3.1252 \(\int \frac{A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=334 \[ \frac{(7 A+31 C) \sin (c+d x)}{16 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{(8 A+39 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac{(11 A+63 C) \sin (c+d x)}{16 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(43 A+219 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

[Out]

((8*A + 39*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(
4*a^(5/2)*d) - ((43*A + 219*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*
x]])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Sin[c + d*x])/(4*d*(a + a*Cos[c
+ d*x])^(5/2)*Sec[c + d*x]^(7/2)) - ((3*A + 19*C)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*
x]^(5/2)) + ((7*A + 31*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)) - ((11*A + 63*C
)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.15423, antiderivative size = 334, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.243, Rules used = {4221, 3042, 2977, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(7 A+31 C) \sin (c+d x)}{16 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{(8 A+39 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac{(11 A+63 C) \sin (c+d x)}{16 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(43 A+219 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)),x]

[Out]

((8*A + 39*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(
4*a^(5/2)*d) - ((43*A + 219*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*
x]])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Sin[c + d*x])/(4*d*(a + a*Cos[c
+ d*x])^(5/2)*Sec[c + d*x]^(7/2)) - ((3*A + 19*C)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*
x]^(5/2)) + ((7*A + 31*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)) - ((11*A + 63*C
)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])

Rule 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2983

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2982

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e
, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 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 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{7}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (\frac{1}{2} a (A-7 C)+2 a (A+3 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{7}{2}}(c+d x)}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (-\frac{5}{4} a^2 (3 A+19 C)+a^2 (7 A+31 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{7}{2}}(c+d x)}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(7 A+31 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} a^3 (7 A+31 C)-a^3 (11 A+63 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{16 a^5}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{7}{2}}(c+d x)}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(7 A+31 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{(11 A+63 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a^4 (11 A+63 C)+2 a^4 (8 A+39 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{16 a^6}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{7}{2}}(c+d x)}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(7 A+31 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{(11 A+63 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{\left ((8 A+39 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{8 a^3}-\frac{\left ((43 A+219 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{7}{2}}(c+d x)}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(7 A+31 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{(11 A+63 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{\left ((8 A+39 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 a^3 d}+\frac{\left ((43 A+219 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac{(8 A+39 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{4 a^{5/2} d}-\frac{(43 A+219 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{7}{2}}(c+d x)}-\frac{(3 A+19 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(7 A+31 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{(11 A+63 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.20035, size = 968, normalized size = 2.9 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{\sec \left (\frac{c}{2}\right ) \left (-A \sin \left (\frac{d x}{2}\right )-C \sin \left (\frac{d x}{2}\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}-\frac{(A+C) \tan \left (\frac{c}{2}\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{\sec \left (\frac{c}{2}\right ) \left (19 A \sin \left (\frac{d x}{2}\right )+35 C \sin \left (\frac{d x}{2}\right )\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{4 d}+\frac{(19 A+35 C) \tan \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{4 d}-\frac{3 (5 A+3 C) \cos \left (\frac{d x}{2}\right ) \sin \left (\frac{c}{2}\right )}{2 d}-\frac{10 C \cos \left (\frac{3 d x}{2}\right ) \sin \left (\frac{3 c}{2}\right )}{d}+\frac{C \cos \left (\frac{5 d x}{2}\right ) \sin \left (\frac{5 c}{2}\right )}{d}-\frac{3 (5 A+3 C) \cos \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{2 d}-\frac{10 C \cos \left (\frac{3 c}{2}\right ) \sin \left (\frac{3 d x}{2}\right )}{d}+\frac{C \cos \left (\frac{5 c}{2}\right ) \sin \left (\frac{5 d x}{2}\right )}{d}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{(a (\cos (c+d x)+1))^{5/2}}-\frac{11 i A e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{4 d (a (\cos (c+d x)+1))^{5/2}}-\frac{63 i C e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{4 d (a (\cos (c+d x)+1))^{5/2}}+\frac{4 i \sqrt{2} A e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (-\sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt{2} \tanh ^{-1}\left (\frac{-1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (a (\cos (c+d x)+1))^{5/2}}+\frac{39 i C e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (-\sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt{2} \tanh ^{-1}\left (\frac{-1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{\sqrt{2} d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)),x]

[Out]

