3.524 \(\int \frac {\cos (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=173 \[ \frac {(9 A-5 B+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(3 A-2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}+\frac {(3 A-B+C) \sin (c+d x)}{2 a d \sqrt {a \sec (c+d x)+a}}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

-(3*A-2*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-1/2*(A-B+C)*sin(d*x+c)/d/(a+a*sec(d*x+c
))^(3/2)+1/4*(9*A-5*B+C)*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)+1/2*(
3*A-B+C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4084, 4022, 3920, 3774, 203, 3795} \[ \frac {(9 A-5 B+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(3 A-2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}+\frac {(3 A-B+C) \sin (c+d x)}{2 a d \sqrt {a \sec (c+d x)+a}}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*A - 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d)) + ((9*A - 5*B + C)*ArcTan
[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x]
)/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((3*A - B + C)*Sin[c + d*x])/(2*a*d*Sqrt[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]

Rule 4084

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

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A-B+C) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos (c+d x) \left (a (3 A-B+C)-\frac {1}{2} a (3 A-3 B-C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-B+C) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {-a^2 (3 A-2 B)+\frac {1}{2} a^2 (3 A-B+C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^3}\\ &=-\frac {(A-B+C) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-B+C) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}-\frac {(3 A-2 B) \int \sqrt {a+a \sec (c+d x)} \, dx}{2 a^2}+\frac {(9 A-5 B+C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A-B+C) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-B+C) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}+\frac {(3 A-2 B) \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 )}{a d}-\frac {(9 A-5 B+C) \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 )}{2 a d}\\ &=-\frac {(3 A-2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {(9 A-5 B+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-B+C) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 2.41, size = 179, normalized size = 1.03 \[ -\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \left (2 \sin ^2\left (\frac {1}{2} (c+d x)\right ) (2 A \cos (c+d x)+3 A-B+C)+\sqrt {2} (9 A-5 B+C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)-1} \tan ^{-1}\left (\frac {\sqrt {\sec (c+d x)-1}}{\sqrt {2}}\right )-4 (3 A-2 B) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)-1} \tan ^{-1}\left (\sqrt {\sec (c+d x)-1}\right )\right )}{2 a d (\cos (c+d x)-1) \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((-4*(3*A - 2*B)*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] + Sqrt[2]*(9*
A - 5*B + C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] + 2*(3*A - B +
C + 2*A*Cos[c + d*x])*Sin[(c + d*x)/2]^2)*Tan[(c + d*x)/2])/(a*d*(-1 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x]
)])

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fricas [A]  time = 21.97, size = 618, normalized size = 3.57 \[ \left [-\frac {\sqrt {2} {\left ({\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right ) + 9 \, A - 5 \, B + C\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 2 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (3 \, A - B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {\sqrt {2} {\left ({\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right ) + 9 \, A - 5 \, B + C\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 ) - 4 \, {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 2 \, B\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 ) - 2 \, {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (3 \, A - B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(sqrt(2)*((9*A - 5*B + C)*cos(d*x + c)^2 + 2*(9*A - 5*B + C)*cos(d*x + c) + 9*A - 5*B + C)*sqrt(-a)*log(
(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2
*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((3*A - 2*B)*cos(d*x + c)^2 + 2*(3*A - 2*B)*co
s(d*x + c) + 3*A - 2*B)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*
cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 4*(2*A*cos(d*x + c)^2 + (3*A - B + C)*co
s(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c)
+ a^2*d), -1/4*(sqrt(2)*((9*A - 5*B + C)*cos(d*x + c)^2 + 2*(9*A - 5*B + C)*cos(d*x + c) + 9*A - 5*B + C)*sqr
t(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))) - 4*((3*A - 2
*B)*cos(d*x + c)^2 + 2*(3*A - 2*B)*cos(d*x + c) + 3*A - 2*B)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*(2*A*cos(d*x + c)^2 + (3*A - B + C)*cos(d*x + c))*sqrt((a*cos(d
*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]

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giac [B]  time = 3.09, size = 474, normalized size = 2.74 \[ \frac {\frac {\sqrt {2} {\left (9 \, A - 5 \, B + 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 )}^{2}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {4 \, {\left (3 \, A - 2 \, B\right )} \log \left (\frac {{\left | -17179869184 \, {\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 )}^{2} - 34359738368 \, \sqrt {2} {\left | a \right |} + 51539607552 \, a \right |}}{{\left | -17179869184 \, {\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 )}^{2} + 34359738368 \, \sqrt {2} {\left | a \right |} + 51539607552 \, a \right |}}\right )}{\sqrt {-a} {\left | a \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {2 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - \sqrt {2} B a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + \sqrt {2} C a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {16 \, \sqrt {2} {\left (3 \, {\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 )}^{2} A - A a\right )}}{{\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 )}^{4} - 6 \, {\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 )}^{2} a + a^{2}\right )} \sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(2)*(9*A - 5*B + C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(sqr
t(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 4*(3*A - 2*B)*log(abs(-17179869184*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 34359738368*sqrt(2)*abs(a) + 51539607552*a)/abs(-17179869184*(sqrt(-
a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 34359738368*sqrt(2)*abs(a) + 51539607552*a)
)/(sqrt(-a)*abs(a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 2*(sqrt(2)*A*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - sqrt(2)
*B*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + sqrt(2)*C*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3 - 16*sqrt(2)*(3*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^2*A - A*a)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*
tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)*sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
)))/d

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maple [B]  time = 1.90, size = 891, normalized size = 5.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*(-1+cos(d*x+c))*(-6*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)+4*B*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin(d*x+c)-9*A*
sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*cos(d*x+c)-6*A*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*
2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+5*B*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*l
n(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*cos(d*x+c)+4*B*(-2*cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin
(d*x+c)-C*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+4*A*cos(d*x+c)^3-9*A*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-
cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+5*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)-C*ln(((-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)+2*A*cos(d*x+c)^2-2*B*cos(d*x+c)^2+2*C*cos(d*x+c)^2-6*A*cos(d*x+c)+2*B*cos(d*x+c)-2*C*cos(d*x+c))*(a*(1+cos(d
*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^3/a^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)/(a*(sec(c + d*x) + 1))**(3/2), x)

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