3.549 \(\int \frac{A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=333 \[ \frac{(79 A-259 B) \sin (c+d x)}{192 a^2 d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}+\frac{(2 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}-\frac{7 (7 A-27 B) \sin (c+d x)}{64 a^3 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(177 A-637 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}+\frac{(A-B) \sin (c+d x)}{6 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

[Out]

((2*A - 7*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(a
^(7/2)*d) - ((177*A - 637*B)*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]])/(64*Sqrt[2]*a^(7/2)*d) + ((A - B)*Sin[c + d*x])/(6*d*(a + a*Cos[c +
d*x])^(7/2)*Sec[c + d*x]^(7/2)) + ((3*A - 7*B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^
(5/2)) + ((79*A - 259*B)*Sin[c + d*x])/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)) - (7*(7*A - 2
7*B)*Sin[c + d*x])/(64*a^3*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.20498, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {2961, 2977, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(79 A-259 B) \sin (c+d x)}{192 a^2 d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}+\frac{(2 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}-\frac{7 (7 A-27 B) \sin (c+d x)}{64 a^3 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(177 A-637 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}+\frac{(A-B) \sin (c+d x)}{6 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(7/2)),x]

[Out]

((2*A - 7*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(a
^(7/2)*d) - ((177*A - 637*B)*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]])/(64*Sqrt[2]*a^(7/2)*d) + ((A - B)*Sin[c + d*x])/(6*d*(a + a*Cos[c +
d*x])^(7/2)*Sec[c + d*x]^(7/2)) + ((3*A - 7*B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^
(5/2)) + ((79*A - 259*B)*Sin[c + d*x])/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)) - (7*(7*A - 2
7*B)*Sin[c + d*x])/(64*a^3*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])

Rule 2961

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{7}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \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{7}{2} a (A-B)-a (A-7 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \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{15}{4} a^2 (3 A-7 B)-\frac{1}{2} a^2 (17 A-77 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(79 A-259 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \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}{8} a^3 (79 A-259 B)-\frac{21}{4} a^3 (7 A-27 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(79 A-259 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x)}-\frac{7 (7 A-27 B) \sin (c+d x)}{64 a^3 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{21}{8} a^4 (7 A-27 B)+24 a^4 (2 A-7 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^7}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(79 A-259 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x)}-\frac{7 (7 A-27 B) \sin (c+d x)}{64 a^3 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{\left ((177 A-637 B) \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}{128 a^3}+\frac{\left ((2 A-7 B) \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}{2 a^4}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(79 A-259 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x)}-\frac{7 (7 A-27 B) \sin (c+d x)}{64 a^3 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{\left ((177 A-637 B) \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 )}{64 a^2 d}-\frac{\left ((2 A-7 B) \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 )}{a^4 d}\\ &=\frac{(2 A-7 B) \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)}}{a^{7/2} d}-\frac{(177 A-637 B) \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)}}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{7}{2}}(c+d x)}+\frac{(3 A-7 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(79 A-259 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac{3}{2}}(c+d x)}-\frac{7 (7 A-27 B) \sin (c+d x)}{64 a^3 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.5768, size = 1017, normalized size = 3.05 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{\sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )-B \sin \left (\frac{d x}{2}\right )\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{(A-B) \tan \left (\frac{c}{2}\right ) \sec ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{\sec \left (\frac{c}{2}\right ) \left (53 B \sin \left (\frac{d x}{2}\right )-41 A \sin \left (\frac{d x}{2}\right )\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{12 d}-\frac{(41 A-53 B) \tan \left (\frac{c}{2}\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{12 d}+\frac{\sec \left (\frac{c}{2}\right ) \left (379 A \sin \left (\frac{d x}{2}\right )-703 B \sin \left (\frac{d x}{2}\right )\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{24 d}+\frac{(379 A-703 B) \tan \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{24 d}+\frac{(427 B-247 A) \cos \left (\frac{d x}{2}\right ) \sin \left (\frac{c}{2}\right )}{12 d}+\frac{8 B \cos \left (\frac{3 d x}{2}\right ) \sin \left (\frac{3 c}{2}\right )}{d}-\frac{(247 A-427 B) \cos \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{12 d}+\frac{8 B \cos \left (\frac{3 c}{2}\right ) \sin \left (\frac{3 d x}{2}\right )}{d}\right ) \cos ^7\left (\frac{c}{2}+\frac{d x}{2}\right )}{(a (\cos (c+d x)+1))^{7/2}}-\frac{49 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 ^7\left (\frac{c}{2}+\frac{d x}{2}\right )}{8 d (a (\cos (c+d x)+1))^{7/2}}+\frac{189 i B 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 ^7\left (\frac{c}{2}+\frac{d x}{2}\right )}{8 d (a (\cos (c+d x)+1))^{7/2}}+\frac{8 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 ^7\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (a (\cos (c+d x)+1))^{7/2}}-\frac{28 i \sqrt{2} B 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 ^7\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (a (\cos (c+d x)+1))^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(7/2)),x]

