3.1190 \(\int \frac{\cos ^4(e+f x)}{\sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=510 \[ \frac{32 b \left (2 a^2-b^2\right ) \cos (e+f x)}{35 a^3 f \left (a^2-b^2\right )^2 \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}+\frac{8 \left (a^2-2 b^2\right ) \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^3 d f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}-\frac{8 \left (5 a^2-3 a b-4 b^2\right ) \tan (e+f x) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right )}{35 a^4 \sqrt{d} f (a-b) (a+b)^{3/2}}-\frac{32 b \left (2 a^2-b^2\right ) \tan (e+f x) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right )}{35 a^5 \sqrt{d} f (a-b) (a+b)^{3/2}}+\frac{12 \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^2 d f (a+b \sin (e+f x))^{5/2}}+\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}} \]

[Out]

(2*Cos[e + f*x]^3*Sqrt[d*Sin[e + f*x]])/(7*a*d*f*(a + b*Sin[e + f*x])^(7/2)) + (12*Cos[e + f*x]*Sqrt[d*Sin[e +
f*x]])/(35*a^2*d*f*(a + b*Sin[e + f*x])^(5/2)) + (8*(a^2 - 2*b^2)*Cos[e + f*x]*Sqrt[d*Sin[e + f*x]])/(35*a^3*
(a^2 - b^2)*d*f*(a + b*Sin[e + f*x])^(3/2)) + (32*b*(2*a^2 - b^2)*Cos[e + f*x])/(35*a^3*(a^2 - b^2)^2*f*Sqrt[d
*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]) - (32*b*(2*a^2 - b^2)*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1
+ Csc[e + f*x]))/(a - b)]*EllipticE[ArcSin[(Sqrt[d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x
]])], -((a + b)/(a - b))]*Tan[e + f*x])/(35*a^5*(a - b)*(a + b)^(3/2)*Sqrt[d]*f) - (8*(5*a^2 - 3*a*b - 4*b^2)*
Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticF[ArcSin[(Sqrt[d]*Sqrt[a + b
*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x]])], -((a + b)/(a - b))]*Tan[e + f*x])/(35*a^4*(a - b)*(a + b)
^(3/2)*Sqrt[d]*f)

________________________________________________________________________________________

Rubi [A]  time = 1.91044, antiderivative size = 510, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2887, 2889, 3056, 2993, 2998, 2816, 2994} \[ \frac{32 b \left (2 a^2-b^2\right ) \cos (e+f x)}{35 a^3 f \left (a^2-b^2\right )^2 \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}+\frac{8 \left (a^2-2 b^2\right ) \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^3 d f \left (a^2-b^2\right ) (a+b \sin (e+f x))^{3/2}}-\frac{8 \left (5 a^2-3 a b-4 b^2\right ) \tan (e+f x) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right )}{35 a^4 \sqrt{d} f (a-b) (a+b)^{3/2}}-\frac{32 b \left (2 a^2-b^2\right ) \tan (e+f x) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right )}{35 a^5 \sqrt{d} f (a-b) (a+b)^{3/2}}+\frac{12 \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^2 d f (a+b \sin (e+f x))^{5/2}}+\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4/(Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^(9/2)),x]

[Out]

