3.115 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=192 \[ \frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}+\frac {2 a^4 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(3 A+22 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (13 A+2 C)-\frac {(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}-\frac {a (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d} \]

[Out]

1/2*a^4*(13*A+2*C)*x+2*a^4*(2*A+3*C)*arctanh(sin(d*x+c))/d+5/2*a^4*(A-2*C)*sin(d*x+c)/d-1/6*a*(3*A-2*C)*(a+a*s
ec(d*x+c))^3*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+a*sec(d*x+c))^4*sin(d*x+c)/d-1/2*(A-2*C)*(a^2+a^2*sec(d*x+c))^2*
sin(d*x+c)/d+1/6*(3*A+22*C)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.53, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4087, 4018, 3996, 3770} \[ \frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}+\frac {2 a^4 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac {(3 A+22 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (13 A+2 C)-\frac {a (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(13*A + 2*C)*x)/2 + (2*a^4*(2*A + 3*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(A - 2*C)*Sin[c + d*x])/(2*d) -
(a*(3*A - 2*C)*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^4*Sin[c + d*x
])/(2*d) - ((A - 2*C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((3*A + 22*C)*(a^4 + a^4*Sec[c + d*x])*
Sin[c + d*x])/(6*d)

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3996

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

Rule 4018

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

Rule 4087

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

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^4 (4 a A-a (3 A-2 C) \sec (c+d x)) \, dx}{2 a}\\ &=-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^3 \left (a^2 (15 A-2 C)-6 a^2 (A-2 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (4 a^3 (9 A-4 C)+2 a^3 (3 A+22 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (30 a^4 (A-2 C)+24 a^4 (2 A+3 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {\int \left (-6 a^5 (13 A+2 C)-24 a^5 (2 A+3 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {1}{2} a^4 (13 A+2 C) x+\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (2 a^4 (2 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (13 A+2 C) x+\frac {2 a^4 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [B]  time = 6.46, size = 1420, normalized size = 7.40 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

((13*A + 2*C)*x*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(16*(A + 2*
C + A*Cos[2*c + 2*d*x])) + ((-2*A - 3*C)*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 +
(d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(4*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((2*A + 3*C)
*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A +
C*Sec[c + d*x]^2))/(4*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a +
a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[c])/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[2*d*x]*Cos[c +
d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2*c])/(32*d*(A + 2*C + A*Cos[2*
c + 2*d*x])) + (A*Cos[c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin
[d*x])/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[2*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d
*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2*d*x])/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^6*Sec[c/2 +
(d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(48*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(C
os[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a
*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*(13*C*Cos[c/2] - 11*C*Sin[c/2]))/(96*d*(A + 2*C + A*Cos[2*c + 2*d*x])*
(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a +
a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*(3*A*Sin[(d*x)/2] + 20*C*Sin[(d*x)/2]))/(24*d*(A + 2*C + A*Cos[2*c +
2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/
2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(48*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2
] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c
+ d*x])^4*(A + C*Sec[c + d*x]^2)*(-13*C*Cos[c/2] - 11*C*Sin[c/2]))/(96*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[
c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Se
c[c + d*x])^4*(A + C*Sec[c + d*x]^2)*(3*A*Sin[(d*x)/2] + 20*C*Sin[(d*x)/2]))/(24*d*(A + 2*C + A*Cos[2*c + 2*d*
x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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fricas [A]  time = 0.46, size = 170, normalized size = 0.89 \[ \frac {3 \, {\left (13 \, A + 2 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (2 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 24 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 12 \, C a^{4} \cos \left (d x + c\right ) + 2 \, C a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*(13*A + 2*C)*a^4*d*x*cos(d*x + c)^3 + 6*(2*A + 3*C)*a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 6*(2*A +
3*C)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (3*A*a^4*cos(d*x + c)^4 + 24*A*a^4*cos(d*x + c)^3 + 2*(3*A +
20*C)*a^4*cos(d*x + c)^2 + 12*C*a^4*cos(d*x + c) + 2*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.65, size = 248, normalized size = 1.29 \[ \frac {3 \, {\left (13 \, A a^{4} + 2 \, C a^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (2 \, A a^{4} + 3 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (2 \, A a^{4} + 3 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {4 \, {\left (3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 38 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(13*A*a^4 + 2*C*a^4)*(d*x + c) + 12*(2*A*a^4 + 3*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*(2*A*a^
4 + 3*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(7*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/
2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 4*(3*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 15*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 6
*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 38*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 27*C*a^4*tan(
1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 1.51, size = 189, normalized size = 0.98 \[ \frac {A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {13 A \,a^{4} x}{2}+\frac {13 A \,a^{4} c}{2 d}+a^{4} C x +\frac {C \,a^{4} c}{d}+\frac {4 A \,a^{4} \sin \left (d x +c \right )}{d}+\frac {6 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {20 a^{4} C \tan \left (d x +c \right )}{3 d}+\frac {4 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

1/2/d*A*a^4*cos(d*x+c)*sin(d*x+c)+13/2*A*a^4*x+13/2/d*A*a^4*c+a^4*C*x+1/d*C*a^4*c+4/d*A*a^4*sin(d*x+c)+6/d*a^4
*C*ln(sec(d*x+c)+tan(d*x+c))+20/3/d*a^4*C*tan(d*x+c)+4/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+2/d*a^4*C*sec(d*x+c)*
tan(d*x+c)+1/d*A*a^4*tan(d*x+c)+1/3/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.35, size = 211, normalized size = 1.10 \[ \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 72 \, {\left (d x + c\right )} A a^{4} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 12 \, {\left (d x + c\right )} C a^{4} - 12 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 12 \, A a^{4} \tan \left (d x + c\right ) + 72 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 72*(d*x + c)*A*a^4 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^
4 + 12*(d*x + c)*C*a^4 - 12*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x +
c) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*C*a^4*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 48*A*a^4*sin(d*x + c) + 12*A*a^4*tan(d*x + c) + 72*C*a^4*tan(d*x + c))/d

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mupad [B]  time = 2.85, size = 252, normalized size = 1.31 \[ \frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^4,x)

[Out]

(4*A*a^4*sin(c + d*x))/d + (13*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*A*a^4*atanh(sin(c/2 +
(d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*C*a^4*atanh(s
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (20*C*a^4*sin(c + d*x))/(3*
d*cos(c + d*x)) + (2*C*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (C*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (A*a
^4*cos(c + d*x)*sin(c + d*x))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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