3.77 \(\int \frac{1}{(a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=590 \[ \frac{9 a b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{7/2} \left (b^2-a^2\right )^{13/4}}-\frac{9 a b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{7/2} \left (b^2-a^2\right )^{13/4}}-\frac{3 \left (\left (-10 a^2 b^2+2 a^4-7 b^4\right ) \cos (c+d x)+15 a b^3\right )}{5 d e^3 \left (a^2-b^2\right )^3 \sqrt{e \sin (c+d x)}}-\frac{3 \left (-10 a^2 b^2+2 a^4-7 b^4\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d e^4 \left (a^2-b^2\right )^3 \sqrt{\sin (c+d x)}}+\frac{9 a^2 b^3 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{2 d e^3 \left (a^2-b^2\right )^3 \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{9 a^2 b^3 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{2 d e^3 \left (a^2-b^2\right )^3 \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}}-\frac{b}{d e \left (a^2-b^2\right ) (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 d e \left (a^2-b^2\right )^2 (e \sin (c+d x))^{5/2}} \]

[Out]

(9*a*b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*(-a^2 + b^2)^(13/4)*d*e^(
7/2)) - (9*a*b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*(-a^2 + b^2)^(13
/4)*d*e^(7/2)) - b/((a^2 - b^2)*d*e*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2)) + (9*a*b - (2*a^2 + 7*b^2)*Co
s[c + d*x])/(5*(a^2 - b^2)^2*d*e*(e*Sin[c + d*x])^(5/2)) - (3*(15*a*b^3 + (2*a^4 - 10*a^2*b^2 - 7*b^4)*Cos[c +
d*x]))/(5*(a^2 - b^2)^3*d*e^3*Sqrt[e*Sin[c + d*x]]) + (9*a^2*b^3*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c
- Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*(a^2 - b^2)^3*(b - Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Sin[c + d*x]]) +
(9*a^2*b^3*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*(a^2 - b^2)^
3*(b + Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Sin[c + d*x]]) - (3*(2*a^4 - 10*a^2*b^2 - 7*b^4)*EllipticE[(c - Pi/2 + d
*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*(a^2 - b^2)^3*d*e^4*Sqrt[Sin[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.68009, antiderivative size = 590, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {2694, 2866, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{9 a b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{7/2} \left (b^2-a^2\right )^{13/4}}-\frac{9 a b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{7/2} \left (b^2-a^2\right )^{13/4}}-\frac{3 \left (\left (-10 a^2 b^2+2 a^4-7 b^4\right ) \cos (c+d x)+15 a b^3\right )}{5 d e^3 \left (a^2-b^2\right )^3 \sqrt{e \sin (c+d x)}}-\frac{3 \left (-10 a^2 b^2+2 a^4-7 b^4\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d e^4 \left (a^2-b^2\right )^3 \sqrt{\sin (c+d x)}}+\frac{9 a^2 b^3 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{2 d e^3 \left (a^2-b^2\right )^3 \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{9 a^2 b^3 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{2 d e^3 \left (a^2-b^2\right )^3 \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}}-\frac{b}{d e \left (a^2-b^2\right ) (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 d e \left (a^2-b^2\right )^2 (e \sin (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[c + d*x])^2*(e*Sin[c + d*x])^(7/2)),x]

[Out]

(9*a*b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*(-a^2 + b^2)^(13/4)*d*e^(
7/2)) - (9*a*b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*(-a^2 + b^2)^(13
/4)*d*e^(7/2)) - b/((a^2 - b^2)*d*e*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2)) + (9*a*b - (2*a^2 + 7*b^2)*Co
s[c + d*x])/(5*(a^2 - b^2)^2*d*e*(e*Sin[c + d*x])^(5/2)) - (3*(15*a*b^3 + (2*a^4 - 10*a^2*b^2 - 7*b^4)*Cos[c +
d*x]))/(5*(a^2 - b^2)^3*d*e^3*Sqrt[e*Sin[c + d*x]]) + (9*a^2*b^3*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c
- Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*(a^2 - b^2)^3*(b - Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Sin[c + d*x]]) +
(9*a^2*b^3*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*(a^2 - b^2)^
3*(b + Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Sin[c + d*x]]) - (3*(2*a^4 - 10*a^2*b^2 - 7*b^4)*EllipticE[(c - Pi/2 + d
*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*(a^2 - b^2)^3*d*e^4*Sqrt[Sin[c + d*x]])

