3.14.80 \(\int \frac {(g \cos (e+f x))^{3/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx\) [1380]

Optimal. Leaf size=439 \[ -\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt [4]{-a^2+b^2} g^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt {b} f}-\frac {g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt [4]{-a^2+b^2} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt {b} f}-\frac {2 g^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.77, antiderivative size = 439, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {2977, 2645, 327, 335, 218, 212, 209, 2774, 2946, 2721, 2720, 2781, 2886, 2884, 214, 211} \begin {gather*} \frac {g^{3/2} \sqrt [4]{b^2-a^2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a \sqrt {b} f}+\frac {g^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a \sqrt {b} f}+\frac {g^2 \left (a^2-b^2\right ) \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 \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {g^2 \left (a^2-b^2\right ) \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 \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {g^{3/2} \text {ArcTan}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {2 g^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((g^(3/2)*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a*f)) + ((-a^2 + b^2)^(1/4)*g^(3/2)*ArcTan[(Sqrt[b]*Sqrt[g*C
os[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*Sqrt[b]*f) - (g^(3/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(
a*f) + ((-a^2 + b^2)^(1/4)*g^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*Sq
rt[b]*f) - (2*g^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(b*f*Sqrt[g*Cos[e + f*x]]) + ((a^2 - b^2)*g^2*
Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b*(a^2 - b*(b - Sqrt[-a^2 + b^2]
))*f*Sqrt[g*Cos[e + f*x]]) + ((a^2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e +
f*x)/2, 2])/(b*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 2774

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

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 2977

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rubi steps

\begin {align*} \int \frac {(g \cos (e+f x))^{3/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx &=\int \left (\frac {(g \cos (e+f x))^{3/2} \csc (e+f x)}{a}-\frac {b (g \cos (e+f x))^{3/2}}{a (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac {\int (g \cos (e+f x))^{3/2} \csc (e+f x) \, dx}{a}-\frac {b \int \frac {(g \cos (e+f x))^{3/2}}{a+b \sin (e+f x)} \, dx}{a}\\ &=-\frac {2 g \sqrt {g \cos (e+f x)}}{a f}-\frac {\text {Subst}\left (\int \frac {x^{3/2}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{a f g}-\frac {g^2 \int \frac {b+a \sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a}\\ &=-\frac {g \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{a f}-\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b}-\frac {\left (\left (-a^2+b^2\right ) g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a b}\\ &=-\frac {(2 g) \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {\left (\sqrt {-a^2+b^2} g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b}+\frac {\left (\sqrt {-a^2+b^2} g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b}+\frac {\left (\left (a^2-b^2\right ) g^3\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 )}{a f}-\frac {\left (g^2 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 g^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {g^2 \text {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}-\frac {g^2 \text {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {\left (2 \left (a^2-b^2\right ) g^3\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 )}{a f}+\frac {\left (\sqrt {-a^2+b^2} g^2 \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 \sqrt {g \cos (e+f x)}}+\frac {\left (\sqrt {-a^2+b^2} g^2 \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 \sqrt {g \cos (e+f x)}}\\ &=-\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {2 g^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {\sqrt {-a^2+b^2} g^2 \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 (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\sqrt {-a^2+b^2} g^2 \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 (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (\sqrt {-a^2+b^2} g^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {\left (\sqrt {-a^2+b^2} g^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}\\ &=-\frac {g^{3/2} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt [4]{-a^2+b^2} g^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt {b} f}-\frac {g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt [4]{-a^2+b^2} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt {b} f}-\frac {2 g^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {\sqrt {-a^2+b^2} g^2 \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 (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\sqrt {-a^2+b^2} g^2 \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 (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 6 vs. order 4 in optimal.
time = 16.45, size = 484, normalized size = 1.10 \begin {gather*} \frac {(g \cos (e+f x))^{3/2} \csc (e+f x) \left (8 a b^{3/2} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {5}{2}}(e+f x)-5 \left (a^2-b^2\right ) \left (2 \sqrt {2} \sqrt [4]{a^2-b^2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \sqrt {2} \sqrt [4]{a^2-b^2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 \sqrt {b} \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right )-2 \sqrt {b} \log \left (1-\sqrt {\cos (e+f x)}\right )+2 \sqrt {b} \log \left (1+\sqrt {\cos (e+f x)}\right )+\sqrt {2} \sqrt [4]{a^2-b^2} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )-\sqrt {2} \sqrt [4]{a^2-b^2} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{20 a \sqrt {b} \left (a^2-b^2\right ) f \cos ^{\frac {3}{2}}(e+f x) (b+a \csc (e+f x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((g*Cos[e + f*x])^(3/2)*Csc[e + f*x]*(8*a*b^(3/2)*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]^(5/2) - 5*(a^2 - b^2)*(2*Sqrt[2]*(a^2 - b^2)^(1/4)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*
Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - 2*Sqrt[2]*(a^2 - b^2)^(1/4)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e +
f*x]])/(a^2 - b^2)^(1/4)] + 4*Sqrt[b]*ArcTan[Sqrt[Cos[e + f*x]]] - 2*Sqrt[b]*Log[1 - Sqrt[Cos[e + f*x]]] + 2*S
qrt[b]*Log[1 + Sqrt[Cos[e + f*x]]] + Sqrt[2]*(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]] - Sqrt[2]*(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]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(20*a*Sqrt[b]*(a^2 -
b^2)*f*Cos[e + f*x]^(3/2)*(b + a*Csc[e + f*x]))

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Maple [A]
time = 18.80, size = 216, normalized size = 0.49

method	result	size
default	\(\frac {-g^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}-g^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}+4 g \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}\, \sqrt {-g}+2 g^{2} \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{2 a \sqrt {-g}\, f}\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-g^(3/2)*ln(2/(cos(1/2*f*x+1/2*e)-1)*(g^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)+2*g*cos(1/2*f*x+1/2*e)-
g))*(-g)^(1/2)-g^(3/2)*ln(2/(cos(1/2*f*x+1/2*e)+1)*(g^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-2*g*cos(1/2*f*
x+1/2*e)-g))*(-g)^(1/2)+4*g*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)*(-g)^(1/2)+2*g^2*ln(2/cos(1/2*f*x+1/2*e)*((-g)
^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-g)))/a/(-g)^(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((g*cos(f*x+e))^(3/2)*csc(f*x+e)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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

[Out]

integrate((g*cos(f*x + e))^(3/2)*csc(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 {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{\sin \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((g*cos(e + f*x))^(3/2)/(sin(e + f*x)*(a + b*sin(e + f*x))),x)

[Out]

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

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