3.115 \(\int \frac{(b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac{15}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=131 \[ \frac{b^2 (3 A+4 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{8 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b^2 (3 A+4 C) \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt{\cos (c+d x)}}+\frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{4 d \cos ^{\frac{9}{2}}(c+d x)} \]

[Out]

(b^2*(3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Sqrt[b*Cos[c + d*x]])/(8*d*Sqrt[Cos[c + d*x]]) + (A*b^2*Sqrt[b*Cos[c +
d*x]]*Sin[c + d*x])/(4*d*Cos[c + d*x]^(9/2)) + (b^2*(3*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d*Cos[c
+ d*x]^(5/2))

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Rubi [A]  time = 0.0636691, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {17, 3012, 3768, 3770} \[ \frac{b^2 (3 A+4 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{8 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b^2 (3 A+4 C) \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt{\cos (c+d x)}}+\frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{4 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(15/2),x]

[Out]

(b^2*(3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Sqrt[b*Cos[c + d*x]])/(8*d*Sqrt[Cos[c + d*x]]) + (A*b^2*Sqrt[b*Cos[c +
d*x]]*Sin[c + d*x])/(4*d*Cos[c + d*x]^(9/2)) + (b^2*(3*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d*Cos[c
+ d*x]^(5/2))

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + 1/2)*b^(n - 1/2)*Sqrt[b*v])/Sqrt[a*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 3012

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{15}{2}}(c+d x)} \, dx &=\frac{\left (b^2 \sqrt{b \cos (c+d x)}\right ) \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{A b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{\left (b^2 (3 A+4 C) \sqrt{b \cos (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{4 \sqrt{\cos (c+d x)}}\\ &=\frac{A b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{b^2 (3 A+4 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{\left (b^2 (3 A+4 C) \sqrt{b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{8 \sqrt{\cos (c+d x)}}\\ &=\frac{b^2 (3 A+4 C) \tanh ^{-1}(\sin (c+d x)) \sqrt{b \cos (c+d x)}}{8 d \sqrt{\cos (c+d x)}}+\frac{A b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{b^2 (3 A+4 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.215726, size = 80, normalized size = 0.61 \[ \frac{(b \cos (c+d x))^{5/2} \left (\sin (c+d x) \left ((3 A+4 C) \cos ^2(c+d x)+2 A\right )+(3 A+4 C) \cos ^4(c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{8 d \cos ^{\frac{13}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(15/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*((3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (2*A + (3*A + 4*C)*Cos[c + d*x]^2)
*Sin[c + d*x]))/(8*d*Cos[c + d*x]^(13/2))

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Maple [A]  time = 0.231, size = 214, normalized size = 1.6 \begin{align*} -{\frac{1}{8\,d} \left ( 3\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -3\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ({\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +4\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -4\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ({\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -3\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4\,C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,A\sin \left ( dx+c \right ) \right ) \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2),x)

[Out]

-1/8/d*(3*A*cos(d*x+c)^4*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-3*A*cos(d*x+c)^4*ln((1-cos(d*x+c)+sin(d*x+
c))/sin(d*x+c))+4*C*cos(d*x+c)^4*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-4*C*cos(d*x+c)^4*ln((1-cos(d*x+c)+
sin(d*x+c))/sin(d*x+c))-3*A*sin(d*x+c)*cos(d*x+c)^2-4*C*sin(d*x+c)*cos(d*x+c)^2-2*A*sin(d*x+c))*(b*cos(d*x+c))
^(5/2)/cos(d*x+c)^(13/2)

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Maxima [B]  time = 2.69141, size = 3594, normalized size = 27.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2),x, algorithm="maxima")

[Out]