Rule 2694

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

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -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 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{1}{(a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}} \, dx &=-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{\int \frac{-a+\frac{7}{2} b \cos (c+d x)}{(a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}} \, dx}{-a^2+b^2}\\ &=-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{5/2}}+\frac{2 \int \frac{\frac{3}{2} a \left (a^2-4 b^2\right )+\frac{3}{4} b \left (2 a^2+7 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}} \, dx}{5 \left (a^2-b^2\right )^2 e^2}\\ &=-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{5/2}}-\frac{3 \left (15 a b^3+\left (2 a^4-10 a^2 b^2-7 b^4\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^3 d e^3 \sqrt{e \sin (c+d x)}}-\frac{4 \int \frac{\left (\frac{3}{4} a \left (a^4-5 a^2 b^2-11 b^4\right )+\frac{3}{8} b \left (2 a^4-10 a^2 b^2-7 b^4\right ) \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{5 \left (a^2-b^2\right )^3 e^4}\\ &=-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{5/2}}-\frac{3 \left (15 a b^3+\left (2 a^4-10 a^2 b^2-7 b^4\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^3 d e^3 \sqrt{e \sin (c+d x)}}+\frac{\left (9 a b^4\right ) \int \frac{\sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3 e^4}-\frac{\left (3 \left (2 a^4-10 a^2 b^2-7 b^4\right )\right ) \int \sqrt{e \sin (c+d x)} \, dx}{10 \left (a^2-b^2\right )^3 e^4}\\ &=-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{5/2}}-\frac{3 \left (15 a b^3+\left (2 a^4-10 a^2 b^2-7 b^4\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^3 d e^3 \sqrt{e \sin (c+d x)}}-\frac{\left (9 a^2 b^3\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{4 \left (a^2-b^2\right )^3 e^3}+\frac{\left (9 a^2 b^3\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{4 \left (a^2-b^2\right )^3 e^3}-\frac{\left (9 a b^5\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d e^3}-\frac{\left (3 \left (2 a^4-10 a^2 b^2-7 b^4\right ) \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{10 \left (a^2-b^2\right )^3 e^4 \sqrt{\sin (c+d x)}}\\ &=-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{5/2}}-\frac{3 \left (15 a b^3+\left (2 a^4-10 a^2 b^2-7 b^4\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^3 d e^3 \sqrt{e \sin (c+d x)}}-\frac{3 \left (2 a^4-10 a^2 b^2-7 b^4\right ) E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 \left (a^2-b^2\right )^3 d e^4 \sqrt{\sin (c+d x)}}-\frac{\left (9 a b^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 \sin (c+d x)}\right )}{\left (a^2-b^2\right )^3 d e^3}-\frac{\left (9 a^2 b^3 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{4 \left (a^2-b^2\right )^3 e^3 \sqrt{e \sin (c+d x)}}+\frac{\left (9 a^2 b^3 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{4 \left (a^2-b^2\right )^3 e^3 \sqrt{e \sin (c+d x)}}\\ &=-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{5/2}}-\frac{3 \left (15 a b^3+\left (2 a^4-10 a^2 b^2-7 b^4\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^3 d e^3 \sqrt{e \sin (c+d x)}}+\frac{9 a^2 b^3 \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{2 \left (a^2-b^2\right )^3 \left (b-\sqrt{-a^2+b^2}\right ) d e^3 \sqrt{e \sin (c+d x)}}+\frac{9 a^2 b^3 \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{2 \left (a^2-b^2\right )^3 \left (b+\sqrt{-a^2+b^2}\right ) d e^3 \sqrt{e \sin (c+d x)}}-\frac{3 \left (2 a^4-10 a^2 b^2-7 b^4\right ) E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 \left (a^2-b^2\right )^3 d e^4 \sqrt{\sin (c+d x)}}+\frac{\left (9 a b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{2 \left (a^2-b^2\right )^3 d e^3}-\frac{\left (9 a b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{2 \left (a^2-b^2\right )^3 d e^3}\\ &=\frac{9 a b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{2 \left (-a^2+b^2\right )^{13/4} d e^{7/2}}-\frac{9 a b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{2 \left (-a^2+b^2\right )^{13/4} d e^{7/2}}-\frac{b}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}+\frac{9 a b-\left (2 a^2+7 b^2\right ) \cos (c+d x)}{5 \left (a^2-b^2\right )^2 d e (e \sin (c+d x))^{5/2}}-\frac{3 \left (15 a b^3+\left (2 a^4-10 a^2 b^2-7 b^4\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^3 d e^3 \sqrt{e \sin (c+d x)}}+\frac{9 a^2 b^3 \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{2 \left (a^2-b^2\right )^3 \left (b-\sqrt{-a^2+b^2}\right ) d e^3 \sqrt{e \sin (c+d x)}}+\frac{9 a^2 b^3 \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{2 \left (a^2-b^2\right )^3 \left (b+\sqrt{-a^2+b^2}\right ) d e^3 \sqrt{e \sin (c+d x)}}-\frac{3 \left (2 a^4-10 a^2 b^2-7 b^4\right ) E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 \left (a^2-b^2\right )^3 d e^4 \sqrt{\sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.84776, size = 950, normalized size = 1.61 \[ \frac{\sin ^4(c+d x) \left (-\frac{\sin (c+d x) b^5}{\left (a^2-b^2\right )^3 (a+b \cos (c+d x))}-\frac{2 \left (\cos (c+d x) a^2-2 b a+b^2 \cos (c+d x)\right ) \csc ^3(c+d x)}{5 \left (a^2-b^2\right )^2}-\frac{2 \left (3 \cos (c+d x) a^4-15 b^2 \cos (c+d x) a^2+20 b^3 a-8 b^4 \cos (c+d x)\right ) \csc (c+d x)}{5 \left (a^2-b^2\right )^3}\right )}{d (e \sin (c+d x))^{7/2}}-\frac{3 \sin ^{\frac{7}{2}}(c+d x) \left (\frac{\left (-7 b^5-10 a^2 b^3+2 a^4 b\right ) \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\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{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \sin (c+d x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (b \sin (c+d x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )\right )\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos ^2(c+d x)}{12 b^{3/2} \left (b^2-a^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac{2 \left (2 a^5-10 b^2 a^3-22 b^4 a\right ) \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\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{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \sin (c+d x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \sin (c+d x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{1-\sin ^2(c+d x)}}\right )}{10 (a-b)^3 (a+b)^3 d (e \sin (c+d x))^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Cos[c + d*x])^2*(e*Sin[c + d*x])^(7/2)),x]

