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

Optimal. Leaf size=288 \[ \frac{2 \left (15 a^2 C+56 a b B+25 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 d}-\frac{2 \left (a^2-b^2\right ) \left (15 a^2 C+56 a b B+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (161 a^2 b B+15 a^3 C+145 a b^2 C+63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (5 a C+7 b B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]

[Out]

(2*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 145*a*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)])/(105*b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(56*a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(105*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(56*
a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*(7*b*B + 5*a*C)*(a + b*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

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Rubi [A]  time = 0.577827, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {3029, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (15 a^2 C+56 a b B+25 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 d}-\frac{2 \left (a^2-b^2\right ) \left (15 a^2 C+56 a b B+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (161 a^2 b B+15 a^3 C+145 a b^2 C+63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (5 a C+7 b B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 145*a*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)])/(105*b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(56*a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(105*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(56*
a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*(7*b*B + 5*a*C)*(a + b*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 2753

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[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2752

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

Rule 2663

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

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\int (a+b \cos (c+d x))^{5/2} (B+C \cos (c+d x)) \, dx\\ &=\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{2}{7} \int (a+b \cos (c+d x))^{3/2} \left (\frac{1}{2} (7 a B+5 b C)+\frac{1}{2} (7 b B+5 a C) \cos (c+d x)\right ) \, dx\\ &=\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{4}{35} \int \sqrt{a+b \cos (c+d x)} \left (\frac{1}{4} \left (35 a^2 B+21 b^2 B+40 a b C\right )+\frac{1}{4} \left (56 a b B+15 a^2 C+25 b^2 C\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{2 \left (56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{8}{105} \int \frac{\frac{1}{8} \left (105 a^3 B+119 a b^2 B+135 a^2 b C+25 b^3 C\right )+\frac{1}{8} \left (161 a^2 b B+63 b^3 B+15 a^3 C+145 a b^2 C\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 \left (56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac{\left (\left (a^2-b^2\right ) \left (56 a b B+15 a^2 C+25 b^2 C\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b}+\frac{\left (161 a^2 b B+63 b^3 B+15 a^3 C+145 a b^2 C\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{105 b}\\ &=\frac{2 \left (56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{\left (\left (161 a^2 b B+63 b^3 B+15 a^3 C+145 a b^2 C\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{105 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (a^2-b^2\right ) \left (56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{105 b \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (161 a^2 b B+63 b^3 B+15 a^3 C+145 a b^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (a^2-b^2\right ) \left (56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 b B+5 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 1.03859, size = 254, normalized size = 0.88 \[ \frac{b \sin (c+d x) (a+b \cos (c+d x)) \left (90 a^2 C+6 b (15 a C+7 b B) \cos (c+d x)+154 a b B+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+2 b \left (135 a^2 b C+105 a^3 B+119 a b^2 B+25 b^3 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 \left (161 a^2 b B+15 a^3 C+145 a b^2 C+63 b^3 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{105 b d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(105*a^3*B + 119*a*b^2*B + 135*a^2*b*C + 25*b^3*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)
/2, (2*b)/(a + b)] + 2*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 145*a*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*((
a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)]) + b*(a + b*Cos[c + d*x
])*(154*a*b*B + 90*a^2*C + 65*b^2*C + 6*b*(7*b*B + 15*a*C)*Cos[c + d*x] + 15*b^2*C*Cos[2*(c + d*x)])*Sin[c + d
*x])/(105*b*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 0.827, size = 1305, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*B*b^4-480*C*a*b^3-360*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(392*B*a*b^3+168*B*b^4+360*
C*a^2*b^2+480*C*a*b^3+280*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-154*B*a^2*b^2-196*B*a*b^3-42*B*b^4-
90*C*a^3*b-180*C*a^2*b^2-170*C*a*b^3-80*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+161*B*EllipticE(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/
2)*a^3*b-161*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1
/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*a^2*b^2+63*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*a*b^3-63*B*EllipticE(cos(1/2*d*x+1/2*c),(-
2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*b^4-56*B*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-
2*b/(a-b))^(1/2))*a^3*b+56*B*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2
)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d
*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-15*C*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a
^3*b+145*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*
d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2-145*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+
b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^3-15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b
/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-10*C*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b
/(a-b))^(1/2))*a^2*b^2+25*C*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/
2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))/b/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^
(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + B a^{2} \cos \left (d x + c\right ) +{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^2*cos(d*x + c)^4 + B*a^2*cos(d*x + c) + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + (C*a^2 + 2*B*a*b)*cos
(d*x + c)^2)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c), x)

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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+b*cos(d*x+c))**(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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