\(\int \frac {(e x)^m}{(a+b x^4)^2 (c+d x^4)^{3/2}} \, dx\) [289]

Optimal result

Integrand size = 26, antiderivative size = 84 \[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\frac {(e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} \operatorname {AppellF1}\left (\frac {1+m}{4},2,\frac {3}{2},\frac {5+m}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 c e (1+m) \sqrt {c+d x^4}} \] Output:

(e*x)^(1+m)*(1+d*x^4/c)^(1/2)*AppellF1(1/4+1/4*m,2,3/2,5/4+1/4*m,-b*x^4/a, 
-d*x^4/c)/a^2/c/e/(1+m)/(d*x^4+c)^(1/2)
 

Mathematica [A] (verified)

Time = 11.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\frac {x (e x)^m \left (1+\frac {d x^4}{c}\right )^{3/2} \operatorname {AppellF1}\left (\frac {1+m}{4},2,\frac {3}{2},\frac {5+m}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 (1+m) \left (c+d x^4\right )^{3/2}} \] Input:

Integrate[(e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)),x]
 

Output:

(x*(e*x)^m*(1 + (d*x^4)/c)^(3/2)*AppellF1[(1 + m)/4, 2, 3/2, (5 + m)/4, -( 
(b*x^4)/a), -((d*x^4)/c)])/(a^2*(1 + m)*(c + d*x^4)^(3/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.

\(\displaystyle \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {\sqrt {\frac {d x^4}{c}+1} \int \frac {(e x)^m}{\left (b x^4+a\right )^2 \left (\frac {d x^4}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^4}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} \operatorname {AppellF1}\left (\frac {m+1}{4},2,\frac {3}{2},\frac {m+5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 c e (m+1) \sqrt {c+d x^4}}\)

Input:

Int[(e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)),x]
 

Output:

((e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*AppellF1[(1 + m)/4, 2, 3/2, (5 + m)/4, 
-((b*x^4)/a), -((d*x^4)/c)])/(a^2*c*e*(1 + m)*Sqrt[c + d*x^4])
 

Defintions of rubi rules used

rule 1012

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])
 

rule 1013

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])
 

Maple [F]

\[\int \frac {\left (e x \right )^{m}}{\left (b \,x^{4}+a \right )^{2} \left (d \,x^{4}+c \right )^{\frac {3}{2}}}d x\]

Input:

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x)
 

Output:

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x)
 

Fricas [F]

\[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {\left (e x\right )^{m}}{{\left (b x^{4} + a\right )}^{2} {\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="fricas")
 

Output:

integral(sqrt(d*x^4 + c)*(e*x)^m/(b^2*d^2*x^16 + 2*(b^2*c*d + a*b*d^2)*x^1 
2 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^8 + 2*(a*b*c^2 + a^2*c*d)*x^4 + a^2* 
c^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x)**m/(b*x**4+a)**2/(d*x**4+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {\left (e x\right )^{m}}{{\left (b x^{4} + a\right )}^{2} {\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate((e*x)^m/((b*x^4 + a)^2*(d*x^4 + c)^(3/2)), x)
 

Giac [F]

\[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {\left (e x\right )^{m}}{{\left (b x^{4} + a\right )}^{2} {\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="giac")
 

Output:

integrate((e*x)^m/((b*x^4 + a)^2*(d*x^4 + c)^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^m}{{\left (b\,x^4+a\right )}^2\,{\left (d\,x^4+c\right )}^{3/2}} \,d x \] Input:

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)),x)
 

Output:

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \left (c+d x^4\right )^{3/2}} \, dx=\text {too large to display} \] Input:

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(3/2),x)
 

Output:

(e**m*(x**m*sqrt(c + d*x**4)*a*d*m*x + 3*x**m*sqrt(c + d*x**4)*a*d*x + x** 
m*sqrt(c + d*x**4)*b*c*m*x + x**m*sqrt(c + d*x**4)*b*c*x - x**m*sqrt(c + d 
*x**4)*b*d*m*x**5 + 5*x**m*sqrt(c + d*x**4)*b*d*x**5 + int((x**m*sqrt(c + 
d*x**4)*x**12)/(a**3*c**2*d*m**2 + 4*a**3*c**2*d*m + 3*a**3*c**2*d + 2*a** 
3*c*d**2*m**2*x**4 + 8*a**3*c*d**2*m*x**4 + 6*a**3*c*d**2*x**4 + a**3*d**3 
*m**2*x**8 + 4*a**3*d**3*m*x**8 + 3*a**3*d**3*x**8 + a**2*b*c**3*m**2 + 2* 
a**2*b*c**3*m + a**2*b*c**3 + 4*a**2*b*c**2*d*m**2*x**4 + 12*a**2*b*c**2*d 
*m*x**4 + 8*a**2*b*c**2*d*x**4 + 5*a**2*b*c*d**2*m**2*x**8 + 18*a**2*b*c*d 
**2*m*x**8 + 13*a**2*b*c*d**2*x**8 + 2*a**2*b*d**3*m**2*x**12 + 8*a**2*b*d 
**3*m*x**12 + 6*a**2*b*d**3*x**12 + 2*a*b**2*c**3*m**2*x**4 + 4*a*b**2*c** 
3*m*x**4 + 2*a*b**2*c**3*x**4 + 5*a*b**2*c**2*d*m**2*x**8 + 12*a*b**2*c**2 
*d*m*x**8 + 7*a*b**2*c**2*d*x**8 + 4*a*b**2*c*d**2*m**2*x**12 + 12*a*b**2* 
c*d**2*m*x**12 + 8*a*b**2*c*d**2*x**12 + a*b**2*d**3*m**2*x**16 + 4*a*b**2 
*d**3*m*x**16 + 3*a*b**2*d**3*x**16 + b**3*c**3*m**2*x**8 + 2*b**3*c**3*m* 
x**8 + b**3*c**3*x**8 + 2*b**3*c**2*d*m**2*x**12 + 4*b**3*c**2*d*m*x**12 + 
 2*b**3*c**2*d*x**12 + b**3*c*d**2*m**2*x**16 + 2*b**3*c*d**2*m*x**16 + b* 
*3*c*d**2*x**16),x)*a**2*b**2*c*d**3*m**4 - 2*int((x**m*sqrt(c + d*x**4)*x 
**12)/(a**3*c**2*d*m**2 + 4*a**3*c**2*d*m + 3*a**3*c**2*d + 2*a**3*c*d**2* 
m**2*x**4 + 8*a**3*c*d**2*m*x**4 + 6*a**3*c*d**2*x**4 + a**3*d**3*m**2*x** 
8 + 4*a**3*d**3*m*x**8 + 3*a**3*d**3*x**8 + a**2*b*c**3*m**2 + 2*a**2*b...