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

Optimal. Leaf size=196 \[ \frac {a \sin (c+d x) \left (a^2 (2 A+3 C)+6 a b B+3 A b^2\right )}{3 d}+\frac {1}{2} x \left (a^3 B+3 a^2 b (A+2 C)+6 a b^2 B+2 A b^3\right )-\frac {b^2 \tan (c+d x) (3 a B+5 A b-6 b C)}{6 d}+\frac {(a B+A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac {b^2 (3 a C+b B) \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

1/2*(2*A*b^3+a^3*B+6*a*b^2*B+3*a^2*b*(A+2*C))*x+b^2*(B*b+3*C*a)*arctanh(sin(d*x+c))/d+1/3*a*(3*A*b^2+6*a*b*B+a
^2*(2*A+3*C))*sin(d*x+c)/d+1/2*(A*b+B*a)*cos(d*x+c)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*se
c(d*x+c))^3*sin(d*x+c)/d-1/6*b^2*(5*A*b+3*B*a-6*C*b)*tan(d*x+c)/d

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Rubi [A]  time = 0.60, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4094, 4076, 4047, 8, 4045, 3770} \[ \frac {a \sin (c+d x) \left (a^2 (2 A+3 C)+6 a b B+3 A b^2\right )}{3 d}+\frac {1}{2} x \left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right )-\frac {b^2 \tan (c+d x) (3 a B+5 A b-6 b C)}{6 d}+\frac {(a B+A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}+\frac {b^2 (3 a C+b B) \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*b^3 + a^3*B + 6*a*b^2*B + 3*a^2*b*(A + 2*C))*x)/2 + (b^2*(b*B + 3*a*C)*ArcTanh[Sin[c + d*x]])/d + (a*(3*
A*b^2 + 6*a*b*B + a^2*(2*A + 3*C))*Sin[c + d*x])/(3*d) + ((A*b + a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[
c + d*x])/(2*d) + (A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) - (b^2*(5*A*b + 3*a*B - 6*b*C)*
Tan[c + d*x])/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3770

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4076

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_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -1]

Rule 4094

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/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 1.32, size = 263, normalized size = 1.34 \[ \frac {a^3 A \sin (3 (c+d x))+3 a \sin (c+d x) \left (a^2 (3 A+4 C)+12 a b B+12 A b^2\right )+3 a^2 (a B+3 A b) \sin (2 (c+d x))+6 (c+d x) \left (a^3 B+3 a^2 b (A+2 C)+6 a b^2 B+2 A b^3\right )-12 b^2 (3 a C+b B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^2 (3 a C+b B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*A*b^3 + a^3*B + 6*a*b^2*B + 3*a^2*b*(A + 2*C))*(c + d*x) - 12*b^2*(b*B + 3*a*C)*Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 12*b^2*(b*B + 3*a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*b^3*C*Sin[(c + d*x)/2])
/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (12*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3
*a*(12*A*b^2 + 12*a*b*B + a^2*(3*A + 4*C))*Sin[c + d*x] + 3*a^2*(3*A*b + a*B)*Sin[2*(c + d*x)] + a^3*A*Sin[3*(
c + d*x)])/(12*d)

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fricas [A]  time = 0.83, size = 201, normalized size = 1.03 \[ \frac {3 \, {\left (B a^{3} + 3 \, {\left (A + 2 \, C\right )} a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 6 \, C b^{3} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{3} + 9 \, B a^{2} b + 9 \, A a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(B*a^3 + 3*(A + 2*C)*a^2*b + 6*B*a*b^2 + 2*A*b^3)*d*x*cos(d*x + c) + 3*(3*C*a*b^2 + B*b^3)*cos(d*x + c)
*log(sin(d*x + c) + 1) - 3*(3*C*a*b^2 + B*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*A*a^3*cos(d*x + c)^3 +
6*C*b^3 + 3*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^2 + 2*((2*A + 3*C)*a^3 + 9*B*a^2*b + 9*A*a*b^2)*cos(d*x + c))*si
n(d*x + c))/(d*cos(d*x + c))

