3.444 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=217 \[ \frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac {1}{8} a^4 x (35 A+48 B+52 C)+\frac {a^4 (B+4 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{8 d}+\frac {a (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d} \]
[Out]
1/8*a^4*(35*A+48*B+52*C)*x+a^4*(B+4*C)*arctanh(sin(d*x+c))/d+5/8*a^4*(7*A+8*B+4*C)*sin(d*x+c)/d+1/3*a*(A+B)*co
s(d*x+c)^2*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^4*sin(d*x+c)/d+1/8*(7*A+8*B+4*C
)*cos(d*x+c)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d-1/24*(35*A+32*B-12*C)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d
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Rubi [A] time = 0.62, 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} \[ \frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac {(7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{8 d}+\frac {1}{8} a^4 x (35 A+48 B+52 C)+\frac {a^4 (B+4 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^4*(35*A + 48*B + 52*C)*x)/8 + (a^4*(B + 4*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(7*A + 8*B + 4*C)*Sin[c + d*
x])/(8*d) + (a*(A + B)*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]^3*(a + a*Se
c[c + d*x])^4*Sin[c + d*x])/(4*d) + ((7*A + 8*B + 4*C)*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(
8*d) - ((35*A + 32*B - 12*C)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(24*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 ^4(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 ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (4 a (A+B)-a (A-4 C) \sec (c+d x)) \, dx}{4 a}\\ &=\frac {a (A+B) \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (3 a^2 (7 A+8 B+4 C)-a^2 (7 A+4 B-12 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {a (A+B) \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {(7 A+8 B+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^3 (35 A+44 B+36 C)-a^3 (35 A+32 B-12 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {a (A+B) \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {(7 A+8 B+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^4 (7 A+8 B+4 C)+24 a^4 (B+4 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {(7 A+8 B+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}-\frac {\int \left (-3 a^5 (35 A+48 B+52 C)-24 a^5 (B+4 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {1}{8} a^4 (35 A+48 B+52 C) x+\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {(7 A+8 B+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^4 (B+4 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} a^4 (35 A+48 B+52 C) x+\frac {a^4 (B+4 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {(7 A+8 B+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end {align*}
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Mathematica [B] time = 6.23, size = 1436, normalized size = 6.62 \[ \text {result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
a^4*(((35*A + 48*B + 52*C)*x*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*
Sec[c + d*x]^2))/(64*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((-B - 4*C)*Cos[c + d*x]^2*(1 + Cos[
c + d*x])^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*Sec[c +
d*x]^2))/(8*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((B + 4*C)*Cos[c + d*x]^2*(1 + Cos[c + d*x]
)^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))
/(8*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((28*A + 27*B + 16*C)*Cos[d*x]*Cos[c + d*x]^2*(1 +
Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(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])) + ((7*A + 4*B + C)*Cos[2*d*x]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d
*x)/2]^8*(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])) + ((4*A + B)*Cos[3*d*x]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(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])) + (A*Cos[4*d*x]*Cos[c + d*
x]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[4*c])/(256*d*(A + 2
*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((28*A + 27*B + 16*C)*Cos[c]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^
4*Sec[c/2 + (d*x)/2]^8*(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])) + ((7*A + 4*B + C)*Cos[2*c]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(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])) + ((4*
