\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\) [521]

Optimal result

Integrand size = 45, antiderivative size = 261 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {(163 A-75 B+19 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/4*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2)-1/16*(17*A-9*B+C)*sin(d*x+c)/a/d/cos(d*x+c)^
(3/2)/(a+a*cos(d*x+c))^(3/2)+1/32*(163*A-75*B+19*C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+
a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+1/48*(95*A-39*B+15*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c)
)^(1/2)-1/48*(299*A-147*B+27*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3120, 3057, 3063, 12, 2861, 211} \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {(163 A-75 B+19 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

((163*A - 75*B + 19*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(
16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)) - ((17*
A - 9*B + C)*Sin[c + d*x])/(16*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((95*A - 39*B + 15*C)*Sin[
c + d*x])/(48*a^2*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((299*A - 147*B + 27*C)*Sin[c + d*x])/(48*a
^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 2861

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 3057

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3063

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*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] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3120

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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)]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\frac {1}{2} a (11 A-3 B+3 C)-a (3 A-3 B-C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{4} a^2 (95 A-39 B+15 C)-a^2 (17 A-9 B+C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {-\frac {1}{8} a^3 (299 A-147 B+27 C)+\frac {1}{4} a^3 (95 A-39 B+15 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{12 a^5} \\ & = -\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {3 a^4 (163 A-75 B+19 C)}{16 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{6 a^6} \\ & = -\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {(163 A-75 B+19 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {(163 A-75 B+19 C) \text {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 {(163 A-75 B+19 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 9.18 (sec) , antiderivative size = 1368, normalized size of antiderivative = 5.24 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=-\frac {C \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (11-31 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+18 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-\frac {19 \text {arctanh}\left (\sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )}{\sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}}\right )}{4 d (a (1+\cos (c+d x)))^{5/2} \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+\frac {2 B \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {8 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (2,2,2,\frac {5}{2};1,1,\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{315 \left (-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {1}{120} \csc ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2 \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (-15 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-343+1465 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-2021 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+824 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (-5145+33980 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-87764 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+109737 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-66122 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+15344 \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{d (a (1+\cos (c+d x)))^{5/2} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{3/2}}-\frac {A \cot ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (640 \cos ^8\left (\frac {1}{2} (c+d x)\right ) \, _5F_4\left (2,2,2,2,\frac {7}{2};1,1,1,\frac {13}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right )-1280 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (2,2,2,\frac {7}{2};1,1,\frac {13}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-6+5 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+33 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3 \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (-105 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-10935+72902 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-188110 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+234156 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-140732 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+33208 \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (-1148175+10333785 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-38990350 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )[phone] \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-96281836 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )[phone] \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-26448512 \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right )+4344400 \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{41580 d (a (1+\cos (c+d x)))^{5/2} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2}} \]

[In]

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

[Out]

