\(\int \frac {A+B x}{(d+e x) \sqrt {a+c x^4}} \, dx\) [228]

Optimal result

Integrand size = 24, antiderivative size = 442 \[ \int \frac {A+B x}{(d+e x) \sqrt {a+c x^4}} \, dx=-\frac {(B d-A e) \text {arctanh}\left (\frac {\sqrt {c d^4+a e^4} x}{d e \sqrt {a+c x^4}}\right )}{2 \sqrt {c d^4+a e^4}}+\frac {(B d-A e) \text {arctanh}\left (\frac {a e^2+c d^2 x^2}{\sqrt {c d^4+a e^4} \sqrt {a+c x^4}}\right )}{2 \sqrt {c d^4+a e^4}}+\frac {\left (A \sqrt {c} d+\sqrt {a} B e\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}}+\frac {(B d-A e) \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}} \] Output:

-1/2*(-A*e+B*d)*arctanh((a*e^4+c*d^4)^(1/2)*x/d/e/(c*x^4+a)^(1/2))/(a*e^4+ 
c*d^4)^(1/2)+1/2*(-A*e+B*d)*arctanh((c*d^2*x^2+a*e^2)/(a*e^4+c*d^4)^(1/2)/ 
(c*x^4+a)^(1/2))/(a*e^4+c*d^4)^(1/2)+1/2*(A*c^(1/2)*d+a^(1/2)*B*e)*(a^(1/2 
)+c^(1/2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2 
*arctan(c^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(1/4)/c^(1/4)/(c^(1/2)*d^2+a^(1/ 
2)*e^2)/(c*x^4+a)^(1/2)+1/4*(-A*e+B*d)*(c^(1/2)*d^2-a^(1/2)*e^2)*(a^(1/2)+ 
c^(1/2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticPi(sin(2*ar 
ctan(c^(1/4)*x/a^(1/4))),1/4*(c^(1/2)*d^2+a^(1/2)*e^2)^2/a^(1/2)/c^(1/2)/d 
^2/e^2,1/2*2^(1/2))/a^(1/4)/c^(1/4)/d/e/(c^(1/2)*d^2+a^(1/2)*e^2)/(c*x^4+a 
)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.82 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x}{(d+e x) \sqrt {a+c x^4}} \, dx=\frac {-\frac {i B \sqrt {1+\frac {c x^4}{a}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right ),-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}+\frac {(B d-A e) \left (-\sqrt [4]{c} d e \sqrt {a+c x^4} \arctan \left (\frac {\sqrt {c} \left (d^2-e^2 x^2\right )+e^2 \sqrt {a+c x^4}}{\sqrt {-c d^4-a e^4}}\right )+\sqrt [4]{-1} \sqrt [4]{a} \sqrt {-c d^4-a e^4} \sqrt {1+\frac {c x^4}{a}} \operatorname {EllipticPi}\left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2},\arcsin \left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )\right )}{\sqrt [4]{c} d \sqrt {-c d^4-a e^4}}}{e \sqrt {a+c x^4}} \] Input:

Integrate[(A + B*x)/((d + e*x)*Sqrt[a + c*x^4]),x]
 

Output:

(((-I)*B*Sqrt[1 + (c*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]] 
*x], -1])/Sqrt[(I*Sqrt[c])/Sqrt[a]] + ((B*d - A*e)*(-(c^(1/4)*d*e*Sqrt[a + 
 c*x^4]*ArcTan[(Sqrt[c]*(d^2 - e^2*x^2) + e^2*Sqrt[a + c*x^4])/Sqrt[-(c*d^ 
4) - a*e^4]]) + (-1)^(1/4)*a^(1/4)*Sqrt[-(c*d^4) - a*e^4]*Sqrt[1 + (c*x^4) 
/a]*EllipticPi[(I*Sqrt[a]*e^2)/(Sqrt[c]*d^2), ArcSin[((-1)^(3/4)*c^(1/4)*x 
)/a^(1/4)], -1]))/(c^(1/4)*d*Sqrt[-(c*d^4) - a*e^4]))/(e*Sqrt[a + c*x^4])
 

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2280, 27, 1577, 488, 219, 2227, 27, 761, 2223}

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.

