2/((c + d*x**4)**(1/4)*c + 2*(c + d*x**4)**(1/4)*d*x**4),x)/a