3.443 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=217 \[ \frac {5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac {a^4 (2 A+8 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(8 A-3 B-18 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (12 A+13 B+8 C)-\frac {(2 A+B-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac {a (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d} \]

[Out]

1/2*a^4*(12*A+13*B+8*C)*x+1/2*a^4*(2*A+8*B+13*C)*arctanh(sin(d*x+c))/d+5/2*a^4*(2*A+B-C)*sin(d*x+c)/d+1/6*a*(4
*A+3*B)*cos(d*x+c)*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+a*sec(d*x+c))^4*sin(d*x+c)/d-1/2*(2*A
+B-C)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d-1/6*(8*A-3*B-18*C)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.60, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4086, 4017, 4018, 3996, 3770} \[ \frac {5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac {a^4 (2 A+8 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(2 A+B-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}-\frac {(8 A-3 B-18 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (12 A+13 B+8 C)+\frac {a (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(12*A + 13*B + 8*C)*x)/2 + (a^4*(2*A + 8*B + 13*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^4*(2*A + B - C)*Si
n[c + d*x])/(2*d) + (a*(4*A + 3*B)*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]^2
*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(3*d) - ((2*A + B - C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) -
((8*A - 3*B - 18*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 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[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 4086

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*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 - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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 ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (a (4 A+3 B)-a (2 A-3 C) \sec (c+d x)) \, dx}{3 a}\\ &=\frac {a (4 A+3 B) \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^3 \left (a^2 (16 A+15 B+6 C)-6 a^2 (2 A+B-C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {a (4 A+3 B) \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A+B-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 (2 a^3 (22 A+18 B+3 C)-2 a^3 (8 A-3 B-18 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {a (4 A+3 B) \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A+B-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B-18 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 (2 A+B-C)+6 a^4 (2 A+8 B+13 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac {a (4 A+3 B) \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A+B-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B-18 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {\int \left (-6 a^5 (12 A+13 B+8 C)-6 a^5 (2 A+8 B+13 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {1}{2} a^4 (12 A+13 B+8 C) x+\frac {5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac {a (4 A+3 B) \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A+B-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B-18 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{2} \left (a^4 (2 A+8 B+13 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (12 A+13 B+8 C) x+\frac {a^4 (2 A+8 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac {a (4 A+3 B) \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A+B-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B-18 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 = 6.31, size = 1518, normalized size = 7.00 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((12*A + 13*B + 8*C)*x*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[
c + d*x]^2))/(16*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((-2*A - 8*B - 13*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 + B*Sec[c + d*x] + C*S
ec[c + d*x]^2))/(16*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((2*A + 8*B + 13*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 + B*Sec[c + d*x] +
C*Sec[c + d*x]^2))/(16*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((27*A + 16*B + 4*C)*Cos[d*x]*C
os[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(32*
d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((4*A + B)*Cos[2*d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]
^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[2*c])/(32*d*(A + 2*C + 2*B*Cos[c + d*x]
+ A*Cos[2*c + 2*d*x])) + (A*Cos[3*d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c
+ d*x] + C*Sec[c + d*x]^2)*Sin[3*c])/(96*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((27*A + 16*B
+ 4*C)*Cos[c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]
^2)*Sin[d*x])/(32*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((4*A + B)*Cos[2*c]*Cos[c + d*x]^6*Se
c[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[2*d*x])/(32*d*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (A*Cos[3*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x]
)^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[3*d*x])/(96*d*(A + 2*C + 2*B*Cos[c + d*x] + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3
2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(B*Sin[(d*x)/2] + 4
*C*Sin[(d*x)/2]))/(8*d*(A + 2*C + 2*B*Cos[c + d*x] + 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 + B*Sec[c + d*x
] + C*Sec[c + d*x]^2))/(32*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(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 + B*Sec[c + d*x] + C*Sec[c + d*
x]^2)*(B*Sin[(d*x)/2] + 4*C*Sin[(d*x)/2]))/(8*d*(A + 2*C + 2*B*Cos[c + d*x] + 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 = 0.48, size = 191, normalized size = 0.88 \[ \frac {6 \, {\left (12 \, A + 13 \, B + 8 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 8 \, B + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + 8 \, B + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (20 \, A + 12 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 6 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*(12*A + 13*B + 8*C)*a^4*d*x*cos(d*x + c)^2 + 3*(2*A + 8*B + 13*C)*a^4*cos(d*x + c)^2*log(sin(d*x + c)
+ 1) - 3*(2*A + 8*B + 13*C)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^4*cos(d*x + c)^4 + 3*(4*A + B
)*a^4*cos(d*x + c)^3 + 2*(20*A + 12*B + 3*C)*a^4*cos(d*x + c)^2 + 6*(B + 4*C)*a^4*cos(d*x + c) + 3*C*a^4)*sin(
d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 0.37, size = 347, normalized size = 1.60 \[ \frac {3 \, {\left (12 \, A a^{4} + 13 \, B a^{4} + 8 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (2 \, A a^{4} + 8 \, B a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{4} + 8 \, B a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {2 \, {\left (30 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 76 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 54 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(12*A*a^4 + 13*B*a^4 + 8*C*a^4)*(d*x + c) + 3*(2*A*a^4 + 8*B*a^4 + 13*C*a^4)*log(abs(tan(1/2*d*x + 1/2*
c) + 1)) - 3*(2*A*a^4 + 8*B*a^4 + 13*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(2*B*a^4*tan(1/2*d*x + 1/2*
c)^3 + 7*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^4*tan(1/2*d*x + 1/2*c) - 9*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*
d*x + 1/2*c)^2 - 1)^2 + 2*(30*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^4*tan(1/2
*d*x + 1/2*c)^5 + 76*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 48*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^4*tan(1/2*d*x + 1
/2*c)^3 + 54*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c) + 6*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 = 1.56, size = 280, normalized size = 1.29 \[ \frac {A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {20 A \,a^{4} \sin \left (d x +c \right )}{3 d}+\frac {a^{4} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {13 a^{4} B x}{2}+\frac {13 a^{4} B c}{2 d}+\frac {a^{4} C \sin \left (d x +c \right )}{d}+\frac {2 A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+6 A \,a^{4} x +\frac {6 A \,a^{4} c}{d}+\frac {4 a^{4} B \sin \left (d x +c \right )}{d}+4 a^{4} C x +\frac {4 C \,a^{4} c}{d}+\frac {13 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{4} C \tan \left (d x +c \right )}{d}+\frac {A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{4} B \tan \left (d x +c \right )}{d}+\frac {a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.37, size = 296, normalized size = 1.36 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 48 \, {\left (d x + c\right )} A a^{4} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 72 \, {\left (d x + c\right )} B a^{4} - 48 \, {\left (d x + c\right )} C a^{4} + 3 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} \sin \left (d x + c\right ) - 48 \, B a^{4} \sin \left (d x + c\right ) - 12 \, C a^{4} \sin \left (d x + c\right ) - 12 \, B a^{4} \tan \left (d x + c\right ) - 48 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 48*(d*x + c)*A*
a^4 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 72*(d*x + c)*B*a^4 - 48*(d*x + c)*C*a^4 + 3*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)) - 6*A*a^4*(log(sin(d*x + c) + 1) -
log(sin(d*x + c) - 1)) - 24*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*C*a^4*(log(sin(d*x + c)
+ 1) - log(sin(d*x + c) - 1)) - 72*A*a^4*sin(d*x + c) - 48*B*a^4*sin(d*x + c) - 12*C*a^4*sin(d*x + c) - 12*B*
a^4*tan(d*x + c) - 48*C*a^4*tan(d*x + c))/d