[Out]

(Sin[c + d*x]^4*((-2*(20*a*b^3 + 3*a^4*Cos[c + d*x] - 15*a^2*b^2*Cos[c + d*x] - 8*b^4*Cos[c + d*x])*Csc[c + d*
x])/(5*(a^2 - b^2)^3) - (2*(-2*a*b + a^2*Cos[c + d*x] + b^2*Cos[c + d*x])*Csc[c + d*x]^3)/(5*(a^2 - b^2)^2) -
(b^5*Sin[c + d*x])/((a^2 - b^2)^3*(a + b*Cos[c + d*x]))))/(d*(e*Sin[c + d*x])^(7/2)) - (3*Sin[c + d*x]^(7/2)*(
((2*a^4*b - 10*a^2*b^3 - 7*b^5)*Cos[c + d*x]^2*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*S
qrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[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[Sin[c + d*x]] + b*Sin[c + d*x]] + Log[Sqrt[a^2 -
b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]) + 8*b^(5/2)*AppellF1[3/4, -1/2
, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))*(a + b*Sqrt[1 - Sin[c + d*x]^
2]))/(12*b^(3/2)*(-a^2 + b^2)*(a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (2*(2*a^5 - 10*a^3*b^2 - 22*a*b^4)*
Cos[c + d*x]*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1
+ ((1 + I)*Sqrt[b]*Sqrt[Sin[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[Sin[c + d*x]] + I*b*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*S
qrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)) + (a*AppellF1[3/4, 1/2, 1, 7/4, Sin[c + d
*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))/(3*(a^2 - b^2)))*(a + b*Sqrt[1 - Sin[c + d*x]^2]
))/((a + b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(10*(a - b)^3*(a + b)^3*d*(e*Sin[c + d*x])^(7/2))

