Integrand size = 24, antiderivative size = 79 \[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=\frac {(e x)^{1+m} \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1+m}{4},-p,2,\frac {5+m}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2 e (1+m)} \] Output:
(e*x)^(1+m)*(b*x^4+a)^p*AppellF1(1/4+1/4*m,-p,2,5/4+1/4*m,-b*x^4/a,-d*x^4/ c)/c^2/e/(1+m)/((1+b*x^4/a)^p)
Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95 \[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=\frac {x (e x)^m \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1+m}{4},-p,2,\frac {5+m}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2 (1+m)} \] Input:
Integrate[((e*x)^m*(a + b*x^4)^p)/(c + d*x^4)^2,x]
Output:
(x*(e*x)^m*(a + b*x^4)^p*AppellF1[(1 + m)/4, -p, 2, (5 + m)/4, -((b*x^4)/a ), -((d*x^4)/c)])/(c^2*(1 + m)*(1 + (b*x^4)/a)^p)
Time = 0.37 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1013, 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.
\(\displaystyle \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \int \frac {(e x)^m \left (\frac {b x^4}{a}+1\right )^p}{\left (d x^4+c\right )^2}dx\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {(e x)^{m+1} \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {m+1}{4},-p,2,\frac {m+5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2 e (m+1)}\) |
Input:
Int[((e*x)^m*(a + b*x^4)^p)/(c + d*x^4)^2,x]
Output:
((e*x)^(1 + m)*(a + b*x^4)^p*AppellF1[(1 + m)/4, -p, 2, (5 + m)/4, -((b*x^ 4)/a), -((d*x^4)/c)])/(c^2*e*(1 + m)*(1 + (b*x^4)/a)^p)
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])
\[\int \frac {\left (e x \right )^{m} \left (b \,x^{4}+a \right )^{p}}{\left (d \,x^{4}+c \right )^{2}}d x\]
Input:
int((e*x)^m*(b*x^4+a)^p/(d*x^4+c)^2,x)
Output:
int((e*x)^m*(b*x^4+a)^p/(d*x^4+c)^2,x)
\[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(b*x^4+a)^p/(d*x^4+c)^2,x, algorithm="fricas")
Output:
integral((b*x^4 + a)^p*(e*x)^m/(d^2*x^8 + 2*c*d*x^4 + c^2), x)
Timed out. \[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate((e*x)**m*(b*x**4+a)**p/(d*x**4+c)**2,x)
Output:
Timed out
\[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(b*x^4+a)^p/(d*x^4+c)^2,x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^p*(e*x)^m/(d*x^4 + c)^2, x)
\[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(b*x^4+a)^p/(d*x^4+c)^2,x, algorithm="giac")
Output:
integrate((b*x^4 + a)^p*(e*x)^m/(d*x^4 + c)^2, x)
Timed out. \[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,{\left (b\,x^4+a\right )}^p}{{\left (d\,x^4+c\right )}^2} \,d x \] Input:
int(((e*x)^m*(a + b*x^4)^p)/(c + d*x^4)^2,x)
Output:
int(((e*x)^m*(a + b*x^4)^p)/(c + d*x^4)^2, x)
\[ \int \frac {(e x)^m \left (a+b x^4\right )^p}{\left (c+d x^4\right )^2} \, dx=e^{m} \left (\int \frac {x^{m} \left (b \,x^{4}+a \right )^{p}}{d^{2} x^{8}+2 c d \,x^{4}+c^{2}}d x \right ) \] Input:
int((e*x)^m*(b*x^4+a)^p/(d*x^4+c)^2,x)
Output:
e**m*int((x**m*(a + b*x**4)**p)/(c**2 + 2*c*d*x**4 + d**2*x**8),x)