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

Optimal. Leaf size=521 \[ \frac {\sqrt {b} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{d \sqrt {a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )}+\frac {\sqrt {-a} \tan ^{-1}\left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )}{2 d}-\frac {\sqrt [4]{b} (a+b)^{3/4} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \Pi \left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{4 \sqrt [4]{b} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}} \]

[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))*(-a)^(1/2)/d+cos(d*x+c)*b^(1/2)*
(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/d/(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))/(a+b)^(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)))*Ell
ipticE(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)/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)*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)/b^(1/4)/d/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)

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Rubi [A]  time = 0.68, antiderivative size = 521, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3215, 1208, 1197, 1103, 1195, 1216, 1706} \[ \frac {\sqrt {b} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{d \sqrt {a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )}+\frac {\sqrt {-a} \tan ^{-1}\left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )}{2 d}-\frac {\sqrt [4]{b} (a+b)^{3/4} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \Pi \left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{4 \sqrt [4]{b} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-a]*ArcTan[(Sqrt[-a]*Cos[c + d*x])/Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4]])/(2*d) + (Sqrt[b
]*Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(Sqrt[a + b]*d*(1 + (Sqrt[b]*Cos[c + d*x]^
2)/Sqrt[a + b])) - (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])/(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[(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])/(4*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 1197

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

Rule 1208

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d -
b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/e^2, Int[(a + b*x^2 + c*x^4
)^(p - 1)/(d + e*x^2), 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] && IGtQ[p + 1/2, 0]

Rule 1216

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 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 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 \csc (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b-2 b x^2+b x^4}}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {-b+b x^2}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{d}-\frac {a \operatorname {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 (\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 )}{d}+\frac {\left ((a+b) \left (-1+\frac {\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 )}{d}\\ &=\frac {\sqrt {-a} \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 )}{2 d}+\frac {\sqrt {b} \cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{\sqrt {a+b} d \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )}-\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 )}{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 )}{4 \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 = 32.00, size = 118912, normalized size = 228.24 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \csc \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \csc \left (d x +c \right ) \sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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