3.1358 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=294 \[ \frac {(3 A-43 B+115 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(5/2)+1/16*(A+7*B-15*C)*sin(d*x+c)/a/d/(a+a*cos(d*
x+c))^(3/2)/sec(d*x+c)^(3/2)+1/16*(3*A-11*B+35*C)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+(2*
B-5*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*(3*A
-43*B+115*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))*cos(d*x+c)^(1/2)*s
ec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)

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Rubi [A]  time = 1.04, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {(3 A-43 B+115 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

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]^(3/2)),x]

[Out]

((2*B - 5*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]])/(a
^(5/2)*d) + ((3*A - 43*B + 115*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]^(5/2)) + ((A + 7*B - 15*C)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)
*Sec[c + d*x]^(3/2)) + ((3*A - 11*B + 35*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 {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{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 {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 {1}{2} a (3 A+5 B-5 C)+a (A-B+5 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 {5}{2}}(c+d x)}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a 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}{4} a^2 (A+7 B-15 C)+\frac {1}{2} a^2 (3 A-11 B+35 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 {5}{2}}(c+d x)}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B+35 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}{4} a^3 (3 A-11 B+35 C)+4 a^3 (2 B-5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac {(A-B+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {\left ((2 B-5 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}{2 a^3}+\frac {\left ((3 A-43 B+115 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 {5}{2}}(c+d x)}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left ((2 B-5 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 )}{a^3 d}-\frac {\left ((3 A-43 B+115 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 {(2 B-5 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)}}{a^{5/2} d}+\frac {(3 A-43 B+115 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 {5}{2}}(c+d x)}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B+35 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 = 28.20, size = 16906, normalized size = 57.50 \[ \text {Result too large to show} \]

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]^(3/2)),x]

[Out]

Result too large to show

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fricas [A]  time = 84.24, size = 329, normalized size = 1.12 \[ -\frac {\sqrt {2} {\left ({\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, B + 115 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 32 \, {\left ({\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 2 \, B - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (16 \, C \cos \left (d x + c\right )^{3} + {\left (7 \, A - 15 \, B + 55 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 11 \, B + 35 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(sqrt(2)*((3*A - 43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^2 + 3*(3*A - 43*B +
115*C)*cos(d*x + c) + 3*A - 43*B + 115*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(
sqrt(a)*sin(d*x + c))) + 32*((2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)*cos(d*x + c)^2 + 3*(2*B - 5*C)*cos(d*x
+ c) + 2*B - 5*C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*(16*
C*cos(d*x + c)^3 + (7*A - 15*B + 55*C)*cos(d*x + c)^2 + (3*A - 11*B + 35*C)*cos(d*x + c))*sqrt(a*cos(d*x + c)
+ a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) +
a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.60, size = 758, normalized size = 2.58 \[ \text {result too large to display} \]

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)^(3/2),x)

[Out]

-1/32/d*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^4*(16*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
cos(d*x+c)^3+7*A*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-15*B*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)-32*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))+39*C*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+80*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)-4*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(
d*x+c)+3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)+4*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*cos(d*x+c)-43*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)-32*B*2^(1/2)*arctan(sin(d*x+c)*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)-20*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+
c)+115*C*sin(d*x+c)*cos(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+80*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)-3*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*A*arcsin((-1+cos(d
*x+c))/sin(d*x+c))*sin(d*x+c)+11*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-43*B*arcsin((-1+cos(d*x+c))/sin(d
*x+c))*sin(d*x+c)-35*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+115*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(
d*x+c))/(1/cos(d*x+c))^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/sin(d*x+c)^9*2^(1/2)/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/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)/((1/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)/(a+a*cos(d*x+c))**(5/2)/sec(d*x+c)**(3/2),x)

[Out]

Timed out

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