\(\displaystyle \int \frac {A+B x}{\sqrt {a+c x^4} (d+e x)} \, dx\)

\(\Big \downarrow \) 2280

\(\displaystyle \int \frac {(B d-A e) x}{\left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx+\int \frac {A d-B e x^2}{\left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle (B d-A e) \int \frac {x}{\left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx+\int \frac {A d-B e x^2}{\left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx\)

\(\Big \downarrow \) 1577

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

\(\displaystyle \int \frac {A d-B e x^2}{\left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx-\frac {(B d-A e) \text {arctanh}\left (\frac {-a e^2-c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 \sqrt {a e^4+c d^4}}\)

\(\Big \downarrow \) 2227

\(\displaystyle \frac {\left (\sqrt {a} B e+A \sqrt {c} d\right ) \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {a} e^2+\sqrt {c} d^2}-\frac {\sqrt {a} d e (B d-A e) \int \frac {\sqrt {c} x^2+\sqrt {a}}{\sqrt {a} \left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx}{\sqrt {a} e^2+\sqrt {c} d^2}-\frac {(B d-A e) \text {arctanh}\left (\frac {-a e^2-c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 \sqrt {a e^4+c d^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (\sqrt {a} B e+A \sqrt {c} d\right ) \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {a} e^2+\sqrt {c} d^2}-\frac {d e (B d-A e) \int \frac {\sqrt {c} x^2+\sqrt {a}}{\left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx}{\sqrt {a} e^2+\sqrt {c} d^2}-\frac {(B d-A e) \text {arctanh}\left (\frac {-a e^2-c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 \sqrt {a e^4+c d^4}}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {d e (B d-A e) \int \frac {\sqrt {c} x^2+\sqrt {a}}{\left (d^2-e^2 x^2\right ) \sqrt {c x^4+a}}dx}{\sqrt {a} e^2+\sqrt {c} d^2}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right ) \left (\sqrt {a} B e+A \sqrt {c} d\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {(B d-A e) \text {arctanh}\left (\frac {-a e^2-c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 \sqrt {a e^4+c d^4}}\)

\(\Big \downarrow \) 2223

\(\displaystyle -\frac {d e (B d-A e) \left (\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {a}}{d^2}-\frac {\sqrt {c}}{e^2}\right ) \operatorname {EllipticPi}\left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {\left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \text {arctanh}\left (\frac {x \sqrt {a e^4+c d^4}}{d e \sqrt {a+c x^4}}\right )}{2 d e \sqrt {a e^4+c d^4}}\right )}{\sqrt {a} e^2+\sqrt {c} d^2}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right ) \left (\sqrt {a} B e+A \sqrt {c} d\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {(B d-A e) \text {arctanh}\left (\frac {-a e^2-c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 \sqrt {a e^4+c d^4}}\)

Input:

Int[(A + B*x)/((d + e*x)*Sqrt[a + c*x^4]),x]
 

Output:

-1/2*((B*d - A*e)*ArcTanh[(-(a*e^2) - c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt 
[a + c*x^4])])/Sqrt[c*d^4 + a*e^4] + ((A*Sqrt[c]*d + Sqrt[a]*B*e)*(Sqrt[a] 
 + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*Ar 
cTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*(Sqrt[c]*d^2 + Sqrt[a] 
*e^2)*Sqrt[a + c*x^4]) - (d*e*(B*d - A*e)*(((Sqrt[c]*d^2 + Sqrt[a]*e^2)*Ar 
cTanh[(Sqrt[c*d^4 + a*e^4]*x)/(d*e*Sqrt[a + c*x^4])])/(2*d*e*Sqrt[c*d^4 + 
a*e^4]) + ((Sqrt[a]/d^2 - Sqrt[c]/e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c 
*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[(Sqrt[c]*d^2 + Sqrt[a]*e^2)^2/ 
(4*Sqrt[a]*Sqrt[c]*d^2*e^2), 2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/ 
4)*c^(1/4)*Sqrt[a + c*x^4])))/(Sqrt[c]*d^2 + Sqrt[a]*e^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1577

