3.1371 \(\int \frac{\sqrt{g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=448 \[ -\frac{a^3 \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt [4]{b^2-a^2}}+\frac{a^3 \sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt [4]{b^2-a^2}}+\frac{2 a^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{b^3 f \sqrt{\cos (e+f x)}}-\frac{a^4 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}-\frac{a^4 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{2 a (g \cos (e+f x))^{3/2}}{3 b^2 f g}-\frac{2 \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b f g}+\frac{4 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b f \sqrt{\cos (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.960273, antiderivative size = 448, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 13, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.394, Rules used = {2898, 2640, 2639, 2565, 30, 2568, 2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac{a^3 \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt [4]{b^2-a^2}}+\frac{a^3 \sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt [4]{b^2-a^2}}+\frac{2 a^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{b^3 f \sqrt{\cos (e+f x)}}-\frac{a^4 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}-\frac{a^4 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^4 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{2 a (g \cos (e+f x))^{3/2}}{3 b^2 f g}-\frac{2 \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b f g}+\frac{4 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b f \sqrt{\cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2898

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])

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 2565

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2568

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

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{\sqrt{g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx &=\int \left (\frac{a^2 \sqrt{g \cos (e+f x)}}{b^3}-\frac{a \sqrt{g \cos (e+f x)} \sin (e+f x)}{b^2}+\frac{\sqrt{g \cos (e+f x)} \sin ^2(e+f x)}{b}-\frac{a^3 \sqrt{g \cos (e+f x)}}{b^3 (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac{a^2 \int \sqrt{g \cos (e+f x)} \, dx}{b^3}-\frac{a^3 \int \frac{\sqrt{g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{b^3}-\frac{a \int \sqrt{g \cos (e+f x)} \sin (e+f x) \, dx}{b^2}+\frac{\int \sqrt{g \cos (e+f x)} \sin ^2(e+f x) \, dx}{b}\\ &=-\frac{2 (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b f g}+\frac{2 \int \sqrt{g \cos (e+f x)} \, dx}{5 b}+\frac{a \operatorname{Subst}\left (\int \sqrt{x} \, dx,x,g \cos (e+f x)\right )}{b^2 f g}+\frac{\left (a^4 g\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^4 g\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^3 g\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 (a^2 \sqrt{g \cos (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \, dx}{b^3 \sqrt{\cos (e+f x)}}\\ &=\frac{2 a (g \cos (e+f x))^{3/2}}{3 b^2 f g}+\frac{2 a^2 \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b^3 f \sqrt{\cos (e+f x)}}-\frac{2 (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b f g}-\frac{\left (2 a^3 g\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^4 g \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^4 g \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 (2 \sqrt{g \cos (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \, dx}{5 b \sqrt{\cos (e+f x)}}\\ &=\frac{2 a (g \cos (e+f x))^{3/2}}{3 b^2 f g}+\frac{2 a^2 \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b^3 f \sqrt{\cos (e+f x)}}+\frac{4 \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 b f \sqrt{\cos (e+f x)}}-\frac{a^4 g \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^4 g \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 \cos (e+f x))^{3/2} \sin (e+f x)}{5 b f g}+\frac{\left (a^3 g\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^3 g\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^3 \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{b^{7/2} \sqrt [4]{-a^2+b^2} f}+\frac{a^3 \sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{b^{7/2} \sqrt [4]{-a^2+b^2} f}+\frac{2 a (g \cos (e+f x))^{3/2}}{3 b^2 f g}+\frac{2 a^2 \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b^3 f \sqrt{\cos (e+f x)}}+\frac{4 \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 b f \sqrt{\cos (e+f x)}}-\frac{a^4 g \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^4 g \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 \cos (e+f x))^{3/2} \sin (e+f x)}{5 b f g}\\ \end{align*}

Mathematica [C]  time = 26.8593, size = 789, normalized size = 1.76 \[ \frac{\sqrt{g \cos (e+f x)} \left (\frac{2 a \cos (e+f x)}{3 b^2}-\frac{\sin (2 (e+f x))}{5 b}\right )}{f}+\frac{\sqrt{g \cos (e+f x)} \left (-\frac{\left (5 a^2+2 b^2\right ) \sin ^2(e+f x) \left (a+b \sqrt{1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \cos ^{\frac{3}{2}}(e+f x) F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (-\log \left (-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+\log \left (\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )\right )}{12 b^{3/2} \left (b^2-a^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}-\frac{4 a b \sin (e+f x) \left (a+b \sqrt{1-\cos ^2(e+f x)}\right ) \left (\frac{a \cos ^{\frac{3}{2}}(e+f x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (-\log \left (-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}+i b \cos (e+f x)\right )+\log \left ((1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}+i b \cos (e+f x)\right )+2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{1-\cos ^2(e+f x)} (a+b \sin (e+f x))}\right )}{5 b^2 f \sqrt{\cos (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 6.549, size = 2363, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x)

[Out]

4/3/f*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*a/b^2*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(
1/2)-2/f*a/b^2*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)-1/2/f*g*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)-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/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)^5-16/3/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/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)*si
n(1/2*f*x+1/2*e)^2)^(1/2)*g/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/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)*EllipticF(cos(1/2*f*x+1/2*e),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)*a^2+4/3/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/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)*EllipticF(cos(1/2*f*x+1/2*e),
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)-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/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)*EllipticE(cos(1/2*f*x+1/2*e),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)+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/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/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/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)*EllipticF(cos(1/2*f*x+1/2*e),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)*a^2-4/3/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/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)*EllipticF(cos(1/2*f*x+1/2*e),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)+
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/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)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(si
n(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(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/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/b^5/
sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*sum((2*sin(1/2*f*x+1/2*e)^2*_alpha^2*b^2-sin(1/2*f*x+1
/2*e)^2*a^2-2*b^2*_alpha^2+a^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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(sin(f*x+e)^3*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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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(sin(f*x+e)**3*(g*cos(f*x+e))**(1/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

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