3.14.91 \(\int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) [1391]

Optimal. Leaf size=380 \[ -\frac {a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \]

[Out]

-a^2*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/f/g^(1/2)-a^2*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)^(3/4)/f/g^(1/2)-2*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^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^2/f/(a^2-b*(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^2/f/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-2*(g*cos(f*x+e))^(1/2)/b/f/g

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Rubi [A]
time = 0.59, antiderivative size = 380, 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 = {2988, 2645, 30, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} -\frac {a^2 \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\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^2 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\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^2 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2/(Sqrt[g*Cos[e + f*x]]*(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])])/(b^(3/2)*(-a^2 + b^2)^(3/4)*f*Sqrt
[g])) - (a^2*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(3/2)*(-a^2 + b^2)^(3/4)
*f*Sqrt[g]) - (2*Sqrt[g*Cos[e + f*x]])/(b*f*g) - (2*a*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(b^2*f*Sqr
t[g*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^2*(a
^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*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^2*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*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 211

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

Rule 214

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 335

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 2645

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 2720

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

Rule 2721

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

Rule 2781

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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 2946

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 2988

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 ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac {\int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{b}-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b}\\ &=-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b^2}+\frac {a^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^2}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,g \cos (e+f x)\right )}{b f g}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{b f 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^2 \sqrt {-a^2+b^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^2 \sqrt {-a^2+b^2}}+\frac {\left (a^2 g\right ) \text {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 f}-\frac {\left (a \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b^2 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {\left (2 a^2 g\right ) \text {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 f}-\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^2 \sqrt {-a^2+b^2} \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^2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \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^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \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^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \sqrt {-a^2+b^2} f}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b \sqrt {-a^2+b^2} f}\\ &=-\frac {a^2 \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 )^{3/4} f \sqrt {g}}-\frac {a^2 \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 )^{3/4} f \sqrt {g}}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}-\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \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^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \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^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 38.27, size = 572, normalized size = 1.51 \begin {gather*} -\frac {\left (a^2-b^2\right ) \left (a^2-b^2+b^2 \cos ^2(e+f x)\right ) \sec (e+f x) \sqrt [4]{\sec ^2(e+f x)} \tan ^3(e+f x) \left (\frac {a^2 \tan ^{-1}\left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{1+\tan ^2(e+f x)}}{\sqrt {b}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{-a^2+b^2} \sqrt [4]{1+\tan ^2(e+f x)}}{\sqrt {b}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}+\frac {a \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)}{a^2-b^2}-\frac {a^3 \cot (e+f x) \Pi \left (-\frac {\sqrt {-a^2+b^2}}{b};\left .\sin ^{-1}\left (\sqrt [4]{1+\tan ^2(e+f x)}\right )\right |-1\right ) \sqrt {-\tan ^2(e+f x)}}{b^2 \left (a^2-b^2\right )}-\frac {a^3 \cot (e+f x) \Pi \left (\frac {\sqrt {-a^2+b^2}}{b};\left .\sin ^{-1}\left (\sqrt [4]{1+\tan ^2(e+f x)}\right )\right |-1\right ) \sqrt {-\tan ^2(e+f x)}}{b^2 \left (a^2-b^2\right )}-\frac {2}{b \sqrt [4]{1+\tan ^2(e+f x)}}\right )}{f \sqrt {g \cos (e+f x)} (b+a \csc (e+f x)) \left (b \left (a^2-b^2\right ) \tan ^5(e+f x)+a^3 \sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}+\tan ^2(e+f x) \left (-a b^2 \sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}+a^3 \left (\sqrt {\sec ^2(e+f x)}+\sqrt {-\tan ^2(e+f x)} \sqrt {-\sec ^2(e+f x) \tan ^2(e+f x)}\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((a^2 - b^2)*(a^2 - b^2 + b^2*Cos[e + f*x]^2)*Sec[e + f*x]*(Sec[e + f*x]^2)^(1/4)*Tan[e + f*x]^3*((a^2*ArcTa
n[((-a^2 + b^2)^(1/4)*(1 + Tan[e + f*x]^2)^(1/4))/Sqrt[b]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - (a^2*ArcTanh[((-a^2
+ b^2)^(1/4)*(1 + Tan[e + f*x]^2)^(1/4))/Sqrt[b]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + (a*Hypergeometric2F1[1/2, 3
/4, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(a^2 - b^2) - (a^3*Cot[e + f*x]*EllipticPi[-(Sqrt[-a^2 + b^2]/b), ArcS
in[(1 + Tan[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2])/(b^2*(a^2 - b^2)) - (a^3*Cot[e + f*x]*EllipticPi[Sq
rt[-a^2 + b^2]/b, ArcSin[(1 + Tan[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2])/(b^2*(a^2 - b^2)) - 2/(b*(1 +
Tan[e + f*x]^2)^(1/4))))/(f*Sqrt[g*Cos[e + f*x]]*(b + a*Csc[e + f*x])*(b*(a^2 - b^2)*Tan[e + f*x]^5 + a^3*Sqr
t[-Tan[e + f*x]^2]*Sqrt[-(Sec[e + f*x]^2*Tan[e + f*x]^2)] + Tan[e + f*x]^2*(-(a*b^2*Sqrt[-Tan[e + f*x]^2]*Sqrt
[-(Sec[e + f*x]^2*Tan[e + f*x]^2)]) + a^3*(Sqrt[Sec[e + f*x]^2] + Sqrt[-Tan[e + f*x]^2]*Sqrt[-(Sec[e + f*x]^2*
Tan[e + f*x]^2)])))))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 39.48, size = 821, normalized size = 2.16

method	result	size
default	\(\text {Expression too large to display}\)	\(821\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-2/b/g*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)+2*a^2/b*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)-g^(1/2)*cos(1/2*f*x+1/2*e)*2^(1/2)-_R),_R=R
ootOf(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))+1/8*(g*(2*cos(1/2*f*x+1/2*e)
^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a*(16*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)*Ellipti
cF(cos(1/2*f*x+1/2*e),2^(1/2))*b^2-sum(1/_alpha/(2*_alpha^2-1)*(8*(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)*_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)*Elliptic
Pi(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*ar
ctanh(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*si
n(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2))/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))/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(f*x + e)^2/(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 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^2}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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