3.599 \(\int \frac{(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=519 \[ \frac{e^{3/2} \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{3/2} d \left (b^2-a^2\right )^{7/4}}+\frac{e^{3/2} \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{3/2} d \left (b^2-a^2\right )^{7/4}}-\frac{a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right ) \sqrt{e \cos (c+d x)}}+\frac{a e^2 \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 d \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{a e^2 \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 d \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2} \]

[Out]

((a^2 + 2*b^2)*e^(3/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*b^(3/2)*(-a^2 +
b^2)^(7/4)*d) + ((a^2 + 2*b^2)*e^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/
(8*b^(3/2)*(-a^2 + b^2)^(7/4)*d) - (a*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)*d*S
qrt[e*Cos[c + d*x]]) + (a*(a^2 + 2*b^2)*e^2*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d
*x)/2, 2])/(8*b^2*(a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) + (a*(a^2 + 2*b^2)*e^2*
Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(8*b^2*(a^2 - b^2)*(a^2 - b*(b +
Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) - (e*Sqrt[e*Cos[c + d*x]])/(2*b*d*(a + b*Sin[c + d*x])^2) + (a*e*Sq
rt[e*Cos[c + d*x]])/(4*b*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.1384, antiderivative size = 519, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {2693, 2864, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac{e^{3/2} \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{3/2} d \left (b^2-a^2\right )^{7/4}}+\frac{e^{3/2} \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{3/2} d \left (b^2-a^2\right )^{7/4}}-\frac{a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right ) \sqrt{e \cos (c+d x)}}+\frac{a e^2 \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 d \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{a e^2 \left (a^2+2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 d \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(3/2)/(a + b*Sin[c + d*x])^3,x]

[Out]

((a^2 + 2*b^2)*e^(3/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*b^(3/2)*(-a^2 +
b^2)^(7/4)*d) + ((a^2 + 2*b^2)*e^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/
(8*b^(3/2)*(-a^2 + b^2)^(7/4)*d) - (a*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)*d*S
qrt[e*Cos[c + d*x]]) + (a*(a^2 + 2*b^2)*e^2*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d
*x)/2, 2])/(8*b^2*(a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) + (a*(a^2 + 2*b^2)*e^2*
Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(8*b^2*(a^2 - b^2)*(a^2 - b*(b +
Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) - (e*Sqrt[e*Cos[c + d*x]])/(2*b*d*(a + b*Sin[c + d*x])^2) + (a*e*Sq
rt[e*Cos[c + d*x]])/(4*b*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

Rule 2693

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

Rule 2864

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a
^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*Sim
p[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[m, -1] && 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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2702

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 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 212

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] &&
!GtQ[a/b, 0]

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]

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]

