\(\int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [492]

Optimal result

Integrand size = 25, antiderivative size = 409 \[ \int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos (c+d x) \sin (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{(a+b) d \sqrt {a+b \sin ^4(c+d x)} \left (\frac {\sqrt {a}}{\sqrt {a+b}}+\tan ^2(c+d x)\right )}-\frac {\sqrt [4]{a} \cos ^2(c+d x) E\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{(a+b)^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt [4]{a} \cos ^2(c+d x) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 (a+b)^{3/4} d \sqrt {a+b \sin ^4(c+d x)}} \] Output:

cos(d*x+c)*sin(d*x+c)*(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a+b)/d/(a+b 
*sin(d*x+c)^4)^(1/2)/(a^(1/2)/(a+b)^(1/2)+tan(d*x+c)^2)-a^(1/4)*cos(d*x+c) 
^2*EllipticE(sin(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4))),1/2*(2-2*a^(1/2 
)/(a+b)^(1/2))^(1/2))*(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)*((a+2*a*tan(d*x+c 
)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)^2)^(1/2)/(a+b)^ 
(3/4)/d/(a+b*sin(d*x+c)^4)^(1/2)+1/2*a^(1/4)*cos(d*x+c)^2*InverseJacobiAM( 
2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)),1/2*(2-2*a^(1/2)/(a+b)^(1/2))^(1/ 
2))*(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)*((a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+ 
c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)^2)^(1/2)/(a+b)^(3/4)/d/(a+b*sin(d 
*x+c)^4)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.43 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {2 i \sqrt {2} \sqrt {a} \cos ^2(c+d x) \left (E\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )\right ) \sqrt {1+\left (1-\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}}{\left (\sqrt {a}+i \sqrt {b}\right ) \sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))}} \] Input:

Integrate[Tan[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

((-2*I)*Sqrt[2]*Sqrt[a]*Cos[c + d*x]^2*(EllipticE[I*ArcSinh[Sqrt[1 - (I*Sq 
rt[b])/Sqrt[a]]*Tan[c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b]) 
] - EllipticF[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], (Sqrt 
[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])])*Sqrt[1 + (1 - (I*Sqrt[b])/Sqrt[a] 
)*Tan[c + d*x]^2]*Sqrt[1 + (1 + (I*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2])/((Sq 
rt[a] + I*Sqrt[b])*Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*d*Sqrt[8*a + 3*b - 4*b*Co 
s[2*(c + d*x)] + b*Cos[4*(c + d*x)]])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3711, 1459, 27, 1416, 1509}

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.

\(\displaystyle \int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^2}{\sqrt {a+b \sin (c+d x)^4}}dx\)

\(\Big \downarrow \) 3711

\(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \int \frac {\tan ^2(c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{d \sqrt {a+b \sin ^4(c+d x)}}\)

\(\Big \downarrow \) 1459

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

\(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {\sqrt [4]{a} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 (a+b)^{3/4} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}-\frac {\int \frac {\sqrt {a}-\sqrt {a+b} \tan ^2(c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{\sqrt {a+b}}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {\sqrt [4]{a} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 (a+b)^{3/4} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{\sqrt [4]{a+b} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}-\frac {\tan (c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}{\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}}}{\sqrt {a+b}}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\)

Input:

Int[Tan[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

(Cos[c + d*x]^2*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4]*((a^ 
(1/4)*EllipticF[2*ArcTan[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)], (1 - Sqrt[ 
a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[(a + 2*a*Ta 
n[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x] 
^2)^2])/(2*(a + b)^(3/4)*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x 
]^4]) - (-((Tan[c + d*x]*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x 
]^4])/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)) + (a^(1/4)*EllipticE[2*ArcTa 
n[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)], (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqr 
t[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[(a + 2*a*Tan[c + d*x]^2 + (a + b)* 
Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/((a + b)^(1/4)* 
Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4]))/Sqrt[a + b]))/(d*S 
qrt[a + b*Sin[c + d*x]^4])
 

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 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1459

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

rule 1509

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 

rule 3042

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

rule 3711

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_)*((d_.)*tan[(e_.) + (f_.)* 
(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff 
*(a + b*Sin[e + f*x]^4)^p*((Sec[e + f*x]^2)^(2*p)/(f*Apart[a*(1 + Tan[e + f 
*x]^2)^2 + b*Tan[e + f*x]^4]^p))   Subst[Int[(d*ff*x)^m*(ExpandToSum[a*(1 + 
 ff^2*x^2)^2 + b*ff^4*x^4, x]^p/(1 + ff^2*x^2)^(2*p + 1)), x], x, Tan[e + f 
*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ[p - 1/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(tan(d*x + c)^2/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b 
), x)
 

Sympy [F]

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

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

Output:

Integral(tan(c + d*x)**2/sqrt(a + b*sin(c + d*x)**4), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(c + d*x)**4*b + a)*tan(c + d*x)**2)/(sin(c + d*x)**4*b + a), 
x)