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giac [B]  time = 0.34, size = 418, normalized size = 2.13 \[ -\frac {\frac {12 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (B a^{3} + 3 \, A a^{2} b + 6 \, C a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*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*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(B*a^3 + 3*A*a^2*b + 6*C*a^2*b + 6*B*a*b^
2 + 2*A*b^3)*(d*x + c) - 6*(3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(3*C*a*b^2 + B*b^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*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 - 9*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*A*a*b^2*tan
(1/2*d*x + 1/2*c)^5 + 4*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*B*a^2*b*tan(1/2*d*
x + 1/2*c)^3 + 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1/2*d*x + 1/2*c) + 3*B*a^3*tan(1/2*d*x + 1/2*c)
+ 6*C*a^3*tan(1/2*d*x + 1/2*c) + 9*A*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a^2*b*tan(1/2*d*x + 1/2*c) + 18*A*a*b^
2*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 = 1.27, size = 278, normalized size = 1.42 \[ \frac {A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {2 a^{3} A \sin \left (d x +c \right )}{3 d}+\frac {a^{3} B \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {a^{3} B x}{2}+\frac {a^{3} B c}{2 d}+\frac {a^{3} C \sin \left (d x +c \right )}{d}+\frac {3 A \,a^{2} b \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {3 a^{2} A b x}{2}+\frac {3 A \,a^{2} b c}{2 d}+\frac {3 a^{2} b B \sin \left (d x +c \right )}{d}+3 C x \,a^{2} b +\frac {3 C \,a^{2} b c}{d}+\frac {3 A a \,b^{2} \sin \left (d x +c \right )}{d}+3 B x a \,b^{2}+\frac {3 B a \,b^{2} c}{d}+\frac {3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+A x \,b^{3}+\frac {A \,b^{3} c}{d}+\frac {b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{3} C \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*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+2/3*a^3*A*sin(d*x+c)/d+1/2/d*a^3*B*sin(d*x+c)*cos(d*x+c)+1/2*a^3*B*x+1/2/d
*a^3*B*c+a^3*C*sin(d*x+c)/d+3/2/d*A*a^2*b*sin(d*x+c)*cos(d*x+c)+3/2*a^2*A*b*x+3/2/d*A*a^2*b*c+3/d*a^2*b*B*sin(
d*x+c)+3*C*x*a^2*b+3/d*C*a^2*b*c+3/d*A*a*b^2*sin(d*x+c)+3*B*x*a*b^2+3/d*B*a*b^2*c+3/d*C*a*b^2*ln(sec(d*x+c)+ta
n(d*x+c))+A*x*b^3+1/d*A*b^3*c+1/d*b^3*B*ln(sec(d*x+c)+tan(d*x+c))+1/d*b^3*C*tan(d*x+c)

