\(\int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx\) [173]

Optimal result

Integrand size = 34, antiderivative size = 194 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {31 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {11 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}} \]

[Out]

31/4*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(3/2)/d-A*cos(d*x+c)*sin(d*x+c)/d/(a-a*sec(d*x+c))^
(3/2)-11/2*A*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+13/4*A*sin(d*x+c)
/a/d/(a-a*sec(d*x+c))^(1/2)+3/2*A*cos(d*x+c)*sin(d*x+c)/a/d/(a-a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4105, 4107, 4005, 3859, 209, 3880} \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {31 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {11 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]

[In]

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

[Out]

(31*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(4*a^(3/2)*d) - (11*A*ArcTan[(Sqrt[a]*Tan[c + d
*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) - (A*Cos[c + d*x]*Sin[c + d*x])/(d*(a - a*Sec[c
+ d*x])^(3/2)) + (13*A*Sin[c + d*x])/(4*a*d*Sqrt[a - a*Sec[c + d*x]]) + (3*A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d
*Sqrt[a - a*Sec[c + d*x]])

Rule 209

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

Rule 3859

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 3880

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
]

Rule 4005

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 4105

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 4107

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) (6 a A+5 a A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-13 a^2 A-9 a^2 A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{4 a^3} \\ & = -\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}+\frac {\int \frac {\frac {31 a^3 A}{2}+\frac {13}{2} a^3 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{4 a^4} \\ & = -\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}+\frac {(31 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{8 a^2}+\frac {(11 A) \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a} \\ & = -\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}+\frac {(31 A) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a d}-\frac {(11 A) \text {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 {31 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {11 A \arctan \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 \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {A \sec (c+d x) \left (-8 \cos ^3(c+d x)-40 (-1+\cos (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1+\sec (c+d x)\right )+\frac {11 (-1+\cos (c+d x)) \left (7 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (1+2 \cos (c+d x)) \sqrt {1+\sec (c+d x)}\right )}{\sqrt {1+\sec (c+d x)}}\right ) \tan (c+d x)}{8 d (a-a \sec (c+d x))^{3/2}} \]

[In]

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

[Out]

(A*Sec[c + d*x]*(-8*Cos[c + d*x]^3 - 40*(-1 + Cos[c + d*x])*Hypergeometric2F1[1/2, 3, 3/2, 1 + Sec[c + d*x]] +
(11*(-1 + Cos[c + d*x])*(7*ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]
] + Cos[c + d*x]*(1 + 2*Cos[c + d*x])*Sqrt[1 + Sec[c + d*x]]))/Sqrt[1 + Sec[c + d*x]])*Tan[c + d*x])/(8*d*(a -
a*Sec[c + d*x])^(3/2))

Maple [A] (verified)

Time = 31.19 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.52

method	result	size
default	\(-\frac {A \sqrt {2}\, \left (2 \cos \left (d x +c \right )^{3} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+9 \cos \left (d x +c \right )^{2} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-6 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \cos \left (d x +c \right )+31 \sqrt {2}\, \cos \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-13 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+44 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-31 \sqrt {2}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-44 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right ) \csc \left (d x +c \right )}{8 a d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {-a \left (\sec \left (d x +c \right )-1\right )}}\)	\(295\)

[In]

int(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*A/a/d*2^(1/2)*(2*cos(d*x+c)^3*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+9*cos(d*x+c)^2*2^(1/2)*(-cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)-6*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)*cos(d*x+c)+31*2^(1/2)*cos(d*x+c)*arctan
((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-13*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+44*cos(d*x+c)*arctan(1/2*2^
(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-31*2^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-44*arctan(1/2*
2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(-a*(sec(d*x+c)-1))^(1/2)*csc(
d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {22 \, \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 ) + 31 \, {\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 ) + 2 \, {\left (2 \, A \cos \left (d x + c\right )^{4} + 9 \, A \cos \left (d x + c\right )^{3} - 6 \, A \cos \left (d x + c\right )^{2} - 13 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{8 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac {22 \, \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 ) - 31 \, {\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 ) - {\left (2 \, A \cos \left (d x + c\right )^{4} + 9 \, A \cos \left (d x + c\right )^{3} - 6 \, A \cos \left (d x + c\right )^{2} - 13 \, 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 )}\right ] \]

[In]

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

[Out]

[-1/8*(22*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)))*si
n(d*x + c) + 31*(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) + 2*(2*A*cos(d*x + c
)^4 + 9*A*cos(d*x + c)^3 - 6*A*cos(d*x + c)^2 - 13*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)), 1/4*(22*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) - 31*(A*cos(d*x + c) - A)*s
qrt(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)^4 + 9*A*cos(d*x + c)^3 - 6*A*cos(d*x + c)^2 - 13*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))]

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=A \left (\int \frac {\cos ^{2}{\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {\cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \]

[In]

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

[Out]

A*(Integral(cos(c + d*x)**2/(-a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a*sqrt(-a*sec(c + d*x) + a)), x) + In
tegral(cos(c + d*x)**2*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))

Maxima [F]

\[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 1.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {22 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {31 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {\sqrt {2} {\left (7 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 18 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{2} a} - \frac {2 \, \sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{4 \, d} \]

[In]

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

[Out]

1/4*(22*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(3/2) - 31*A*arctan(1/2*sqrt(2)*sqrt(a*
tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(3/2) - sqrt(2)*(7*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 18*sqrt(a*t
an(1/2*d*x + 1/2*c)^2 - a)*A*a)/((a*tan(1/2*d*x + 1/2*c)^2 + a)^2*a) - 2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2
- a)*A/(a^2*tan(1/2*d*x + 1/2*c)^2))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {A}{\cos \left (c+d\,x\right )}\right )}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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