\(\int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx\) [169]

Optimal result

Integrand size = 34, antiderivative size = 155 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {11 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \sqrt {2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}} \]

[Out]

11/4*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/d/a^(1/2)-2*A*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)+5/4*A*sin(d*x+c)/d/(a-a*sec(d*x+c))^(1/2)+1/2*A*cos(d*x+c)*sin(d*x+c)
/d/(a-a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4107, 4005, 3859, 209, 3880} \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {11 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \sqrt {2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}} \]

[In]

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

[Out]

(11*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(4*Sqrt[a]*d) - (2*Sqrt[2]*A*ArcTan[(Sqrt[a]*Ta
n[c + d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[a]*d) + (5*A*Sin[c + d*x])/(4*d*Sqrt[a - a*Sec[c + d*x]
]) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*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 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)}{2 d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-\frac {5 a A}{2}-\frac {3}{2} a A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a} \\ & = \frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}+\frac {\int \frac {\frac {11 a^2 A}{4}+\frac {5}{4} a^2 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a^2} \\ & = \frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}+(2 A) \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx+\frac {(11 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{8 a} \\ & = \frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}}+\frac {(11 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 d}-\frac {(4 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 )}{d} \\ & = \frac {11 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \sqrt {2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {5 A \sin (c+d x)}{4 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a-a \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {A \left (\sqrt {1+\sec (c+d x)} (5 \sin (c+d x)+\sin (2 (c+d x)))+11 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \tan (c+d x)-8 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right ) \tan (c+d x)\right )}{4 d \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \]

[In]

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

[Out]

(A*(Sqrt[1 + Sec[c + d*x]]*(5*Sin[c + d*x] + Sin[2*(c + d*x)]) + 11*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*Tan[c + d*
x] - 8*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]]*Tan[c + d*x]))/(4*d*Sqrt[1 + Sec[c + d*x]]*Sqrt[a - a*S
ec[c + d*x]])

Maple [A] (verified)

Time = 28.83 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {A \sqrt {2}\, \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+7 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \cos \left (d x +c \right )+5 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+11 \sqrt {2}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+16 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-a \left (\sec \left (d x +c \right )-1\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(203\)

[In]

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

[Out]

1/8*A/d*2^(1/2)*sin(d*x+c)*(2*cos(d*x+c)^2*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+7*(-cos(d*x+c)/(cos(d*x+
c)+1))^(1/2)*2^(1/2)*cos(d*x+c)+5*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+11*2^(1/2)*arctan((-cos(d*x+c)/(c
os(d*x+c)+1))^(1/2))+16*arctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))/(cos(d*x+c)+1)/(-a*(sec(d*x+c)
-1))^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 462, normalized size of antiderivative = 2.98 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\left [\frac {8 \, \sqrt {2} A a \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 11 \, A \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} + 5 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{8 \, a d \sin \left (d x + c\right )}, \frac {8 \, \sqrt {2} A \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 11 \, A \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - {\left (2 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} + 5 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \, a d \sin \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[1/8*(8*sqrt(2)*A*a*sqrt(-1/a)*log(-(2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d
*x + c))*sqrt(-1/a) - (3*cos(d*x + c) + 1)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin(d*x + c) - 11*
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)^3 + 7*A*cos(d*x + c)^2 + 5*A*cos(
d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c)), 1/4*(8*sqrt(2)*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) - 11*A*sqrt(a)*arct
an(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 + 7*A*cos(d*x + c)^2 + 5*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.99 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {\frac {8 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {11 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{\sqrt {a}} - \frac {\sqrt {2} {\left (3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 10 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{2}}}{4 \, d} \]

[In]

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

[Out]

1/4*(8*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) - 11*A*arctan(1/2*sqrt(2)*sqrt(a*t
an(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) - sqrt(2)*(3*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 10*sqrt(a*ta
n(1/2*d*x + 1/2*c)^2 - a)*A*a)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^2)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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