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

Optimal. Leaf size=192 \[ \frac {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {23 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {a} d}-\frac {2 \sqrt {2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \sin (c+d x) \cos (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}} \]

[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)

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Rubi [A]  time = 0.52, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4022, 3920, 3774, 203, 3795} \[ \frac {9 A \sin (c+d x)}{8 d \sqrt {a-a \sec (c+d x)}}+\frac {23 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {a} d}-\frac {2 \sqrt {2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a-a \sec (c+d x)}}+\frac {7 A \sin (c+d x) \cos (c+d x)}{12 d \sqrt {a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[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 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 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 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
]

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 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]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx &=\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) \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 )}{8 d}-\frac {(4 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 )}{d}\\ &=\frac {23 A \tan ^{-1}\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 \tan ^{-1}\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*}

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Mathematica [C]  time = 1.95, size = 330, normalized size = 1.72 \[ \frac {A e^{-4 i (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+i \sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (9 e^{i (c+d x)}+40 e^{2 i (c+d x)}+47 e^{3 i (c+d x)}+47 e^{4 i (c+d x)}+40 e^{5 i (c+d x)}+9 e^{6 i (c+d x)}+2 e^{7 i (c+d x)}+69 e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-96 \sqrt {2} e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+69 e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )+2\right )}{48 d \sqrt {a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(2 + 9*E^(I*(c + d*x)) + 40*E^((2*I)*(c + d*x)) + 47*E^((3*I)*(c + d*x)) + 47*E^((4*I)*(c + d*x)) + 40*E^((
5*I)*(c + d*x)) + 9*E^((6*I)*(c + d*x)) + 2*E^((7*I)*(c + d*x)) + 69*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c
+ d*x))]*ArcSinh[E^(I*(c + d*x))] - 96*Sqrt[2]*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 +
E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + 69*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x)
)]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]*(Cos[(c + d*x)/2] + I*Sin[(c + d*x)/2])*Sin[(c + d*x)/
2])/(48*d*E^((4*I)*(c + d*x))*Sqrt[a - a*Sec[c + d*x]])

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fricas [A]  time = 0.47, size = 484, normalized size = 2.52 \[ \left [\frac {48 \, \sqrt {2} A a \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 69 \, A \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{48 \, a d \sin \left (d x + c\right )}, \frac {48 \, \sqrt {2} A \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 69 \, A \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - {\left (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{24 \, a d \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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giac [C]  time = 4.36, size = 290, normalized size = 1.51 \[ -\frac {\frac {{\left (-48 i \, \sqrt {2} A \arctan \left (-i\right ) + 69 i \, A \arctan \left (-\frac {1}{2} i \, \sqrt {2}\right ) + 49 \, \sqrt {2} A\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\sqrt {-a}} + \frac {48 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{\sqrt {a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {69 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{\sqrt {a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (21 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {5}{2}} A + 80 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A a + 108 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*I*sqrt(2)*A*arctan(-I) + 69*I*A*arctan(-1/2*I*sqrt(2)) + 49*sqrt(2)*A)*sgn(tan(1/2*d*x + 1/2*c))/s
qrt(-a) + 48*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(sqrt(a)*sgn(tan(1/2*d*x + 1/2*c)^2
- 1)*sgn(tan(1/2*d*x + 1/2*c))) - 69*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(sqrt(a)
*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - 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*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))))/d

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maple [B]  time = 2.12, size = 625, normalized size = 3.26 \[ -\frac {A \left (-1+\cos \left (d x +c \right )\right )^{3} \left (96 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+192 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right ) \sqrt {2}+40 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}+96 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}-160 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+190 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-320 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \sqrt {2}+465 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+480 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-160 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}-49 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+960 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right ) \sqrt {2}+155 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sqrt {2}+690 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+480 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}+135 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+1380 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \cos \left (d x +c \right )+690 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )\right ) \sqrt {2}}{240 d \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (-1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*A/d*(-1+cos(d*x+c))^3*(96*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^2*2^(1/2)+192*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(5/2)*cos(d*x+c)*2^(1/2)+40*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^5*2^(1/2)+96*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)-160*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*2^(1/2)+190*(-2*
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*2^(1/2)-320*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*2^(1
/2)+465*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*2^(1/2)+480*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2))*cos(d*x+c)^2*2^(1/2)-160*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)-49*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*cos(d*x+c)^2*2^(1/2)+960*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)*2^(1/2)+155*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*2^(1/2)+690*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*c
os(d*x+c)^2+480*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*2^(1/2)+135*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*2^(1/2)+1380*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)+690*arctan(1/2*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)))/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))^(1/2)/s
in(d*x+c)^5*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {-a \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+\frac {A}{\cos \left (c+d\,x\right )}\right )}{\sqrt {a-\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ A \left (\int \frac {\cos ^{3}{\left (c + d x \right )}}{\sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {\cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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