Integrand size = 21, antiderivative size = 59 \[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \sqrt [3]{a+b x^4} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{3},2,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2 \sqrt [3]{1+\frac {b x^4}{a}}} \] Output:
x*(b*x^4+a)^(1/3)*AppellF1(1/4,-1/3,2,5/4,-b*x^4/a,-d*x^4/c)/c^2/(1+b*x^4/ a)^(1/3)
Leaf count is larger than twice the leaf count of optimal. \(233\) vs. \(2(59)=118\).
Time = 10.43 (sec) , antiderivative size = 233, normalized size of antiderivative = 3.95 \[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \left (\frac {b x^4 \left (1+\frac {b x^4}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {5}{4},\frac {2}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2}+\frac {3 \left (\frac {a+b x^4}{c}+\frac {45 a^2 \operatorname {AppellF1}\left (\frac {1}{4},\frac {2}{3},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{15 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {2}{3},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-4 x^4 \left (3 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {2}{3},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {5}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}\right )}{c+d x^4}\right )}{12 \left (a+b x^4\right )^{2/3}} \] Input:
Integrate[(a + b*x^4)^(1/3)/(c + d*x^4)^2,x]
Output:
(x*((b*x^4*(1 + (b*x^4)/a)^(2/3)*AppellF1[5/4, 2/3, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])/c^2 + (3*((a + b*x^4)/c + (45*a^2*AppellF1[1/4, 2/3, 1, 5/4 , -((b*x^4)/a), -((d*x^4)/c)])/(15*a*c*AppellF1[1/4, 2/3, 1, 5/4, -((b*x^4 )/a), -((d*x^4)/c)] - 4*x^4*(3*a*d*AppellF1[5/4, 2/3, 2, 9/4, -((b*x^4)/a) , -((d*x^4)/c)] + 2*b*c*AppellF1[5/4, 5/3, 1, 9/4, -((b*x^4)/a), -((d*x^4) /c)]))))/(c + d*x^4)))/(12*(a + b*x^4)^(2/3))
Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {937, 936}
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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt [3]{a+b x^4} \int \frac {\sqrt [3]{\frac {b x^4}{a}+1}}{\left (d x^4+c\right )^2}dx}{\sqrt [3]{\frac {b x^4}{a}+1}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \sqrt [3]{a+b x^4} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{3},2,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2 \sqrt [3]{\frac {b x^4}{a}+1}}\) |
Input:
Int[(a + b*x^4)^(1/3)/(c + d*x^4)^2,x]
Output:
(x*(a + b*x^4)^(1/3)*AppellF1[1/4, -1/3, 2, 5/4, -((b*x^4)/a), -((d*x^4)/c )])/(c^2*(1 + (b*x^4)/a)^(1/3))
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{3}}}{\left (d \,x^{4}+c \right )^{2}}d x\]
Input:
int((b*x^4+a)^(1/3)/(d*x^4+c)^2,x)
Output:
int((b*x^4+a)^(1/3)/(d*x^4+c)^2,x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate((b*x^4+a)^(1/3)/(d*x^4+c)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {\sqrt [3]{a + b x^{4}}}{\left (c + d x^{4}\right )^{2}}\, dx \] Input:
integrate((b*x**4+a)**(1/3)/(d*x**4+c)**2,x)
Output:
Integral((a + b*x**4)**(1/3)/(c + d*x**4)**2, x)
\[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{3}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:
integrate((b*x^4+a)^(1/3)/(d*x^4+c)^2,x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(1/3)/(d*x^4 + c)^2, x)
\[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{3}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:
integrate((b*x^4+a)^(1/3)/(d*x^4+c)^2,x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(1/3)/(d*x^4 + c)^2, x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{1/3}}{{\left (d\,x^4+c\right )}^2} \,d x \] Input:
int((a + b*x^4)^(1/3)/(c + d*x^4)^2,x)
Output:
int((a + b*x^4)^(1/3)/(c + d*x^4)^2, x)
\[ \int \frac {\sqrt [3]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{3}}}{d^{2} x^{8}+2 c d \,x^{4}+c^{2}}d x \] Input:
int((b*x^4+a)^(1/3)/(d*x^4+c)^2,x)
Output:
int((a + b*x**4)**(1/3)/(c**2 + 2*c*d*x**4 + d**2*x**8),x)