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

Optimal. Leaf size=192 \[ \frac {b \tan (c+d x) \left (-\left (a^2 (6 A-8 C)\right )+9 a b B+b^2 (3 A+2 C)\right )}{3 d}+a^2 x (a B+3 A b)+\frac {\left (2 a^3 C+6 a^2 b B+3 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^2 \tan (c+d x) \sec (c+d x) (6 a A-5 a C-3 b B)}{6 d}-\frac {b (3 A-C) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]

[Out]

a^2*(3*A*b+B*a)*x+1/2*(6*a^2*b*B+b^3*B+2*a^3*C+3*a*b^2*(2*A+C))*arctanh(sin(d*x+c))/d+A*(a+b*sec(d*x+c))^3*sin
(d*x+c)/d+1/3*b*(9*a*b*B-a^2*(6*A-8*C)+b^2*(3*A+2*C))*tan(d*x+c)/d-1/6*b^2*(6*A*a-3*B*b-5*C*a)*sec(d*x+c)*tan(
d*x+c)/d-1/3*b*(3*A-C)*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]  time = 0.37, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4094, 4056, 4048, 3770, 3767, 8} \[ \frac {b \tan (c+d x) \left (a^2 (-(6 A-8 C))+9 a b B+b^2 (3 A+2 C)\right )}{3 d}+\frac {\left (6 a^2 b B+2 a^3 C+3 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (a B+3 A b)-\frac {b^2 \tan (c+d x) \sec (c+d x) (6 a A-5 a C-3 b B)}{6 d}-\frac {b (3 A-C) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*(3*A*b + a*B)*x + ((6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*(2*A + C))*ArcTanh[Sin[c + d*x]])/(2*d) + (A*(a
+ b*Sec[c + d*x])^3*Sin[c + d*x])/d + (b*(9*a*b*B - a^2*(6*A - 8*C) + b^2*(3*A + 2*C))*Tan[c + d*x])/(3*d) - (
b^2*(6*a*A - 3*b*B - 5*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) - (b*(3*A - C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x
])/(3*d)

Rule 8

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

Rule 3767

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 3770

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

Rule 4048

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 4056

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

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 (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 (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\int (a+b \sec (c+d x))^2 \left (3 A b+a B+(b B+a C) \sec (c+d x)-b (3 A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \sec (c+d x)) \left (3 a (3 A b+a B)+\left (3 A b^2+6 a b B+3 a^2 C+2 b^2 C\right ) \sec (c+d x)-b (6 a A-3 b B-5 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b^2 (6 a A-3 b B-5 a C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (6 a^2 (3 A b+a B)+3 \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \sec (c+d x)+2 b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^2 (3 A b+a B) x+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b^2 (6 a A-3 b B-5 a C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{2} \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \int \sec (c+d x) \, dx+\frac {1}{3} \left (b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=a^2 (3 A b+a B) x+\frac {\left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b^2 (6 a A-3 b B-5 a C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac {\left (b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=a^2 (3 A b+a B) x+\frac {\left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}-\frac {b^2 (6 a A-3 b B-5 a C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 6.37, size = 509, normalized size = 2.65 \[ \frac {\cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (12 a^3 A \sin (c+d x)+\frac {4 b \sin \left (\frac {1}{2} (c+d x)\right ) \left (9 a^2 C+9 a b B+3 A b^2+2 b^2 C\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 b \sin \left (\frac {1}{2} (c+d x)\right ) \left (9 a^2 C+9 a b B+3 A b^2+2 b^2 C\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+12 a^2 (c+d x) (a B+3 A b)-6 \left (2 a^3 C+6 a^2 b B+3 a b^2 (2 A+C)+b^3 B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \left (2 a^3 C+6 a^2 b B+3 a b^2 (2 A+C)+b^3 B\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2 (9 a C+b (3 B+C))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^2 (9 a C+b (3 B+C))}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}\right )}{6 d (a \cos (c+d x)+b)^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(12*a^2*(3*A*b + a*B)*(c + d*x)
- 6*(6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*(2*A + C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(6*a^2*b*B
+ b^3*B + 2*a^3*C + 3*a*b^2*(2*A + C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^2*(9*a*C + b*(3*B + C)))
/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
+ (4*b*(3*A*b^2 + 9*a*b*B + 9*a^2*C + 2*b^2*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*b^
3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (b^2*(9*a*C + b*(3*B + C)))/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^2 + (4*b*(3*A*b^2 + 9*a*b*B + 9*a^2*C + 2*b^2*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2]) + 12*a^3*A*Sin[c + d*x]))/(6*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c
+ d*x)]))

