3.9.92 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [892]

Optimal. Leaf size=303 \[ \frac {1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \tan (c+d x)}{6 d}-\frac {b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d} \]

[Out]

1/2*a*(8*A*b^3+a^3*B+12*a*b^2*B+4*a^2*b*(A+2*C))*x+1/2*b^2*(2*A*b^2+8*B*a*b+12*C*a^2+C*b^2)*arctanh(sin(d*x+c)
)/d+1/6*(12*A*b^2+15*a*b*B+a^2*(4*A+6*C))*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/6*(4*A*b+3*B*a)*cos(d*x+c)*(a+b*se
c(d*x+c))^3*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*sec(d*x+c))^4*sin(d*x+c)/d-1/6*b*(39*a^2*b*B-6*b^3*B+4*a*b^2*
(11*A-6*C)+4*a^3*(2*A+3*C))*tan(d*x+c)/d-1/6*b^2*(18*a*b*B+3*b^2*(6*A-C)+a^2*(4*A+6*C))*sec(d*x+c)*tan(d*x+c)/
d

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Rubi [A]
time = 0.59, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4179, 4133, 3855, 3852, 8} \begin {gather*} \frac {b^2 \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\sin (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \sec (c+d x))^2}{6 d}-\frac {b^2 \tan (c+d x) \sec (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{6 d}-\frac {b \tan (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{6 d}+\frac {1}{2} a x \left (a^3 B+4 a^2 b (A+2 C)+12 a b^2 B+8 A b^3\right )+\frac {(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{6 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*x)/2 + (b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*ArcTa
nh[Sin[c + d*x]])/(2*d) + ((12*A*b^2 + 15*a*b*B + a^2*(4*A + 6*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(6*d)
+ ((4*A*b + 3*a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]^2*(a + b*Sec[c +
d*x])^4*Sin[c + d*x])/(3*d) - (b*(39*a^2*b*B - 6*b^3*B + 4*a*b^2*(11*A - 6*C) + 4*a^3*(2*A + 3*C))*Tan[c + d*x
])/(6*d) - (b^2*(18*a*b*B + 3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(6*d)

Rule 8

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 4133

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

Rule 4179

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))^4 \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))^4 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (4 A b+3 a B+(2 a A+3 b B+3 a C) \sec (c+d x)-b (2 A-3 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (12 A b^2+15 a b B+a^2 (4 A+6 C)+\left (3 a^2 B+6 b^2 B+4 a b (A+3 C)\right ) \sec (c+d x)-6 b (2 A b+a B-b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {1}{6} \int (a+b \sec (c+d x)) \left (3 \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )-b \left (8 a A b+3 a^2 B-6 b^2 B-18 a b C\right ) \sec (c+d x)-2 b \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{12} \int \left (6 a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right )+6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \sec (c+d x)-2 b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{2} \left (b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right )\right ) \int \sec (c+d x) \, dx-\frac {1}{6} \left (b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {\left (b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac {1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \tan (c+d x)}{6 d}-\frac {b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 4.70, size = 370, normalized size = 1.22 \begin {gather*} \frac {6 a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) (c+d x)-6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 b^3 (b B+4 a 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 {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 b^3 (b B+4 a 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 )}+3 a^2 \left (24 A b^2+16 a b B+a^2 (3 A+4 C)\right ) \sin (c+d x)+3 a^3 (4 A b+a B) \sin (2 (c+d x))+a^4 A \sin (3 (c+d x))}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*(c + d*x) - 6*b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*
C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] + (3*b^4*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (12*b^3*(b*B + 4*a*C)*Sin[(c + d*x
)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (3*b^4*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (12*b^3*(b*B
+ 4*a*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*a^2*(24*A*b^2 + 16*a*b*B + a^2*(3*A + 4*
C))*Sin[c + d*x] + 3*a^3*(4*A*b + a*B)*Sin[2*(c + d*x)] + a^4*A*Sin[3*(c + d*x)])/(12*d)

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Maple [A]
time = 0.17, size = 284, normalized size = 0.94 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 311, normalized size = 1.03 \begin {gather*} -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 48 \, {\left (d x + c\right )} C a^{3} b - 72 \, {\left (d x + c\right )} B a^{2} b^{2} - 48 \, {\left (d x + c\right )} A a b^{3} + 3 \, C b^{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 )} - 36 \, C a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A b^{4} {\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^{4} \sin \left (d x + c\right ) - 48 \, B a^{3} b \sin \left (d x + c\right ) - 72 \, A a^{2} b^{2} \sin \left (d x + c\right ) - 48 \, C a b^{3} \tan \left (d x + c\right ) - 12 \, B b^{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)^3*(a+b*sec(d*x+c))^4*(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^4 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 12*(2*d*x + 2*c
+ sin(2*d*x + 2*c))*A*a^3*b - 48*(d*x + c)*C*a^3*b - 72*(d*x + c)*B*a^2*b^2 - 48*(d*x + c)*A*a*b^3 + 3*C*b^4*(
2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 36*C*a^2*b^2*(log(sin(d
*x + c) + 1) - log(sin(d*x + c) - 1)) - 24*B*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*A*b^4*(
log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 12*C*a^4*sin(d*x + c) - 48*B*a^3*b*sin(d*x + c) - 72*A*a^2*b^
2*sin(d*x + c) - 48*C*a*b^3*tan(d*x + c) - 12*B*b^4*tan(d*x + c))/d