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mupad [B]  time = 4.53, size = 373, normalized size = 1.72 \[ \frac {2\,\left (6\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}+\frac {13\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}+4\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,13{}\mathrm {i}}{2}\right )}{d}+\frac {\frac {A\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {83\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{4}+\frac {A\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{48}+\frac {5\,B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{8}+B\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {B\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{16}+2\,C\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {C\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {41\,A\,a^4\,\sin \left (c+d\,x\right )}{24}+B\,a^4\,\sin \left (c+d\,x\right )+\frac {3\,C\,a^4\,\sin \left (c+d\,x\right )}{4}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*(6*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*a^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2
))*1i + (13*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - B*a^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2
+ (d*x)/2))*4i + 4*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (C*a^4*atan((sin(c/2 + (d*x)/2)*1i)/cos
(c/2 + (d*x)/2))*13i)/2))/d + ((A*a^4*sin(2*c + 2*d*x))/2 + (83*A*a^4*sin(3*c + 3*d*x))/48 + (A*a^4*sin(4*c +
4*d*x))/4 + (A*a^4*sin(5*c + 5*d*x))/48 + (5*B*a^4*sin(2*c + 2*d*x))/8 + B*a^4*sin(3*c + 3*d*x) + (B*a^4*sin(4
*c + 4*d*x))/16 + 2*C*a^4*sin(2*c + 2*d*x) + (C*a^4*sin(3*c + 3*d*x))/4 + (41*A*a^4*sin(c + d*x))/24 + B*a^4*s
in(c + d*x) + (3*C*a^4*sin(c + d*x))/4)/(d*(cos(2*c + 2*d*x)/2 + 1/2))

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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)**3*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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