\(\int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx\) [1385]

Optimal result

Integrand size = 31, antiderivative size = 413 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a \left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}+\frac {a \left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}-\frac {2 \left (5 a^2-3 b^2\right ) g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f} \]

[Out]

-a*(-a^2+b^2)^(3/4)*g^(5/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/f+a*(-a^2+b^
2)^(3/4)*g^(5/2)*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/f-2/15*g*(g*cos(f*x+e)
)^(3/2)*(5*a-3*b*sin(f*x+e))/b^2/f+a^2*(a^2-b^2)*g^3*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticP
i(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/(b-(-a^2+b^2)^(1/2))/(g*cos(f*x+
e))^(1/2)+a^2*(a^2-b^2)*g^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^4/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-2/5*(5*a^2-3*b^
2)*g^2*(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)/b^3/f/cos(f*x+e)^(1/2)

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {2944, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a g^{5/2} \left (b^2-a^2\right )^{3/4} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}+\frac {a g^{5/2} \left (b^2-a^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}+\frac {a^2 g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a^2 g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {2 g^2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{5 b^3 f \sqrt {\cos (e+f x)}}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f} \]

[In]

Int[((g*Cos[e + f*x])^(5/2)*Sin[e + f*x])/(a + b*Sin[e + f*x]),x]

[Out]

-((a*(-a^2 + b^2)^(3/4)*g^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(7/2)*
f)) + (a*(-a^2 + b^2)^(3/4)*g^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(
7/2)*f) - (2*(5*a^2 - 3*b^2)*g^2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(5*b^3*f*Sqrt[Cos[e + f*x]])
+ (a^2*(a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^4*(b -
Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + (a^2*(a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt
[-a^2 + b^2]), (e + f*x)/2, 2])/(b^4*(b + Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (2*g*(g*Cos[e + f*x])^(3
/2)*(5*a - 3*b*Sin[e + f*x]))/(15*b^2*f)

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 304

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 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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 2780

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 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 2944

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}+\frac {\left (2 g^2\right ) \int \frac {\sqrt {g \cos (e+f x)} \left (-a b-\frac {1}{2} \left (5 a^2-3 b^2\right ) \sin (e+f x)\right )}{a+b \sin (e+f x)} \, dx}{5 b^2} \\ & = -\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}-\frac {\left (\left (5 a^2-3 b^2\right ) g^2\right ) \int \sqrt {g \cos (e+f x)} \, dx}{5 b^3}+\frac {\left (a \left (a^2-b^2\right ) g^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{b^3} \\ & = -\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}-\frac {\left (a^2 \left (a^2-b^2\right ) g^3\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^4}+\frac {\left (a^2 \left (a^2-b^2\right ) g^3\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^4}+\frac {\left (a \left (a^2-b^2\right ) g^3\right ) \text {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 )}{b^2 f}-\frac {\left (\left (5 a^2-3 b^2\right ) g^2 \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 b^3 \sqrt {\cos (e+f x)}} \\ & = -\frac {2 \left (5 a^2-3 b^2\right ) g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^3 f \sqrt {\cos (e+f x)}}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}+\frac {\left (2 a \left (a^2-b^2\right ) g^3\right ) \text {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 )}{b^2 f}-\frac {\left (a^2 \left (a^2-b^2\right ) g^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^4 \sqrt {g \cos (e+f x)}}+\frac {\left (a^2 \left (a^2-b^2\right ) g^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^4 \sqrt {g \cos (e+f x)}} \\ & = -\frac {2 \left (5 a^2-3 b^2\right ) g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f}-\frac {\left (a \left (a^2-b^2\right ) g^3\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 )}{b^3 f}+\frac {\left (a \left (a^2-b^2\right ) g^3\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 )}{b^3 f} \\ & = -\frac {a \left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}+\frac {a \left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}-\frac {2 \left (5 a^2-3 b^2\right ) g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^2 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 17.14 (sec) , antiderivative size = 737, normalized size of antiderivative = 1.78 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\frac {(g \cos (e+f x))^{5/2} \left (\frac {2 \cos ^{\frac {3}{2}}(e+f x) (-5 a+3 b \sin (e+f x))}{3 b^2}+\frac {\left (5 a^2-3 b^2\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{12 b^{7/2} \left (-a^2+b^2\right ) (a+b \sin (e+f x))}+\frac {4 a \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{b \sqrt {\sin ^2(e+f x)} (a+b \sin (e+f x))}\right )}{5 f \cos ^{\frac {5}{2}}(e+f x)} \]