-1/16*((12*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*c
os(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 44*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^
2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(b^2*si
n(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*sin(4*d*x + 4*c)
+ 4*b^2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(b^2*cos(8*d*x + 8*c)^2 +
16*b^2*cos(6*d*x + 6*c)^2 + 36*b^2*cos(4*d*x + 4*c)^2 + 16*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(8*d*x + 8*c)^2 + 1
6*b^2*sin(6*d*x + 6*c)^2 + 36*b^2*sin(4*d*x + 4*c)^2 + 48*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*b^2*sin(2
*d*x + 2*c)^2 + 8*b^2*cos(2*d*x + 2*c) + b^2 + 2*(4*b^2*cos(6*d*x + 6*c) + 6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(
2*d*x + 2*c) + b^2)*cos(8*d*x + 8*c) + 8*(6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*cos(6*d*x + 6
*c) + 12*(4*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c) + 4*(2*b^2*sin(6*d*x + 6*c) + 3*b^2*sin(4*d*x + 4*c)
+ 2*b^2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*sin(6*d*x +
6*c))*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 3*(b^2*cos(8*d*x + 8*c)^2 + 16*b^2
*cos(6*d*x + 6*c)^2 + 36*b^2*cos(4*d*x + 4*c)^2 + 16*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(8*d*x + 8*c)^2 + 16*b^2*
sin(6*d*x + 6*c)^2 + 36*b^2*sin(4*d*x + 4*c)^2 + 48*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*b^2*sin(2*d*x +
2*c)^2 + 8*b^2*cos(2*d*x + 2*c) + b^2 + 2*(4*b^2*cos(6*d*x + 6*c) + 6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x
+ 2*c) + b^2)*cos(8*d*x + 8*c) + 8*(6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*cos(6*d*x + 6*c) +
12*(4*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c) + 4*(2*b^2*sin(6*d*x + 6*c) + 3*b^2*sin(4*d*x + 4*c) + 2*b^
2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*
log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
)))^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 12*(b^2*cos(8*d*x + 8*c) + 4*b^2*cos(6*d
*x + 6*c) + 6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c))) - 44*(b^2*cos(8*d*x + 8*c) + 4*b^2*cos(6*d*x + 6*c) + 6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c
) + b^2)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 44*(b^2*cos(8*d*x + 8*c) + 4*b^2*cos(6*d*x + 6
*c) + 6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
))) + 12*(b^2*cos(8*d*x + 8*c) + 4*b^2*cos(6*d*x + 6*c) + 6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^
2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A*sqrt(b)/(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c)
+ 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c)
+ 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x +
4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x +
8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c
)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c
) + 1) + 4*(4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c))) - 4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
- (b^2*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d
*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d
*x + 4*c))*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (b^2*cos(4*d*x + 4*c)^2 + 4*b
^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2
*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c))*log(cos(1/2*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2*c) + b^2)*sin(
3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2*c) + b^2)*sin
(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(b)/(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + co
s(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d
*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1))/d

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Fricas [A]  time = 1.73427, size = 714, normalized size = 5.45 \begin{align*} \left [\frac{{\left (3 \, A + 4 \, C\right )} b^{\frac{5}{2}} \cos \left (d x + c\right )^{5} \log \left (-\frac{b \cos \left (d x + c\right )^{3} - 2 \, \sqrt{b \cos \left (d x + c\right )} \sqrt{b} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \,{\left ({\left (3 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 2 \, A b^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{5}}, -\frac{{\left (3 \, A + 4 \, C\right )} \sqrt{-b} b^{2} \arctan \left (\frac{\sqrt{b \cos \left (d x + c\right )} \sqrt{-b} \sin \left (d x + c\right )}{b \sqrt{\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} -{\left ({\left (3 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 2 \, A b^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2),x, algorithm="fricas")

[Out]

[1/16*((3*A + 4*C)*b^(5/2)*cos(d*x + c)^5*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x
+ c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*((3*A + 4*C)*b^2*cos(d*x + c)^2 + 2*A*b^2)*sqrt(b*
cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5), -1/8*((3*A + 4*C)*sqrt(-b)*b^2*arctan(sqrt(
b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c)^5 - ((3*A + 4*C)*b^2*cos(d*x + c)^2
+ 2*A*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)]

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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((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(15/2),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 )} \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}{\cos \left (d x + c\right )^{\frac{15}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)/cos(d*x + c)^(15/2), x)