(2*Cos[e + f*x]^3*Sqrt[d*Sin[e + f*x]])/(7*a*d*f*(a + b*Sin[e + f*x])^(7/2)) + (12*Cos[e + f*x]*Sqrt[d*Sin[e +
f*x]])/(35*a^2*d*f*(a + b*Sin[e + f*x])^(5/2)) + (8*(a^2 - 2*b^2)*Cos[e + f*x]*Sqrt[d*Sin[e + f*x]])/(35*a^3*
(a^2 - b^2)*d*f*(a + b*Sin[e + f*x])^(3/2)) + (32*b*(2*a^2 - b^2)*Cos[e + f*x])/(35*a^3*(a^2 - b^2)^2*f*Sqrt[d
*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]) - (32*b*(2*a^2 - b^2)*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1
+ Csc[e + f*x]))/(a - b)]*EllipticE[ArcSin[(Sqrt[d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x
]])], -((a + b)/(a - b))]*Tan[e + f*x])/(35*a^5*(a - b)*(a + b)^(3/2)*Sqrt[d]*f) - (8*(5*a^2 - 3*a*b - 4*b^2)*
Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticF[ArcSin[(Sqrt[d]*Sqrt[a + b
*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x]])], -((a + b)/(a - b))]*Tan[e + f*x])/(35*a^4*(a - b)*(a + b)
^(3/2)*Sqrt[d]*f)

Rule 2887

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) +
(f_.)*(x_)]], x_Symbol] :> -Simp[(g*(g*Cos[e + f*x])^(p - 1)*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^(m + 1)
)/(a*d*f*(m + 1)), x] + Dist[(g^2*(2*m + 3))/(2*a*(m + 1)), Int[((g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])
^(m + 1))/Sqrt[d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && E
qQ[m + p + 1/2, 0]

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 2993

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)])^(3/2)), x_Symbol] :> Simp[(2*(A*b - a*B)*Cos[e + f*x])/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[
d*Sin[e + f*x]]), x] + Dist[d/(a^2 - b^2), Int[(A*b - a*B + (a*A - b*B)*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]
]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]

Rule 2998

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]

Rule 2816

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 2994

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*Tan[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c
- d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[
(c + d)/b, 2])], -((c + d)/(c - d))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]

Rubi steps

\begin{align*} \int \frac{\cos ^4(e+f x)}{\sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{9/2}} \, dx &=\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}}+\frac{6 \int \frac{\cos ^2(e+f x)}{\sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{7/2}} \, dx}{7 a}\\ &=\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}}+\frac{6 \int \frac{1-\sin ^2(e+f x)}{\sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{7/2}} \, dx}{7 a}\\ &=\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}}+\frac{12 \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^2 d f (a+b \sin (e+f x))^{5/2}}+\frac{12 \int \frac{2 \left (a^2-b^2\right ) d-\left (a^2-b^2\right ) d \sin ^2(e+f x)}{\sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx}{35 a^2 \left (a^2-b^2\right ) d}\\ &=\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}}+\frac{12 \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^2 d f (a+b \sin (e+f x))^{5/2}}+\frac{8 \left (a^2-2 b^2\right ) \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^3 \left (a^2-b^2\right ) d f (a+b \sin (e+f x))^{3/2}}+\frac{8 \int \frac{\frac{1}{2} \left (5 a^4-9 a^2 b^2+4 b^4\right ) d^2-\frac{3}{2} a b \left (a^2-b^2\right ) d^2 \sin (e+f x)}{\sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^{3/2}} \, dx}{35 a^3 \left (a^2-b^2\right )^2 d^2}\\ &=\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}}+\frac{12 \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^2 d f (a+b \sin (e+f x))^{5/2}}+\frac{8 \left (a^2-2 b^2\right ) \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^3 \left (a^2-b^2\right ) d f (a+b \sin (e+f x))^{3/2}}+\frac{32 b \left (2 a^2-b^2\right ) \cos (e+f x)}{35 a^3 \left (a^2-b^2\right )^2 f \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}+\frac{8 \int \frac{\frac{3}{2} a^2 b \left (a^2-b^2\right ) d^2+\frac{1}{2} b \left (5 a^4-9 a^2 b^2+4 b^4\right ) d^2+\left (\frac{3}{2} a b^2 \left (a^2-b^2\right ) d^2+\frac{1}{2} a \left (5 a^4-9 a^2 b^2+4 b^4\right ) d^2\right ) \sin (e+f x)}{(d \sin (e+f x))^{3/2} \sqrt{a+b \sin (e+f x)}} \, dx}{35 a^3 \left (a^2-b^2\right )^3 d}\\ &=\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}}+\frac{12 \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^2 d f (a+b \sin (e+f x))^{5/2}}+\frac{8 \left (a^2-2 b^2\right ) \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^3 \left (a^2-b^2\right ) d f (a+b \sin (e+f x))^{3/2}}+\frac{32 b \left (2 a^2-b^2\right ) \cos (e+f x)}{35 a^3 \left (a^2-b^2\right )^2 f \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}+\frac{\left (4 \left (5 a^2-3 a b-4 b^2\right )\right ) \int \frac{1}{\sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}} \, dx}{35 a^3 (a-b) (a+b)^2}+\frac{\left (16 b \left (2 a^2-b^2\right ) d\right ) \int \frac{1+\sin (e+f x)}{(d \sin (e+f x))^{3/2} \sqrt{a+b \sin (e+f x)}} \, dx}{35 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 \cos ^3(e+f x) \sqrt{d \sin (e+f x)}}{7 a d f (a+b \sin (e+f x))^{7/2}}+\frac{12 \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^2 d f (a+b \sin (e+f x))^{5/2}}+\frac{8 \left (a^2-2 b^2\right ) \cos (e+f x) \sqrt{d \sin (e+f x)}}{35 a^3 \left (a^2-b^2\right ) d f (a+b \sin (e+f x))^{3/2}}+\frac{32 b \left (2 a^2-b^2\right ) \cos (e+f x)}{35 a^3 \left (a^2-b^2\right )^2 f \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}-\frac{32 b \left (2 a^2-b^2\right ) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (1+\csc (e+f x))}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right ) \tan (e+f x)}{35 a^5 (a-b) (a+b)^{3/2} \sqrt{d} f}-\frac{8 \left (5 a^2-3 a b-4 b^2\right ) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (1+\csc (e+f x))}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right ) \tan (e+f x)}{35 a^4 (a-b) (a+b)^{3/2} \sqrt{d} f}\\ \end{align*}