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^3} \, dx &=-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}-\frac{e^2 \int \frac{\sin (c+d x)}{\sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2} \, dx}{4 b}\\ &=-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{e^2 \int \frac{b-\frac{1}{2} a \sin (c+d x)}{\sqrt{e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{4 b \left (a^2-b^2\right )}\\ &=-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\left (a e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )}+\frac{\left (\left (a^2+2 b^2\right ) e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{8 b^2 \left (a^2-b^2\right )}\\ &=-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\left (a \left (a^2+2 b^2\right ) e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{16 b^2 \left (-a^2+b^2\right )^{3/2}}+\frac{\left (a \left (a^2+2 b^2\right ) e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{16 b^2 \left (-a^2+b^2\right )^{3/2}}+\frac{\left (\left (a^2+2 b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \cos (c+d x)\right )}{8 b \left (a^2-b^2\right ) d}-\frac{\left (a e^2 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right ) \sqrt{e \cos (c+d x)}}\\ &=-\frac{a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right ) d \sqrt{e \cos (c+d x)}}-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\left (\left (a^2+2 b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{4 b \left (a^2-b^2\right ) d}+\frac{\left (a \left (a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \left (\sqrt{-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{16 b^2 \left (-a^2+b^2\right )^{3/2} \sqrt{e \cos (c+d x)}}+\frac{\left (a \left (a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \left (\sqrt{-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{16 b^2 \left (-a^2+b^2\right )^{3/2} \sqrt{e \cos (c+d x)}}\\ &=-\frac{a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right ) d \sqrt{e \cos (c+d x)}}-\frac{a \left (a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}+\frac{a \left (a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\left (\left (a^2+2 b^2\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{8 b \left (-a^2+b^2\right )^{3/2} d}+\frac{\left (\left (a^2+2 b^2\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{8 b \left (-a^2+b^2\right )^{3/2} d}\\ &=\frac{\left (a^2+2 b^2\right ) e^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{3/2} \left (-a^2+b^2\right )^{7/4} d}+\frac{\left (a^2+2 b^2\right ) e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{3/2} \left (-a^2+b^2\right )^{7/4} d}-\frac{a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right ) d \sqrt{e \cos (c+d x)}}-\frac{a \left (a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}+\frac{a \left (a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^2 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{e \sqrt{e \cos (c+d x)}}{2 b d (a+b \sin (c+d x))^2}+\frac{a e \sqrt{e \cos (c+d x)}}{4 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 24.2999, size = 1211, normalized size = 2.33 \[ \frac{(e \cos (c+d x))^{3/2} \sec (c+d x) \left (-\frac{a}{4 b \left (b^2-a^2\right ) (a+b \sin (c+d x))}-\frac{1}{2 b (a+b \sin (c+d x))^2}\right )}{d}-\frac{(e \cos (c+d x))^{3/2} \left (\frac{4 b \left (a+b \sqrt{1-\cos ^2(c+d x)}\right ) \left (\frac{5 a \left (a^2-b^2\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) \sqrt{\cos (c+d x)}}{\sqrt{1-\cos ^2(c+d x)} \left (5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )-2 \left (2 F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (b^2-a^2\right ) F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (\cos ^2(c+d x)-1\right )\right )}-\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \sqrt{b} \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )+\log \left (i b \cos (c+d x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}\right )-\log \left (i b \cos (c+d x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}\right )\right )}{\left (b^2-a^2\right )^{3/4}}\right ) \sin (c+d x)}{\sqrt{1-\cos ^2(c+d x)} (a+b \sin (c+d x))}-\frac{2 a \left (a+b \sqrt{1-\cos ^2(c+d x)}\right ) \left (\frac{5 b \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{1-\cos ^2(c+d x)} F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )}{\left (2 \left (2 F_1\left (\frac{5}{4};-\frac{1}{2},2;\frac{9}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (a^2-b^2\right ) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right ) \cos ^2(c+d x)-5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right ) \left (a^2+b^2 \left (\cos ^2(c+d x)-1\right )\right )}+\frac{a \left (-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \cos (c+d x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (b \cos (c+d x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}\right )\right )}{4 \sqrt{2} \sqrt{b} \left (a^2-b^2\right )^{3/4}}\right ) \sin ^2(c+d x)}{\left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}\right )}{8 (a-b) b (a+b) d \cos ^{\frac{3}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^(3/2)/(a + b*Sin[c + d*x])^3,x]

[Out]

((e*Cos[c + d*x])^(3/2)*Sec[c + d*x]*(-1/(2*b*(a + b*Sin[c + d*x])^2) - a/(4*b*(-a^2 + b^2)*(a + b*Sin[c + d*x
]))))/d - ((e*Cos[c + d*x])^(3/2)*((4*b*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2,
1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d*x]])/(Sqrt[1 - Cos[c + d*x]^2]*(5*(a
^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/
4, 1/2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, C
os[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) - ((1/8
- I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*
Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sq
rt[Cos[c + d*x]] + I*b*Cos[c + d*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c +
d*x]] + I*b*Cos[c + d*x]]))/(-a^2 + b^2)^(3/4))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x]))
- (2*a*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos
[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d*x]]*Sqrt[1 - Cos[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1
, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Cos[c + d*x]^
2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*
x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) + (a*(-2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*S
qrt[Cos[c + d*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*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[c + d*x]] + b*Cos[c + d*x]] + Log[Sqrt[a^2 -
b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2
)^(3/4)))*Sin[c + d*x]^2)/((1 - Cos[c + d*x]^2)*(a + b*Sin[c + d*x]))))/(8*(a - b)*b*(a + b)*d*Cos[c + d*x]^(3
/2))

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Maple [C]  time = 27.009, size = 45147, normalized size = 87. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*cos(d*x + c))^(3/2)/(b*sin(d*x + c) + a)^3, 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((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^3,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((e*cos(d*x+c))**(3/2)/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*cos(d*x + c))^(3/2)/(b*sin(d*x + c) + a)^3, x)