3.15.7 \(\int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\) [1407]
Optimal. Leaf size=651 \[ \frac {b \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a \left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}} \]
[Out]
b*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(5/2)-b^(9/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1
/4)/g^(1/2))/a^2/(-a^2+b^2)^(7/4)/f/g^(5/2)+b*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(5/2)-b^(9/2)*arct
anh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^2/(-a^2+b^2)^(7/4)/f/g^(5/2)-2/3*b/a^2/f/g/(g*cos
(f*x+e))^(3/2)-csc(f*x+e)/a/f/g/(g*cos(f*x+e))^(3/2)+5/3*sin(f*x+e)/a/f/g/(g*cos(f*x+e))^(3/2)-2/3*b^2*(b-a*si
n(f*x+e))/a^2/(a^2-b^2)/f/g/(g*cos(f*x+e))^(3/2)+5/3*(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)/a/f/g^2/(g*cos(f*x+e))^(1/2)+2/3*b^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)/a/(a^2-b^2)/f/g^2/(g*cos(f*x+e))^(
1/2)-b^4*(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)/a/(a^2-b^2)/f/g^2/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-b^4*(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)/a/(a^2-b^2)/f/g^2/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)
________________________________________________________________________________________
Rubi [A]
time = 1.01, antiderivative size = 651, normalized size of antiderivative = 1.00, number of steps
used = 25, number of rules used = 18, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {2977, 2645,
331, 335, 218, 212, 209, 2650, 2716, 2721, 2720, 2775, 2946, 2781, 2886, 2884, 214, 211}
\begin {gather*} -\frac {b^{9/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b \text {ArcTan}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}+\frac {2 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac {b^{9/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f g^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac {b^4 \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 )}{a 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 {b^4 \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 )}{a 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 {b \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^2/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]
[Out]
(b*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f*g^(5/2)) - (b^(9/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a
^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(7/4)*f*g^(5/2)) + (b*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2
*f*g^(5/2)) - (b^(9/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)
^(7/4)*f*g^(5/2)) - (2*b)/(3*a^2*f*g*(g*Cos[e + f*x])^(3/2)) - Csc[e + f*x]/(a*f*g*(g*Cos[e + f*x])^(3/2)) + (
5*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*a*f*g^2*Sqrt[g*Cos[e + f*x]]) + (2*b^2*Sqrt[Cos[e + f*x]]*E
llipticF[(e + f*x)/2, 2])/(3*a*(a^2 - b^2)*f*g^2*Sqrt[g*Cos[e + f*x]]) - (b^4*Sqrt[Cos[e + f*x]]*EllipticPi[(2
*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*(a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[
e + f*x]]) - (b^4*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*(a^2 - b^2)*
(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) + (5*Sin[e + f*x])/(3*a*f*g*(g*Cos[e + f*x])^(3/2
)) - (2*b^2*(b - a*Sin[e + f*x]))/(3*a^2*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2))
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 331
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 2650
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*Cos[e + f*
x])^(n + 1)*((a*Sin[e + f*x])^(m + 1)/(a*b*f*(m + 1))), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
Rule 2716
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 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 2775
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
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 {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx &=\int \left (-\frac {b \csc (e+f x)}{a^2 (g \cos (e+f x))^{5/2}}+\frac {\csc ^2(e+f x)}{a (g \cos (e+f x))^{5/2}}+\frac {b^2}{a^2 (g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac {\int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2}} \, dx}{a}-\frac {b \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{5/2}} \, dx}{a^2}+\frac {b^2 \int \frac {1}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx}{a^2}\\ &=-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {5 \int \frac {1}{(g \cos (e+f x))^{5/2}} \, dx}{2 a}-\frac {\left (2 b^2\right ) \int \frac {-\frac {a^2}{2}+\frac {3 b^2}{2}-\frac {1}{2} a b \sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{3 a^2 \left (a^2-b^2\right ) g^2}+\frac {b \text {Subst}\left (\int \frac {1}{x^{5/2} \left (1-\frac {x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{a^2 f g}\\ &=-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {b \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^2 f g^3}+\frac {5 \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{6 a g^2}+\frac {b^2 \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{3 a \left (a^2-b^2\right ) g^2}-\frac {b^4 \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^2 \left (a^2-b^2\right ) g^2}\\ &=-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^2 f g^3}-\frac {b^4 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 a \left (-a^2+b^2\right )^{3/2} g^2}-\frac {b^4 \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 a \left (-a^2+b^2\right )^{3/2} g^2}-\frac {b^5 \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^2 \left (a^2-b^2\right ) f g}+\frac {\left (5 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{6 a g^2 \sqrt {g \cos (e+f x)}}+\frac {\left (b^2 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{3 a \left (a^2-b^2\right ) g^2 \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac {b \text {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^2 f g^2}+\frac {b \text {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^2 f g^2}-\frac {\left (2 b^5\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^2 \left (a^2-b^2\right ) f g}-\frac {\left (b^4 \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 a \left (-a^2+b^2\right )^{3/2} g^2 \sqrt {g \cos (e+f x)}}-\frac {\left (b^4 \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 a \left (-a^2+b^2\right )^{3/2} g^2 \sqrt {g \cos (e+f x)}}\\ &=\frac {b \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {b^4 \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 )}{a \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 {b^4 \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 )}{a \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 {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\frac {b^5 \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^2 \left (-a^2+b^2\right )^{3/2} f g^2}-\frac {b^5 \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^2 \left (-a^2+b^2\right )^{3/2} f g^2}\\ &=\frac {b \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{3 a \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {b^4 \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 )}{a \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 {b^4 \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 )}{a \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 {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in
optimal.