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fricas [A]  time = 0.81, size = 225, normalized size = 1.17 \[ \frac {12 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, {\left (2 \, A + C\right )} a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, {\left (2 \, A + C\right )} a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 2 \, C b^{3} + 2 \, {\left (9 \, C a^{2} b + 9 \, B a b^{2} + {\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.37, size = 438, normalized size = 2.28 \[ \frac {\frac {12 \, A 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} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} {\left (d x + c\right )} + 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2} + 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 ) - 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2} + 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 (18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C 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 )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(B*a^3 + 3*A*a^2*b)*(d*x + c) + 3*(2*C*a^3
+ 6*B*a^2*b + 6*A*a*b^2 + 3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*C*a^3 + 6*B*a^2*b + 6*
A*a*b^2 + 3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(18*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*
a*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^3*tan
(1/2*d*x + 1/2*c)^5 + 6*C*b^3*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*b^2*tan(1/2*
d*x + 1/2*c)^3 - 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b*tan(1/2*d*x + 1
/2*c) + 18*B*a*b^2*tan(1/2*d*x + 1/2*c) + 9*C*a*b^2*tan(1/2*d*x + 1/2*c) + 6*A*b^3*tan(1/2*d*x + 1/2*c) + 3*B*
b^3*tan(1/2*d*x + 1/2*c) + 6*C*b^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 = 1.24, size = 294, normalized size = 1.53 \[ \frac {a^{3} A \sin \left (d x +c \right )}{d}+a^{3} B x +\frac {a^{3} B c}{d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+3 a^{2} A b x +\frac {3 A \,a^{2} b c}{d}+\frac {3 a^{2} b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 B a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {3 C a \,b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {A \,b^{3} \tan \left (d x +c \right )}{d}+\frac {b^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 b^{3} C \tan \left (d x +c \right )}{3 d}+\frac {b^{3} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 280, normalized size = 1.46 \[ \frac {12 \, {\left (d x + c\right )} B a^{3} + 36 \, {\left (d x + c\right )} A a^{2} b + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{3} - 9 \, C a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{2} b \tan \left (d x + c\right ) + 36 \, B a b^{2} \tan \left (d x + c\right ) + 12 \, A b^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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*(12*(d*x + c)*B*a^3 + 36*(d*x + c)*A*a^2*b + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b^3 - 9*C*a*b^2*(2*sin
(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*b^3*(2*sin(d*x + c)/(sin
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d
*x + c) - 1)) + 18*B*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*A*a*b^2*(log(sin(d*x + c) + 1)
- log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 36*C*a^2*b*tan(d*x + c) + 36*B*a*b^2*tan(d*x + c) + 12*A*b
^3*tan(d*x + c))/d

