3.592 \(\int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=514 \[ \frac {\left (2 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \left (a^2-b^2\right )^2 \sqrt {e \cos (c+d x)}}-\frac {7 a^2 b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 d e^2 \left (a^2-b^2\right )^2 \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {7 a^2 b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 d e^2 \left (a^2-b^2\right )^2 \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {b}{d e \left (a^2-b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 d e \left (a^2-b^2\right )^2 (e \cos (c+d x))^{3/2}}+\frac {7 a b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{5/2} \left (b^2-a^2\right )^{11/4}}+\frac {7 a b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{5/2} \left (b^2-a^2\right )^{11/4}} \]

[Out]

7/2*a*b^(5/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(11/4)/d/e^(5/2)+7/2*a*
b^(5/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(11/4)/d/e^(5/2)+b/(a^2-b^2)
/d/e/(e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))+1/3*(-7*a*b+(2*a^2+5*b^2)*sin(d*x+c))/(a^2-b^2)^2/d/e/(e*cos(d*x+c)
)^(3/2)+1/3*(2*a^2+5*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)
)*cos(d*x+c)^(1/2)/(a^2-b^2)^2/d/e^2/(e*cos(d*x+c))^(1/2)-7/2*a^2*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x
+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/(a^2-b^2)^2/d/e^2/(a^
2-b*(b-(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)-7/2*a^2*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticPi(sin(1/2*d*x+1/2*c),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/(a^2-b^2)^2/d/e^2/(a^2-b*(b+(-a^
2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.31, antiderivative size = 514, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2694, 2866, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac {7 a b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{5/2} \left (b^2-a^2\right )^{11/4}}+\frac {7 a b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d e^{5/2} \left (b^2-a^2\right )^{11/4}}+\frac {\left (2 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \left (a^2-b^2\right )^2 \sqrt {e \cos (c+d x)}}-\frac {7 a^2 b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 d e^2 \left (a^2-b^2\right )^2 \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {7 a^2 b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 d e^2 \left (a^2-b^2\right )^2 \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {b}{d e \left (a^2-b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 d e \left (a^2-b^2\right )^2 (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])^2),x]

[Out]

(7*a*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*(-a^2 + b^2)^(11/4)*d*e^(
5/2)) + (7*a*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*(-a^2 + b^2)^(11
/4)*d*e^(5/2)) + ((2*a^2 + 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*(a^2 - b^2)^2*d*e^2*Sqrt[e*
Cos[c + d*x]]) - (7*a^2*b^2*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*(a
^2 - b^2)^2*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*e^2*Sqrt[e*Cos[c + d*x]]) - (7*a^2*b^2*Sqrt[Cos[c + d*x]]*Ellip
ticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*(a^2 - b^2)^2*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*e^2*S
qrt[e*Cos[c + d*x]]) + b/((a^2 - b^2)*d*e*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x])) - (7*a*b - (2*a^2 + 5*b
^2)*Sin[c + d*x])/(3*(a^2 - b^2)^2*d*e*(e*Cos[c + d*x])^(3/2))

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]

