3.1385 \(\int \frac{(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=413 \[ -\frac{a g^{5/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}+\frac{a g^{5/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}-\frac{2 g^2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b^3 f \sqrt{\cos (e+f x)}}+\frac{a^2 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}+\frac{a^2 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}-\frac{2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f} \]

[Out]

-((a*(-a^2 + b^2)^(3/4)*g^(5/2)*ArcTan[(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)

________________________________________________________________________________________

Rubi [A]  time = 0.972255, antiderivative size = 413, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.355, Rules used = {2865, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac{a g^{5/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}+\frac{a g^{5/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}-\frac{2 g^2 \left (5 a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b^3 f \sqrt{\cos (e+f x)}}+\frac{a^2 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}+\frac{a^2 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}-\frac{2 g (g \cos (e+f x))^{3/2} (5 a-3 b \sin (e+f x))}{15 b^2 f} \]

Antiderivative was successfully verified.

[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 2865

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 2867

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 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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(g \cos (e+f x))^{5/2} \sin (e+f x)}{a+b \sin (e+f x)} \, dx &=-\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 ) \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 )}{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 ) \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 )}{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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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 ) \operatorname{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 ) \operatorname{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} \tan ^{-1}\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} \tanh ^{-1}\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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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]  time = 24.8467, size = 737, normalized size = 1.78 \[ \frac{(g \cos (e+f x))^{5/2} \left (\frac{\left (5 a^2-3 b^2\right ) \left (a+b \sqrt{\sin ^2(e+f x)}\right ) \left (8 b^{5/2} \cos ^{\frac{3}{2}}(e+f x) F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (-\log \left (-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+\log \left (\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )\right )}{12 b^{7/2} \left (b^2-a^2\right ) (a+b \sin (e+f x))}+\frac{4 a \sin (e+f x) \left (a+b \sqrt{\sin ^2(e+f x)}\right ) \left (\frac{a \cos ^{\frac{3}{2}}(e+f x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (-\log \left (-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}+i b \cos (e+f x)\right )+\log \left ((1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}+i b \cos (e+f x)\right )+2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right )}{b \sqrt{\sin ^2(e+f x)} (a+b \sin (e+f x))}+\frac{2 \cos ^{\frac{3}{2}}(e+f x) (3 b \sin (e+f x)-5 a)}{3 b^2}\right )}{5 f \cos ^{\frac{5}{2}}(e+f x)} \]

Warning: Unable to verify antiderivative.

[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]  time = 6.952, size = 2612, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3/f*g^2*a/b^2*cos(1/2*f*x+1/2*e)^2*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)-4/3/f*g^2*a/b^2*(2*cos(1/2*f*x+1/2*e)
^2*g-g)^(1/2)+2/f*g^2*a/b^2*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)+1/2/f*g^3*a^3/b^2*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)-co
s(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^3*a*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))-16/3/f*(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/b/(-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*co
s(1/2*f*x+1/2*e)^2-1))^(1/2)*cos(1/2*f*x+1/2*e)^5+16/3/f*(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/b/(-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)*cos(1/2*f*x+1/2*e)^5+8/f*(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/b/(-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)*cos(
1/2*f*x+1/2*e)^3+4/f*(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/b^3/(-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)*(sin(1/2*f*x+1/2*e)^
2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*a^2-16/3/f*(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/b/(-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)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2
)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))+4/f*(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/b/(-
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)
*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))-8/f*(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/b/(-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)*cos(1/2*f*x+1/2*e)^3-8/3/f*(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/b/(-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)*cos(1/2*f*x+1/2*e)-4/f*(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/b^3/(-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*co
s(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)*EllipticF(cos(1/2*
f*x+1/2*e),2^(1/2))*a^2+16/3/f*(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/b/(-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)*(sin(1/2*f*x
+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))-4/f*(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/b/(-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)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/
2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))+8/3/f*(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/b
/(-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)*cos(1/2*f*x+1/2*e)+1/4/f*(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/b^5/a^2/sin(1/2*f*x
+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*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)*(-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))*_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))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)