\(\int \frac {(e x)^m}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\) [283]

Optimal result

Integrand size = 26, antiderivative size = 81 \[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx=\frac {(e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} \operatorname {AppellF1}\left (\frac {1+m}{4},2,\frac {1}{2},\frac {5+m}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2 e (1+m) \sqrt {c+d x^4}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(179\) vs. \(2(81)=162\).

Time = 10.01 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.21 \[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx=\frac {x (e x)^m \sqrt {c+d x^4} \left (-a b c d \operatorname {AppellF1}\left (\frac {1+m}{4},-\frac {1}{2},1,\frac {5+m}{4},-\frac {d x^4}{c},-\frac {b x^4}{a}\right )+b c (b c-a d) \operatorname {AppellF1}\left (\frac {1+m}{4},2,-\frac {1}{2},\frac {5+m}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+a^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{4},\frac {5+m}{4},-\frac {d x^4}{c}\right )\right )}{a^2 c (b c-a d)^2 (1+m) \sqrt {1+\frac {d x^4}{c}}} \] Input:

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

Output:

(x*(e*x)^m*Sqrt[c + d*x^4]*(-(a*b*c*d*AppellF1[(1 + m)/4, -1/2, 1, (5 + m) 
/4, -((d*x^4)/c), -((b*x^4)/a)]) + b*c*(b*c - a*d)*AppellF1[(1 + m)/4, 2, 
-1/2, (5 + m)/4, -((b*x^4)/a), -((d*x^4)/c)] + a^2*d^2*Hypergeometric2F1[1 
/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)]))/(a^2*c*(b*c - a*d)^2*(1 + m)*Sqr 
t[1 + (d*x^4)/c])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 81, 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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

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

Input:

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

Output:

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

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

Input:

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)
 

Output:

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)
 

Fricas [F]

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

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(d*x^4 + c)*(e*x)^m/(b^2*d*x^12 + (b^2*c + 2*a*b*d)*x^8 + (2* 
a*b*c + a^2*d)*x^4 + a^2*c), x)
 

Sympy [F]

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

integrate((e*x)**m/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)
 

Output:

Integral((e*x)**m/((a + b*x**4)**2*sqrt(c + d*x**4)), x)
 

Maxima [F]

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

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(1/2)),x)
 

Output:

int((e*x)^m/((a + b*x^4)^2*(c + d*x^4)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {(e x)^m}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx=e^{m} \left (\int \frac {x^{m} \sqrt {d \,x^{4}+c}}{b^{2} d \,x^{12}+2 a b d \,x^{8}+b^{2} c \,x^{8}+a^{2} d \,x^{4}+2 a b c \,x^{4}+a^{2} c}d x \right ) \] Input:

int((e*x)^m/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)
 

Output:

e**m*int((x**m*sqrt(c + d*x**4))/(a**2*c + a**2*d*x**4 + 2*a*b*c*x**4 + 2* 
a*b*d*x**8 + b**2*c*x**8 + b**2*d*x**12),x)