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Fricas [A]
time = 3.42, size = 262, normalized size = 0.86 \begin {gather*} \frac {6 \, {\left (B a^{4} + 4 \, {\left (A + 2 \, C\right )} a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, C b^{4} + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 12 \, B a^{3} b + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*(B*a^4 + 4*(A + 2*C)*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*d*x*cos(d*x + c)^2 + 3*(12*C*a^2*b^2 + 8*B*a*b^
3 + (2*A + C)*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(12*C*a^2*b^2 + 8*B*a*b^3 + (2*A + C)*b^4)*cos(d*x
+ c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^4*cos(d*x + c)^4 + 3*C*b^4 + 3*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^3 +
2*((2*A + 3*C)*a^4 + 12*B*a^3*b + 18*A*a^2*b^2)*cos(d*x + c)^2 + 6*(4*C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c)^2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [A]
time = 0.56, size = 543, normalized size = 1.79 \begin {gather*} \frac {3 \, {\left (B a^{4} + 4 \, A a^{3} b + 8 \, C a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + 2 \, A b^{4} + C b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + 2 \, A b^{4} + C b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{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} - 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a^{2} b^{2} \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 ) + 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, A a^{2} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(B*a^4 + 4*A*a^3*b + 8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*(d*x + c) + 3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*
A*b^4 + C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*A*b^4 + C*b^4)*log(abs(tan
(1/2*d*x + 1/2*c) - 1)) - 6*(8*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^4*tan(1/2*d*x + 1/2*c)^3 - C*b^4*tan(1/2
*d*x + 1/2*c)^3 - 8*C*a*b^3*tan(1/2*d*x + 1/2*c) - 2*B*b^4*tan(1/2*d*x + 1/2*c) - C*b^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 - 3*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^4*
tan(1/2*d*x + 1/2*c)^5 - 12*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 24*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 36*A*a^2*b^2*
tan(1/2*d*x + 1/2*c)^5 + 4*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 48*B*a^3*b*tan(1/2
*d*x + 1/2*c)^3 + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*tan(1/2*d*x + 1
/2*c) + 6*C*a^4*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b*tan(1/2*d*x + 1/2*c) + 24*B*a^3*b*tan(1/2*d*x + 1/2*c) + 36*
A*a^2*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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Mupad [B]
time = 8.39, size = 2500, normalized size = 8.25 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + b/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)^3*((8*A*a^4)/3 + 2*B*a^4 - 4*B*b^4 - 4*C*b^4 + 8*A*a^3*b - 16*C*a*b^3) + tan(c/2 + (d*x)
/2)^7*((8*A*a^4)/3 - 2*B*a^4 + 4*B*b^4 - 4*C*b^4 - 8*A*a^3*b + 16*C*a*b^3) - tan(c/2 + (d*x)/2)^9*(2*A*a^4 - B
*a^4 - 2*B*b^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 - 4*A*a^3*b + 8*B*a^3*b - 8*C*a*b^3) - tan(c/2 + (d*x)/2)*(2*A
*a^4 + B*a^4 + 2*B*b^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 + 4*A*a^3*b + 8*B*a^3*b + 8*C*a*b^3) + tan(c/2 + (d*x)
/2)^5*(4*C*a^4 - (4*A*a^4)/3 - 6*C*b^4 + 24*A*a^2*b^2 + 16*B*a^3*b))/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d
*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (atan((((A*b^4 + (C*b
^4)/2 + 6*C*a^2*b^2 + 4*B*a*b^3)*(32*A*b^4 + 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 128*A*a*b^3
+ 64*A*a^3*b + 128*B*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*
