3.1398 \(\int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=453 \[ \frac {a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f g^{3/2} \left (b^2-a^2\right )^{5/4}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f g^{3/2} \left (b^2-a^2\right )^{5/4}}-\frac {2 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)}}-\frac {2 b}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f g \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f g \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}} \]

[Out]

a^2*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/(-a^2+b^2)^(5/4)/f/g^(3/2)/b^(1/2)-a^2*arcta
nh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/(-a^2+b^2)^(5/4)/f/g^(3/2)/b^(1/2)-2*b/(a^2-b^2)/f/g
/(g*cos(f*x+e))^(1/2)+2*a*sin(f*x+e)/(a^2-b^2)/f/g/(g*cos(f*x+e))^(1/2)-a^3*(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/(a^2-b^2)/f/g
/(b-(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-a^3*(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/(a^2-b^2)/f/g/(b+(-a^2+b^2)^(1/2))/(g*cos(
f*x+e))^(1/2)-2*a*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(g*cos
(f*x+e))^(1/2)/(a^2-b^2)/f/g^2/cos(f*x+e)^(1/2)

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Rubi [A]  time = 0.88, antiderivative size = 453, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {2902, 2636, 2640, 2639, 2565, 30, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac {a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f g^{3/2} \left (b^2-a^2\right )^{5/4}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {b} f g^{3/2} \left (b^2-a^2\right )^{5/4}}-\frac {2 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)}}-\frac {2 b}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f g \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b f g \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]

[Out]

(a^2*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(Sqrt[b]*(-a^2 + b^2)^(5/4)*f*g^(3/2
)) - (a^2*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(Sqrt[b]*(-a^2 + b^2)^(5/4)*f*
g^(3/2)) - (2*b)/((a^2 - b^2)*f*g*Sqrt[g*Cos[e + f*x]]) - (2*a*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])
/((a^2 - b^2)*f*g^2*Sqrt[Cos[e + f*x]]) - (a^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e
+ f*x)/2, 2])/(b*(a^2 - b^2)*(b - Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[e + f*x]]) - (a^3*Sqrt[Cos[e + f*x]]*Ellipt
icPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b*(a^2 - b^2)*(b + Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[e + f
*x]]) + (2*a*Sin[e + f*x])/((a^2 - b^2)*f*g*Sqrt[g*Cos[e + f*x]])

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), 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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

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

Rule 2640

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

Rule 2701

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

Rubi steps

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

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Mathematica [C]  time = 17.01, size = 785, normalized size = 1.73 \[ \frac {2 \cos (e+f x) (a \sin (e+f x)-b)}{f \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac {a \cos ^{\frac {3}{2}}(e+f x) \left (-\frac {\sin ^2(e+f x) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \cos ^{\frac {3}{2}}(e+f x) F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (-\log \left (-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )+\log \left (\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )+2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )\right )}{12 \sqrt {b} \left (b^2-a^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}-\frac {4 a \sin (e+f x) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \cos ^{\frac {3}{2}}(e+f x) F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (-\log \left (-(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\cos (e+f x)}+\sqrt {b^2-a^2}+i b \cos (e+f x)\right )+\log \left ((1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\cos (e+f x)}+\sqrt {b^2-a^2}+i b \cos (e+f x)\right )+2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt {b} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}\right )}{f (a-b) (a+b) (g \cos (e+f x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^2/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]

[Out]

(2*Cos[e + f*x]*(-b + a*Sin[e + f*x]))/((a^2 - b^2)*f*(g*Cos[e + f*x])^(3/2)) - (a*Cos[e + f*x]^(3/2)*((-4*a*(
a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*AppellF1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^
2)]*Cos[e + f*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(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]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[e + f*x]
)/(Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) - ((a + b*Sqrt[1 - Cos[e + f*x]^2])*(8*b^(5/2)*AppellF1[3/4,
-1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)
^(3/4)*(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]*S
qrt[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]]))*Sin[e + f*x]^2)/(12*Sqrt[b]*(-a^2 + b^2)*(1 - Cos[e + f*x]^2)*(a + b*Sin[e + f*x]))))/((a - b)*(a +
b)*f*(g*Cos[e + f*x])^(3/2))

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fricas [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(sin(f*x+e)^2/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^2/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)

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maple [C]  time = 8.01, size = 1105, normalized size = 2.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x)

[Out]

1/2/f/g^2*b/(a^2-b^2)*2^(1/2)/(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)-1/2/f/g*b*a
^2/(a-b)/(a+b)*sum((_R^6-_R^4*g-_R^2*g^2+g^3)/(_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))-1/2/f/g^2*b/(a^2-b^2)*2^(1/2)/(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)-4/f/g*a/(a+b)/(a-b)/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)^3-2/f/g*a/(a+b)/(a-b)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)/si
n(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*(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*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))+4/f/g*a
/(a+b)/(a-b)/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)+1/8/f/g*a/b^2/(a+b)/(a
-b)/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*(8*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*c
os(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))*(g*(2*_alpha^2*b^2
+a^2-2*b^2)/b^2)^(1/2)*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(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))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+2^(1
/2)*a^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*
_alpha^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))*(-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))
*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(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 \[ \int \frac {\sin \left (f x + e\right )^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^2/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (e+f\,x\right )}^2}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^2/((g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))),x)

[Out]

int(sin(e + f*x)^2/((g*cos(e + f*x))^(3/2)*(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)**2/(g*cos(f*x+e))**(3/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

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