3.585 \(\int \frac{(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=459 \[ \frac{7 a e^{9/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{9/2} d}-\frac{7 a e^{9/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{9/2} d}+\frac{7 e^4 \left (5 a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 b^4 d \sqrt{\cos (c+d x)}}-\frac{7 a^2 e^5 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^5 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}-\frac{7 a^2 e^5 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^5 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}}+\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))} \]

[Out]

(7*a*(-a^2 + b^2)^(3/4)*e^(9/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*b^(9/2
)*d) - (7*a*(-a^2 + b^2)^(3/4)*e^(9/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(
2*b^(9/2)*d) + (7*(5*a^2 - 3*b^2)*e^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b^4*d*Sqrt[Cos[c + d*
x]]) - (7*a^2*(a^2 - b^2)*e^5*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*
b^5*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) - (7*a^2*(a^2 - b^2)*e^5*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b
)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*b^5*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) + (7*e^3*(e*C
os[c + d*x])^(3/2)*(5*a - 3*b*Sin[c + d*x]))/(15*b^3*d) - (e*(e*Cos[c + d*x])^(7/2))/(b*d*(a + b*Sin[c + d*x])
)

________________________________________________________________________________________

Rubi [A]  time = 1.12032, antiderivative size = 459, 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, 2865, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{7 a e^{9/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{9/2} d}-\frac{7 a e^{9/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{9/2} d}+\frac{7 e^4 \left (5 a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 b^4 d \sqrt{\cos (c+d x)}}-\frac{7 a^2 e^5 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^5 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}-\frac{7 a^2 e^5 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^5 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}}+\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(9/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(7*a*(-a^2 + b^2)^(3/4)*e^(9/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*b^(9/2
)*d) - (7*a*(-a^2 + b^2)^(3/4)*e^(9/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(
2*b^(9/2)*d) + (7*(5*a^2 - 3*b^2)*e^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b^4*d*Sqrt[Cos[c + d*
x]]) - (7*a^2*(a^2 - b^2)*e^5*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*
b^5*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) - (7*a^2*(a^2 - b^2)*e^5*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b
)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*b^5*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) + (7*e^3*(e*C
os[c + d*x])^(3/2)*(5*a - 3*b*Sin[c + d*x]))/(15*b^3*d) - (e*(e*Cos[c + d*x])^(7/2))/(b*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 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{(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^2} \, dx &=-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}-\frac{\left (7 e^2\right ) \int \frac{(e \cos (c+d x))^{5/2} \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b}\\ &=\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}-\frac{\left (7 e^4\right ) \int \frac{\sqrt{e \cos (c+d x)} \left (-a b-\frac{1}{2} \left (5 a^2-3 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 b^3}\\ &=\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}+\frac{\left (7 \left (5 a^2-3 b^2\right ) e^4\right ) \int \sqrt{e \cos (c+d x)} \, dx}{10 b^4}-\frac{\left (7 a \left (a^2-b^2\right ) e^4\right ) \int \frac{\sqrt{e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx}{2 b^4}\\ &=\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}+\frac{\left (7 a^2 \left (a^2-b^2\right ) e^5\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^5}-\frac{\left (7 a^2 \left (a^2-b^2\right ) e^5\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^5}-\frac{\left (7 a \left (a^2-b^2\right ) e^5\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \cos (c+d x)\right )}{2 b^3 d}+\frac{\left (7 \left (5 a^2-3 b^2\right ) e^4 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{10 b^4 \sqrt{\cos (c+d x)}}\\ &=\frac{7 \left (5 a^2-3 b^2\right ) e^4 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt{\cos (c+d x)}}+\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}-\frac{\left (7 a \left (a^2-b^2\right ) e^5\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{b^3 d}+\frac{\left (7 a^2 \left (a^2-b^2\right ) e^5 \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}{4 b^5 \sqrt{e \cos (c+d x)}}-\frac{\left (7 a^2 \left (a^2-b^2\right ) e^5 \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}{4 b^5 \sqrt{e \cos (c+d x)}}\\ &=\frac{7 \left (5 a^2-3 b^2\right ) e^4 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt{\cos (c+d x)}}-\frac{7 a^2 \left (a^2-b^2\right ) e^5 \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 )}{2 b^5 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{7 a^2 \left (a^2-b^2\right ) e^5 \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 )}{2 b^5 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}+\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}+\frac{\left (7 a \left (a^2-b^2\right ) e^5\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 )}{2 b^4 d}-\frac{\left (7 a \left (a^2-b^2\right ) e^5\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 )}{2 b^4 d}\\ &=\frac{7 a \left (-a^2+b^2\right )^{3/4} e^{9/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{2 b^{9/2} d}-\frac{7 a \left (-a^2+b^2\right )^{3/4} e^{9/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{2 b^{9/2} d}+\frac{7 \left (5 a^2-3 b^2\right ) e^4 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt{\cos (c+d x)}}-\frac{7 a^2 \left (a^2-b^2\right ) e^5 \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 )}{2 b^5 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{7 a^2 \left (a^2-b^2\right ) e^5 \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 )}{2 b^5 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}+\frac{7 e^3 (e \cos (c+d x))^{3/2} (5 a-3 b \sin (c+d x))}{15 b^3 d}-\frac{e (e \cos (c+d x))^{7/2}}{b d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 26.893, size = 835, normalized size = 1.82 \[ \frac{7 \left (-\frac{\left (5 a^2-3 b^2\right ) \left (a+b \sqrt{1-\cos ^2(c+d x)}\right ) \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(c+d x) b^{5/2}+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \cos (c+d x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (b \cos (c+d x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}\right )\right )\right ) \sin ^2(c+d x)}{12 b^{3/2} \left (b^2-a^2\right ) \left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}-\frac{4 a b \left (a+b \sqrt{1-\cos ^2(c+d x)}\right ) \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \cos (c+d x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \cos (c+d x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \sin (c+d x)}{\sqrt{1-\cos ^2(c+d x)} (a+b \sin (c+d x))}\right ) (e \cos (c+d x))^{9/2}}{10 b^3 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{\sec ^4(c+d x) \left (\frac{4 a \cos (c+d x)}{3 b^3}-\frac{\sin (2 (c+d x))}{5 b^2}+\frac{a^2 \cos (c+d x)-b^2 \cos (c+d x)}{b^3 (a+b \sin (c+d x))}\right ) (e \cos (c+d x))^{9/2}}{d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^(9/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(7*(e*Cos[c + d*x])^(9/2)*((-4*a*b*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((a*AppellF1[3/4, 1/2, 1, 7/4, Cos[c + d*x
]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos[c + d*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(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)*Sqrt[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]]))/(Sqrt[
b]*(-a^2 + b^2)^(1/4)))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x])) - ((5*a^2 - 3*b^2)*(a +
b*Sqrt[1 - Cos[c + d*x]^2])*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2
+ b^2)]*Cos[c + d*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[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]*Sqr
t[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))*Sin[c + d*x]^2)/(12*b^(3/2)*(-a^2 + b^2)*(1 - Co
s[c + d*x]^2)*(a + b*Sin[c + d*x]))))/(10*b^3*d*Cos[c + d*x]^(9/2)) + ((e*Cos[c + d*x])^(9/2)*Sec[c + d*x]^4*(
(4*a*Cos[c + d*x])/(3*b^3) + (a^2*Cos[c + d*x] - b^2*Cos[c + d*x])/(b^3*(a + b*Sin[c + d*x])) - Sin[2*(c + d*x
)]/(5*b^2)))/d

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Maple [C]  time = 8.373, size = 20346, normalized size = 44.3 \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))^(9/2)/(a+b*sin(d*x+c))^2,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{9}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(9/2)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(9/2)/(b*sin(d*x + c) + a)^2, 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))^(9/2)/(a+b*sin(d*x+c))^2,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))**(9/2)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [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))^(9/2)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out