3.172 \(\int \frac{\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{5 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac{7 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}+\frac{2 A \sin (c+d x)}{a d \sqrt{a-a \sec (c+d x)}}-\frac{A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.354706, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{5 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac{7 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}+\frac{2 A \sin (c+d x)}{a d \sqrt{a-a \sec (c+d x)}}-\frac{A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2),x]

[Out]

(5*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(a^(3/2)*d) - (7*A*ArcTan[(Sqrt[a]*Tan[c + d*x])
/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) - (A*Sin[c + d*x])/(d*(a - a*Sec[c + d*x])^(3/2)) +
(2*A*Sin[c + d*x])/(a*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))^{3/2}} \, dx &=-\frac{A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos (c+d x) (4 a A+3 a A \sec (c+d x))}{\sqrt{a-a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{2 A \sin (c+d x)}{a d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{-5 a^2 A-2 a^2 A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{2 a^3}\\ &=-\frac{A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{2 A \sin (c+d x)}{a d \sqrt{a-a \sec (c+d x)}}+\frac{(5 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{2 a^2}+\frac{(7 A) \int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{2 a}\\ &=-\frac{A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{2 A \sin (c+d x)}{a d \sqrt{a-a \sec (c+d x)}}+\frac{(5 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 d}-\frac{(7 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 )}{a d}\\ &=\frac{5 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac{7 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}-\frac{A \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{2 A \sin (c+d x)}{a d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.61696, size = 361, normalized size = 2.47 \[ A \left (\frac{\sin ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^2(c+d x) \left (-\frac{2 \sin \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{d}+\frac{2 \sin \left (\frac{3 c}{2}\right ) \sin \left (\frac{3 d x}{2}\right )}{d}+\frac{2 \cos \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )}{d}-\frac{2 \cos \left (\frac{3 c}{2}\right ) \cos \left (\frac{3 d x}{2}\right )}{d}-\frac{2 \cot \left (\frac{c}{2}\right ) \csc \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{2 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}\right )}{(a-a \sec (c+d x))^{3/2}}+\frac{\sqrt{2} e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sin ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) \left (-5 \sinh ^{-1}\left (e^{i (c+d x)}\right )+7 \sqrt{2} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-5 \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{d (a-a \sec (c+d x))^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2),x]

[Out]

A*((Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-5*ArcSinh[E^(I*(c
+ d*x))] + 7*Sqrt[2]*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - 5*ArcTanh[Sqrt[1
 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]^(3/2)*Sin[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a - a*Sec[c + d*x])
^(3/2)) + (Sec[c + d*x]^2*((2*Cos[c/2]*Cos[(d*x)/2])/d - (2*Cos[(3*c)/2]*Cos[(3*d*x)/2])/d - (2*Cot[c/2]*Csc[c
/2 + (d*x)/2])/d + (2*Csc[c/2]*Csc[c/2 + (d*x)/2]^2*Sin[(d*x)/2])/d - (2*Sin[c/2]*Sin[(d*x)/2])/d + (2*Sin[(3*
c)/2]*Sin[(3*d*x)/2])/d)*Sin[c/2 + (d*x)/2]^3)/(a - a*Sec[c + d*x])^(3/2))

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Maple [B]  time = 0.283, size = 462, normalized size = 3.2 \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))^(3/2),x)

[Out]

1/3*A/d*2^(1/2)*(-1+cos(d*x+c))^3*(-3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*2^(1/2)-6*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*2^(1/2)-7*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^2*2^(1/2)-3*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)+3*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+21*arctan
(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*2^(1/2)-2*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)+7*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+30*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2))*cos(d*x+c)^2+5*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-21*2^(1/2)*arctan(1/(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2))-6*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-30*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))^(3/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))^(3/2)/sin(d*x+c)
^5

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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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((A*sec(d*x + c) + A)*cos(d*x + c)/(-a*sec(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 0.541201, size = 1345, normalized size = 9.21 \begin{align*} \left [-\frac{7 \, \sqrt{2}{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{-a} \log \left (\frac{2 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} +{\left (3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 10 \,{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{-a} \log \left (\frac{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} -{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \,{\left (A \cos \left (d x + c\right )^{3} - A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \,{\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac{7 \, \sqrt{2}{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 10 \,{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \,{\left (A \cos \left (d x + c\right )^{3} - A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{2 \,{\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}\right ] \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))^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(7*sqrt(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) + 10*(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*(A*cos(d*x + c)^3
 - A*cos(d*x + c)^2 - 2*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)
*sin(d*x + c)), 1/2*(7*sqrt(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) - 10*(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*(A*cos(d*x + c)^3 - A*cos(d*x
+ c)^2 - 2*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c)
)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} A \left (\int \frac{\cos{\left (c + d x \right )}}{- a \sqrt{- a \sec{\left (c + d x \right )} + a} \sec{\left (c + d x \right )} + a \sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx + \int \frac{\cos{\left (c + d x \right )} \sec{\left (c + d x \right )}}{- a \sqrt{- a \sec{\left (c + d x \right )} + a} \sec{\left (c + d x \right )} + a \sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx\right ) \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))**(3/2),x)

[Out]

A*(Integral(cos(c + d*x)/(-a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a*sqrt(-a*sec(c + d*x) + a)), x) + Integ
ral(cos(c + d*x)*sec(c + d*x)/(-a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a*sqrt(-a*sec(c + d*x) + a)), x))

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Giac [B]  time = 2.06882, size = 344, normalized size = 2.36 \begin{align*} -\frac{A{\left (\frac{7 \, \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{3}{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 (3 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} + 4 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a\right )}}{{\left ({\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} + 3 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} a + 2 \, a^{2}\right )} a \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{10 \, \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{3}{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 )}\right )}}{2 \, 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))^(3/2),x, algorithm="giac")

[Out]

-1/2*A*(7*sqrt(2)*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*
sgn(tan(1/2*d*x + 1/2*c))) - sqrt(2)*(3*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2) + 4*sqrt(a*tan(1/2*d*x + 1/2*c)^2
 - a)*a)/(((a*tan(1/2*d*x + 1/2*c)^2 - a)^2 + 3*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a + 2*a^2)*a*sgn(tan(1/2*d*x +
1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - 10*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(
a^(3/2)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))))/d