\(\int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx\) [388]

Optimal result

Integrand size = 16, antiderivative size = 348 \[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a+b \tan ^4(c+d x)}}\right )}{2 \sqrt {a+b} d}-\frac {\sqrt [4]{b} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right ) \left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+b \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )^2}}}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) d \sqrt {a+b \tan ^4(c+d x)}}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2}{4 \sqrt {a} \sqrt {b}},2 \arctan \left (\frac {\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right ) \left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+b \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) \sqrt [4]{b} d \sqrt {a+b \tan ^4(c+d x)}} \] Output:

1/2*arctan((a+b)^(1/2)*tan(d*x+c)/(a+tan(d*x+c)^4*b)^(1/2))/(a+b)^(1/2)/d- 
1/2*b^(1/4)*InverseJacobiAM(2*arctan(b^(1/4)*tan(d*x+c)/a^(1/4)),1/2*2^(1/ 
2))*(a^(1/2)+b^(1/2)*tan(d*x+c)^2)*((a+tan(d*x+c)^4*b)/(a^(1/2)+b^(1/2)*ta 
n(d*x+c)^2)^2)^(1/2)/a^(1/4)/(a^(1/2)-b^(1/2))/d/(a+tan(d*x+c)^4*b)^(1/2)+ 
1/4*(a^(1/2)+b^(1/2))*EllipticPi(sin(2*arctan(b^(1/4)*tan(d*x+c)/a^(1/4))) 
,-1/4*(a^(1/2)-b^(1/2))^2/a^(1/2)/b^(1/2),1/2*2^(1/2))*(a^(1/2)+b^(1/2)*ta 
n(d*x+c)^2)*((a+tan(d*x+c)^4*b)/(a^(1/2)+b^(1/2)*tan(d*x+c)^2)^2)^(1/2)/a^ 
(1/4)/(a^(1/2)-b^(1/2))/b^(1/4)/d/(a+tan(d*x+c)^4*b)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.49 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=-\frac {i \operatorname {EllipticPi}\left (-\frac {i \sqrt {a}}{\sqrt {b}},i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),-1\right ) \sqrt {1+\frac {b \tan ^4(c+d x)}{a}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {a+b \tan ^4(c+d x)}} \] Input:

Integrate[1/Sqrt[a + b*Tan[c + d*x]^4],x]
 

Output:

((-I)*EllipticPi[((-I)*Sqrt[a])/Sqrt[b], I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a 
]]*Tan[c + d*x]], -1]*Sqrt[1 + (b*Tan[c + d*x]^4)/a])/(Sqrt[(I*Sqrt[b])/Sq 
rt[a]]*d*Sqrt[a + b*Tan[c + d*x]^4])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4144, 1541, 27, 761, 2221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4144

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

\(\Big \downarrow \) 1541

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 2221

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

Input:

Int[1/Sqrt[a + b*Tan[c + d*x]^4],x]
 

Output:

(-1/2*(b^(1/4)*EllipticF[2*ArcTan[(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(S 
qrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sq 
rt[b]*Tan[c + d*x]^2)^2])/(a^(1/4)*(Sqrt[a] - Sqrt[b])*Sqrt[a + b*Tan[c + 
d*x]^4]) + (((Sqrt[a] - Sqrt[b])*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a 
+ b*Tan[c + d*x]^4]])/(2*Sqrt[a + b]) + ((Sqrt[a] + Sqrt[b])*EllipticPi[-1 
/4*(Sqrt[a] - Sqrt[b])^2/(Sqrt[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[c + d*x] 
)/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d* 
x]^4)/(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)^2])/(4*a^(1/4)*b^(1/4)*Sqrt[a + b 
*Tan[c + d*x]^4]))/(Sqrt[a] - Sqrt[b]))/d
 

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 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 2221

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))*(ArcTan[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*ArcTan[q*x 
], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && Po 
sQ[c/a] && EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && PosQ[c*(d/e) + a*(e/d)]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4144

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> 
With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f)   Subst[Int[(a + b* 
(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, 
 b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || 
EqQ[n^2, 16])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.87 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.35

method	result	size
derivativedivides	\(\frac {\sqrt {1-\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tan \left (d x +c \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, \frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{d \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tan \left (d x +c \right )^{4}}}\)	\(123\)
default	\(\frac {\sqrt {1-\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tan \left (d x +c \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, \frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{d \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tan \left (d x +c \right )^{4}}}\)	\(123\)

Input:

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

Output:

1/d/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)*(1+ 
I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)/(a+b*tan(d*x+c)^4)^(1/2)*EllipticPi( 
tan(d*x+c)*(I/a^(1/2)*b^(1/2))^(1/2),I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/2) 
)^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2))
 

Fricas [F]

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

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

Output:

integral(1/sqrt(b*tan(d*x + c)^4 + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(tan(c + d*x)**4*b + a)/(tan(c + d*x)**4*b + a),x)