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

Optimal. Leaf size=182 \[ \frac {b \left (a^2 (4 A+6 C)+A b^2\right ) \sin (c+d x)}{2 d}+\frac {a \left (a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac {A b \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac {b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

1/8*a*(12*b^2*(A+2*C)+a^2*(3*A+4*C))*x+b^3*C*arctanh(sin(d*x+c))/d+1/2*b*(A*b^2+a^2*(4*A+6*C))*sin(d*x+c)/d+1/
8*a*(2*A*b^2+a^2*(3*A+4*C))*cos(d*x+c)*sin(d*x+c)/d+1/4*A*b*cos(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/4*A
*cos(d*x+c)^3*(a+b*sec(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.56, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4095, 4094, 4074, 4047, 8, 4045, 3770} \[ \frac {b \left (a^2 (4 A+6 C)+A b^2\right ) \sin (c+d x)}{2 d}+\frac {a \left (a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right )+\frac {A b \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac {b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*x)/8 + (b^3*C*ArcTanh[Sin[c + d*x]])/d + (b*(A*b^2 + a^2*(4*A + 6*C))*
Sin[c + d*x])/(2*d) + (a*(2*A*b^2 + a^2*(3*A + 4*C))*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (A*b*Cos[c + d*x]^2*(a
+ b*Sec[c + d*x])^2*Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*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 4074

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[(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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, 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]

Rule 4095

Int[((A_.) + 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] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - 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, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (3 A+4 C) \sec (c+d x)+4 b C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{12} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (3 \left (2 A b^2+a^2 (3 A+4 C)\right )+3 a b (5 A+8 C) \sec (c+d x)+12 b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {1}{24} \int \cos (c+d x) \left (-12 b \left (A b^2+a^2 (4 A+6 C)\right )-3 a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \sec (c+d x)-24 b^3 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {1}{24} \int \cos (c+d x) \left (-12 b \left (A b^2+a^2 (4 A+6 C)\right )-24 b^3 C \sec ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right )\right ) \int 1 \, dx\\ &=\frac {1}{8} a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac {b \left (A b^2+a^2 (4 A+6 C)\right ) \sin (c+d x)}{2 d}+\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\left (b^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac {b^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \left (A b^2+a^2 (4 A+6 C)\right ) \sin (c+d x)}{2 d}+\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.64, size = 177, normalized size = 0.97 \[ \frac {a^3 A \sin (4 (c+d x))+4 a (c+d x) \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right )+8 a \left (a^2 (A+C)+3 A b^2\right ) \sin (2 (c+d x))+8 b \left (3 a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x)+8 a^2 A b \sin (3 (c+d x))-32 b^3 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 b^3 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*(c + d*x) - 32*b^3*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*b
^3*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*b*(4*A*b^2 + 3*a^2*(3*A + 4*C))*Sin[c + d*x] + 8*a*(3*A*b^2
+ a^2*(A + C))*Sin[2*(c + d*x)] + 8*a^2*A*b*Sin[3*(c + d*x)] + a^3*A*Sin[4*(c + d*x)])/(32*d)

