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

Optimal. Leaf size=211 \[ -\frac {(75 A-19 B-5 C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/32*(75*A-19*B-5*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-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2)-1/16*(13*A-5*B-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-9*B+C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1
/2)

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Rubi [A]  time = 0.64, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3041, 2978, 2984, 12, 2782, 205} \[ \frac {(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(75 A-19 B-5 C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((75*A - 19*B - 5*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(1
6*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) - ((13*A
- 5*B - 3*C)*Sin[c + d*x])/(16*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((49*A - 9*B + C)*Sin[c +
d*x])/(16*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 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 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 2978

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 2984

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 3041

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*} \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {(A-B+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-B+C)-2 a (A-B-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-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-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-9 B+C)-\frac {1}{2} a^2 (13 A-5 B-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-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-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-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {\int -\frac {a^3 (75 A-19 B-5 C)}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a^5}\\ &=-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-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-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {(75 A-19 B-5 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 \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-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-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {(75 A-19 B-5 C) \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 {(75 A-19 B-5 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 )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-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-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 2.44, size = 225, normalized size = 1.07 \[ \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) (2 (85 A-13 B+5 C) \cos (c+d x)+(49 A-9 B+C) \cos (2 (c+d x))+113 A-9 B+C)}{4 \sqrt {\cos (c+d x)}}-\frac {i (75 A-19 B-5 C) e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{4 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(((-I)*(75*A - 19*B - 5*C)*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*
x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[1 + E^((2*I)*(c + d*x))] + (
(113*A - 9*B + C + 2*(85*A - 13*B + 5*C)*Cos[c + d*x] + (49*A - 9*B + C)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*
Tan[(c + d*x)/2])/(4*Sqrt[Cos[c + d*x]])))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2))

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fricas [A]  time = 0.96, size = 264, normalized size = 1.25 \[ -\frac {\sqrt {2} {\left ({\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (49 \, A - 9 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (85 \, A - 13 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 32 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+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*(sqrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^4 + 3*(75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5
*C)*cos(d*x + c)^2 + (75*A - 19*B - 5*C)*cos(d*x + c))*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqr
t(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*((49*A - 9*B + C)*cos(d*x + c)^2
+ (85*A - 13*B + 5*C)*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*c
os(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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.37, size = 675, normalized size = 3.20 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right )^{2} \left (98 A \left (\cos ^{5}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+268 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{4}\left (d x +c \right )\right )+136 A \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-18 B \left (\cos ^{5}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}-75 A \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-204 A \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+19 B \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-26 B \left (\cos ^{4}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+5 C \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-75 A \sqrt {2}\, \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right )-234 A \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+19 B \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+18 B \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+2 C \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+5 C \sqrt {2}\, \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right )-64 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+26 B \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+8 C \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-10 C \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 d \cos \left (d x +c \right )^{\frac {5}{2}} \sin \left (d x +c \right )^{5} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/d*(-1+cos(d*x+c))^2*(98*A*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+268*A*(cos(d*x+c)/(1+cos(d*x+c)
))^(5/2)*cos(d*x+c)^4+136*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-18*B*cos(d*x+c)^5*(cos(d*x+c)/(1+co
s(d*x+c)))^(3/2)-75*A*cos(d*x+c)^4*2^(1/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-204*A*cos(d*x+c)^2*(c
os(d*x+c)/(1+cos(d*x+c)))^(5/2)+19*B*cos(d*x+c)^4*2^(1/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-26*B*c
os(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+5*C*cos(d*x+c)^4*2^(1/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d
*x+c))-75*A*2^(1/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3-234*A*cos(d*x+c)*(cos(d*x+c)/(1
+cos(d*x+c)))^(5/2)+19*B*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+18*B*cos(d*x+c)^3*
(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+2*C*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+5*C*2^(1/2)*sin(d*x+c)*ar
csin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3-64*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+26*B*cos(d*x+c)^2*(cos(d*
x+c)/(1+cos(d*x+c)))^(3/2)+8*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-10*C*cos(d*x+c)^3*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2))*(a*(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(5/2)/sin(d*x+c)^5/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/
a^3

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maxima [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((A+B*cos(d*x+c)+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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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