________________________________________________________________________________________

Maple [B]  time = 13.2, size = 2888, normalized size = 4.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x)

[Out]

-9/4/d/e^3*a*b^3/(a-b)^3/(a+b)^3/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e
*sin(d*x+c))^(1/2)+1)-9/4/d/e^3*a*b^3/(a-b)^3/(a+b)^3/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2*(a
^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)-1)-9/8/d/e^3*a*b^3/(a-b)^3/(a+b)^3/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*l
n((e*sin(d*x+c)-(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*sin(d*x+c
)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))-1/d/e^3*a*b^5/(a-b)^3/(a+
b)^3*(e*sin(d*x+c))^(3/2)/(-b^2*cos(d*x+c)^2*e^2+e^2*a^2)+1/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4/(a^2-b^2
)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))+1
/d/e^3*sin(d*x+c)^2*cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^6/(a-b)^2/(a+b)^2/(a^2-b^2)/(-b^2*cos(d*x+c)^2+a^2)+4/5/
d/e*a*b/(a+b)^2/(a-b)^2/(e*sin(d*x+c))^(5/2)-6/5/d/e^3*cos(d*x+c)^3/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2/(cos(d*x+
c)^2-1)*a^2-6/5/d/e^3*cos(d*x+c)^3/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2/(cos(d*x+c)^2-1)*b^2-2/d/e^3/cos(d*x+c)/(e
*sin(d*x+c))^(1/2)*b^4/(a^2-b^2)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-s
in(d*x+c))^(1/2),1/2*2^(1/2))-3/4/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*b^2/(a-b)^2/(a+b)^2/(a^2-b^2)*(1-s
in(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2
),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*b^2/(a-b)^2/(a+b)^2/(a^2
-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*
x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+1/2/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4/(a-b)^2/(a+b)^
2/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-
sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/2/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^2/(a-b)^3/
(a+b)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(
d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))*a^2-1/2/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4/(a-b)^3/
(a+b)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(
d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-1/2/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4/(a-b)^3/(a+b
)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+
c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-1/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4/(a-b)^2/(a+b)^2/(a
^2-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2
))+1/2/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4/(a-b)^2/(a+b)^2/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c
))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))+2/d/e^3*cos(d*x+c)/(e*sin(d*x+c))^(1/2)*
b^4/(a^2-b^2)^3-6/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^2/(a^2-b^2)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^
(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+3/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)
*b^2/(a^2-b^2)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1
/2*2^(1/2))*a^2+3/5/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*sin(d*x+c)^(5/2)/(cos(d*x+c)^2-1)*(1-sin
(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+3/5/d/e^3/cos(d*x+c)/(e*
sin(d*x+c))^(1/2)/(a^2-b^2)^2*sin(d*x+c)^(5/2)/(cos(d*x+c)^2-1)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*El
lipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-6/5/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*sin(d*x+c)
^(5/2)/(cos(d*x+c)^2-1)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2)
)*a^2-6/5/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*sin(d*x+c)^(5/2)/(cos(d*x+c)^2-1)*(1-sin(d*x+c))^(
1/2)*(2+2*sin(d*x+c))^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2+8/5/d/e^3*cos(d*x+c)/(e*sin(d*x+c)
)^(1/2)/(a^2-b^2)^2/(cos(d*x+c)^2-1)*a^2+8/5/d/e^3*cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2/(cos(d*x+c)^2-1
)*b^2+6/d/e^3*cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^2/(a^2-b^2)^3*a^2-8/d/e^3*a*b^3/(a-b)^3/(a+b)^3/(e*sin(d*x+c))
^(1/2)-3/2/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^2/(a-b)^3/(a+b)^3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/
2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2
))*a^2+1/2/d/e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4/(a-b)^2/(a+b)^2/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d
*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),
1/2*2^(1/2))

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Maxima [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(1/(a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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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(1/(a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),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(1/(a+b*cos(d*x+c))**2/(e*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(1/((b*cos(d*x + c) + a)^2*(e*sin(d*x + c))^(7/2)), x)