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fricas [A]  time = 0.48, size = 146, normalized size = 0.80 \[ \frac {4 \, C b^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, C b^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, {\left (A + 2 \, C\right )} a b^{2}\right )} d x + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, A a^{2} b \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, A + 3 \, C\right )} a^{2} b + 8 \, A b^{3} + {\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, A a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.30, size = 503, normalized size = 2.76 \[ \frac {8 \, C b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, C b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (3 \, A a^{3} + 4 \, C a^{3} + 12 \, A a b^{2} + 24 \, C a b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, A b^{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 )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*C*b^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*C*b^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (3*A*a^3 + 4*C*
a^3 + 12*A*a*b^2 + 24*C*a*b^2)*(d*x + c) - 2*(5*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 4*C*a^3*tan(1/2*d*x + 1/2*c)^7
- 24*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 -
8*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 4*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*A*a^2*b*
tan(1/2*d*x + 1/2*c)^5 - 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 24*A*b^3*tan(
1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*A*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 - 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*A*b^3*tan(1/2*d*x + 1/
2*c)^3 - 5*A*a^3*tan(1/2*d*x + 1/2*c) - 4*C*a^3*tan(1/2*d*x + 1/2*c) - 24*A*a^2*b*tan(1/2*d*x + 1/2*c) - 24*C*
a^2*b*tan(1/2*d*x + 1/2*c) - 12*A*a*b^2*tan(1/2*d*x + 1/2*c) - 8*A*b^3*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 = 1.12, size = 252, normalized size = 1.38 \[ \frac {A \,a^{3} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 A \,a^{3} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{8 d}+\frac {3 a^{3} A x}{8}+\frac {3 A \,a^{3} c}{8 d}+\frac {C \,a^{3} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {a^{3} C x}{2}+\frac {C \,a^{3} c}{2 d}+\frac {A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{2} b}{d}+\frac {2 A \,a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 C \,a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 A a \,b^{2} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {3 A x a \,b^{2}}{2}+\frac {3 A a \,b^{2} c}{2 d}+3 a \,b^{2} C x +\frac {3 C a \,b^{2} c}{d}+\frac {A \,b^{3} \sin \left (d x +c \right )}{d}+\frac {b^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*A*a^3*sin(d*x+c)*cos(d*x+c)^3+3/8/d*A*a^3*sin(d*x+c)*cos(d*x+c)+3/8*a^3*A*x+3/8/d*A*a^3*c+1/2/d*C*a^3*si
n(d*x+c)*cos(d*x+c)+1/2*a^3*C*x+1/2/d*C*a^3*c+1/d*A*sin(d*x+c)*cos(d*x+c)^2*a^2*b+2/d*A*a^2*b*sin(d*x+c)+3/d*C
*a^2*b*sin(d*x+c)+3/2/d*A*a*b^2*sin(d*x+c)*cos(d*x+c)+3/2*A*x*a*b^2+3/2/d*A*a*b^2*c+3*a*b^2*C*x+3/d*C*a*b^2*c+
1/d*A*b^3*sin(d*x+c)+1/d*b^3*C*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.35, size = 174, normalized size = 0.96 \[ \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} + 96 \, {\left (d x + c\right )} C a b^{2} + 16 \, C b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, C a^{2} b \sin \left (d x + c\right ) + 32 \, A b^{3} \sin \left (d x + c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*((12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 + 8*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3
- 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b^2 + 96*(d*x + c)*C
*a*b^2 + 16*C*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*C*a^2*b*sin(d*x + c) + 32*A*b^3*sin(d*x
+ c))/d