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 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 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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2} \, dx &=\frac {b}{\left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}+\frac {\int \frac {-a+\frac {5}{2} b \sin (c+d x)}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx}{-a^2+b^2}\\ &=\frac {b}{\left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d e (e \cos (c+d x))^{3/2}}+\frac {2 \int \frac {\frac {1}{2} a \left (a^2-8 b^2\right )+\frac {1}{4} b \left (2 a^2+5 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{3 \left (a^2-b^2\right )^2 e^2}\\ &=\frac {b}{\left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d e (e \cos (c+d x))^{3/2}}-\frac {\left (7 a b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{2 \left (a^2-b^2\right )^2 e^2}+\frac {\left (2 a^2+5 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{6 \left (a^2-b^2\right )^2 e^2}\\ &=\frac {b}{\left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d e (e \cos (c+d x))^{3/2}}+\frac {\left (7 a^2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 \left (-a^2+b^2\right )^{5/2} e^2}+\frac {\left (7 a^2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 \left (-a^2+b^2\right )^{5/2} e^2}-\frac {\left (7 a b^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 )}{2 \left (a^2-b^2\right )^2 d e}+\frac {\left (\left (2 a^2+5 b^2\right ) \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 \left (a^2-b^2\right )^2 e^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {\left (2 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 \left (a^2-b^2\right )^2 d e^2 \sqrt {e \cos (c+d x)}}+\frac {b}{\left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d e (e \cos (c+d x))^{3/2}}-\frac {\left (7 a b^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 )}{\left (a^2-b^2\right )^2 d e}+\frac {\left (7 a^2 b^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}{4 \left (-a^2+b^2\right )^{5/2} e^2 \sqrt {e \cos (c+d x)}}+\frac {\left (7 a^2 b^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}{4 \left (-a^2+b^2\right )^{5/2} e^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {\left (2 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 \left (a^2-b^2\right )^2 d e^2 \sqrt {e \cos (c+d x)}}-\frac {7 a^2 b^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 )}{2 \left (-a^2+b^2\right )^{5/2} \left (b-\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \cos (c+d x)}}+\frac {7 a^2 b^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 )}{2 \left (-a^2+b^2\right )^{5/2} \left (b+\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \cos (c+d x)}}+\frac {b}{\left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d e (e \cos (c+d x))^{3/2}}+\frac {\left (7 a b^3\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 \left (-a^2+b^2\right )^{5/2} d e^2}+\frac {\left (7 a b^3\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 \left (-a^2+b^2\right )^{5/2} d e^2}\\ &=\frac {7 a b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \left (-a^2+b^2\right )^{11/4} d e^{5/2}}+\frac {7 a b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \left (-a^2+b^2\right )^{11/4} d e^{5/2}}+\frac {\left (2 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 \left (a^2-b^2\right )^2 d e^2 \sqrt {e \cos (c+d x)}}-\frac {7 a^2 b^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 )}{2 \left (-a^2+b^2\right )^{5/2} \left (b-\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \cos (c+d x)}}+\frac {7 a^2 b^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 )}{2 \left (-a^2+b^2\right )^{5/2} \left (b+\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \cos (c+d x)}}+\frac {b}{\left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}-\frac {7 a b-\left (2 a^2+5 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d e (e \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 24.34, size = 1258, normalized size = 2.45 \[ \frac {\left (\frac {2 \sec ^2(c+d x) \left (\sin (c+d x) a^2-2 b a+b^2 \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2}-\frac {b^3}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}\right ) \cos ^3(c+d x)}{d (e \cos (c+d x))^{5/2}}+\frac {\left (-\frac {2 \left (5 b^3+2 a^2 b\right ) \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))}-\frac {2 \left (2 a^3-16 a b^2\right ) \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))}\right ) \cos ^{\frac {5}{2}}(c+d x)}{6 (a-b)^2 (a+b)^2 d (e \cos (c+d x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])^2),x]

[Out]

(Cos[c + d*x]^(5/2)*((-2*(2*a^3 - 16*a*b^2)*(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*AppellF
1[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, 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))) - ((
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
)*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]]))/(-a^2 + b^2)^(3/4))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x
])) - (2*(2*a^2*b + 5*b^3)*(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)*A
ppellF1[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]*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]]))/(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]))))/(6*(a - b)^2*(a +
b)^2*d*(e*Cos[c + d*x])^(5/2)) + (Cos[c + d*x]^3*(-(b^3/((a^2 - b^2)^2*(a + b*Sin[c + d*x]))) + (2*Sec[c + d*x
]^2*(-2*a*b + a^2*Sin[c + d*x] + b^2*Sin[c + d*x]))/(3*(a^2 - b^2)^2)))/(d*(e*Cos[c + d*x])^(5/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)^2), x)

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maple [C]  time = 17.26, size = 6022, normalized size = 11.72 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c))^2,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate(1/((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))^2),x)

[Out]

int(1/((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))**(5/2)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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