A + B)*Cos[3*c]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(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])) + (A*Cos[4*c]*Cos[c + d*x]^2*(1 + Cos[
c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[4*d*x])/(256*d*(A + 2*C + 2*B*Cos
[c + d*x] + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(A + B*Sec[c +
d*x] + C*Sec[c + d*x]^2)*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]^2*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8
*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*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 = 179, normalized size = 0.82 \[ \frac {3 \, {\left (35 \, A + 48 \, B + 52 \, C\right )} a^{4} d x \cos \left (d x + c\right ) + 12 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (27 \, A + 16 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, C a^{4}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
1/24*(3*(35*A + 48*B + 52*C)*a^4*d*x*cos(d*x + c) + 12*(B + 4*C)*a^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 12*(
B + 4*C)*a^4*cos(d*x + c)*log(-sin(d*x + c) + 1) + (6*A*a^4*cos(d*x + c)^4 + 8*(4*A + B)*a^4*cos(d*x + c)^3 +
3*(27*A + 16*B + 4*C)*a^4*cos(d*x + c)^2 + 32*(5*A + 5*B + 3*C)*a^4*cos(d*x + c) + 24*C*a^4)*sin(d*x + c))/(d*
cos(d*x + c))
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giac [A] time = 0.34, size = 332, normalized size = 1.53 \[ -\frac {\frac {48 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4} + 52 \, C a^{4}\right )} {\left (d x + c\right )} - 24 \, {\left (B a^{4} + 4 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 24 \, {\left (B a^{4} + 4 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 84 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 385 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 276 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 300 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 279 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 108 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/24*(48*C*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(35*A*a^4 + 48*B*a^4 + 52*C*a^4)*(d*x +
c) - 24*(B*a^4 + 4*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 24*(B*a^4 + 4*C*a^4)*log(abs(tan(1/2*d*x + 1/2*
c) - 1)) - 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 84*C*a^4*tan(1/2*d*x + 1/2
*c)^7 + 385*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 424*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 276*C*a^4*tan(1/2*d*x + 1/2*c)^5
+ 511*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 520*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 300*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 27
9*A*a^4*tan(1/2*d*x + 1/2*c) + 216*B*a^4*tan(1/2*d*x + 1/2*c) + 108*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x +
1/2*c)^2 + 1)^4)/d
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maple [A] time = 1.48, size = 289, normalized size = 1.33 \[ \frac {4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {20 a^{4} B \sin \left (d x +c \right )}{3 d}+\frac {35 A \,a^{4} x}{8}+\frac {A \,a^{4} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {27 A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {a^{4} C \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {2 a^{4} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+6 a^{4} B x +\frac {20 A \,a^{4} \sin \left (d x +c \right )}{3 d}+\frac {4 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {35 A \,a^{4} c}{8 d}+\frac {a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {13 a^{4} C x}{2}+\frac {13 C \,a^{4} c}{2 d}+\frac {4 a^{4} C \sin \left (d x +c \right )}{d}+\frac {a^{4} C \tan \left (d x +c \right )}{d}+\frac {6 a^{4} B c}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
4/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+1/3/d*B*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*a^4*B*sin(d*x+c)+35/8*A*a^4*x+1
/4/d*A*a^4*sin(d*x+c)*cos(d*x+c)^3+27/8/d*A*a^4*cos(d*x+c)*sin(d*x+c)+1/2/d*a^4*C*sin(d*x+c)*cos(d*x+c)+2/d*a^
4*B*cos(d*x+c)*sin(d*x+c)+6*a^4*B*x+20/3/d*A*a^4*sin(d*x+c)+4/3/d*A*sin(d*x+c)*cos(d*x+c)^2*a^4+35/8/d*A*a^4*c
