3.5.42 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [442]

Optimal. Leaf size=209 \[ \frac {1}{2} a^4 (13 A+8 B+2 C) x+\frac {a^4 (8 A+13 B+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {(A-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+18 B+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

1/2*a^4*(13*A+8*B+2*C)*x+1/2*a^4*(8*A+13*B+12*C)*arctanh(sin(d*x+c))/d+5/2*a^4*(A-B-2*C)*sin(d*x+c)/d-1/6*a*(3
*A-2*C)*(a+a*sec(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-B-2*C)*(a^2+a^
2*sec(d*x+c))^2*sin(d*x+c)/d+1/6*(3*A+18*B+22*C)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]
time = 0.40, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4171, 4103, 4081, 3855} \begin {gather*} \frac {5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}+\frac {a^4 (8 A+13 B+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(3 A+18 B+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+8 B+2 C)-\frac {(A-B-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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3855

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

Rule 4081

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 4103

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] &&
NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rule 4171

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 ^2(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 (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 (2 a (2 A+B)-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+6 B-2 C)-6 a^2 (A-B-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-B-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 (2 a^3 (18 A+3 B-8 C)+2 a^3 (3 A+18 B+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-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+18 B+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-B-2 C)+6 a^4 (8 A+13 B+12 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {5 a^4 (A-B-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-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+18 B+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+8 B+2 C)-6 a^5 (8 A+13 B+12 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {1}{2} a^4 (13 A+8 B+2 C) x+\frac {5 a^4 (A-B-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-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+18 B+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{2} \left (a^4 (8 A+13 B+12 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (13 A+8 B+2 C) x+\frac {a^4 (8 A+13 B+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^4 (A-B-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-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+18 B+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] Leaf count is larger than twice the leaf count of optimal. \(524\) vs. \(2(209)=418\).
time = 4.82, size = 524, normalized size = 2.51 \begin {gather*} \frac {a^4 (1+\cos (c+d x))^4 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-96 (8 A+13 B+12 C) \cos ^3(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec (c) (36 (13 A+8 B+2 C) d x \cos (d x)+36 (13 A+8 B+2 C) d x \cos (2 c+d x)+156 A d x \cos (2 c+3 d x)+96 B d x \cos (2 c+3 d x)+24 C d x \cos (2 c+3 d x)+156 A d x \cos (4 c+3 d x)+96 B d x \cos (4 c+3 d x)+24 C d x \cos (4 c+3 d x)+102 A \sin (d x)+384 B \sin (d x)+672 C \sin (d x)-42 A \sin (2 c+d x)-192 B \sin (2 c+d x)-288 C \sin (2 c+d x)+96 A \sin (c+2 d x)+48 B \sin (c+2 d x)+96 C \sin (c+2 d x)+96 A \sin (3 c+2 d x)+48 B \sin (3 c+2 d x)+96 C \sin (3 c+2 d x)+57 A \sin (2 c+3 d x)+192 B \sin (2 c+3 d x)+320 C \sin (2 c+3 d x)+9 A \sin (4 c+3 d x)+48 A \sin (3 c+4 d x)+12 B \sin (3 c+4 d x)+48 A \sin (5 c+4 d x)+12 B \sin (5 c+4 d x)+3 A \sin (4 c+5 d x)+3 A \sin (6 c+5 d x))\right )}{1536 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^4*(1 + Cos[c + d*x])^4*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^8*Sec[c + d*x]^3*(-96*(8*A
+ 13*B + 12*C)*Cos[c + d*x]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]]) + Sec[c]*(36*(13*A + 8*B + 2*C)*d*x*Cos[d*x] + 36*(13*A + 8*B + 2*C)*d*x*Cos[2*c + d*x] + 156*A*d*x*Cos[
2*c + 3*d*x] + 96*B*d*x*Cos[2*c + 3*d*x] + 24*C*d*x*Cos[2*c + 3*d*x] + 156*A*d*x*Cos[4*c + 3*d*x] + 96*B*d*x*C
os[4*c + 3*d*x] + 24*C*d*x*Cos[4*c + 3*d*x] + 102*A*Sin[d*x] + 384*B*Sin[d*x] + 672*C*Sin[d*x] - 42*A*Sin[2*c
+ d*x] - 192*B*Sin[2*c + d*x] - 288*C*Sin[2*c + d*x] + 96*A*Sin[c + 2*d*x] + 48*B*Sin[c + 2*d*x] + 96*C*Sin[c
+ 2*d*x] + 96*A*Sin[3*c + 2*d*x] + 48*B*Sin[3*c + 2*d*x] + 96*C*Sin[3*c + 2*d*x] + 57*A*Sin[2*c + 3*d*x] + 192
*B*Sin[2*c + 3*d*x] + 320*C*Sin[2*c + 3*d*x] + 9*A*Sin[4*c + 3*d*x] + 48*A*Sin[3*c + 4*d*x] + 12*B*Sin[3*c + 4
*d*x] + 48*A*Sin[5*c + 4*d*x] + 12*B*Sin[5*c + 4*d*x] + 3*A*Sin[4*c + 5*d*x] + 3*A*Sin[6*c + 5*d*x])))/(1536*d
*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]
time = 1.00, size = 280, normalized size = 1.34 Too large to display