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maxima [A]  time = 0.37, size = 216, normalized size = 1.10 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 36 \, {\left (d x + c\right )} C a^{2} b - 36 \, {\left (d x + c\right )} B a b^{2} - 12 \, {\left (d x + c\right )} A b^{3} - 18 \, C a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, B b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{3} \sin \left (d x + c\right ) - 36 \, B a^{2} b \sin \left (d x + c\right ) - 36 \, A a b^{2} \sin \left (d x + c\right ) - 12 \, C b^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 6.41, size = 2470, normalized size = 12.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(2*A*a^3 + B*a^3 + 2*C*a^3 + 2*C*b^3 + 6*A*a*b^2 + 3*A*a^2*b + 6*B*a^2*b) - tan(c/2 + (d*x
)/2)^7*(2*A*a^3 - B*a^3 + 2*C*a^3 - 2*C*b^3 + 6*A*a*b^2 - 3*A*a^2*b + 6*B*a^2*b) + tan(c/2 + (d*x)/2)^3*(2*C*a
^3 - B*a^3 - (2*A*a^3)/3 + 6*C*b^3 + 6*A*a*b^2 - 3*A*a^2*b + 6*B*a^2*b) - tan(c/2 + (d*x)/2)^5*(B*a^3 - (2*A*a
^3)/3 + 2*C*a^3 - 6*C*b^3 + 6*A*a*b^2 + 3*A*a^2*b + 6*B*a^2*b))/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)
/2)^6 - tan(c/2 + (d*x)/2)^8 + 1)) - (atan((((B*b^3 + 3*C*a*b^2)*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b
+ 96*B*a*b^2 + 96*C*a*b^2 + 96*C*a^2*b) + tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2
*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 288*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 192*A*B*a*b^5 +
48*A*B*a^5*b + 192*B*C*a*b^5 + 96*B*C*a^5*b + 320*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 288*A*C*a^4*b^2 + 576*B*C*a
^3*b^3))*(B*b^3 + 3*C*a*b^2)*1i - ((B*b^3 + 3*C*a*b^2)*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b
^2 + 96*C*a*b^2 + 96*C*a^2*b) - tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*
A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 288*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^
5*b + 192*B*C*a*b^5 + 96*B*C*a^5*b + 320*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 288*A*C*a^4*b^2 + 576*B*C*a^3*b^3))*(
B*b^3 + 3*C*a*b^2)*1i)/(((B*b^3 + 3*C*a*b^2)*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2 + 96*C*
a*b^2 + 96*C*a^2*b) + tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*A^2*a^4*b^
2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 288*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b + 192*
B*C*a*b^5 + 96*B*C*a^5*b + 320*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 288*A*C*a^4*b^2 + 576*B*C*a^3*b^3))*(B*b^3 + 3*
C*a*b^2) + ((B*b^3 + 3*C*a*b^2)*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a^2*b + 96*B*a*b^2 + 96*C*a*b^2 + 96*C*
a^2*b) - tan(c/2 + (d*x)/2)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*A^2*a^4*b^2 + 288*B^2*a
^2*b^4 + 96*B^2*a^4*b^2 + 288*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b + 192*B*C*a*b^5 + 9
6*B*C*a^5*b + 320*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 288*A*C*a^4*b^2 + 576*B*C*a^3*b^3))*(B*b^3 + 3*C*a*b^2) - 64
*A*B^2*b^9 + 64*A^2*B*b^9 - 192*B^3*a*b^8 + 576*B^3*a^2*b^7 - 32*B^3*a^3*b^6 + 192*B^3*a^4*b^5 + 16*B^3*a^6*b^
3 - 1728*C^3*a^4*b^5 + 1728*C^3*a^5*b^4 + 384*A*B^2*a*b^8 + 192*A^2*C*a*b^8 - 96*A*B^2*a^2*b^7 + 640*A*B^2*a^3
*b^6 + 96*A*B^2*a^5*b^4 + 192*A^2*B*a^2*b^7 + 144*A^2*B*a^4*b^5 - 576*A*C^2*a^2*b^7 + 1152*A*C^2*a^3*b^6 - 864
*A*C^2*a^4*b^5 + 1728*A*C^2*a^5*b^4 + 576*A^2*C*a^3*b^6 + 432*A^2*C*a^5*b^4 - 2880*B*C^2*a^3*b^6 + 4032*B*C^2*
a^4*b^5 - 288*B*C^2*a^5*b^4 + 576*B*C^2*a^6*b^3 - 1344*B^2*C*a^2*b^7 + 2880*B^2*C*a^3*b^6 - 192*B^2*C*a^4*b^5
+ 768*B^2*C*a^5*b^4 + 48*B^2*C*a^7*b^2 - 384*A*B*C*a*b^8 + 1536*A*B*C*a^2*b^7 - 576*A*B*C*a^3*b^6 + 2496*A*B*C
*a^4*b^5 + 288*A*B*C*a^6*b^3))*(B*b^3*2i + C*a*b^2*6i))/d + (atanh((2*tan(c/2 + (d*x)/2)*(A*b^3*1i + (B*a^3*1i
)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i + C*a^2*b*3i)*(32*A^2*b^6 + 8*B^2*a^6 + 32*B^2*b^6 + 96*A^2*a^2*b^4 + 72*A^2
*a^4*b^2 + 288*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 288*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 192*A*B*a*b^5 + 48*A*B*a^5*b
+ 192*B*C*a*b^5 + 96*B*C*a^5*b + 320*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 288*A*C*a^4*b^2 + 576*B*C*a^3*b^3))/(2*(
A*b^3*1i + (B*a^3*1i)/2 + (A*a^2*b*3i)/2 + B*a*b^2*3i + C*a^2*b*3i)^2*(32*A*b^3 + 16*B*a^3 + 32*B*b^3 + 48*A*a
^2*b + 96*B*a*b^2 + 96*C*a*b^2 + 96*C*a^2*b) - 64*A*B^2*b^9 + 64*A^2*B*b^9 - 192*B^3*a*b^8 + 576*B^3*a^2*b^7 -
32*B^3*a^3*b^6 + 192*B^3*a^4*b^5 + 16*B^3*a^6*b^3 - 1728*C^3*a^4*b^5 + 1728*C^3*a^5*b^4 + 384*A*B^2*a*b^8 + 1
92*A^2*C*a*b^8 - 96*A*B^2*a^2*b^7 + 640*A*B^2*a^3*b^6 + 96*A*B^2*a^5*b^4 + 192*A^2*B*a^2*b^7 + 144*A^2*B*a^4*b
^5 - 576*A*C^2*a^2*b^7 + 1152*A*C^2*a^3*b^6 - 864*A*C^2*a^4*b^5 + 1728*A*C^2*a^5*b^4 + 576*A^2*C*a^3*b^6 + 432
*A^2*C*a^5*b^4 - 2880*B*C^2*a^3*b^6 + 4032*B*C^2*a^4*b^5 - 288*B*C^2*a^5*b^4 + 576*B*C^2*a^6*b^3 - 1344*B^2*C*
a^2*b^7 + 2880*B^2*C*a^3*b^6 - 192*B^2*C*a^4*b^5 + 768*B^2*C*a^5*b^4 + 48*B^2*C*a^7*b^2 - 384*A*B*C*a*b^8 + 15
36*A*B*C*a^2*b^7 - 576*A*B*C*a^3*b^6 + 2496*A*B*C*a^4*b^5 + 288*A*B*C*a^6*b^3))*(A*b^3*2i + B*a^3*1i + A*a^2*b
*3i + B*a*b^2*6i + C*a^2*b*6i))/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)**3*(a+b*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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