+1/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+13/2*a^4*C*x+13/2/d*C*a^4*c+4/d*a^4*C*sin(d*x+c)+1/d*a^4*C*tan(d*x+c)+6/d
*a^4*B*c
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maxima [A] time = 0.35, size = 290, normalized size = 1.34 \[ -\frac {128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 96 \, {\left (d x + c\right )} A a^{4} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 384 \, {\left (d x + c\right )} B a^{4} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 576 \, {\left (d x + c\right )} C a^{4} - 48 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 192 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 384 \, A a^{4} \sin \left (d x + c\right ) - 576 \, B a^{4} \sin \left (d x + c\right ) - 384 \, C a^{4} \sin \left (d x + c\right ) - 96 \, C a^{4} \tan \left (d x + c\right )}{96 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
-1/96*(128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*A*a^4 - 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 96*(d*x + c)*A*a^4 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*B*a^4 - 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 384*(d*x + c)*B*a^4 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c)
)*C*a^4 - 576*(d*x + c)*C*a^4 - 48*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 192*C*a^4*(log(sin(
d*x + c) + 1) - log(sin(d*x + c) - 1)) - 384*A*a^4*sin(d*x + c) - 576*B*a^4*sin(d*x + c) - 384*C*a^4*sin(d*x +
c) - 96*C*a^4*tan(d*x + c))/d
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mupad [B] time = 4.60, size = 1244, normalized size = 5.73 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^4*(a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
- (tan(c/2 + (d*x)/2)^5*((21*A*a^4)/2 + 8*B*a^4 - 10*C*a^4) + tan(c/2 + (d*x)/2)^9*((35*A*a^4)/4 + 10*B*a^4 +
5*C*a^4) - tan(c/2 + (d*x)/2)^3*((58*A*a^4)/3 + (76*B*a^4)/3 + 24*C*a^4) + tan(c/2 + (d*x)/2)^7*((70*A*a^4)/3
+ (76*B*a^4)/3 + 8*C*a^4) - tan(c/2 + (d*x)/2)*((93*A*a^4)/4 + 18*B*a^4 + 11*C*a^4))/(d*(3*tan(c/2 + (d*x)/2)^
2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(c/2 + (d*x)/2)^10 + 1)) - (
a^4*atan(((a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 1680*A*B*a^8 + 1820*A*C*a
^8 + 2752*B*C*a^8) - (a^4*(35*A + 48*B + 52*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4)*1i)/8)*(35*A + 48*B + 52*C)
)/8 + (a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 1680*A*B*a^8 + 1820*A*C*a^8 +
2752*B*C*a^8) + (a^4*(35*A + 48*B + 52*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4)*1i)/8)*(35*A + 48*B + 52*C))/8)
/(1920*B^3*a^12 + 4160*C^3*a^12 + 3080*A*B^2*a^12 + 1225*A^2*B*a^12 + 10080*A*C^2*a^12 + 4900*A^2*C*a^12 + 132
00*B*C^2*a^12 + 10720*B^2*C*a^12 - (a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 +
1680*A*B*a^8 + 1820*A*C*a^8 + 2752*B*C*a^8) - (a^4*(35*A + 48*B + 52*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4)*1i
)/8)*(35*A + 48*B + 52*C)*1i)/8 + (a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 1
680*A*B*a^8 + 1820*A*C*a^8 + 2752*B*C*a^8) + (a^4*(35*A + 48*B + 52*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4)*1i)
/8)*(35*A + 48*B + 52*C)*1i)/8 + 14840*A*B*C*a^12))*(35*A + 48*B + 52*C))/(4*d) - (a^4*atan((a^4*(tan(c/2 + (d
*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 1680*A*B*a^8 + 1820*A*C*a^8 + 2752*B*C*a^8) + a^4*(B
+ 4*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4))*(B + 4*C)*1i + a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^
2*a^8 + 1864*C^2*a^8 + 1680*A*B*a^8 + 1820*A*C*a^8 + 2752*B*C*a^8) - a^4*(B + 4*C)*(140*A*a^4 + 224*B*a^4 + 33
6*C*a^4))*(B + 4*C)*1i)/(1920*B^3*a^12 + 4160*C^3*a^12 + 3080*A*B^2*a^12 + 1225*A^2*B*a^12 + 10080*A*C^2*a^12
+ 4900*A^2*C*a^12 + 13200*B*C^2*a^12 + 10720*B^2*C*a^12 + a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2
*a^8 + 1864*C^2*a^8 + 1680*A*B*a^8 + 1820*A*C*a^8 + 2752*B*C*a^8) + a^4*(B + 4*C)*(140*A*a^4 + 224*B*a^4 + 336
*C*a^4))*(B + 4*C) - a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 1680*A*B*a^8 +
1820*A*C*a^8 + 2752*B*C*a^8) - a^4*(B + 4*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4))*(B + 4*C) + 14840*A*B*C*a^12
))*(B + 4*C)*2i)/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)**4*(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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