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mupad [B]  time = 5.79, size = 2008, normalized size = 11.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((5*A*a^3)/4 + 2*A*b^3 + C*a^3 + 3*A*a*b^2 + 6*A*a^2*b + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^7
*((5*A*a^3)/4 - 2*A*b^3 + C*a^3 + 3*A*a*b^2 - 6*A*a^2*b - 6*C*a^2*b) + tan(c/2 + (d*x)/2)^3*(6*A*b^3 - (3*A*a^
3)/4 + C*a^3 + 3*A*a*b^2 + 10*A*a^2*b + 18*C*a^2*b) + tan(c/2 + (d*x)/2)^5*((3*A*a^3)/4 + 6*A*b^3 - C*a^3 - 3*
A*a*b^2 + 10*A*a^2*b + 18*C*a^2*b))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)
^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (C*b^3*atan((C*b^3*(tan(c/2 + (d*x)/2)*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b
^6 + 72*A^2*a^2*b^4 + 36*A^2*a^4*b^2 + 288*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A
*C*a^4*b^2) + C*b^3*(12*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A*a*b^2 + 96*C*a*b^2))*1i + C*b^3*(tan(c/2 + (d*x)/2)
*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b^6 + 72*A^2*a^2*b^4 + 36*A^2*a^4*b^2 + 288*C^2*a^2*b^4 + 96*C^2*a^4*b^2
+ 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A*C*a^4*b^2) - C*b^3*(12*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A*a*b^2 + 96*C*
a*b^2))*1i)/(576*C^3*a^2*b^7 - 192*C^3*a*b^8 - 32*C^3*a^3*b^6 + 192*C^3*a^4*b^5 + 16*C^3*a^6*b^3 + C*b^3*(tan(
c/2 + (d*x)/2)*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b^6 + 72*A^2*a^2*b^4 + 36*A^2*a^4*b^2 + 288*C^2*a^2*b^4 + 9
6*C^2*a^4*b^2 + 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A*C*a^4*b^2) + C*b^3*(12*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A
*a*b^2 + 96*C*a*b^2)) - C*b^3*(tan(c/2 + (d*x)/2)*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b^6 + 72*A^2*a^2*b^4 + 3
6*A^2*a^4*b^2 + 288*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A*C*a^4*b^2) - C*b^3*(12
*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A*a*b^2 + 96*C*a*b^2)) - 96*A*C^2*a*b^8 + 576*A*C^2*a^2*b^7 - 24*A*C^2*a^3*b
^6 + 240*A*C^2*a^4*b^5 + 24*A*C^2*a^6*b^3 + 144*A^2*C*a^2*b^7 + 72*A^2*C*a^4*b^5 + 9*A^2*C*a^6*b^3))*2i)/d - (
a*atan(((a*(tan(c/2 + (d*x)/2)*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b^6 + 72*A^2*a^2*b^4 + 36*A^2*a^4*b^2 + 288
*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A*C*a^4*b^2) - (a*(3*A*a^2 + 12*A*b^2 + 4*C
*a^2 + 24*C*b^2)*(12*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A*a*b^2 + 96*C*a*b^2)*1i)/8)*(3*A*a^2 + 12*A*b^2 + 4*C*a
^2 + 24*C*b^2))/8 + (a*(tan(c/2 + (d*x)/2)*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b^6 + 72*A^2*a^2*b^4 + 36*A^2*a
^4*b^2 + 288*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A*C*a^4*b^2) + (a*(3*A*a^2 + 12
*A*b^2 + 4*C*a^2 + 24*C*b^2)*(12*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A*a*b^2 + 96*C*a*b^2)*1i)/8)*(3*A*a^2 + 12*A
*b^2 + 4*C*a^2 + 24*C*b^2))/8)/((a*(tan(c/2 + (d*x)/2)*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b^6 + 72*A^2*a^2*b^
4 + 36*A^2*a^4*b^2 + 288*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A*C*a^4*b^2) + (a*(
3*A*a^2 + 12*A*b^2 + 4*C*a^2 + 24*C*b^2)*(12*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A*a*b^2 + 96*C*a*b^2)*1i)/8)*(3*
A*a^2 + 12*A*b^2 + 4*C*a^2 + 24*C*b^2)*1i)/8 - (a*(tan(c/2 + (d*x)/2)*((9*A^2*a^6)/2 + 8*C^2*a^6 + 32*C^2*b^6
+ 72*A^2*a^2*b^4 + 36*A^2*a^4*b^2 + 288*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 12*A*C*a^6 + 288*A*C*a^2*b^4 + 120*A*C*
a^4*b^2) - (a*(3*A*a^2 + 12*A*b^2 + 4*C*a^2 + 24*C*b^2)*(12*A*a^3 + 16*C*a^3 + 32*C*b^3 + 48*A*a*b^2 + 96*C*a*
b^2)*1i)/8)*(3*A*a^2 + 12*A*b^2 + 4*C*a^2 + 24*C*b^2)*1i)/8 - 192*C^3*a*b^8 + 576*C^3*a^2*b^7 - 32*C^3*a^3*b^6
+ 192*C^3*a^4*b^5 + 16*C^3*a^6*b^3 - 96*A*C^2*a*b^8 + 576*A*C^2*a^2*b^7 - 24*A*C^2*a^3*b^6 + 240*A*C^2*a^4*b^
5 + 24*A*C^2*a^6*b^3 + 144*A^2*C*a^2*b^7 + 72*A^2*C*a^4*b^5 + 9*A^2*C*a^6*b^3))*(3*A*a^2 + 12*A*b^2 + 4*C*a^2
+ 24*C*b^2))/(4*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+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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