\(\int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [244]

Optimal result

Integrand size = 23, antiderivative size = 469 \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}\right )}{2 \sqrt {-a} d}+\frac {\sqrt [4]{b} \sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 a d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{4 a \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \]

[Out]

-1/2*arctan(cos(d*x+c)*(-a)^(1/2)/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2))/d/(-a)^(1/2)+1/2*b^(1/4)*(a+b)^
(1/4)*(cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))*El
lipticF(sin(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4))),1/2*(2+2*b^(1/2)/(a+b)^(1/2))^(1/2))*(1+cos(d*x+c)^2*b^(
1/2)/(a+b)^(1/2))*(b^(1/2)-(a+b)^(1/2))*((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/(1+cos(d*x+c)^2*b^(1/2)/(
a+b)^(1/2))^2)^(1/2)/a/d/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)-1/4*(a+b)^(1/4)*(cos(2*arctan(b^(1/4)*cos
(d*x+c)/(a+b)^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))*EllipticPi(sin(2*arctan(b^(1/4)*c
os(d*x+c)/(a+b)^(1/4))),1/4*(b^(1/2)+(a+b)^(1/2))^2/b^(1/2)/(a+b)^(1/2),1/2*(2+2*b^(1/2)/(a+b)^(1/2))^(1/2))*(
1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))*(b^(1/2)-(a+b)^(1/2))^2*((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/(1+co
s(d*x+c)^2*b^(1/2)/(a+b)^(1/2))^2)^(1/2)/a/b^(1/4)/d/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3294, 1230, 1117, 1720} \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )}{2 \sqrt {-a} d}+\frac {\sqrt [4]{b} \sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 a d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{4 a \sqrt [4]{b} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}} \]

[In]

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

[Out]

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

Rule 1117

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 1230

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

Rule 1720

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

Rule 3294

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {\left ((a+b) \left (-1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right ) \text {Subst}\left (\int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {a+b}}}{\left (1-x^2\right ) \sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{a d}+\frac {\left (\sqrt {b} \left (\sqrt {b}-\sqrt {a+b}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\arctan \left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}\right )}{2 \sqrt {-a} d}+\frac {\sqrt [4]{b} \sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 a d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{4 a \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 14.57 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.04 \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos ^3(c+d x) \sqrt {\left (1-\frac {i \sqrt {a}}{\sqrt {b}}\right ) \sec ^2(c+d x)} \left (\operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \left (a-i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right ) \sqrt {-\frac {i \left (a+i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right )}{\sqrt {a} \sqrt {b}}}+i \left (\sqrt {a}+i \sqrt {b}\right ) \sqrt {b} \operatorname {EllipticPi}\left (-\frac {2 i \sqrt {b}}{\sqrt {a}-i \sqrt {b}},\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}} \sqrt {\frac {(a+b) \left (a \sec ^4(c+d x)+b \tan ^4(c+d x)\right )}{a b}}\right )}{(a+b) d \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}} \sqrt {\cos ^4(c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}} \]

[In]

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

[Out]

(Cos[c + d*x]^3*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2]*(EllipticF[ArcSin[Sqrt[1 + (I*Sqrt[a])/Sqrt[b]
+ (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]/Sqrt[2]], (2*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])]*(a - I*Sqrt[a]*Sq
rt[b] + (a + b)*Tan[c + d*x]^2)*Sqrt[((-I)*(a + I*Sqrt[a]*Sqrt[b] + (a + b)*Tan[c + d*x]^2))/(Sqrt[a]*Sqrt[b])
] + I*(Sqrt[a] + I*Sqrt[b])*Sqrt[b]*EllipticPi[((-2*I)*Sqrt[b])/(Sqrt[a] - I*Sqrt[b]), ArcSin[Sqrt[1 + (I*Sqrt
[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]/Sqrt[2]], (2*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])]*Sqrt
[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[((a + b)*(a*Sec[c + d*x]^4 + b*T
an[c + d*x]^4))/(a*b)]))/((a + b)*d*Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b]
)]*Sqrt[Cos[c + d*x]^4*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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