3.1359 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=352 \[ \frac {(8 A-20 B+39 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(43 A-115 B+219 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(7 A-15 B+31 C) \sin (c+d x)}{16 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {(11 A-35 B+63 C) \sin (c+d x)}{16 a^2 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(3 A-11 B+19 C) \sin (c+d x)}{16 a d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(7/2)-1/16*(3*A-11*B+19*C)*sin(d*x+c)/a/d/(a+a*cos
(d*x+c))^(3/2)/sec(d*x+c)^(5/2)+1/16*(7*A-15*B+31*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)-
1/16*(11*A-35*B+63*C)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+1/4*(8*A-20*B+39*C)*arcsin(sin(
d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d-1/32*(43*A-115*B+219*C)*arc
tan(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/
a^(5/2)/d*2^(1/2)

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Rubi [A]  time = 1.27, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4221, 3041, 2977, 2983, 2982, 2782, 205, 2774, 216} \[ \frac {(7 A-15 B+31 C) \sin (c+d x)}{16 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {(8 A-20 B+39 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(11 A-35 B+63 C) \sin (c+d x)}{16 a^2 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(43 A-115 B+219 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(3 A-11 B+19 C) \sin (c+d x)}{16 a d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*A - 20*B + 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 - 115*B + 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 - B + C)*Sin[c + d*x]
)/(4*d*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(7/2)) - ((3*A - 11*B + 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 - 15*B + 31*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]*Se
c[c + d*x]^(3/2)) - ((11*A - 35*B + 63*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*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 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]

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 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 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 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 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 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)]

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]

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)+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+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac {(A-B+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 B-7 C)+2 a (A-B+3 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B+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-11 B+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-11 B+19 C)+a^2 (7 A-15 B+31 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B+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-11 B+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-15 B+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-15 B+31 C)-a^3 (11 A-35 B+63 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{16 a^5}\\ &=-\frac {(A-B+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-11 B+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-15 B+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-35 B+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-35 B+63 C)+2 a^4 (8 A-20 B+39 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{16 a^6}\\ &=-\frac {(A-B+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-11 B+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-15 B+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-35 B+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-20 B+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-115 B+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-B+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-11 B+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-15 B+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-35 B+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-20 B+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-115 B+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-20 B+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-115 B+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-B+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-11 B+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-15 B+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-35 B+63 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 28.49, size = 17727, normalized size = 50.36 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.54, size = 924, normalized size = 2.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+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*(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)-16*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3+28*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*cos(d*x+c)^3+15*A*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+32*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)*cos(d*x+c)-39*B*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)-80*B*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)/c
os(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*cos(d*x+c)*arctan(sin(d*x+c)*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)*2^(1/2)+43*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c
)*cos(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)-115*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)
+20*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-80*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)+219*C*sin(d*x+c)*cos(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-32*C*2^(1/
2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+156*C*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)+43*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-11*A*2^(1/2)*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)-115*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+35*B*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)-63*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/(1/cos
(d*x+c))^(5/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/sin(d*x+c)^11*2^(1/2)/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/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)/((1/cos(c + d*x))^(5/2)*(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)/(a+a*cos(d*x+c))**(5/2)/sec(d*x+c)**(5/2),x)

[Out]

Timed out

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