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

Optimal. Leaf size=184 \[ \frac{23 A \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{79 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}-\frac{11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]

[Out]

(7*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(a^(5/2)*d) - (79*A*ArcTan[(Sqrt[a]*Tan[c + d*x]
)/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(8*Sqrt[2]*a^(5/2)*d) - (A*Sin[c + d*x])/(2*d*(a - a*Sec[c + d*x])^(5/2
)) - (11*A*Sin[c + d*x])/(8*a*d*(a - a*Sec[c + d*x])^(3/2)) + (23*A*Sin[c + d*x])/(8*a^2*d*Sqrt[a - a*Sec[c +
d*x]])

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Rubi [A]  time = 0.505332, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {4020, 4022, 3920, 3774, 203, 3795} \[ \frac{23 A \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{79 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}-\frac{11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(a^(5/2)*d) - (79*A*ArcTan[(Sqrt[a]*Tan[c + d*x]
)/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(8*Sqrt[2]*a^(5/2)*d) - (A*Sin[c + d*x])/(2*d*(a - a*Sec[c + d*x])^(5/2
)) - (11*A*Sin[c + d*x])/(8*a*d*(a - a*Sec[c + d*x])^(3/2)) + (23*A*Sin[c + d*x])/(8*a^2*d*Sqrt[a - a*Sec[c +
d*x]])

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4022

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

Rule 3920

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

Rule 3774

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

Rule 203

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

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx &=-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}+\frac{\int \frac{\cos (c+d x) (6 a A+5 a A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos (c+d x) \left (23 a^2 A+\frac{33}{2} a^2 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{23 A \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{-28 a^3 A-\frac{23}{2} a^3 A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{23 A \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{(7 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{2 a^3}+\frac{(79 A) \int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{16 a^2}\\ &=-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{23 A \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{(7 A) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^2 d}-\frac{(79 A) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^2 d}\\ &=\frac{7 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{79 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}-\frac{A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{23 A \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.79081, size = 423, normalized size = 2.3 \[ A \left (\frac{\sin ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) \left (\frac{15 \sin \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{d}-\frac{4 \sin \left (\frac{3 c}{2}\right ) \sin \left (\frac{3 d x}{2}\right )}{d}-\frac{15 \cos \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )}{d}+\frac{4 \cos \left (\frac{3 c}{2}\right ) \cos \left (\frac{3 d x}{2}\right )}{d}-\frac{\cot \left (\frac{c}{2}\right ) \csc ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{23 \cot \left (\frac{c}{2}\right ) \csc \left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{\csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}-\frac{23 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}\right )}{(a-a \sec (c+d x))^{5/2}}+\frac{e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sin ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^{\frac{5}{2}}(c+d x) \left (28 \sinh ^{-1}\left (e^{i (c+d x)}\right )-\frac{79 \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{2}}+28 \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{\sqrt{2} d (a-a \sec (c+d x))^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

A*((Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(28*ArcSinh[E^(I*(c + d*x))]
- (79*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[2] + 28*ArcTanh[Sqrt[1 + E
^((2*I)*(c + d*x))]])*Sec[c + d*x]^(5/2)*Sin[c/2 + (d*x)/2]^5)/(Sqrt[2]*d*E^((I/2)*(c + d*x))*(a - a*Sec[c + d
*x])^(5/2)) + (Sec[c + d*x]^3*((-15*Cos[c/2]*Cos[(d*x)/2])/d + (4*Cos[(3*c)/2]*Cos[(3*d*x)/2])/d + (23*Cot[c/2
]*Csc[c/2 + (d*x)/2])/(2*d) - (Cot[c/2]*Csc[c/2 + (d*x)/2]^3)/d - (23*Csc[c/2]*Csc[c/2 + (d*x)/2]^2*Sin[(d*x)/
2])/(2*d) + (Csc[c/2]*Csc[c/2 + (d*x)/2]^4*Sin[(d*x)/2])/d + (15*Sin[c/2]*Sin[(d*x)/2])/d - (4*Sin[(3*c)/2]*Si
n[(3*d*x)/2])/d)*Sin[c/2 + (d*x)/2]^5)/(a - a*Sec[c + d*x])^(5/2))

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Maple [B]  time = 0.342, size = 788, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*A/d*2^(1/2)*(-1+cos(d*x+c))^5*(195*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)+450*(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^3*2^(1/2)+237*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/
2)+180*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^2*2^(1/2)-210*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(
d*x+c)*2^(1/2)-395*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)-474*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(5/2)*cos(d*x+c)^2*2^(1/2)-135*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+120*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)*cos(d*x+c)^5*2^(1/2)-343*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^4*2^(1/2)+1185*2^(1/2)*cos(d
*x+c)^4*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+790*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^2*2
^(1/2)+237*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)+1680*cos(d*x+c)^4*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2))+736*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-578*cos(d*x+c)^2*2^(1/2)*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-2370*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*2^(1/2)-39
5*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)-3360*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*c
os(d*x+c)^2-280*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+345*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)+1185*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+1680*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)))/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))^(5/2)/sin(d*x+c)^9

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.55588, size = 1609, normalized size = 8.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*(79*sqrt(2)*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(-a)*log((2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x
+ c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) + (3*a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c)
- 1)*sin(d*x + c)))*sin(d*x + c) + 112*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(-a)*log((2*(cos(d*x + c)
^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) - (2*a*cos(d*x + c) + a)*sin(d*x + c))/sin
(d*x + c))*sin(d*x + c) + 4*(8*A*cos(d*x + c)^4 - 27*A*cos(d*x + c)^3 - 12*A*cos(d*x + c)^2 + 23*A*cos(d*x + c
))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) + a^3*d)*sin(d*x + c
)), 1/16*(79*sqrt(2)*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) - a
)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) - 112*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) +
A)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) -
2*(8*A*cos(d*x + c)^4 - 27*A*cos(d*x + c)^3 - 12*A*cos(d*x + c)^2 + 23*A*cos(d*x + c))*sqrt((a*cos(d*x + c) -
a)/cos(d*x + c)))/((a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) + a^3*d)*sin(d*x + c))]

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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(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 2.26015, size = 393, normalized size = 2.14 \begin{align*} -\frac{A{\left (\frac{79 \, \sqrt{2} \arctan \left (\frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{\sqrt{a}}\right )}{a^{\frac{5}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{112 \, \arctan \left (\frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{2 \, \sqrt{a}}\right )}{a^{\frac{5}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{16 \, \sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{\sqrt{2}{\left (17 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} + 15 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a\right )}}{a^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*A*(79*sqrt(2)*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(5/2)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
)*sgn(tan(1/2*d*x + 1/2*c))) - 112*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(5/2)*sgn
(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - 16*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/((a*ta
n(1/2*d*x + 1/2*c)^2 + a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - sqrt(2)*(17*(a*tan(
1/2*d*x + 1/2*c)^2 - a)^(3/2) + 15*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*a)/(a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*
sgn(tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c)^4))/d