3.1233 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^{\frac{7}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx\)
Optimal. Leaf size=201 \[ \frac{2 (13 A+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}-\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}} \]
[Out]
-((Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[C
os[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (2*(13*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a
+ a*Cos[c + d*x]]) - (2*A*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.640555, antiderivative size = 201, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used =
{4221, 3044, 2984, 12, 2782, 205} \[ \frac{2 (13 A+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}-\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/Sqrt[a + a*Cos[c + d*x]],x]
[Out]
-((Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[C
os[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (2*(13*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a
+ a*Cos[c + d*x]]) - (2*A*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]])
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]
Rule 3044
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[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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]
Rubi steps
\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{a A}{2}+\frac{1}{2} a (4 A+5 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=-\frac{2 A \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^2 (13 A+15 C)-\frac{1}{2} a^2 A \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 (13 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int -\frac{15 a^3 (A+C)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^3}\\ &=\frac{2 (13 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\left ((A+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\\ &=\frac{2 (13 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (2 a (A+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 )}{d}\\ &=-\frac{\sqrt{2} (A+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)}}{\sqrt{a} d}+\frac{2 (13 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [C] time = 7.98825, size = 1757, normalized size = 8.74 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/Sqrt[a + a*Cos[c + d*x]],x]
[Out]
(2*Cos[c/2 + (d*x)/2]*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(-(C*Sin[c/2 +
(d*x)/2])/(2*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) - ((A + C)*Csc[c/2 + (d*x)/2]^7*(4725*Sin[c/2 + (d*x)/2]^2 -
48825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8 + 655812*Sin[c/2 + (d*x
)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d
*x)/2]^10 - 40*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {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]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2,
11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 226656*Sin[c/2 + (d*x)/2]^14
- 1500*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]
^14 - 42048*Sin[c/2 + (d*x)/2]^16 + 440*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 4725*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*S
qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[
c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 291060*Arc
Tanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*Sqrt[Sin[c/2 + (d*x)/2]^2/
(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[
c/2 + (d*x)/2]^6*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1458000*ArcTanh[Sqrt[Sin[c/2 + (d*
x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^8*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)
/2]^2)] - 1598400*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^10*Sqrt
[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1080000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c
/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^12*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 414720*Arc
Tanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*Sqrt[Sin[c/2 + (d*x)/2]^2
/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 69120*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[
c/2 + (d*x)/2]^16*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 60*Cos[(c + d*x)/2]^4*Hypergeomet
ricPFQ[{2, 2, 9/2}, {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]^10*(-5 +
4*Sin[c/2 + (d*x)/2]^2)))/(675*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) + (C*((3*Sin
[c/2 + (d*x)/2])/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2) + 4*(Sin[c/2 + (d*x)/2]/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)
+ (2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])))/30))/(d*Sqrt[a*(1 + Cos[c + d*x])])
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Maple [B] time = 0.203, size = 503, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x)
[Out]
1/15/d*2^(1/2)/a*(15*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3+15*C*
(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3+45*A*(cos(d*x+c)/(1+cos(d*x+
c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+45*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+c
os(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+45*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*
cos(d*x+c)+45*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)+15*A*(cos(d*x+
c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+15*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+
cos(d*x+c))/sin(d*x+c))+13*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+15*C*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-A*2^(1/2)*co
s(d*x+c)*sin(d*x+c)+3*A*2^(1/2)*sin(d*x+c))*cos(d*x+c)*(1/cos(d*x+c))^(7/2)*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)^4/(-1+cos(d*x+c))^2/(1+cos(d*x+c))^3
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.90142, size = 433, normalized size = 2.15 \begin{align*} \frac{\frac{15 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{3} +{\left (A + C\right )} a \cos \left (d x + c\right )^{2}\right )} \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 )}{\sqrt{a}} + \frac{2 \,{\left ({\left (13 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/15*(15*sqrt(2)*((A + C)*a*cos(d*x + c)^3 + (A + C)*a*cos(d*x + c)^2)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)
*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*((13*A + 15*C)*cos(d*x + c)^2 - A*cos(d*x + c) + 3*A)*
sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2)/(a+a*cos(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac{7}{2}}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(7/2)/sqrt(a*cos(d*x + c) + a), x)