[In]

Integrate[((g*Cos[e + f*x])^(5/2)*Sin[e + f*x])/(a + b*Sin[e + f*x]),x]

[Out]

((g*Cos[e + f*x])^(5/2)*((2*Cos[e + f*x]^(3/2)*(-5*a + 3*b*Sin[e + f*x]))/(3*b^2) + ((5*a^2 - 3*b^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 + (Sq
rt[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]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(12*b^(7/2)*(-a^2 + b^2)*(a + b*Sin[e + f*x])) + (4*a*(
(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[C
os[e + f*x]] + I*b*Cos[e + f*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[e + f*x]*(a + b*Sqrt[Sin[e + f*x]^2]))/(b
*Sqrt[Sin[e + f*x]^2]*(a + b*Sin[e + f*x]))))/(5*f*Cos[e + f*x]^(5/2))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 7.29 (sec) , antiderivative size = 1533, normalized size of antiderivative = 3.71

method	result	size
default	\(\text {Expression too large to display}\)	\(1533\)

[In]

int((g*cos(f*x+e))^(5/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

(-4*g^3*a*(1/6/b^2*(-2*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*sin(1/2*f*x+1/2*e)^2+(-2*g*sin(1/2*f*x+1/2*e)^2+g)^
(1/2))/g-1/16/b^4*(a^2-b^2)/(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b
^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2))/(2*g*cos(1/2*f*x+1/2*e)^2-g+(g
^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/
2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*
(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4))))-1/12*(g*(2*cos(1/2*
f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*(64*cos(1/2*f*x+1/2*e)^5*sin(1/2*f*x+1/2*e)^2*a^2*b^4-64*cos(1
/2*f*x+1/2*e)^5*a^2*b^4-96*cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x+1/2*e)^2*a^2*b^4-48*EllipticF(cos(1/2*f*x+1/2*e),2
^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+1/2*e)^2*a^4*b^2+64*Elliptic
F(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+1/2*e)
^2*a^2*b^4-48*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1
/2)*sin(1/2*f*x+1/2*e)^2*a^2*b^4+96*cos(1/2*f*x+1/2*e)^3*a^2*b^4+32*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2*a^
2*b^4+48*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*a
^4*b^2-64*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*
a^2*b^4+48*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)
*a^2*b^4-32*a^2*b^4*cos(1/2*f*x+1/2*e)-3*sum((sin(1/2*f*x+1/2*e)^2*(2*_alpha^2*a^2*b^2-2*_alpha^2*b^4-a^4+a^2*
b^2)-2*_alpha^2*a^2*b^2+2*_alpha^2*b^4+a^4-a^2*b^2)/_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)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2
*(_alpha^2-1),2^(1/2))*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(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)+a^2*2
^(1/2)*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(b^2*_alpha^2+4*a^2*cos(1/2*f*x+1/2*e)^2-3*b^2*cos(1/2*f*x+1
/2*e)^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))*(-g*sin(1/2*f*x+1/2*e)^2*(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)/(-g*sin(1/2*f*x+1/2*e)^2*(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*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2))/b^5/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2
))^(1/2)/a^2/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2))/f

Fricas [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

integrate((g*cos(f*x+e))^(5/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

integrate((g*cos(f*x+e))**(5/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(5/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^(5/2)*sin(f*x + e)/(b*sin(f*x + e) + a), x)

Giac [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(5/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^(5/2)*sin(f*x + e)/(b*sin(f*x + e) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {\sin \left (e+f\,x\right )\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]

[In]

int((sin(e + f*x)*(g*cos(e + f*x))^(5/2))/(a + b*sin(e + f*x)),x)

[Out]

int((sin(e + f*x)*(g*cos(e + f*x))^(5/2))/(a + b*sin(e + f*x)), x)