Integrand size = 19, antiderivative size = 401 \[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\frac {e \text {arctanh}\left (\frac {\sqrt {c d^4+a e^4} x}{d e \sqrt {a+c x^4}}\right )}{2 \sqrt {c d^4+a e^4}}-\frac {e \text {arctanh}\left (\frac {a e^2+c d^2 x^2}{\sqrt {c d^4+a e^4} \sqrt {a+c x^4}}\right )}{2 \sqrt {c d^4+a e^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}}-\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}} \] Output:
1/2*e*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*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*(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^(1/2)*d^2+a^(1/2)*e^2)/(c*x^4+a)^(1/2)-1/4*(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*arctan(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/(c^(1/2)*d^2+a^( 1/2)*e^2)/(c*x^4+a)^(1/2)
Result contains complex when optimal does not.
Time = 10.38 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.50 \[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\frac {\sqrt {1+\frac {c x^4}{a}} \left (-2 \sqrt [4]{-1} \sqrt [4]{a} \sqrt {1+\frac {c d^4}{a e^4}} e \operatorname {EllipticPi}\left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2},\arcsin \left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )+\sqrt [4]{c} d \log \left (\frac {-d^2+e^2 x^2}{c d^2 x^2+a e^2 \left (1+\sqrt {1+\frac {c d^4}{a e^4}} \sqrt {1+\frac {c x^4}{a}}\right )}\right )\right )}{2 \sqrt [4]{c} d \sqrt {1+\frac {c d^4}{a e^4}} e \sqrt {a+c x^4}} \] 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])
Time = 1.10 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.07, 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))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
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]
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]
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)]
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]
Result contains complex when optimal does not.
Time = 0.41 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.42
| method | result | size |
| default | \(\frac {-\frac {\operatorname {arctanh}\left (\frac {\frac {2 c \,x^{2} d^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}+\frac {e \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, d \sqrt {c \,x^{4}+a}}}{e}\) | \(169\) |
| elliptic | \(\frac {-\frac {\operatorname {arctanh}\left (\frac {\frac {2 c \,x^{2} d^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}+\frac {e \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, d \sqrt {c \,x^{4}+a}}}{e}\) | \(169\) |
Input:
int(1/(e*x+d)/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/e*(-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)*EllipticP i(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)))
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
\[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int \frac {1}{\sqrt {a + c x^{4}} \left (d + e x\right )}\, dx \] 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)
\[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}} \,d x } \] 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)
\[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}} \,d x } \] 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)
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)
\[ \int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx=\int \frac {\sqrt {c \,x^{4}+a}}{c e \,x^{5}+c d \,x^{4}+a e x +a d}d x \] Input:
int(1/(e*x+d)/(c*x^4+a)^(1/2),x)
Output:
int(sqrt(a + c*x**4)/(a*d + a*e*x + c*d*x**4 + c*e*x**5),x)