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

Optimal. Leaf size=486 \[ \frac{2 \left (a \left (3 a^2-8 b^2\right ) \sin (c+d x)+5 b^3\right )}{5 d e^3 \left (a^2-b^2\right )^2 \sqrt{e \cos (c+d x)}}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d e^{7/2} \left (b^2-a^2\right )^{9/4}}-\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d e^{7/2} \left (b^2-a^2\right )^{9/4}}-\frac{2 a \left (3 a^2-8 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}}+\frac{a b^3 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{d e^3 \left (a^2-b^2\right )^2 \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}+\frac{a b^3 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{d e^3 \left (a^2-b^2\right )^2 \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}}-\frac{2 (b-a \sin (c+d x))}{5 d e \left (a^2-b^2\right ) (e \cos (c+d x))^{5/2}} \]

[Out]

(b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(9/4)*d*e^(7/2)) -
(b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(9/4)*d*e^(7/2))
- (2*a*(3*a^2 - 8*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*(a^2 - b^2)^2*d*e^4*Sqrt[Cos[c + d*
x]]) + (a*b^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/((a^2 - b^2)^2*(b -
Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Cos[c + d*x]]) + (a*b^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b
^2]), (c + d*x)/2, 2])/((a^2 - b^2)^2*(b + Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Cos[c + d*x]]) - (2*(b - a*Sin[c + d
*x]))/(5*(a^2 - b^2)*d*e*(e*Cos[c + d*x])^(5/2)) + (2*(5*b^3 + a*(3*a^2 - 8*b^2)*Sin[c + d*x]))/(5*(a^2 - b^2)
^2*d*e^3*Sqrt[e*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.32848, antiderivative size = 486, 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 = {2696, 2866, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{2 \left (a \left (3 a^2-8 b^2\right ) \sin (c+d x)+5 b^3\right )}{5 d e^3 \left (a^2-b^2\right )^2 \sqrt{e \cos (c+d x)}}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d e^{7/2} \left (b^2-a^2\right )^{9/4}}-\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d e^{7/2} \left (b^2-a^2\right )^{9/4}}-\frac{2 a \left (3 a^2-8 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}}+\frac{a b^3 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{d e^3 \left (a^2-b^2\right )^2 \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}+\frac{a b^3 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{d e^3 \left (a^2-b^2\right )^2 \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}}-\frac{2 (b-a \sin (c+d x))}{5 d e \left (a^2-b^2\right ) (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(9/4)*d*e^(7/2)) -
(b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(9/4)*d*e^(7/2))
- (2*a*(3*a^2 - 8*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*(a^2 - b^2)^2*d*e^4*Sqrt[Cos[c + d*
x]]) + (a*b^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/((a^2 - b^2)^2*(b -
Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Cos[c + d*x]]) + (a*b^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b
^2]), (c + d*x)/2, 2])/((a^2 - b^2)^2*(b + Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Cos[c + d*x]]) - (2*(b - a*Sin[c + d
*x]))/(5*(a^2 - b^2)*d*e*(e*Cos[c + d*x])^(5/2)) + (2*(5*b^3 + a*(3*a^2 - 8*b^2)*Sin[c + d*x]))/(5*(a^2 - b^2)
^2*d*e^3*Sqrt[e*Cos[c + d*x]])

Rule 2696

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

Mathematica [C]  time = 6.75686, size = 881, normalized size = 1.81 \[ \frac{\cos ^4(c+d x) \left (\frac{2 (a \sin (c+d x)-b) \sec ^3(c+d x)}{5 \left (a^2-b^2\right )}+\frac{2 \left (3 \sin (c+d x) a^3-8 b^2 \sin (c+d x) a+5 b^3\right ) \sec (c+d x)}{5 \left (a^2-b^2\right )^2}\right )}{d (e \cos (c+d x))^{7/2}}-\frac{\cos ^{\frac{7}{2}}(c+d x) \left (-\frac{\left (3 a^3 b-8 a b^3\right ) \left (a+b \sqrt{1-\cos ^2(c+d x)}\right ) \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(c+d x) b^{5/2}+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \cos (c+d x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (b \cos (c+d x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}\right )\right )\right ) \sin ^2(c+d x)}{12 b^{3/2} \left (b^2-a^2\right ) \left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}-\frac{2 \left (3 a^4-8 b^2 a^2-5 b^4\right ) \left (a+b \sqrt{1-\cos ^2(c+d x)}\right ) \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \cos (c+d x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \cos (c+d x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \sin (c+d x)}{\sqrt{1-\cos ^2(c+d x)} (a+b \sin (c+d x))}\right )}{5 (a-b)^2 (a+b)^2 d (e \cos (c+d x))^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Cos[c + d*x]^(7/2)*((-2*(3*a^4 - 8*a^2*b^2 - 5*b^4)*(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[Co
s[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])) - ((3
*a^3*b - 8*a*b^3)*(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]*Sqrt[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 - Cos[c + d*x]^2)*(a + b*Sin[c + d*x]))))/(5*(a - b)^2*(a + b)^2*d*(e*Cos[c + d*x])^(7/2))
+ (Cos[c + d*x]^4*((2*Sec[c + d*x]^3*(-b + a*Sin[c + d*x]))/(5*(a^2 - b^2)) + (2*Sec[c + d*x]*(5*b^3 + 3*a^3*S
in[c + d*x] - 8*a*b^2*Sin[c + d*x]))/(5*(a^2 - b^2)^2)))/(d*(e*Cos[c + d*x])^(7/2))

