\(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\) [214]

Optimal result

Integrand size = 37, antiderivative size = 199 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=-\frac {5 (15 A-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

-5/32*(15*A-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
)-1/4*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2)-1/16*(13*A-3*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c
))^(3/2)/cos(d*x+c)^(1/2)+1/16*(49*A+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 = 0.67 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3121, 3057, 3063, 12, 2861, 211} \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=-\frac {5 (15 A-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[In]

Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(-5*(15*A - 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 + C)*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) - ((13*A - 3*C)*Sin
[c + d*x])/(16*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((49*A + C)*Sin[c + d*x])/(16*a^2*d*Sqrt[C
os[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 3121

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)]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\frac {1}{2} a (9 A+C)-2 a (A-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{4} a^2 (49 A+C)-\frac {1}{2} a^2 (13 A-3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {\int -\frac {5 a^3 (15 A-C)}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a^5} \\ & = -\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {(5 (15 A-C)) \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 \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {(5 (15 A-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 {5 (15 A-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+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 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 = 7.13 (sec) , antiderivative size = 726, normalized size of antiderivative = 3.65 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{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 ) \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \left (3-\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-5 \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 ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\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}}+\frac {2 A \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}} \]

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(C*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(3 - Sin[c/2 +
(d*x)/2]^2 - 5*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4*Csc[c/2
+ (d*x)/2]^2*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)) +
(2*A*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*HypergeometricPFQ[{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)))/120))/(d*(a*(1 +
Cos[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(402\) vs. \(2(168)=336\).

Time = 11.91 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.03

method	result	size
default	\(\frac {\left (75 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )-5 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )+150 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+98 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-10 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+2 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+170 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-5 C \cos \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+10 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+64 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 a^{3} d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\cos \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(403\)
parts	\(\frac {A \left (75 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+49 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+225 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+85 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}+225 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+32 \sqrt {2}\, \sin \left (d x +c \right )+75 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\cos \left (d x +c \right )}\, a^{3}}+\frac {C \left (\sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+5 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-10 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}}\)	\(457\)

[In]

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+cos(d*x+c)*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/32/a^3/d*(75*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^3-5*C*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*
cos(d*x+c)^3+150*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2+98*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)-10*C*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2+2*C*cos(d*x+c)^2*sin(d*x+c)*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)+75*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)+170*A*cos(d*x+c)*sin(d*x+
c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-5*C*cos(d*x+c)*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+10*C*cos(d*x+c)*sin(
d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+64*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))*(a*(1+cos(d*x+c)))
^(1/2)/(1+cos(d*x+c))^3/cos(d*x+c)^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.24 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=-\frac {5 \, \sqrt {2} {\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + {\left (15 \, A - C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (49 \, A + C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 32 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(5*sqrt(2)*((15*A - C)*cos(d*x + c)^4 + 3*(15*A - C)*cos(d*x + c)^3 + 3*(15*A - C)*cos(d*x + c)^2 + (15*
A - C)*cos(d*x + c))*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*((49*A + C)*cos(d*x + c)^2 + 5*(17*A + C)*cos(d*x + c) + 32*A)*sqr
t(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*
d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(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)*cos(d*x + c)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a*cos(c + d*x))^(5/2)),x)

[Out]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a*cos(c + d*x))^(5/2)), x)