\(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [185]

Optimal result

Integrand size = 35, antiderivative size = 152 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d} \]

[Out]

-(A+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+2/15*(15*A+14*C)*tan(d*
x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/5*C*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/15*C*(a+a*sec(d*x+c))^(
1/2)*tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4174, 4095, 4086, 3880, 209} \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 a d} \]

[In]

Int[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

-((Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (2*(15*A
+ 14*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[
c + d*x]]) - (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*a*d)

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 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 4086

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

Rule 4095

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

Rule 4174

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\sec ^2(c+d x) \left (\frac {1}{2} a (5 A+4 C)-\frac {1}{2} a C \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a} \\ & = \frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+\frac {4 \int \frac {\sec (c+d x) \left (-\frac {a^2 C}{4}+\frac {1}{4} a^2 (15 A+14 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^2} \\ & = \frac {2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+(-A-C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx \\ & = \frac {2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+\frac {(2 (A+C)) \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 )}{d} \\ & = -\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (-15 \sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+2 \sqrt {1-\sec (c+d x)} \left (15 A+13 C-C \sec (c+d x)+3 C \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

((-15*Sqrt[2]*(A + C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + 2*Sqrt[1 - Sec[c + d*x]]*(15*A + 13*C - C*Sec[
c + d*x] + 3*C*Sec[c + d*x]^2))*Tan[c + d*x])/(15*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(306\) vs. \(2(131)=262\).

Time = 0.38 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.02

method	result	size
parts	\(-\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\ln \left (\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )\right )}{d a}-\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-34 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+40 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+30 \cot \left (d x +c \right )-30 \csc \left (d x +c \right )\right )}{15 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\)	\(307\)
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (15 A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+15 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-30 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-34 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+60 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+40 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-30 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-30 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{15 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\)	\(331\)

[In]

int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-A/d/a*(a*(1+sec(d*x+c)))^(1/2)*(ln((cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)-cot(d*x+c)+csc
(d*x+c))*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*cot(d*x+c)-2*csc(d*x+c))-1/15*C/d/a*(-2*a/((1-cos(d*x+c)
)^2*csc(d*x+c)^2-1))^(1/2)*(15*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c
))^2*csc(d*x+c)^2-1)^(5/2)-34*(1-cos(d*x+c))^5*csc(d*x+c)^5+40*(1-cos(d*x+c))^3*csc(d*x+c)^3+30*cot(d*x+c)-30*
csc(d*x+c))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 377, normalized size of antiderivative = 2.48 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {15 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{3} + {\left (A + C\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (15 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left ({\left (15 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{3} + {\left (A + C\right )} a \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

[In]

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/30*(15*sqrt(2)*((A + C)*a*cos(d*x + c)^3 + (A + C)*a*cos(d*x + c)^2)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos
(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((15*A + 13*C)*cos(d*x + c)^2 - C*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2), 1/15*(2*((15*A + 13*C)*cos(d*x
+ c)^2 - C*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 15*sqrt(2)*((A + C)*a*c
os(d*x + c)^3 + (A + C)*a*cos(d*x + c)^2)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/
(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

[In]

integrate(sec(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**2/sqrt(a*(sec(c + d*x) + 1)), x)

Maxima [F]

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

[In]

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^2/sqrt(a*sec(d*x + c) + a), x)

Giac [A] (verification not implemented)

none

Time = 1.35 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {15 \, {\left (\sqrt {2} A + \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {2 \, {\left (15 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (30 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 20 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (15 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 17 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{15 \, d} \]

[In]

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/15*(15*(sqrt(2)*A + sqrt(2)*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))
)/(sqrt(-a)*sgn(cos(d*x + c))) + 2*(15*sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^2*sgn(cos(d*x + c)) -
(30*sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 20*sqrt(2)*C*a^2*sgn(cos(d*x + c)) - (15*sqrt(2)*A*a^2*sgn(cos(d*x + c))
+ 17*sqrt(2)*C*a^2*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((
a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

Mupad [F(-1)]

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

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(1/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(1/2)), x)