3.245 \(\int \frac {\csc ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

Optimal. Leaf size=776 \[ -\frac {\sqrt {b} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{2 a d \sqrt {a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )}{4 \sqrt {-a} d}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{2 a d}-\frac {\sqrt [4]{b} \left (-\sqrt {b} \sqrt {a+b}+a+b\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 a d \sqrt [4]{a+b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}+\frac {\sqrt [4]{b} (a+b)^{3/4} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 a d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \Pi \left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{8 a \sqrt [4]{b} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}} \]

[Out]

-1/4*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*cot(d*x+c)*csc
(d*x+c)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/a/d-1/2*cos(d*x+c)*b^(1/2)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x
+c)^4)^(1/2)/a/d/(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))/(a+b)^(1/2)+1/2*b^(1/4)*(a+b)^(3/4)*(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)))*EllipticE(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))*((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)-1/8*(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)*cos(d*x+c)/(a+b)^(1/4))),1/4*((a+b)^(1/2)+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+cos(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)-1/2*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)))*EllipticF(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))*(a+b-b^(1/2)*(a+b)^(1/2))*((a+b-2*b*c
os(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/(a+b)^(1/4)/d/(a+b-2*b*cos(d
*x+c)^2+b*cos(d*x+c)^4)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.04, antiderivative size = 776, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3215, 1223, 1714, 1195, 1708, 1103, 1706} \[ -\frac {\sqrt {b} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{2 a d \sqrt {a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )}{4 \sqrt {-a} d}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{2 a d}-\frac {\sqrt [4]{b} \left (-\sqrt {b} \sqrt {a+b}+a+b\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 a d \sqrt [4]{a+b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}+\frac {\sqrt [4]{b} (a+b)^{3/4} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 a d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \Pi \left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{8 a \sqrt [4]{b} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-ArcTan[(Sqrt[-a]*Cos[c + d*x])/Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4]]/(4*Sqrt[-a]*d) - (Sqrt[b]
*Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(2*a*Sqrt[a + b]*d*(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]*Cot[c + d*x]*Csc[c + d*x])/(2*a*d)
+ (b^(1/4)*(a + b)^(3/4)*(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)]*EllipticE[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]) - (b^
(1/4)*(a + b - 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*(a + b)^(1/4)*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)]*Ellip
ticPi[(Sqrt[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]/Sqrt[a + b])/2])/(8*a*b^(1/4)*d*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])

Rule 1103

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)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> 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)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), 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 1223

Int[((d_) + (e_.)*(x_)^2)^(q_)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Simp[(e^2*x*(d + e*x^2)
^(q + 1)*Sqrt[a + b*x^2 + c*x^4])/(2*d*(q + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(2*d*(q + 1)*(c*d^2 - b*d
*e + a*e^2)), Int[((d + e*x^2)^(q + 1)*Simp[a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1) - 2*e*(c*d*(q + 1) - b*e
*(q + 2))*x^2 + c*e^2*(2*q + 5)*x^4, x])/Sqrt[a + b*x^2 + c*x^4], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b
^2 - 4*a*c, 0] && ILtQ[q, -1]

Rule 1706

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)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), 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 1708

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

Rule 1714

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]
, A = Coeff[P4x, x, 0], B = Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Dist[C/(e*q), Int[(1 - q*x^2)/Sqrt[a + b
*x^2 + c*x^4], x], x] + Dist[1/(c*e), Int[(A*c*e + a*C*d*q + (B*c*e - C*(c*d - a*e*q))*x^2)/((d + e*x^2)*Sqrt[
a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[P4x, x^2, 2] && 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] &&  !GtQ[b^2 - 4*a*c, 0]

Rule 3215

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*} \int \frac {\csc ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)} \cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\operatorname {Subst}\left (\int \frac {-a+b-2 b x^2+b x^4}{\left (1-x^2\right ) \sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{2 a d}\\ &=-\frac {\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)} \cot (c+d x) \csc (c+d x)}{2 a d}-\frac {\operatorname {Subst}\left (\int \frac {-b (-a+b)+b^{3/2} \sqrt {a+b}+\left (2 b^2-b \left (b+\sqrt {b} \sqrt {a+b}\right )\right ) x^2}{\left (1-x^2\right ) \sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{2 a b d}+\frac {\left (\sqrt {b} \sqrt {a+b}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a+b}}}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{2 a d}\\ &=-\frac {\sqrt {b} \cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{2 a \sqrt {a+b} d \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )}-\frac {\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)} \cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\sqrt [4]{b} (a+b)^{3/4} \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}} E\left (2 \tan ^{-1}\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 {\left (\sqrt {a+b} \left (\sqrt {b}-\sqrt {a+b}\right )\right ) \operatorname {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 )}{2 a d}-\frac {\left (\sqrt {b} \left (a+b-\sqrt {b} \sqrt {a+b}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{a \sqrt {a+b} d}\\ &=-\frac {\tan ^{-1}\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 )}{4 \sqrt {-a} d}-\frac {\sqrt {b} \cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{2 a \sqrt {a+b} d \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )}-\frac {\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)} \cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\sqrt [4]{b} (a+b)^{3/4} \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}} E\left (2 \tan ^{-1}\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]{b} \left (a+b-\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}} F\left (2 \tan ^{-1}\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 \sqrt [4]{a+b} 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}} \Pi \left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}};2 \tan ^{-1}\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 )}{8 a \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 32.63, size = 119171, normalized size = 153.57 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

Result too large to show

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\csc \left (d x + c\right )^{3}}{\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

integral(csc(d*x + c)^3/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 2.48Not invertible Error: Bad Argument Value

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maple [F]  time = 1.62, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}\left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)

[Out]

int(csc(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (d x + c\right )^{3}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(csc(d*x + c)^3/sqrt(b*sin(d*x + c)^4 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\sin \left (c+d\,x\right )}^3\,\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^3*(a + b*sin(c + d*x)^4)^(1/2)),x)

[Out]

int(1/(sin(c + d*x)^3*(a + b*sin(c + d*x)^4)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3/(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(csc(c + d*x)**3/sqrt(a + b*sin(c + d*x)**4), x)

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