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mupad [B]  time = 6.49, size = 2437, normalized size = 12.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*atan((a^2*(tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A^2*a^4*b^2
+ 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B*C*a
*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3) - a^2*(3*A*b + B
*a)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2)*1i)*(3*A*b + B*a) + a
^2*(tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 96*B^2*a^2*b
^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B*C*a*b^5 + 192*B*C
*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3) + a^2*(3*A*b + B*a)*(32*B*a^3
+ 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2)*1i)*(3*A*b + B*a))/(64*B*C^2*a^9 -
64*B^2*C*a^9 - 192*B^3*a^8*b + 1728*A^3*a^4*b^5 - 1728*A^3*a^5*b^4 + 16*B^3*a^3*b^6 + 192*B^3*a^5*b^4 - 32*B^3
*a^6*b^3 + 576*B^3*a^7*b^2 + a^2*(tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 +
288*A^2*a^4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*
a^5*b + 48*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3)
- a^2*(3*A*b + B*a)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2)*1i)*(
3*A*b + B*a)*1i - a^2*(tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A^2*a^4
*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*
B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3) + a^2*(3*A*
b + B*a)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2)*1i)*(3*A*b + B*a
)*1i + 192*A*C^2*a^8*b + 384*B^2*C*a^8*b + 48*A*B^2*a^2*b^7 + 768*A*B^2*a^4*b^5 - 192*A*B^2*a^5*b^4 + 2880*A*B
^2*a^6*b^3 - 1344*A*B^2*a^7*b^2 + 576*A^2*B*a^3*b^6 - 288*A^2*B*a^4*b^5 + 4032*A^2*B*a^5*b^4 - 2880*A^2*B*a^6*
b^3 + 432*A*C^2*a^4*b^5 + 576*A*C^2*a^6*b^3 + 1728*A^2*C*a^4*b^5 - 864*A^2*C*a^5*b^4 + 1152*A^2*C*a^6*b^3 - 57
6*A^2*C*a^7*b^2 + 144*B*C^2*a^5*b^4 + 192*B*C^2*a^7*b^2 + 96*B^2*C*a^4*b^5 + 640*B^2*C*a^6*b^3 - 96*B^2*C*a^7*
b^2 - 384*A*B*C*a^8*b + 288*A*B*C*a^3*b^6 + 2496*A*B*C*a^5*b^4 - 576*A*B*C*a^6*b^3 + 1536*A*B*C*a^7*b^2))*(3*A
*b + B*a))/d - (atanh((2*tan(c/2 + (d*x)/2)*((B*b^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b + (3*C*a*b^2)/2)*(32*B^
2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2
*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 28
8*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3))/(64*B*C^2*a^9 - 2*((B*b^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2
*b + (3*C*a*b^2)/2)^2*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2) - 6
4*B^2*C*a^9 - 192*B^3*a^8*b + 1728*A^3*a^4*b^5 - 1728*A^3*a^5*b^4 + 16*B^3*a^3*b^6 + 192*B^3*a^5*b^4 - 32*B^3*
a^6*b^3 + 576*B^3*a^7*b^2 + 192*A*C^2*a^8*b + 384*B^2*C*a^8*b + 48*A*B^2*a^2*b^7 + 768*A*B^2*a^4*b^5 - 192*A*B
^2*a^5*b^4 + 2880*A*B^2*a^6*b^3 - 1344*A*B^2*a^7*b^2 + 576*A^2*B*a^3*b^6 - 288*A^2*B*a^4*b^5 + 4032*A^2*B*a^5*
b^4 - 2880*A^2*B*a^6*b^3 + 432*A*C^2*a^4*b^5 + 576*A*C^2*a^6*b^3 + 1728*A^2*C*a^4*b^5 - 864*A^2*C*a^5*b^4 + 11
52*A^2*C*a^6*b^3 - 576*A^2*C*a^7*b^2 + 144*B*C^2*a^5*b^4 + 192*B*C^2*a^7*b^2 + 96*B^2*C*a^4*b^5 + 640*B^2*C*a^
6*b^3 - 96*B^2*C*a^7*b^2 - 384*A*B*C*a^8*b + 288*A*B*C*a^3*b^6 + 2496*A*B*C*a^5*b^4 - 576*A*B*C*a^6*b^3 + 1536
*A*B*C*a^7*b^2))*(B*b^3 + 2*C*a^3 + 6*A*a*b^2 + 6*B*a^2*b + 3*C*a*b^2))/d - (tan(c/2 + (d*x)/2)*(2*A*a^3 + 2*A
*b^3 + B*b^3 + 2*C*b^3 + 6*B*a*b^2 + 3*C*a*b^2 + 6*C*a^2*b) + tan(c/2 + (d*x)/2)^7*(2*A*b^3 - 2*A*a^3 - B*b^3
+ 2*C*b^3 + 6*B*a*b^2 - 3*C*a*b^2 + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^3*(6*A*a^3 + 2*A*b^3 - B*b^3 - (2*C*b^3)/3
+ 6*B*a*b^2 - 3*C*a*b^2 + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^5*(2*A*b^3 - 6*A*a^3 + B*b^3 - (2*C*b^3)/3 + 6*B*a*
b^2 + 3*C*a*b^2 + 6*C*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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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