\(\int \frac {\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx\) [170]

Optimal result

Integrand size = 34, antiderivative size = 192 \[ \int \frac {\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {23 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {a} d}-\frac {2 \sqrt {2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a-a \sec (c+d x)}} \]

[Out]

23/8*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)+9/8*A*sin(d*x+c)/d/(a-a*sec(d*x+c))^(1/2)+7/12*A*cos(d*x+c)*sin(d*x+c
)/d/(a-a*sec(d*x+c))^(1/2)+1/3*A*cos(d*x+c)^2*sin(d*x+c)/d/(a-a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4107, 4005, 3859, 209, 3880} \[ \int \frac {\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {23 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {a} d}-\frac {2 \sqrt {2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \sin (c+d x) \cos (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}} \]

[In]

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

[Out]

(23*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(8*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) + (9*A*Sin[c + d*x])/(8*d*Sqrt[a - a*Sec[c + d*x]
]) + (7*A*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a - a*Sec[c + d*x]]) + (A*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sq
rt[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 ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {\cos ^2(c+d x) \left (-\frac {7 a A}{2}-\frac {5}{2} a A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{3 a} \\ & = \frac {7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a-a \sec (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (\frac {27 a^2 A}{4}+\frac {21}{4} a^2 A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{6 a^2} \\ & = \frac {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {-\frac {69 a^3 A}{8}-\frac {27}{8} a^3 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{6 a^3} \\ & = \frac {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 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 {(23 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{16 a} \\ & = \frac {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a-a \sec (c+d x)}}+\frac {(23 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 )}{8 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 {23 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 \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 {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a-a \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {A \left (69 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-48 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+(7+33 \cos (c+d x)+7 \cos (2 (c+d x))+2 \cos (3 (c+d x))) \sqrt {1+\sec (c+d x)}\right ) \tan (c+d x)}{24 d \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \]

[In]

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

[Out]

(A*(69*ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 48*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]] + (7 + 33*Cos[c +
d*x] + 7*Cos[2*(c + d*x)] + 2*Cos[3*(c + d*x)])*Sqrt[1 + Sec[c + d*x]])*Tan[c + d*x])/(24*d*Sqrt[1 + Sec[c + d
*x]]*Sqrt[a - a*Sec[c + d*x]])

Maple [A] (verified)

Time = 29.78 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {A \sqrt {2}\, \sin \left (d x +c \right ) \left (8 \cos \left (d x +c \right )^{3} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+22 \cos \left (d x +c \right )^{2} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+41 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \cos \left (d x +c \right )+27 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+69 \sqrt {2}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+96 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right )}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-a \left (\sec \left (d x +c \right )-1\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(236\)

[In]

int(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/48*A/d*2^(1/2)*sin(d*x+c)*(8*cos(d*x+c)^3*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+22*cos(d*x+c)^2*2^(1/2)
*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+41*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)*cos(d*x+c)+27*2^(1/2)*(-cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)+69*2^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+96*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 = 0.29 (sec) , antiderivative size = 484, normalized size of antiderivative = 2.52 \[ \int \frac {\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\left [\frac {48 \, \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 ) - 69 \, 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 (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{48 \, a d \sin \left (d x + c\right )}, \frac {48 \, \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 ) - 69 \, 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 (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{24 \, a d \sin \left (d x + c\right )}\right ] \]

[In]

integrate(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/48*(48*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) - 6
9*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*co
s(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 2*(8*A*cos(d*x + c)^4 + 22*A*cos(d*x + c)^3 + 41*A*
cos(d*x + c)^2 + 27*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c)), 1/24*(48*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) - 69*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)))*si
n(d*x + c) - (8*A*cos(d*x + c)^4 + 22*A*cos(d*x + c)^3 + 41*A*cos(d*x + c)^2 + 27*A*cos(d*x + c))*sqrt((a*cos(
d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c))]

Sympy [F]

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

[In]

integrate(cos(d*x+c)**3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(1/2),x)

[Out]

A*(Integral(cos(c + d*x)**3/sqrt(-a*sec(c + d*x) + a), x) + Integral(cos(c + d*x)**3*sec(c + d*x)/sqrt(-a*sec(
c + d*x) + a), x))

Maxima [F]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 1.03 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {\frac {48 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {69 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{\sqrt {a}} - \frac {\sqrt {2} {\left (21 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {5}{2}} A + 80 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A a + 108 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{3}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/24*(48*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) - 69*A*arctan(1/2*sqrt(2)*sqrt(a
*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) - sqrt(2)*(21*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(5/2)*A + 80*(a*tan
(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A*a + 108*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A*a^2)/(a*tan(1/2*d*x + 1/2*c)^2 +
a)^3)/d

Mupad [F(-1)]

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

[In]

int((cos(c + d*x)^3*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)^3*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(1/2), x)