3.1389 \(\int \frac {\sin ^4(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)
Optimal. Leaf size=509 \[ -\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b^3 f g}+\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {2 a \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 b^2 f g}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g} \]
[Out]
2/5*(g*cos(f*x+e))^(5/2)/b/f/g^3-a^4*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/(-a
^2+b^2)^(3/4)/f/g^(1/2)-a^4*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/(-a^2+b^2)^
(3/4)/f/g^(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))*co
s(f*x+e)^(1/2)/b^4/f/(g*cos(f*x+e))^(1/2)-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)/b^2/f/(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^4/f/(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^4/f/(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^3/f/g-2*(g*cos(f*x+e))^(1/2)/b/f/g+2/3*a*sin(f*x+e)*(g*cos(f*x+e))^(1/2)/b^
2/f/g
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Rubi [A] time = 1.51, antiderivative size = 509, normalized size of antiderivative = 1.00,
number of steps used = 23, number of rules used = 15, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used
= {2909, 2565, 14, 2568, 2642, 2641, 30, 2867, 2702, 2807, 2805, 329, 212, 208, 205}
\[ -\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b^3 f g}-\frac {a^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {g \cos (e+f x)}}+\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^5 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 b^2 f g}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g} \]
Antiderivative was successfully verified.
[In]
Int[Sin[e + f*x]^4/(Sqrt[g*Cos[e + f*x]]*(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^(7/2)*(-a^2 + b^2)^(3/4)*f*Sqrt
[g])) - (a^4*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(7/2)*(-a^2 + b^2)^(3/4)
*f*Sqrt[g]) - (2*a^2*Sqrt[g*Cos[e + f*x]])/(b^3*f*g) - (2*Sqrt[g*Cos[e + f*x]])/(b*f*g) + (2*(g*Cos[e + f*x])^
(5/2))/(5*b*f*g^3) - (2*a^3*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(b^4*f*Sqrt[g*Cos[e + f*x]]) - (4*a*
Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*b^2*f*Sqrt[g*Cos[e + f*x]]) + (a^5*Sqrt[Cos[e + f*x]]*Ellipti
cPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^4*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*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^4*(a^2 - b*(b + Sqrt
[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) + (2*a*Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(3*b^2*f*g)
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 2568
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]
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 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)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac {\int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{b}-\frac {a \int \frac {\sin ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b}\\ &=-\frac {a \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{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}{b^2}-\frac {\operatorname {Subst}\left (\int \frac {1-\frac {x^2}{g^2}}{\sqrt {x}} \, dx,x,g \cos (e+f x)\right )}{b f g}\\ &=\frac {2 a \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f g}+\frac {a^2 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{b^3}-\frac {a^3 \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^3}-\frac {(2 a) \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{3 b^2}-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{\sqrt {x}}-\frac {x^{3/2}}{g^2}\right ) \, dx,x,g \cos (e+f x)\right )}{b f g}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}+\frac {2 a \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f g}-\frac {a^3 \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b^4}+\frac {a^4 \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^4}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,g \cos (e+f x)\right )}{b^3 f g}-\frac {\left (2 a \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{3 b^2 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b^3 f g}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}-\frac {4 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}+\frac {2 a \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f g}-\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^4 \sqrt {-a^2+b^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^4 \sqrt {-a^2+b^2}}+\frac {\left (a^4 g\right ) \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^3 f}-\frac {\left (a^3 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b^4 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b^3 f g}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}-\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \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 b^2 f \sqrt {g \cos (e+f x)}}+\frac {2 a \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f g}+\frac {\left (2 a^4 g\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^3 f}-\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^4 \sqrt {-a^2+b^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^4 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b^3 f g}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}-\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \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 b^2 f \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^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \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^4 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f g}-\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^3 \sqrt {-a^2+b^2} f}-\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^3 \sqrt {-a^2+b^2} f}\\ &=-\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^{7/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\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^{7/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b^3 f g}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}-\frac {2 a^3 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \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 b^2 f \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^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \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^4 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f g}\\ \end {align*}
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Mathematica [C] time = 26.92, size = 1953, normalized size = 3.84 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sin[e + f*x]^4/(Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x])),x]
[Out]
(Cos[e + f*x]*(Cos[2*(e + f*x)]/(5*b) + (2*a*Sin[e + f*x])/(3*b^2)))/(f*Sqrt[g*Cos[e + f*x]]) - (Sqrt[Cos[e +
f*x]]*((-2*(10*a^2 - 27*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)*A
ppellF1[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]*Sqr
t[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])) + ((30*a^2
+ 27*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)*S
qrt[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)*Arc
Tan[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[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))) + ((1/4 - 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])) + (28*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)*App
ellF1[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]*Sqrt[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]))))/(60*b^2*f*Sqrt[g*Cos
[e + f*x]])
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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/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(sin(f*x + e)^4/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)), x)
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maple [C] time = 6.87, size = 1455, normalized size = 2.86 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(f*x+e)^4/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x)
[Out]
8/5/f/b*cos(1/2*f*x+1/2*e)^4/g*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)-8/5/f/b*cos(1/2*f*x+1/2*e)^2/g*(2*cos(1/2*f*
x+1/2*e)^2*g-g)^(1/2)-8/5/f/b/g*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)-2/f/b^3*a^2/g*(g*(2*cos(1/2*f*x+1/2*e)^2-1)
)^(1/2)+2/f*a^4/b^3*g*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))-8/3/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/b^2/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)*sin(1/2*f*x+1/2*e)^3/(g*(2*cos(1/2*f*x+1/2*e
)^2-1))^(1/2)*cos(1/2*f*x+1/2*e)+4/3/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/b^2/(-g*(2*
sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)*sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*cos(
1/2*f*x+1/2*e)+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^3/b^4/(-g*(2*sin(1/2*f*x+1/2*e)
^4-sin(1/2*f*x+1/2*e)^2))^(1/2)/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*EllipticF(cos(1/2*f*x+
1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)+4/3/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/b^2/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)/sin(1/2*f*x+1/2
*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2
*sin(1/2*f*x+1/2*e)^2-1)^(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^3/b^6/(-g*(2*
sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*sum(
1/_alpha/(2*_alpha^2-1)*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*
x+1/2*e)^2-1)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^2+4*b^2)/a^2,2^(1/2))*_alpha^3*b^2-8*b^2*_alp
ha*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^
2+4*b^2)/a^2,2^(1/2))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+2^(1/2)*a^2*arctanh(1/2/(g*(2*_alpha^2*b^2+a^2-
2*b^2)/b^2)^(1/2)/(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(1/2)/(4*a^2-3*b^2)*g*2^(1/2)*(-16*sin(1/
2*f*x+1/2*e)^2*_alpha^2*a^2+12*sin(1/2*f*x+1/2*e)^2*_alpha^2*b^2+4*_alpha^4*b^2+12*sin(1/2*f*x+1/2*e)^2*a^2-9*
sin(1/2*f*x+1/2*e)^2*b^2+4*_alpha^2*a^2-7*b^2*_alpha^2-3*a^2+3*b^2))*(sin(1/2*f*x+1/2*e)^2*g*(-2*sin(1/2*f*x+1
/2*e)^2+1))^(1/2))/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(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),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))*(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(sin(f*x + e)^4/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (e+f\,x\right )}^4}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(e + f*x)^4/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))),x)
[Out]
int(sin(e + f*x)^4/((g*cos(e + f*x))^(1/2)*(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/(a+b*sin(f*x+e))/(g*cos(f*x+e))**(1/2),x)
[Out]
Timed out
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