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+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 320, normalized size = 1.53 \begin {gather*} \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} + 48 \, {\left (d x + c\right )} B 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} - 3 \, B 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 )} - 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 )} + 36 \, 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 )} + 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 \, B a^{4} \sin \left (d x + c\right ) + 12 \, A a^{4} \tan \left (d x + c\right ) + 48 \, B a^{4} \tan \left (d x + c\right ) + 72 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \end {gather*}

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

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Fricas [A]
time = 2.43, size = 191, normalized size = 0.91 \begin {gather*} \frac {6 \, {\left (13 \, A + 8 \, B + 2 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, A + 13 \, B + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, A + 13 \, B + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 6 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 12 \, B + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 2 \, C a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

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+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*(13*A + 8*B + 2*C)*a^4*d*x*cos(d*x + c)^3 + 3*(8*A + 13*B + 12*C)*a^4*cos(d*x + c)^3*log(sin(d*x + c)
+ 1) - 3*(8*A + 13*B + 12*C)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(3*A*a^4*cos(d*x + c)^4 + 6*(4*A +
B)*a^4*cos(d*x + c)^3 + 2*(3*A + 12*B + 20*C)*a^4*cos(d*x + c)^2 + 3*(B + 4*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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]
time = 0.56, size = 347, normalized size = 1.66 \begin {gather*} \frac {3 \, {\left (13 \, A a^{4} + 8 \, B a^{4} + 2 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (8 \, A a^{4} + 13 \, B a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, A a^{4} + 13 \, B a^{4} + 12 \, 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} + 2 \, B 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 ) + 2 \, B 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 (6 \, 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} + 30 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, 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} - 76 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, 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 ) + 54 \, 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} \end {gather*}

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+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(13*A*a^4 + 8*B*a^4 + 2*C*a^4)*(d*x + c) + 3*(8*A*a^4 + 13*B*a^4 + 12*C*a^4)*log(abs(tan(1/2*d*x + 1/2*
c) + 1)) - 3*(8*A*a^4 + 13*B*a^4 + 12*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 + 2*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) + 2*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2
*d*x + 1/2*c)^2 + 1)^2 - 2*(6*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 + 30*C*a^4*tan(1/
2*d*x + 1/2*c)^5 - 12*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 - 76*C*a^4*tan(1/2*d*x +
1/2*c)^3 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c) + 54*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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Mupad [B]
time = 5.06, size = 625, normalized size = 2.99 \begin {gather*} \frac {3\,A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {33\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{32}+\frac {3\,A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{2}+\frac {3\,A\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{32}+\frac {3\,B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{2}+3\,B\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {3\,B\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+3\,C\,a^4\,\sin \left (2\,c+2\,d\,x\right )+5\,C\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {15\,A\,a^4\,\sin \left (c+d\,x\right )}{16}+3\,B\,a^4\,\sin \left (c+d\,x\right )+6\,C\,a^4\,\sin \left (c+d\,x\right )+\frac {117\,A\,a^4\,\cos \left (c+d\,x\right )\,\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 )}{4}+18\,A\,a^4\,\cos \left (c+d\,x\right )\,\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 )+18\,B\,a^4\,\cos \left (c+d\,x\right )\,\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 {117\,B\,a^4\,\cos \left (c+d\,x\right )\,\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 )}{4}+\frac {9\,C\,a^4\,\cos \left (c+d\,x\right )\,\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}+27\,C\,a^4\,\cos \left (c+d\,x\right )\,\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 )+\frac {39\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{4}+6\,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 )\,\cos \left (3\,c+3\,d\,x\right )+6\,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 )\,\cos \left (3\,c+3\,d\,x\right )+\frac {39\,B\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}+9\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{3\,d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*A*a^4*sin(2*c + 2*d*x) + (33*A*a^4*sin(3*c + 3*d*x))/32 + (3*A*a^4*sin(4*c + 4*d*x))/2 + (3*A*a^4*sin(5*c +
5*d*x))/32 + (3*B*a^4*sin(2*c + 2*d*x))/2 + 3*B*a^4*sin(3*c + 3*d*x) + (3*B*a^4*sin(4*c + 4*d*x))/8 + 3*C*a^4
*sin(2*c + 2*d*x) + 5*C*a^4*sin(3*c + 3*d*x) + (15*A*a^4*sin(c + d*x))/16 + 3*B*a^4*sin(c + d*x) + 6*C*a^4*sin
(c + d*x) + (117*A*a^4*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + 18*A*a^4*cos(c + d*x)*ata
nh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 18*B*a^4*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
+ (117*B*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (9*C*a^4*cos(c + d*x)*atan(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + 27*C*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (39*
A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + 6*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c
/2 + (d*x)/2))*cos(3*c + 3*d*x) + 6*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x) + (39*B
*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + (3*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c
/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 + 9*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/(3
*d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))

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