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

Optimal. Leaf size=169 \[ -\frac {(5 A+3 B-6 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {1}{2} a^3 x (5 A+7 B+6 C)+\frac {a^3 (B+3 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(A+B) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 a d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

[Out]

1/2*a^3*(5*A+7*B+6*C)*x+a^3*(B+3*C)*arctanh(sin(d*x+c))/d+5/2*a^3*(A+B)*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+a*s
ec(d*x+c))^3*sin(d*x+c)/d+1/2*(A+B)*cos(d*x+c)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d-1/6*(5*A+3*B-6*C)*(a^3+a^
3*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.44, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {(5 A+3 B-6 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {(A+B) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 a d}+\frac {1}{2} a^3 x (5 A+7 B+6 C)+\frac {a^3 (B+3 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(5*A + 7*B + 6*C)*x)/2 + (a^3*(B + 3*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(A + B)*Sin[c + d*x])/(2*d) + (
A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((A + B)*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])^2
*Sin[c + d*x])/(2*a*d) - ((5*A + 3*B - 6*C)*(a^3 + a^3*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))^3 \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))^3 \sin (c+d x)}{3 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 (3 a (A+B)-a (A-3 C) \sec (c+d x)) \, dx}{3 a}\\ &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (5 A+6 B+3 C)-a^2 (5 A+3 B-6 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A+3 B-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (A+B)+6 a^3 (B+3 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A+3 B-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {\int \left (-3 a^4 (5 A+7 B+6 C)-6 a^4 (B+3 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {1}{2} a^3 (5 A+7 B+6 C) x+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A+3 B-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^3 (B+3 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^3 (5 A+7 B+6 C) x+\frac {a^3 (B+3 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A+3 B-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [B]  time = 2.27, size = 379, normalized size = 2.24 \[ \frac {a^3 \cos ^2(c+d x) (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {3 \sin (c) (15 A+4 (3 B+C)) \cos (d x)}{d}+\frac {3 \cos (c) (15 A+4 (3 B+C)) \sin (d x)}{d}+\frac {3 (3 A+B) \sin (2 c) \cos (2 d x)}{d}+\frac {3 (3 A+B) \cos (2 c) \sin (2 d x)}{d}+6 x (5 A+7 B+6 C)+\frac {A \sin (3 c) \cos (3 d x)}{d}+\frac {A \cos (3 c) \sin (3 d x)}{d}-\frac {12 (B+3 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {12 (B+3 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {12 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 C \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{48 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(6*(5*A +
7*B + 6*C)*x - (12*(B + 3*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (12*(B + 3*C)*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]])/d + (3*(15*A + 4*(3*B + C))*Cos[d*x]*Sin[c])/d + (3*(3*A + B)*Cos[2*d*x]*Sin[2*c])/d + (A
*Cos[3*d*x]*Sin[3*c])/d + (3*(15*A + 4*(3*B + C))*Cos[c]*Sin[d*x])/d + (3*(3*A + B)*Cos[2*c]*Sin[2*d*x])/d + (
A*Cos[3*c]*Sin[3*d*x])/d + (12*C*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))
+ (12*C*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(48*(A + 2*C + 2*B*Co
s[c + d*x] + A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.49, size = 156, normalized size = 0.92 \[ \frac {3 \, {\left (5 \, A + 7 \, B + 6 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 3 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 9 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(5*A + 7*B + 6*C)*a^3*d*x*cos(d*x + c) + 3*(B + 3*C)*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - 3*(B + 3*
C)*a^3*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*A*a^3*cos(d*x + c)^3 + 3*(3*A + B)*a^3*cos(d*x + c)^2 + 2*(11*
A + 9*B + 3*C)*a^3*cos(d*x + c) + 6*C*a^3)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.35, size = 281, normalized size = 1.66 \[ -\frac {\frac {12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (5 \, A a^{3} + 7 \, B a^{3} + 6 \, C a^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*C*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(5*A*a^3 + 7*B*a^3 + 6*C*a^3)*(d*x + c) -
6*(B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 4
0*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*A*a^3*
tan(1/2*d*x + 1/2*c) + 21*B*a^3*tan(1/2*d*x + 1/2*c) + 6*C*a^3*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 = 1.44, size = 221, normalized size = 1.31 \[ \frac {A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {11 a^{3} A \sin \left (d x +c \right )}{3 d}+\frac {a^{3} B \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {7 a^{3} B x}{2}+\frac {7 a^{3} B c}{2 d}+\frac {a^{3} C \sin \left (d x +c \right )}{d}+\frac {3 A \,a^{3} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {5 a^{3} A x}{2}+\frac {5 A \,a^{3} c}{2 d}+\frac {3 a^{3} B \sin \left (d x +c \right )}{d}+3 a^{3} C x +\frac {3 C \,a^{3} c}{d}+\frac {3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3} C \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d*A*cos(d*x+c)^2*sin(d*x+c)*a^3+11/3*a^3*A*sin(d*x+c)/d+1/2/d*a^3*B*sin(d*x+c)*cos(d*x+c)+7/2*a^3*B*x+7/2/
d*a^3*B*c+a^3*C*sin(d*x+c)/d+3/2/d*A*a^3*sin(d*x+c)*cos(d*x+c)+5/2*a^3*A*x+5/2/d*A*a^3*c+3*a^3*B*sin(d*x+c)/d+
3*a^3*C*x+3/d*C*a^3*c+3/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+a^3*C*tan(d*x+c)
/d

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maxima [A]  time = 0.37, size = 210, normalized size = 1.24 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 12 \, {\left (d x + c\right )} A a^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 36 \, {\left (d x + c\right )} B a^{3} - 36 \, {\left (d x + c\right )} C a^{3} - 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 18 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, A a^{3} \sin \left (d x + c\right ) - 36 \, B a^{3} \sin \left (d x + c\right ) - 12 \, C a^{3} \sin \left (d x + c\right ) - 12 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(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^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 12*(d*x + c)*A*a
^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 36*(d*x + c)*B*a^3 - 36*(d*x + c)*C*a^3 - 6*B*a^3*(log(sin(d*x
+ c) + 1) - log(sin(d*x + c) - 1)) - 18*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*A*a^3*sin(
d*x + c) - 36*B*a^3*sin(d*x + c) - 12*C*a^3*sin(d*x + c) - 12*C*a^3*tan(d*x + c))/d

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mupad [B]  time = 3.80, size = 290, normalized size = 1.72 \[ \frac {5\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+7\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+6\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {23\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{12}+\frac {3\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{24}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{8}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{8}+C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 7*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B
*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i + 6*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
- C*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*6i)/d + ((23*A*a^3*sin(2*c + 2*d*x))/12 + (3*A*a^3*s
in(3*c + 3*d*x))/8 + (A*a^3*sin(4*c + 4*d*x))/24 + (3*B*a^3*sin(2*c + 2*d*x))/2 + (B*a^3*sin(3*c + 3*d*x))/8 +
(C*a^3*sin(2*c + 2*d*x))/2 + (3*A*a^3*sin(c + d*x))/8 + (B*a^3*sin(c + d*x))/8 + C*a^3*sin(c + d*x))/(d*cos(c
+ d*x))

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

[Out]

Timed out

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