[Out]

(((-49*I)/8)*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]^7)/(d*E^((I/2)*(c + d*x))*(a*(1 + Cos
[c + d*x]))^(7/2)) + (((189*I)/8)*B*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]^7)/(d*E^((I/2)*(
c + d*x))*(a*(1 + Cos[c + d*x]))^(7/2)) + ((8*I)*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqr
t[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]^7)/(d*E^((I/2)*(c + d*x
))*(a*(1 + Cos[c + d*x]))^(7/2)) - ((28*I)*Sqrt[2]*B*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]^7)/(d*E^((I/2)*(c + d*x))*(a*
(1 + Cos[c + d*x]))^(7/2)) + (Cos[c/2 + (d*x)/2]^7*Sqrt[Sec[c + d*x]]*(((-247*A + 427*B)*Cos[(d*x)/2]*Sin[c/2]
)/(12*d) + (8*B*Cos[(3*d*x)/2]*Sin[(3*c)/2])/d - ((247*A - 427*B)*Cos[c/2]*Sin[(d*x)/2])/(12*d) + (Sec[c/2]*Se
c[c/2 + (d*x)/2]^2*(379*A*Sin[(d*x)/2] - 703*B*Sin[(d*x)/2]))/(24*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^6*(A*Sin[(
d*x)/2] - B*Sin[(d*x)/2]))/(3*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^4*(-41*A*Sin[(d*x)/2] + 53*B*Sin[(d*x)/2]))/(1
2*d) + (8*B*Cos[(3*c)/2]*Sin[(3*d*x)/2])/d + ((379*A - 703*B)*Sec[c/2 + (d*x)/2]*Tan[c/2])/(24*d) - ((41*A - 5
3*B)*Sec[c/2 + (d*x)/2]^3*Tan[c/2])/(12*d) + ((A - B)*Sec[c/2 + (d*x)/2]^5*Tan[c/2])/(3*d)))/(a*(1 + Cos[c + d
*x]))^(7/2)

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Maple [B]  time = 0.71, size = 855, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(7/2)/sec(d*x+c)^(7/2),x)

[Out]

-1/384/d*2^(1/2)/a^4*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^7*cos(d*x+c)*(-192*B*cos(d*x+c)^4*2^(1/2)*(cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)+384*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)/cos(d*x+c))+247*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3-1344*B*cos(d*x+c)^2*sin(d*x+c
)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*2^(1/2)-907*B*cos(d*x+c)^3*2^(1/2)*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)+768*A*cos(d*x+c)*2^(1/2)*sin(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)/cos(d*x+c))+115*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+531*A*arcsin((-1+cos(d*x+c))/sin(d
*x+c))*cos(d*x+c)^2*sin(d*x+c)-2688*B*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)/cos(d*x+c))-343*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-1911*B*arcsin((-1+cos(d*x+c
))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+384*A*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*2
^(1/2)*sin(d*x+c)-215*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+1062*A*arcsin((-1+cos(d*x+c))/sin
(d*x+c))*sin(d*x+c)*cos(d*x+c)-1344*B*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)+875*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-3822*B*arcsin((-1+cos(d*x+c))/sin(d*x+c)
)*sin(d*x+c)*cos(d*x+c)-147*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+531*A*arcsin((-1+cos(d*x+c))/sin(d*x+c
))*sin(d*x+c)+567*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-1911*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*
x+c))/(1/cos(d*x+c))^(7/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)/sin(d*x+c)^15

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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((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 172.239, size = 1041, normalized size = 3.13 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right ) + 177 \, A - 637 \, B\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 ) - 384 \,{\left ({\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 7 \, B\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 (192 \, B \cos \left (d x + c\right )^{4} -{\left (247 \, A - 1099 \, B\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (181 \, A - 721 \, B\right )} \cos \left (d x + c\right )^{2} - 21 \,{\left (7 \, A - 27 \, B\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 )}}}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/384*(3*sqrt(2)*((177*A - 637*B)*cos(d*x + c)^4 + 4*(177*A - 637*B)*cos(d*x + c)^3 + 6*(177*A - 637*B)*cos(d*
x + c)^2 + 4*(177*A - 637*B)*cos(d*x + c) + 177*A - 637*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqr
t(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 384*((2*A - 7*B)*cos(d*x + c)^4 + 4*(2*A - 7*B)*cos(d*x + c)^3 + 6*(
2*A - 7*B)*cos(d*x + c)^2 + 4*(2*A - 7*B)*cos(d*x + c) + 2*A - 7*B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sq
rt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*(192*B*cos(d*x + c)^4 - (247*A - 1099*B)*cos(d*x + c)^3 - 2*(181*
A - 721*B)*cos(d*x + c)^2 - 21*(7*A - 27*B)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x +
c)))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*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+B*cos(d*x+c))/(a+a*cos(d*x+c))**(7/2)/sec(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{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(7/2)*sec(d*x + c)^(7/2)), x)