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

Optimal. Leaf size=212 \[ \frac {(19 A+163 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {5 (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{48 a^3 d}-\frac {(21 A+197 C) \sin (c+d x)}{24 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {(A+17 C) \sin (c+d x) \cos ^2(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

-1/4*(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(A+17*C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos
(d*x+c))^(3/2)+1/32*(19*A+163*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1
/2)-1/24*(21*A+197*C)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)+5/48*(3*A+19*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2
)/a^3/d

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Rubi [A]  time = 0.61, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 2977, 2968, 3023, 2751, 2649, 206} \[ \frac {5 (3 A+19 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{48 a^3 d}-\frac {(21 A+197 C) \sin (c+d x)}{24 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(19 A+163 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {(A+17 C) \sin (c+d x) \cos ^2(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((19*A + 163*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (
(A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((A + 17*C)*Cos[c + d*x]^2*Sin[c + d*x
])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) - ((21*A + 197*C)*Sin[c + d*x])/(24*a^2*d*Sqrt[a + a*Cos[c + d*x]]) + (
5*(3*A + 19*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(48*a^3*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3042

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*} \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (a (A-3 C)+\frac {1}{2} a (3 A+11 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(A+17 C) \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos (c+d x) \left (-a^2 (A+17 C)+\frac {5}{4} a^2 (3 A+19 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(A+17 C) \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {-a^2 (A+17 C) \cos (c+d x)+\frac {5}{4} a^2 (3 A+19 C) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(A+17 C) \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {5 (3 A+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}+\frac {\int \frac {\frac {5}{8} a^3 (3 A+19 C)-\frac {1}{4} a^3 (21 A+197 C) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(A+17 C) \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(21 A+197 C) \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {5 (3 A+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}+\frac {(19 A+163 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(A+17 C) \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(21 A+197 C) \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {5 (3 A+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}-\frac {(19 A+163 C) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {(19 A+163 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(A+17 C) \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(21 A+197 C) \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {5 (3 A+19 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.86, size = 112, normalized size = 0.53 \[ \frac {6 (19 A+163 C) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\tan \left (\frac {1}{2} (c+d x)\right ) ((39 A+479 C) \cos (c+d x)+27 A+80 C \cos (2 (c+d x))-8 C \cos (3 (c+d x))+379 C)}{48 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(19*A + 163*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 - (27*A + 379*C + (39*A + 479*C)*Cos[c + d*x] +
80*C*Cos[2*(c + d*x)] - 8*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(48*a*d*(a*(1 + Cos[c + d*x]))^(3/2))

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fricas [A]  time = 0.42, size = 249, normalized size = 1.17 \[ \frac {3 \, \sqrt {2} {\left ({\left (19 \, A + 163 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 163 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 163 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 163 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (32 \, C \cos \left (d x + c\right )^{3} - 160 \, C \cos \left (d x + c\right )^{2} - {\left (39 \, A + 503 \, C\right )} \cos \left (d x + c\right ) - 27 \, A - 299 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*sqrt(2)*((19*A + 163*C)*cos(d*x + c)^3 + 3*(19*A + 163*C)*cos(d*x + c)^2 + 3*(19*A + 163*C)*cos(d*x +
c) + 19*A + 163*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) -
2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(32*C*cos(d*x + c)^3 - 160*C*cos(d*x + c)^
2 - (39*A + 503*C)*cos(d*x + c) - 27*A - 299*C)*sqrt(a*cos(d*x + c) + a)*sin(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 [A]  time = 4.92, size = 203, normalized size = 0.96 \[ \frac {\frac {{\left ({\left (3 \, {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} + C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{6}} - \frac {\sqrt {2} {\left (7 \, A a^{5} + 23 \, C a^{5}\right )}}{a^{6}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {4 \, \sqrt {2} {\left (15 \, A a^{5} + 167 \, C a^{5}\right )}}{a^{6}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, \sqrt {2} {\left (11 \, A a^{5} + 155 \, C a^{5}\right )}}{a^{6}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {3 \, \sqrt {2} {\left (19 \, A + 163 \, C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(((3*(2*sqrt(2)*(A*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/a^6 - sqrt(2)*(7*A*a^5 + 23*C*a^5)/a^6)*tan(1/2*d*
x + 1/2*c)^2 - 4*sqrt(2)*(15*A*a^5 + 167*C*a^5)/a^6)*tan(1/2*d*x + 1/2*c)^2 - 3*sqrt(2)*(11*A*a^5 + 155*C*a^5)
/a^6)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 3*sqrt(2)*(19*A + 163*C)*log(abs(-sqrt(a)*ta
n(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/a^(5/2))/d

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maple [A]  time = 1.32, size = 362, normalized size = 1.71 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (128 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+57 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +489 C \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -512 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-39 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-87 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+6 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(128*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6
+57*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a+4
89*C*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-51
2*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-39*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/
2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2-87*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^2+6*
A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+6*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/cos(1/2*d
*x+1/2*c)^3/a^(7/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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