3.1402 \(\int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=601 \[ \frac {2 a^2 \sqrt {g \cos (e+f x)}}{b f g^3 \left (a^2-b^2\right )}-\frac {2 b \sqrt {g \cos (e+f x)}}{f g^3 \left (a^2-b^2\right )}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {2 b}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}+\frac {2 a \sin (e+f x)}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f g^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac {a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}} \]

[Out]

-a^4*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(3/2)/(-a^2+b^2)^(7/4)/f/g^(5/2)-a^4*arct
anh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(3/2)/(-a^2+b^2)^(7/4)/f/g^(5/2)-2/3*b/(a^2-b^2)/
f/g/(g*cos(f*x+e))^(3/2)+2/3*a*sin(f*x+e)/(a^2-b^2)/f/g/(g*cos(f*x+e))^(3/2)-4/3*a*(cos(1/2*f*x+1/2*e)^2)^(1/2
)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)/(a^2-b^2)/f/g^2/(g*cos(f*x+e))^(1/
2)+2*a^3*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2
)/b^2/(a^2-b^2)/f/g^2/(g*cos(f*x+e))^(1/2)-a^5*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(
1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^2/(a^2-b^2)/f/g^2/(a^2-b*(b-(-a^2+b^2)^(1/
2)))/(g*cos(f*x+e))^(1/2)-a^5*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*
b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^2/(a^2-b^2)/f/g^2/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e
))^(1/2)+2*a^2*(g*cos(f*x+e))^(1/2)/b/(a^2-b^2)/f/g^3-2*b*(g*cos(f*x+e))^(1/2)/(a^2-b^2)/f/g^3

________________________________________________________________________________________

Rubi [A]  time = 1.40, antiderivative size = 601, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 16, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {2902, 2566, 2642, 2641, 2565, 14, 2909, 30, 2867, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac {2 a^2 \sqrt {g \cos (e+f x)}}{b f g^3 \left (a^2-b^2\right )}-\frac {2 b \sqrt {g \cos (e+f x)}}{f g^3 \left (a^2-b^2\right )}-\frac {a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f g^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac {a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 b}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}+\frac {2 a \sin (e+f x)}{3 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]

[Out]

-((a^4*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(3/2)*(-a^2 + b^2)^(7/4)*f*g^(5
/2))) - (a^4*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(3/2)*(-a^2 + b^2)^(7/4)
*f*g^(5/2)) - (2*b)/(3*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2)) + (2*a^2*Sqrt[g*Cos[e + f*x]])/(b*(a^2 - b^2)*f
*g^3) - (2*b*Sqrt[g*Cos[e + f*x]])/((a^2 - b^2)*f*g^3) - (4*a*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3
*(a^2 - b^2)*f*g^2*Sqrt[g*Cos[e + f*x]]) + (2*a^3*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(b^2*(a^2 - b^
2)*f*g^2*Sqrt[g*Cos[e + f*x]]) - (a^5*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2,
 2])/(b^2*(a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) - (a^5*Sqrt[Cos[e + f*x]]*E
llipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^2*(a^2 - b^2)*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*g
^2*Sqrt[g*Cos[e + f*x]]) + (2*a*Sin[e + f*x])/(3*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
 !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
 &&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2566

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(a*Sin[e
+ f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Sin[e +
 f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Integ
ersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2702

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, -Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[(b*g)/f, Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
 + f*x]]*(q - b*Cos[e + f*x])), x], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
 b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2867

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

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2909

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a*d)/b, Int[(
(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1))/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && N
eQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1, p, 1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx &=\frac {a \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{5/2}} \, dx}{a^2-b^2}-\frac {b \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{5/2}} \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{\left (a^2-b^2\right ) g^2}\\ &=\frac {2 a \sin (e+f x)}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\frac {(2 a) \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{3 \left (a^2-b^2\right ) g^2}-\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{b \left (a^2-b^2\right ) g^2}+\frac {a^3 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b \left (a^2-b^2\right ) g^2}+\frac {b \operatorname {Subst}\left (\int \frac {1-\frac {x^2}{g^2}}{x^{5/2}} \, dx,x,g \cos (e+f x)\right )}{\left (a^2-b^2\right ) f g}\\ &=\frac {2 a \sin (e+f x)}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,g \cos (e+f x)\right )}{b \left (a^2-b^2\right ) f g^3}+\frac {a^3 \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b^2 \left (a^2-b^2\right ) g^2}-\frac {a^4 \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^2 \left (a^2-b^2\right ) g^2}+\frac {b \operatorname {Subst}\left (\int \left (\frac {1}{x^{5/2}}-\frac {1}{g^2 \sqrt {x}}\right ) \, dx,x,g \cos (e+f x)\right )}{\left (a^2-b^2\right ) f g}-\frac {\left (2 a \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{3 \left (a^2-b^2\right ) g^2 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 b}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b \left (a^2-b^2\right ) f g^3}-\frac {2 b \sqrt {g \cos (e+f x)}}{\left (a^2-b^2\right ) f g^3}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\frac {a^5 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2 \left (-a^2+b^2\right )^{3/2} g^2}-\frac {a^5 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2 \left (-a^2+b^2\right )^{3/2} g^2}-\frac {a^4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) g^2+b^2 x^2\right )} \, dx,x,g \cos (e+f x)\right )}{b \left (a^2-b^2\right ) f g}+\frac {\left (a^3 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b^2 \left (a^2-b^2\right ) g^2 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 b}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b \left (a^2-b^2\right ) f g^3}-\frac {2 b \sqrt {g \cos (e+f x)}}{\left (a^2-b^2\right ) f g^3}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\frac {\left (2 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) g^2+b^2 x^4} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \left (a^2-b^2\right ) f g}-\frac {\left (a^5 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2 \left (-a^2+b^2\right )^{3/2} g^2 \sqrt {g \cos (e+f x)}}-\frac {\left (a^5 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2 \left (-a^2+b^2\right )^{3/2} g^2 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 b}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b \left (a^2-b^2\right ) f g^3}-\frac {2 b \sqrt {g \cos (e+f x)}}{\left (a^2-b^2\right ) f g^3}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\frac {a^4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \left (-a^2+b^2\right )^{3/2} f g^2}-\frac {a^4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \left (-a^2+b^2\right )^{3/2} f g^2}\\ &=-\frac {a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {2 b}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b \left (a^2-b^2\right ) f g^3}-\frac {2 b \sqrt {g \cos (e+f x)}}{\left (a^2-b^2\right ) f g^3}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{3 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 26.84, size = 1958, normalized size = 3.26 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^4/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]