Mathematica [C]  time = 6.5265, size = 1670, normalized size = 3.27 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[e + f*x]^4/(Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^(9/2)),x]

[Out]

(Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((-2*(a^2*Cos[e + f*x] - b^2*Cos[e + f*x]))/(7*a*b^2*(a + b*Sin[e + f*x
])^4) + (4*(5*a^2*Cos[e + f*x] + 3*b^2*Cos[e + f*x]))/(35*a^2*b^2*(a + b*Sin[e + f*x])^3) - (2*(5*a^4*Cos[e +
f*x] - 9*a^2*b^2*Cos[e + f*x] + 8*b^4*Cos[e + f*x]))/(35*a^3*b^2*(a^2 - b^2)*(a + b*Sin[e + f*x])^2) - (32*(2*
a^2*b^2*Cos[e + f*x] - b^4*Cos[e + f*x]))/(35*a^4*(a^2 - b^2)^2*(a + b*Sin[e + f*x]))))/(f*Sqrt[d*Sin[e + f*x]
]) + (4*Sqrt[Sin[e + f*x]]*((4*a*(5*a^4 - 9*a^2*b^2 + 4*b^4)*Sqrt[((a + b)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-a + b
)]*EllipticF[ArcSin[Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sec[e
+ f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[-(((a + b)*Csc[(-e + Pi/2 - f*x)/2]^2*Sin[e + f*x])/a)]*Sqrt[(Csc[(-e
+ Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a])/((a + b)*Sqrt[Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]) + 4*a*(-8*a
^3*b + 4*a*b^3)*((Sqrt[((a + b)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-a + b)]*EllipticF[ArcSin[Sqrt[(Csc[(-e + Pi/2 -
f*x)/2]^2*(a + b*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[-((
(a + b)*Csc[(-e + Pi/2 - f*x)/2]^2*Sin[e + f*x])/a)]*Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a]
)/((a + b)*Sqrt[Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]) - (Sqrt[((a + b)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-a + b)]
*EllipticPi[-(a/b), ArcSin[Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/(-a + b)
]*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[-(((a + b)*Csc[(-e + Pi/2 - f*x)/2]^2*Sin[e + f*x])/a)]*Sqrt[(C
sc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a])/(b*Sqrt[Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])) + 2*(8*a^
2*b^2 - 4*b^4)*((Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]])/(b*Sqrt[Sin[e + f*x]]) + (I*Cos[(-e + Pi/2 - f*x)/2]*C
sc[e + f*x]*EllipticE[I*ArcSinh[Sin[(-e + Pi/2 - f*x)/2]/Sqrt[Sin[e + f*x]]], (-2*a)/(-a - b)]*Sqrt[a + b*Sin[
e + f*x]])/(b*Sqrt[Cos[(-e + Pi/2 - f*x)/2]^2*Csc[e + f*x]]*Sqrt[(Csc[e + f*x]*(a + b*Sin[e + f*x]))/(a + b)])
+ (2*a*((a*Sqrt[((a + b)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-a + b)]*EllipticF[ArcSin[Sqrt[(Csc[(-e + Pi/2 - f*x)/2
]^2*(a + b*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[-(((a + b
)*Csc[(-e + Pi/2 - f*x)/2]^2*Sin[e + f*x])/a)]*Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a])/((a
+ b)*Sqrt[Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]) - (a*Sqrt[((a + b)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-a + b)]*Ell
ipticPi[-(a/b), ArcSin[Sqrt[(Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a]/Sqrt[2]], (-2*a)/(-a + b)]*Se
c[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[-(((a + b)*Csc[(-e + Pi/2 - f*x)/2]^2*Sin[e + f*x])/a)]*Sqrt[(Csc[(
-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/a])/(b*Sqrt[Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])))/b)))/(35*a^4
*(a - b)^2*(a + b)^2*f*Sqrt[d*Sin[e + f*x]])

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Maple [B]  time = 0.898, size = 24365, normalized size = 47.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4/(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac{9}{2}} \sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4/(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(f*x + e)^4/((b*sin(f*x + e) + a)^(9/2)*sqrt(d*sin(f*x + e))), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (f x + e\right ) + a} \sqrt{d \sin \left (f x + e\right )} \cos \left (f x + e\right )^{4}}{b^{5} d \cos \left (f x + e\right )^{6} -{\left (10 \, a^{2} b^{3} + 3 \, b^{5}\right )} d \cos \left (f x + e\right )^{4} +{\left (5 \, a^{4} b + 20 \, a^{2} b^{3} + 3 \, b^{5}\right )} d \cos \left (f x + e\right )^{2} -{\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} d -{\left (5 \, a b^{4} d \cos \left (f x + e\right )^{4} - 10 \,{\left (a^{3} b^{2} + a b^{4}\right )} d \cos \left (f x + e\right )^{2} +{\left (a^{5} + 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} d\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4/(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e))*cos(f*x + e)^4/(b^5*d*cos(f*x + e)^6 - (10*a^2*b^3 + 3
*b^5)*d*cos(f*x + e)^4 + (5*a^4*b + 20*a^2*b^3 + 3*b^5)*d*cos(f*x + e)^2 - (5*a^4*b + 10*a^2*b^3 + b^5)*d - (5
*a*b^4*d*cos(f*x + e)^4 - 10*(a^3*b^2 + a*b^4)*d*cos(f*x + e)^2 + (a^5 + 10*a^3*b^2 + 5*a*b^4)*d)*sin(f*x + e)
), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**4/(a+b*sin(f*x+e))**(9/2)/(d*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac{9}{2}} \sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4/(a+b*sin(f*x+e))^(9/2)/(d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(f*x + e)^4/((b*sin(f*x + e) + a)^(9/2)*sqrt(d*sin(f*x + e))), x)