3.3.48 \(\int \frac {\csc ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [248]

3.3.48.1 Optimal result
3.3.48.2 Mathematica [C] (verified)
3.3.48.3 Rubi [A] (verified)
3.3.48.4 Maple [F]
3.3.48.5 Fricas [F]
3.3.48.6 Sympy [F]
3.3.48.7 Maxima [F]
3.3.48.8 Giac [F]
3.3.48.9 Mupad [F(-1)]

3.3.48.1 Optimal result

Integrand size = 25, antiderivative size = 493 \[ \int \frac {\csc ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\cos ^2(c+d x) \cot (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt {a+b} \cos (c+d x) \sin (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )}-\frac {\sqrt [4]{a+b} \cos ^2(c+d x) E\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (a+b+\sqrt {a} \sqrt {a+b}\right ) \cos ^2(c+d x) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 a^{3/4} (a+b)^{3/4} d \sqrt {a+b \sin ^4(c+d x)}} \]

output
-(a+b)^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))^2 
)^(1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticE(sin(2*arct 
an((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+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^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)/a^(3/4)/d/(a+b*sin(d*x+c)^4)^ 
(1/2)+1/2*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))^2)^( 
1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticF(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+b 
+a^(1/2)*(a+b)^(1/2))*((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^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)/a^(3/4 
)/(a+b)^(3/4)/d/(a+b*sin(d*x+c)^4)^(1/2)-cos(d*x+c)^2*cot(d*x+c)*(a+2*a*ta 
n(d*x+c)^2+(a+b)*tan(d*x+c)^4)/a/d/(a+b*sin(d*x+c)^4)^(1/2)+cos(d*x+c)*sin 
(d*x+c)*(a+b)^(1/2)*(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/a/d/(a+b*sin(d 
*x+c)^4)^(1/2)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)
 
3.3.48.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 13.20 (sec) , antiderivative size = 464, normalized size of antiderivative = 0.94 \[ \int \frac {\csc ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))} \cot (c+d x)}{2 \sqrt {2} a d}-\frac {2 \sqrt {2} \left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \left (a+b \sin ^4(c+d x)\right ) \tan (c+d x)+\sqrt {a} \left (i \sqrt {a}+\sqrt {b}\right ) \cos ^2(c+d x) E\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {1+\left (1-\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}-\sqrt {a} \sqrt {b} \cos ^2(c+d x) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {1+\left (1-\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}\right )}{a \sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))}} \]

input
Integrate[Csc[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]^4],x]
 
output
-1/2*(Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]*Cot[c + 
d*x])/(Sqrt[2]*a*d) - (2*Sqrt[2]*(Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*(a + b*Sin 
[c + d*x]^4)*Tan[c + d*x] + Sqrt[a]*(I*Sqrt[a] + Sqrt[b])*Cos[c + d*x]^2*E 
llipticE[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] - Sqrt[a]*S 
qrt[b]*Cos[c + d*x]^2*EllipticF[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Ta 
n[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]))/(a*Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*d*Sqrt[8*a + 3*b - 4*b*Cos[ 
2*(c + d*x)] + b*Cos[4*(c + d*x)]])
 
3.3.48.3 Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3698, 1604, 25, 1511, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (c+d x)^2 \sqrt {a+b \sin (c+d x)^4}}dx\)

\(\Big \downarrow \) 3698

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

\(\Big \downarrow \) 1604

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1511

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 1509

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

input
Int[Csc[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]*(-(( 
Cot[c + d*x]*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4])/a) + ( 
(a^(1/4)*(a + b + Sqrt[a]*Sqrt[a + b])*EllipticF[2*ArcTan[((a + b)^(1/4)*T 
an[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*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(S 
qrt[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]) - Sqrt[a + b]*(-((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*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*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(Sqrt[a] + Sq 
rt[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])))/a))/(d*Sqrt[a + b*Sin[c + d*x]^4])
 

3.3.48.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 1511
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

rule 1604
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) 
/(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1))   Int[(f*x)^(m + 2)*(a + b*x^2 
 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x 
], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ 
m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3698
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( 
p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1) 
*(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[x^m*(ExpandToSum[a*(1 + ff^2*x 
^2)^2 + b*ff^4*x^4, x]^p/(1 + ff^2*x^2)^(m/2 + 2*p + 1)), x], x, Tan[e + f* 
x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[p - 1 
/2]
 
3.3.48.4 Maple [F]

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

input
int(csc(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x)
 
output
int(csc(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x)
 
3.3.48.5 Fricas [F]

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

input
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")
 
output
integral(csc(d*x + c)^2/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b 
), x)
 
3.3.48.6 Sympy [F]

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

input
integrate(csc(d*x+c)**2/(a+b*sin(d*x+c)**4)**(1/2),x)
 
output
Integral(csc(c + d*x)**2/sqrt(a + b*sin(c + d*x)**4), x)
 
3.3.48.7 Maxima [F]

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

input
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")
 
output
integrate(csc(d*x + c)^2/sqrt(b*sin(d*x + c)^4 + a), x)
 
3.3.48.8 Giac [F]

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

input
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")
 
output
sage0*x
 
3.3.48.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {1}{{\sin \left (c+d\,x\right )}^2\,\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

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