time = 61.25, size = 2183, normalized size = 3.35 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Csc[e + f*x]^2/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]
[Out]
(Cos[e + f*x]^(5/2)*((-2*(10*a^3 - 18*a*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*Appell
F1[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)))/(Sqrt[1 - Cos[e + f*x]^2]*(b + a*Csc[e + f*x])) - ((-5*a
^2*b + 3*b^3)*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*Csc[e + f*x]*((-10*Sqrt[
2]*(2*a^2 - b^2)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3
/4)) + (10*Sqrt[2]*(2*a^2 - b^2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b
]*(a^2 - b^2)^(3/4)) - (20*ArcTan[Sqrt[Cos[e + f*x]]])/a - (16*b*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))/(-a^2 + b^2) - (200*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, 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))) + (10*Log[1
- Sqrt[Cos[e + f*x]]])/a - (10*Log[1 + Sqrt[Cos[e + f*x]]])/a - (5*Sqrt[2]*(2*a^2 - b^2)*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*Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*Sq
rt[2]*(2*a^2 - b^2)*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*Sqrt[b]*(a^2 - b^2)^(3/4))))/(20*(1 - Cos[e + f*x]^2)*(-1 + 2*Cos[e + f*x]^2)*(b + a*Csc[e + f*x])) - (
2*(-7*a^2*b + 9*b^3)*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Csc[e + f*x]*((5*b*(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)]*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))) - (-2*Sqrt[2]*b^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*Sqrt[2]*b
^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 4*(a^2 - b^2)^(3/4)*ArcTan[Sqrt[Co
s[e + f*x]]] - 2*(a^2 - b^2)^(3/4)*Log[1 - Sqrt[Cos[e + f*x]]] + 2*(a^2 - b^2)^(3/4)*Log[1 + Sqrt[Cos[e + f*x]
]] - Sqrt[2]*b^(3/2)*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]*b^(3/2)*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]])/(8*a*(a^2 - b^2)^(3/4))))/((1 - Cos[e + f*x]^2)*(b + a*Csc[e + f*x]))))/(12*a*(a - b)*(a + b)*f*(g*Cos[e
+ f*x])^(5/2)) + (Cos[e + f*x]^3*(-(Csc[e + f*x]/a) + (2*Sec[e + f*x]^2*(-b + a*Sin[e + f*x]))/(3*(a^2 - b^2)
)))/(f*(g*Cos[e + f*x])^(5/2))
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 127.35, size = 1571, normalized size = 2.41
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method |
result |
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\(\text {Expression too large to display}\) |
\(1571\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)^2/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)
[Out]
(1/6/g^3*b/(2+2^(1/2))/(2^(1/2)-2)/(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)-1/6/g^3*b*2^(1/2)/(2+2^(1/2))/(2^(1/2)-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/g*b^5/(a-b)/(a+b)/a^2*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=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))+1/6/g^3*b/(2+2^(1/2))/(2^(1/2)-2)/(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)+1/6/g^3*b*2^(1/2)/(2+2^(1/2))/(2^(1/2
)-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/g^2*b/a^2/(-g)^(1/2)*ln(
(-2*g+2*(-g)^(1/2)*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2))/cos(1/2*f*x+1/2*e))+2/g^(5/2)*b/(2+2^(1/2))^2/(2^(1/2)-
2)^2/a^2*ln((4*g*cos(1/2*f*x+1/2*e)+2*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)-1))
+2/g^(5/2)*b/(2+2^(1/2))^2/(2^(1/2)-2)^2/a^2*ln((-4*g*cos(1/2*f*x+1/2*e)+2*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)+1))+1/24*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)/g^2/a/cos
(1/2*f*x+1/2*e)/(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(3/2)/(a^2-b^2)*(40*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)^(3/2)*a^2*g*cos(1/2*f*x+1/2*e)*sin(1/2*
f*x+1/2*e)^2-24*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)^
(3/2)*b^2*g*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2-80*a^2*g*sin(1/2*f*x+1/2*e)^6+48*b^2*g*sin(1/2*f*x+1/2*e)^
6+3/a^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*_alpha*(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/(-2*sin(1/2
*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(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*s
in(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)^(3/2)*b^2*cos(1/2*f*x+1/2*e)+80*a^2*g*sin(1/2*f*x+1/2*e)^4-48*b^2*g*sin(1/2*f*x+1/2*e)^4-12*a^2*g*sin(1/2*
f*x+1/2*e)^2+12*b^2*g*sin(1/2*f*x+1/2*e)^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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^2/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.
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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^2/(g*cos(f*x+e))^(5/2)/(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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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(csc(f*x+e)**2/(g*cos(f*x+e))**(5/2)/(a+b*sin(f*x+e)),x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 5008 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(csc(f*x+e)^2/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="giac")
[Out]
integrate(csc(f*x + e)^2/((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 \begin {gather*} \int \frac {1}{{\sin \left (e+f\,x\right )}^2\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\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(1/(sin(e + f*x)^2*(g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))),x)
[Out]
int(1/(sin(e + f*x)^2*(g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))), x)
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