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Maple [C]  time = 5.316, size = 2399, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/e^4*b^3/(a^2-b^2)^2*2^(1/2)/(cos(1/2*d*x+1/2*c)-1/2*2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)+3/80/d/
e^4*b/(a^2-b^2)/(cos(1/2*d*x+1/2*c)+1/2*2^(1/2))^2*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)+3/80/d/e^4*b*2^(1/2)/(a
^2-b^2)/(cos(1/2*d*x+1/2*c)+1/2*2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-1/2/d/e^4*b^3/(a^2-b^2)^2*2^(1/2)
/(cos(1/2*d*x+1/2*c)+1/2*2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)+3/80/d/e^4*b/(a^2-b^2)/(cos(1/2*d*x+1/2*
c)-1/2*2^(1/2))^2*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-3/80/d/e^4*b*2^(1/2)/(a^2-b^2)/(cos(1/2*d*x+1/2*c)-1/2*2
^(1/2))*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-1/80/d/e^4*b/(a^2-b^2)*2^(1/2)/(cos(1/2*d*x+1/2*c)-1/2*2^(1/2))^3*
(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)+1/80/d/e^4*b/(a^2-b^2)*2^(1/2)/(cos(1/2*d*x+1/2*c)+1/2*2^(1/2))^3*(-2*sin(
1/2*d*x+1/2*c)^2*e+e)^(1/2)+1/2/d/e^3*b^5/(a-b)^2/(a+b)^2*sum((_R^6-_R^4*e-_R^2*e^2+e^3)/(_R^7*b^2-3*_R^5*b^2*
e+8*_R^3*a^2*e^2-5*_R^3*b^2*e^2-_R*b^2*e^3)*ln((-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-e^(1/2)*cos(1/2*d*x+1/2*c)*
2^(1/2)-_R),_R=RootOf(b^2*_Z^8-4*b^2*e*_Z^6+(16*a^2*e^2-10*b^2*e^2)*_Z^4-4*b^2*e^3*_Z^2+b^2*e^4))+4/d*(e*(2*co
s(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e^4*a/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2
)/(a^2-b^2)^2*b^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4*e+sin(1/2*d*x+1/2*c)^2*e)^(1/2)*cos(1/2*
d*x+1/2*c)-2/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e^4*a/sin(1/2*d*x+1/2*c)^3/(e*(2*cos(
1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b^2)^2*b^2/(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*s
in(1/2*d*x+1/2*c)^4*e+sin(1/2*d*x+1/2*c)^2*e)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-48/5/d*(e*(2*cos(1/2
*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e^4*a*sin(1/2*d*x+1/2*c)^3/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(
a^2-b^2)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4*e+
sin(1/2*d*x+1/2*c)^2*e)^(1/2)*cos(1/2*d*x+1/2*c)+24/5/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1
/2)/e^4*a*sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b^2)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2
*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4*e+sin(1/2*d*x+1/2*c)^2*e)^(1/2)*EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)+48/5/d*(e*(2*cos(1/2*d*
x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e^4*a*sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b
^2)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4*e+sin(1
/2*d*x+1/2*c)^2*e)^(1/2)*cos(1/2*d*x+1/2*c)-24/5/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e
^4*a/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b^2)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+
1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4*e+sin(1/2*d*x+1/2*c)^2*e)^(1/2)*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-16/5/d*(e*(2*cos(1/2*d*x+1/2
*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e^4*a/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b^2)/(
8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4*e+sin(1/2*d*
x+1/2*c)^2*e)^(1/2)*cos(1/2*d*x+1/2*c)+6/5/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e^4*a/s
in(1/2*d*x+1/2*c)^3/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b^2)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*
c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4*e+sin(1/2*d*x+1/2*c)^2*e)^(1/2)*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/8/d*(e*(2*cos(1/2*d*x+1/2*c)^2
-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/e^3*a/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)*b^2/(a-b)^2/(a+b
)^2*sum(1/_alpha*(2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*e*(4*_alpha^2-3)/(4*a^2-3*b^2)*
(4*cos(1/2*d*x+1/2*c)^2*a^2-3*b^2*cos(1/2*d*x+1/2*c)^2+b^2*_alpha^2-3*a^2+2*b^2)*2^(1/2)/(e*(2*_alpha^2*b^2+a^
2-2*b^2)/b^2)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-sin(1/2*d*x+1/2*c)^2*e*(2*sin(1/2*d*x+1/2*c)^2-
1))^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(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{1}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}{\left (b \sin \left (d x + c\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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