a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 192*C^2*a
^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 256*A*B*a*b^7 + 64*A*B*a^7*b + 128*B*C*a*b^7 + 128*
B*C*a^7*b + 1536*A*B*a^3*b^5 + 896*A*B*a^5*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 + 1536*B
*C*a^3*b^5 + 1536*B*C*a^5*b^3))*(A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2 + 4*B*a*b^3)*1i - ((A*b^4 + (C*b^4)/2 + 6*C*a
^2*b^2 + 4*B*a*b^3)*(32*A*b^4 + 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b
+ 128*B*a*b^3 + 128*C*a^3*b) - tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512
*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152
*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 256*A*B*a*b^7 + 64*A*B*a^7*b + 128*B*C*a*b^7 + 128*B*C*a^7*b + 1
536*A*B*a^3*b^5 + 896*A*B*a^5*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 + 1536*B*C*a^3*b^5 +
1536*B*C*a^5*b^3))*(A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2 + 4*B*a*b^3)*1i)/(((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2 + 4*B*
a*b^3)*(32*A*b^4 + 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*B*a*b^
3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4
+ 128*A^2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4
+ 512*C^2*a^6*b^2 + 32*A*C*b^8 + 256*A*B*a*b^7 + 64*A*B*a^7*b + 128*B*C*a*b^7 + 128*B*C*a^7*b + 1536*A*B*a^3*b
^5 + 896*A*B*a^5*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 + 1536*B*C*a^3*b^5 + 1536*B*C*a^5*
b^3))*(A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2 + 4*B*a*b^3) + ((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2 + 4*B*a*b^3)*(32*A*b^4
+ 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*B*a*b^3 + 128*C*a^3*b)
- tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^
2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^
2 + 32*A*C*b^8 + 256*A*B*a*b^7 + 64*A*B*a^7*b + 128*B*C*a*b^7 + 128*B*C*a^7*b + 1536*A*B*a^3*b^5 + 896*A*B*a^5
*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 + 1536*B*C*a^3*b^5 + 1536*B*C*a^5*b^3))*(A*b^4 + (
C*b^4)/2 + 6*C*a^2*b^2 + 4*B*a*b^3) - 256*A^3*a*b^11 + 1024*A^3*a^2*b^10 - 128*A^3*a^3*b^9 + 1024*A^3*a^4*b^8
+ 256*A^3*a^6*b^6 - 6144*B^3*a^4*b^8 + 9216*B^3*a^5*b^7 - 512*B^3*a^6*b^6 + 1536*B^3*a^7*b^5 + 64*B^3*a^9*b^3
- 64*C^3*a^3*b^9 - 1536*C^3*a^5*b^7 + 512*C^3*a^6*b^6 - 9216*C^3*a^7*b^5 + 6144*C^3*a^8*b^4 - 64*A*C^2*a*b^11
- 256*A^2*C*a*b^11 - 7168*A*B^2*a^3*b^9 + 14592*A*B^2*a^4*b^8 - 2304*A*B^2*a^5*b^7 + 7552*A*B^2*a^6*b^6 + 528*
A*B^2*a^8*b^4 - 2432*A^2*B*a^2*b^10 + 7168*A^2*B*a^3*b^9 - 1056*A^2*B*a^4*b^8 + 5888*A^2*B*a^5*b^7 + 1152*A^2*
B*a^7*b^5 - 1824*A*C^2*a^3*b^9 + 1024*A*C^2*a^4*b^8 - 13056*A*C^2*a^5*b^7 + 13824*A*C^2*a^6*b^6 - 4608*A*C^2*a
^7*b^5 + 6144*A*C^2*a^8*b^4 + 512*A^2*C*a^2*b^10 - 3456*A^2*C*a^3*b^9 + 8704*A^2*C*a^4*b^8 - 1536*A^2*C*a^5*b^
7 + 7296*A^2*C*a^6*b^6 + 1536*A^2*C*a^8*b^4 - 96*B*C^2*a^2*b^10 - 3336*B*C^2*a^4*b^8 + 1536*B*C^2*a^5*b^7 - 26
304*B*C^2*a^6*b^6 + 22656*B*C^2*a^7*b^5 - 1152*B*C^2*a^8*b^4 + 1536*B*C^2*a^9*b^3 - 1536*B^2*C*a^3*b^9 + 1152*
B^2*C*a^4*b^8 - 22656*B^2*C*a^5*b^7 + 26304*B^2*C*a^6*b^6 - 1536*B^2*C*a^7*b^5 + 3336*B^2*C*a^8*b^4 + 96*B^2*C
*a^10*b^2 - 1408*A*B*C*a^2*b^10 + 1536*A*B*C*a^3*b^9 - 19488*A*B*C*a^4*b^8 + 30592*A*B*C*a^5*b^7 - 6528*A*B*C*
a^6*b^6 + 15168*A*B*C*a^7*b^5 + 768*A*B*C*a^9*b^3))*(A*b^4*2i + C*b^4*1i + C*a^2*b^2*12i + B*a*b^3*8i))/d - (a
*atan(((a*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^
2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^
2*a^6*b^2 + 32*A*C*b^8 + 256*A*B*a*b^7 + 64*A*B*a^7*b + 128*B*C*a*b^7 + 128*B*C*a^7*b + 1536*A*B*a^3*b^5 + 896
*A*B*a^5*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 + 1536*B*C*a^3*b^5 + 1536*B*C*a^5*b^3) - (
a*(8*A*b^3 + B*a^3 + 4*A*a^2*b + 12*B*a*b^2 + 8...

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