(((-11*I)/4)*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 - E^(I
*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])]*Cos[c/2 + (d*x)/2]^5)/(d*E^((I/2)*(c + d*x))*(a*(1 + Cos
[c + d*x]))^(5/2)) - (((63*I)/4)*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])]*Cos[c/2 + (d*x)/2]^5)/(d*E^((I/2)*(c
+ d*x))*(a*(1 + Cos[c + d*x]))^(5/2)) + ((4*I)*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt
[1 + E^((2*I)*(c + d*x))]*(-ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1
+ E^((2*I)*(c + d*x))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Cos[c/2 + (d*x)/2]^5)/(d*E^((I/2)*(c + d*x)
)*(a*(1 + Cos[c + d*x]))^(5/2)) + ((39*I)*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*
(c + d*x))]*(-ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c
+ d*x))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Cos[c/2 + (d*x)/2]^5)/(Sqrt[2]*d*E^((I/2)*(c + d*x))*(a*(
1 + Cos[c + d*x]))^(5/2)) + (Cos[c/2 + (d*x)/2]^5*Sqrt[Sec[c + d*x]]*((-3*(5*A + 3*C)*Cos[(d*x)/2]*Sin[c/2])/(
2*d) - (10*C*Cos[(3*d*x)/2]*Sin[(3*c)/2])/d + (C*Cos[(5*d*x)/2]*Sin[(5*c)/2])/d - (3*(5*A + 3*C)*Cos[c/2]*Sin[
(d*x)/2])/(2*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^4*(-(A*Sin[(d*x)/2]) - C*Sin[(d*x)/2]))/(2*d) + (Sec[c/2]*Sec[c
/2 + (d*x)/2]^2*(19*A*Sin[(d*x)/2] + 35*C*Sin[(d*x)/2]))/(4*d) - (10*C*Cos[(3*c)/2]*Sin[(3*d*x)/2])/d + (C*Cos
[(5*c)/2]*Sin[(5*d*x)/2])/d + ((19*A + 35*C)*Sec[c/2 + (d*x)/2]*Tan[c/2])/(4*d) - ((A + C)*Sec[c/2 + (d*x)/2]^
3*Tan[c/2])/(2*d)))/(a*(1 + Cos[c + d*x]))^(5/2)

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Maple [B]  time = 0.211, size = 642, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(5/2),x)

[Out]

-1/32/d*2^(1/2)/a^3*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^5*(-8*C*cos(d*x+c)^4*2^(1/2)*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)+28*C*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+15*A*2^(1/2)*(cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+32*A*sin(d*x+c)*cos(d*x+c)*2^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)/cos(d*x+c))+75*C*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+156*C*2^(1/2)*arctan(sin(d
*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)*sin(d*x+c)-4*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*cos(d*x+c)+32*A*2^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)+4
3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)-32*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*co
s(d*x+c)+156*C*2^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)+219*C*arcsin
((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)-11*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+43*A*arcsin(
(-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-63*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+219*C*arcsin((-1+cos(d*x
+c))/sin(d*x+c))*sin(d*x+c))/(1/cos(d*x+c))^(5/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/sin(d*x+c)^11

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(5/2)), x)

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Fricas [A]  time = 132.385, size = 886, normalized size = 2.65 \begin{align*} \frac{\sqrt{2}{\left ({\left (43 \, A + 219 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (43 \, A + 219 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (43 \, A + 219 \, C\right )} \cos \left (d x + c\right ) + 43 \, A + 219 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) - 8 \,{\left ({\left (8 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (8 \, A + 39 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 39 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) + \frac{2 \,{\left (8 \, C \cos \left (d x + c\right )^{4} - 20 \, C \cos \left (d x + c\right )^{3} - 5 \,{\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (11 \, A + 63 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*((43*A + 219*C)*cos(d*x + c)^3 + 3*(43*A + 219*C)*cos(d*x + c)^2 + 3*(43*A + 219*C)*cos(d*x + c)
+ 43*A + 219*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) -
8*((8*A + 39*C)*cos(d*x + c)^3 + 3*(8*A + 39*C)*cos(d*x + c)^2 + 3*(8*A + 39*C)*cos(d*x + c) + 8*A + 39*C)*sqr
t(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*(8*C*cos(d*x + c)^4 - 20*C
*cos(d*x + c)^3 - 5*(3*A + 19*C)*cos(d*x + c)^2 - (11*A + 63*C)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x
+ c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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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((A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(5/2)/sec(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(5/2)), x)