-1/4*(C*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*(11 - 31*Sin[c/2 + (d*x)/2]^2 + 18*Sin[c/2
+ (d*x)/2]^4 - (19*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4)/Sqr
t[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(d*(a*(1 + Cos[c + d*x]))^(5/2)*Sqrt[1 - 2*Sin[c/2 +
(d*x)/2]^2]) + (2*B*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*((8*Cos[(c + d*x)/2]^6*Hyperge
ometricPFQ[{2, 2, 2, 5/2}, {1, 1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2
]^2)/(315*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) + (Csc[c/2 + (d*x)/2]^8*(1 - 2*Sin[c/2 + (d*x)/2]^2)^2*Sqrt[Sin[c/2 +
(d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-15*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2
)]]*Cos[(c + d*x)/2]^4*(-343 + 1465*Sin[c/2 + (d*x)/2]^2 - 2021*Sin[c/2 + (d*x)/2]^4 + 824*Sin[c/2 + (d*x)/2]^
6) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-5145 + 33980*Sin[c/2 + (d*x)/2]^2 - 87764*Sin[
c/2 + (d*x)/2]^4 + 109737*Sin[c/2 + (d*x)/2]^6 - 66122*Sin[c/2 + (d*x)/2]^8 + 15344*Sin[c/2 + (d*x)/2]^10)))/1
20))/(d*(a*(1 + Cos[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - (A*Cot[c/2 + (d*x)/2]^5*Csc[c/2 + (
d*x)/2]^4*Sec[(c + d*x)/2]^4*(640*Cos[(c + d*x)/2]^8*HypergeometricPFQ[{2, 2, 2, 2, 7/2}, {1, 1, 1, 13/2}, Sin
[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 - 1280*Cos[(c + d*x)/2]^6*Hypergeometri
cPFQ[{2, 2, 2, 7/2}, {1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(
-6 + 5*Sin[c/2 + (d*x)/2]^2) + 33*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (
d*x)/2]^2)]*(-105*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-10935
+ 72902*Sin[c/2 + (d*x)/2]^2 - 188110*Sin[c/2 + (d*x)/2]^4 + 234156*Sin[c/2 + (d*x)/2]^6 - 140732*Sin[c/2 + (
d*x)/2]^8 + 33208*Sin[c/2 + (d*x)/2]^10) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-1148175
+ 10333785*Sin[c/2 + (d*x)/2]^2 - 38990350*Sin[c/2 + (d*x)/2]^4 + 79946462*Sin[c/2 + (d*x)/2]^6 - 96281836*Sin
[c/2 + (d*x)/2]^8 + 68243596*Sin[c/2 + (d*x)/2]^10 - 26448512*Sin[c/2 + (d*x)/2]^12 + 4344400*Sin[c/2 + (d*x)/
2]^14))))/(41580*d*(a*(1 + Cos[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(762\) vs. \(2(224)=448\).

Time = 13.63 (sec) , antiderivative size = 763, normalized size of antiderivative = 2.92

method	result	size
default	\(\text {Expression too large to display}\)	\(763\)
parts	\(\text {Expression too large to display}\)	\(778\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/96/a^3/d*(489*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^4-225*B*
2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^4+57*2^(1/2)*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*C*cos(d*x+c)^4*arcsin(cot(d*x+c)-csc(d*x+c))+1467*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-675*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*ar
csin(cot(d*x+c)-csc(d*x+c))+171*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*arcsin(cot(d*x+c)-csc
(d*x+c))+1467*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)^2*arcsin(cot(d*x+c)-csc(d*x+c))-675*B*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)^2*arcsin(cot(d*x+c)-csc(d*x+c))+171*C*cos(d*x+c)^2*2^(1/2)*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+489*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x
+c)*arcsin(cot(d*x+c)-csc(d*x+c))*2^(1/2)+598*A*cos(d*x+c)^3*sin(d*x+c)-225*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*cos(d*x+c)*arcsin(cot(d*x+c)-csc(d*x+c))*2^(1/2)-294*B*sin(d*x+c)*cos(d*x+c)^3+57*C*cos(d*x+c)*2^(1/2)*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+54*C*cos(d*x+c)^3*sin(d*x+c)+1006*A*sin(d*x+c)*cos(
d*x+c)^2-510*B*sin(d*x+c)*cos(d*x+c)^2+78*C*cos(d*x+c)^2*sin(d*x+c)+320*A*sin(d*x+c)*cos(d*x+c)-192*B*sin(d*x+
c)*cos(d*x+c)-64*A*sin(d*x+c))*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^3/cos(d*x+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.10 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (299 \, A - 147 \, B + 27 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (503 \, A - 255 \, B + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 32 \, {\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 32 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/96*(3*sqrt(2)*((163*A - 75*B + 19*C)*cos(d*x + c)^5 + 3*(163*A - 75*B + 19*C)*cos(d*x + c)^4 + 3*(163*A - 75
*B + 19*C)*cos(d*x + c)^3 + (163*A - 75*B + 19*C)*cos(d*x + c)^2)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x +
c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*((299*A - 147*B + 27*
C)*cos(d*x + c)^3 + (503*A - 255*B + 39*C)*cos(d*x + c)^2 + 32*(5*A - 3*B)*cos(d*x + c) - 32*A)*sqrt(a*cos(d*x
+ c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x +
c)^3 + a^3*d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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