[Out]

(2*Cos[e + f*x]*(-b + a*Sin[e + f*x]))/(3*(a^2 - b^2)*f*(g*Cos[e + f*x])^(5/2)) + (Cos[e + f*x]^(5/2)*((-2*(-7
*a^2 + 3*b^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b
^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e + f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/
2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + f*
x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e
+ f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1
 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]]
)/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos
[e + f*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]])
)/(-a^2 + b^2)^(3/4))*Sin[e + f*x])/(Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) + ((3*a^2 - 3*b^2)*(a + b*
Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*(((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e
 + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 + ((1 + I)*
Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + (4*Sqrt[Cos[e + f*x]])/b - (4*
a*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(5/2))/(5*(a^2 -
b^2)) + (10*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[C
os[e + f*x]])/(Sqrt[1 - Cos[e + f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e +
 f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)
] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2
)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) + ((1/4 - I/4)*(-2*a^2 + b^2)*Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^
2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/4 - I/4)*(-2*a^2 + b
^2)*Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]])/(b^(3/2)
*(-a^2 + b^2)^(3/4)))*Sin[e + f*x])/(Sqrt[1 - Cos[e + f*x]^2]*(-1 + 2*Cos[e + f*x]^2)*(a + b*Sin[e + f*x])) -
(4*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos
[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]]*Sqrt[1 - Cos[e + f*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1
, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Cos[e + f*x]^
2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*
x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) + (a*(-2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*S
qrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] -
 Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]] + Log[Sqrt[a^2 -
 b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2
)^(3/4)))*Sin[e + f*x]^2)/((1 - Cos[e + f*x]^2)*(a + b*Sin[e + f*x]))))/(6*(a - b)*(a + b)*f*(g*Cos[e + f*x])^
(5/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^4/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)

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maple [C]  time = 11.51, size = 1268, normalized size = 2.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x)

[Out]

2/f/g^3/b*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)-2/f/g/b*a^4/(a-b)/(a+b)*sum((_R^4+_R^2*g)/(_R^7*b^2-3*_R^5*b^2*
g+8*_R^3*a^2*g^2-5*_R^3*b^2*g^2-_R*b^2*g^3)*ln((-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-cos(1/2*f*x+1/2*e)*g^(1/2)*
2^(1/2)-_R),_R=RootOf(b^2*_Z^8-4*b^2*g*_Z^6+(16*a^2*g^2-10*b^2*g^2)*_Z^4-4*b^2*g^3*_Z^2+b^2*g^4))-1/12/f/g^3*b
/(a^2-b^2)/(cos(1/2*f*x+1/2*e)-1/2*2^(1/2))^2*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)+1/12/f/g^3*b*2^(1/2)/(a^2-b^
2)/(cos(1/2*f*x+1/2*e)-1/2*2^(1/2))*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-1/12/f/g^3*b/(a^2-b^2)/(cos(1/2*f*x+1/
2*e)+1/2*2^(1/2))^2*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-1/12/f/g^3*b*2^(1/2)/(a^2-b^2)/(cos(1/2*f*x+1/2*e)+1/2
*2^(1/2))*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-2/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a/
g^2/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)/b^2*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1
/2*e)^2+1)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2)
)+1/8/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a^5/g^2/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x
+1/2*e)^2-1))^(1/2)/(a-b)/(a+b)/b^4*sum(1/_alpha/(2*_alpha^2-1)*(2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1
/2)*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*cos(1/2*f*x+1/2*e)^2*a^2-3*b^2*cos(1/2*f*x+1/2*e)^2+b^2*_alp
ha^2-3*a^2+2*b^2)*2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2
*e)^2))^(1/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)/(-
sin(1/2*f*x+1/2*e)^2*g*(2*sin(1/2*f*x+1/2*e)^2-1))^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1)
,2^(1/2))),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))+1/3/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)
^(1/2)*a/g^3/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)/(a^2-b^2)*cos(1/2*f*x+1/2*e)*(-g*(2*sin(1
/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)/(cos(1/2*f*x+1/2*e)^2-1/2)^2-2/3/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*
sin(1/2*f*x+1/2*e)^2)^(1/2)*a/g^2/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)/(a^2-b^2)*(sin(1/2*f
*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)*
EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^4/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (e+f\,x\right )}^4}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^4/((g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))),x)

[Out]

int(sin(e + f*x)^4/((g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4/(g*cos(f*x+e))**(5/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

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