3.1.1 \(\int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx\) [1]

3.1.1.1 Optimal result

Integrand size = 19, antiderivative size = 405 \[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\frac {e \arctan \left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {a+c x^4}}\right )}{2 \sqrt {-c d^4-a e^4}}-\frac {e \text {arctanh}\left (\frac {a e^2+c d^2 x^2}{\sqrt {c d^4+a e^4} \sqrt {a+c x^4}}\right )}{2 \sqrt {c d^4+a e^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}}-\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}} \]

output

1/2*e*arctan(x*(-a*e^4-c*d^4)^(1/2)/d/e/(c*x^4+a)^(1/2))/(-a*e^4-c*d^4)^(1 
/2)-1/2*e*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*c^(1/4)*d*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2 
)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4 
))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2) 
^(1/2)/a^(1/4)/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)-1/4*(cos(2*arctan 
(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticPi( 
sin(2*arctan(c^(1/4)*x/a^(1/4))),1/4*(e^2*a^(1/2)+d^2*c^(1/2))^2/d^2/e^2/a 
^(1/2)/c^(1/2),1/2*2^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2))*(a^(1/2)+x^2*c^(1/2 
))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(1/4)/d/(e^2*a^(1/2 
)+d^2*c^(1/2))/(c*x^4+a)^(1/2)
 

3.1.1.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.22 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\frac {\sqrt {1+\frac {c x^4}{a}} \left (-2 \sqrt [4]{-1} \sqrt [4]{a} \sqrt {1+\frac {c d^4}{a e^4}} e \operatorname {EllipticPi}\left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2},\arcsin \left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )+\sqrt [4]{c} d \log \left (\frac {-d^2+e^2 x^2}{c d^2 x^2+a e^2 \left (1+\sqrt {1+\frac {c d^4}{a e^4}} \sqrt {1+\frac {c x^4}{a}}\right )}\right )\right )}{2 \sqrt [4]{c} d \sqrt {1+\frac {c d^4}{a e^4}} e \sqrt {a+c x^4}} \]

input

Integrate[1/((d + e*x)*Sqrt[a + c*x^4]),x]
 

output

(Sqrt[1 + (c*x^4)/a]*(-2*(-1)^(1/4)*a^(1/4)*Sqrt[1 + (c*d^4)/(a*e^4)]*e*El 
lipticPi[(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*Log[(-d^2 + e^2*x^2)/(c*d^2*x^2 + a*e^2*(1 + Sqrt[1 
+ (c*d^4)/(a*e^4)]*Sqrt[1 + (c*x^4)/a]))]))/(2*c^(1/4)*d*Sqrt[1 + (c*d^4)/ 
(a*e^4)]*e*Sqrt[a + c*x^4])
 

3.1.1.3 Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2263, 1541, 27, 761, 1577, 488, 219, 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.

\(\displaystyle \int \frac {1}{\sqrt {a+c x^4} (d+e x)} \, dx\)

\(\Big \downarrow \) 2263

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

\(\Big \downarrow \) 1541

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1577

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 2223

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

input

Int[1/((d + e*x)*Sqrt[a + c*x^4]),x]
 

output

(e*ArcTanh[(-(a*e^2) - c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])])/ 
(2*Sqrt[c*d^4 + a*e^4]) + d*((c^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c* 
x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1 
/2])/(2*a^(1/4)*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + c*x^4]) + (e^2*(((Sqr 
t[c]*d^2 + Sqrt[a]*e^2)*ArcTanh[(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[(Sqr 
t[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 + Sqr 
t[a]*e^2))
 

3.1.1.3.1 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 1541

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[c/a, 2]}, Simp[(c*d + a*e*q)/(c*d^2 - a*e^2)   Int[1/Sqrt[a + c*x^4 
], x], x] - Simp[(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}, x] && NeQ[c*d^2 + a*e 
^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/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 2263

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Simp[d 
  Int[1/((d^2 - e^2*x^2)*Sqrt[a + c*x^4]), x], x] - Simp[e   Int[x/((d^2 - 
e^2*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x]
 

3.1.1.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.64 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.42

method	result	size
default	\(\frac {-\frac {\operatorname {arctanh}\left (\frac {\frac {2 c \,x^{2} d^{2}}{e^{2}}+2 a}{2 \sqrt {\frac {c \,d^{4}}{e^{4}}+a}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {\frac {c \,d^{4}}{e^{4}}+a}}+\frac {e \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \Pi \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, d \sqrt {c \,x^{4}+a}}}{e}\)	\(169\)
elliptic	\(\frac {-\frac {\operatorname {arctanh}\left (\frac {\frac {2 c \,x^{2} d^{2}}{e^{2}}+2 a}{2 \sqrt {\frac {c \,d^{4}}{e^{4}}+a}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {\frac {c \,d^{4}}{e^{4}}+a}}+\frac {e \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \Pi \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, d \sqrt {c \,x^{4}+a}}}{e}\)	\(169\)

input

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

output

1/e*(-1/2/(c/e^4*d^4+a)^(1/2)*arctanh(1/2*(2*c*x^2/e^2*d^2+2*a)/(c/e^4*d^4 
+a)^(1/2)/(c*x^4+a)^(1/2))+1/(I/a^(1/2)*c^(1/2))^(1/2)*e/d*(1-I/a^(1/2)*c^ 
(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticP 
i(x*(I/a^(1/2)*c^(1/2))^(1/2),-I*a^(1/2)/c^(1/2)*e^2/d^2,(-I/a^(1/2)*c^(1/ 
2))^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)))
 

3.1.1.5 Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.1.1.6 Sympy [F]

\[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int \frac {1}{\sqrt {a + c x^{4}} \left (d + e x\right )}\, dx \]

input

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

output

Integral(1/(sqrt(a + c*x**4)*(d + e*x)), x)
 

3.1.1.7 Maxima [F]

\[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}} \,d x } \]

input

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

output

integrate(1/(sqrt(c*x^4 + a)*(e*x + d)), x)
 

3.1.1.8 Giac [F]

\[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}} \,d x } \]

input

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

output

integrate(1/(sqrt(c*x^4 + a)*(e*x + d)), x)
 

3.1.1.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int \frac {1}{\sqrt {c\,x^4+a}\,\left (d+e\,x\right )} \,d x \]

input

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

output

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