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, c, d, e, p, q}, x]
 

rule 2223

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]) 
, x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(ArcTanh[Rt[(-c)* 
(d/e) - a*(e/d), 2]*(x/Sqrt[a + c*x^4])]/(2*d*e*Rt[(-c)*(d/e) - a*(e/d), 2] 
)), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^ 
2)]/(4*d*e*q*Sqrt[a + c*x^4]))*EllipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*Arc 
Tan[q*x], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0 
] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && NegQ[c*(d/e) + a*(e 
/d)]
 

rule 2227

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]) 
, x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(A*(c*d + a*e*q) - a*B*(e + d*q) 
)/(c*d^2 - a*e^2)   Int[1/Sqrt[a + c*x^4], x], x] + Simp[a*(B*d - A*e)*((e 
+ d*q)/(c*d^2 - a*e^2))   Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x] 
, x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] 
 && NeQ[c*A^2 - a*B^2, 0]
 

rule 2280

Int[(Px_)/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Wit 
h[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2], D = Coeff 
[Px, x, 3]}, Int[(x*(B*d - A*e + (d*D - C*e)*x^2))/((d^2 - e^2*x^2)*Sqrt[a 
+ c*x^4]), x] + Int[(A*d + (C*d - B*e)*x^2 - D*e*x^4)/((d^2 - e^2*x^2)*Sqrt 
[a + c*x^4]), x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Px, x] && LeQ[Expon[Px 
, x], 3] && NeQ[c*d^4 + a*e^4, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.76 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {B \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{e \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {\left (A e -B d \right ) \left (-\frac {\operatorname {arctanh}\left (\frac {\frac {2 c \,x^{2} d^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}+\frac {e \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, d \sqrt {c \,x^{4}+a}}\right )}{e^{2}}\)	\(251\)
elliptic	\(\frac {B \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{e \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {\left (A e -B d \right ) \left (-\frac {\operatorname {arctanh}\left (\frac {\frac {2 c \,x^{2} d^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}+\frac {e \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, d \sqrt {c \,x^{4}+a}}\right )}{e^{2}}\)	\(251\)

Input:

int((B*x+A)/(e*x+d)/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

B/e/(I*c^(1/2)/a^(1/2))^(1/2)*(1-I*c^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*c^(1/2) 
*x^2/a^(1/2))^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I*c^(1/2)/a^(1/2))^(1/2), 
I)+(A*e-B*d)/e^2*(-1/2/(a+c*d^4/e^4)^(1/2)*arctanh(1/2*(2*c*x^2*d^2/e^2+2* 
a)/(a+c*d^4/e^4)^(1/2)/(c*x^4+a)^(1/2))+1/(I*c^(1/2)/a^(1/2))^(1/2)/d*e*(1 
-I*c^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*c^(1/2)*x^2/a^(1/2))^(1/2)/(c*x^4+a)^(1 
/2)*EllipticPi(x*(I*c^(1/2)/a^(1/2))^(1/2),-I/c^(1/2)*a^(1/2)/d^2*e^2,(-I/ 
a^(1/2)*c^(1/2))^(1/2)/(I*c^(1/2)/a^(1/2))^(1/2)))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B x}{(d+e x) \sqrt {a+c x^4}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

integrate((B*x+A)/(e*x+d)/(c*x**4+a)**(1/2),x)
 

Output:

Integral((A + B*x)/(sqrt(a + c*x**4)*(d + e*x)), x)
 

Maxima [F]

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

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

Output:

integrate((B*x + A)/(sqrt(c*x^4 + a)*(e*x + d)), x)
 

Giac [F]

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

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

Output:

integrate((B*x + A)/(sqrt(c*x^4 + a)*(e*x + d)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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