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

Optimal. Leaf size=219 \[ -\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac {2 a b \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^2 x \left (a^2 (A+2 C)+12 A b^2\right )-\frac {a b^3 (9 A-4 C) \tan (c+d x) \sec (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^2}{6 d}+\frac {2 A b \sin (c+d x) (a+b \sec (c+d x))^3}{d}+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^4}{2 d} \]

[Out]

1/2*a^2*(12*A*b^2+a^2*(A+2*C))*x+2*a*b*(2*A*b^2+(2*a^2+b^2)*C)*arctanh(sin(d*x+c))/d+2*A*b*(a+b*sec(d*x+c))^3*
sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+b*sec(d*x+c))^4*sin(d*x+c)/d-1/6*b^2*(a^2*(39*A-34*C)-2*b^2*(3*A+2*C))*tan(d*
x+c)/d-1/3*a*b^3*(9*A-4*C)*sec(d*x+c)*tan(d*x+c)/d-1/6*b^2*(15*A-2*C)*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]  time = 0.61, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4095, 4094, 4056, 4048, 3770, 3767, 8} \[ -\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac {2 a b \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^2 x \left (a^2 (A+2 C)+12 A b^2\right )-\frac {b^2 (15 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^2}{6 d}-\frac {a b^3 (9 A-4 C) \tan (c+d x) \sec (c+d x)}{3 d}+\frac {2 A b \sin (c+d x) (a+b \sec (c+d x))^3}{d}+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^4}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*b^2 + a^2*(A + 2*C))*x)/2 + (2*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/d + (2*A*b*(a
+ b*Sec[c + d*x])^3*Sin[c + d*x])/d + (A*Cos[c + d*x]*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(2*d) - (b^2*(a^2*
(39*A - 34*C) - 2*b^2*(3*A + 2*C))*Tan[c + d*x])/(6*d) - (a*b^3*(9*A - 4*C)*Sec[c + d*x]*Tan[c + d*x])/(3*d) -
(b^2*(15*A - 2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6*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]

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 ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (4 A b+a (A+2 C) \sec (c+d x)-b (3 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac {1}{2} \int (a+b \sec (c+d x))^2 \left (12 A b^2+a^2 (A+2 C)-2 a b (A-2 C) \sec (c+d x)-b^2 (15 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+b \sec (c+d x)) \left (3 a \left (12 A b^2+a^2 (A+2 C)\right )-b \left (3 a^2 (A-6 C)-2 b^2 (3 A+2 C)\right ) \sec (c+d x)-4 a b^2 (9 A-4 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{12} \int \left (6 a^2 \left (12 A b^2+a^2 (A+2 C)\right )+24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \sec (c+d x)-2 b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {1}{2} a^2 \left (12 A b^2+a^2 (A+2 C)\right ) x+\frac {2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\left (2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right )\right ) \int \sec (c+d x) \, dx-\frac {1}{6} \left (b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {1}{2} a^2 \left (12 A b^2+a^2 (A+2 C)\right ) x+\frac {2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {\left (b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac {1}{2} a^2 \left (12 A b^2+a^2 (A+2 C)\right ) x+\frac {2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}-\frac {a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [B]  time = 6.32, size = 864, normalized size = 3.95 \[ -\frac {2 \left (2 b C a^3+2 A b^3 a+b^3 C a\right ) \cos ^4(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^4}{d (b+a \cos (c+d x))^4}+\frac {2 \left (2 b C a^3+2 A b^3 a+b^3 C a\right ) \cos ^4(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^4}{d (b+a \cos (c+d x))^4}+\frac {b^4 C \cos ^4(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^4}{6 d (b+a \cos (c+d x))^4 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\cos ^4(c+d x) \left (3 A \sin \left (\frac {1}{2} (c+d x)\right ) b^4+2 C \sin \left (\frac {1}{2} (c+d x)\right ) b^4+18 a^2 C \sin \left (\frac {1}{2} (c+d x)\right ) b^2\right ) (a+b \sec (c+d x))^4}{3 d (b+a \cos (c+d x))^4 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\cos ^4(c+d x) \left (3 A \sin \left (\frac {1}{2} (c+d x)\right ) b^4+2 C \sin \left (\frac {1}{2} (c+d x)\right ) b^4+18 a^2 C \sin \left (\frac {1}{2} (c+d x)\right ) b^2\right ) (a+b \sec (c+d x))^4}{3 d (b+a \cos (c+d x))^4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 a^3 A b \cos ^4(c+d x) \sin (c+d x) (a+b \sec (c+d x))^4}{d (b+a \cos (c+d x))^4}+\frac {a^4 A \cos ^4(c+d x) \sin (2 (c+d x)) (a+b \sec (c+d x))^4}{4 d (b+a \cos (c+d x))^4}+\frac {\left (C b^4+12 a C b^3\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{12 d (b+a \cos (c+d x))^4 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\left (-C b^4-12 a C b^3\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{12 d (b+a \cos (c+d x))^4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^4 C \cos ^4(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^4}{6 d (b+a \cos (c+d x))^4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {a^2 \left (A a^2+2 C a^2+12 A b^2\right ) (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{2 d (b+a \cos (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a^2*A + 12*A*b^2 + 2*a^2*C)*(c + d*x)*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4)/(2*d*(b + a*Cos[c + d*x])^4
) - (2*(2*a*A*b^3 + 2*a^3*b*C + a*b^3*C)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c
+ d*x])^4)/(d*(b + a*Cos[c + d*x])^4) + (2*(2*a*A*b^3 + 2*a^3*b*C + a*b^3*C)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^4)/(d*(b + a*Cos[c + d*x])^4) + ((12*a*b^3*C + b^4*C)*Cos[c + d*x]
^4*(a + b*Sec[c + d*x])^4)/(12*d*(b + a*Cos[c + d*x])^4*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (b^4*C*Cos[
c + d*x]^4*(a + b*Sec[c + d*x])^4*Sin[(c + d*x)/2])/(6*d*(b + a*Cos[c + d*x])^4*(Cos[(c + d*x)/2] - Sin[(c + d
*x)/2])^3) + (b^4*C*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*Sin[(c + d*x)/2])/(6*d*(b + a*Cos[c + d*x])^4*(Cos[(
c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((-12*a*b^3*C - b^4*C)*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4)/(12*d*(b +
a*Cos[c + d*x])^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*(3*A*b^4*S
in[(c + d*x)/2] + 18*a^2*b^2*C*Sin[(c + d*x)/2] + 2*b^4*C*Sin[(c + d*x)/2]))/(3*d*(b + a*Cos[c + d*x])^4*(Cos[
(c + d*x)/2] - Sin[(c + d*x)/2])) + (Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*(3*A*b^4*Sin[(c + d*x)/2] + 18*a^2*
b^2*C*Sin[(c + d*x)/2] + 2*b^4*C*Sin[(c + d*x)/2]))/(3*d*(b + a*Cos[c + d*x])^4*(Cos[(c + d*x)/2] + Sin[(c + d
*x)/2])) + (4*a^3*A*b*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(d*(b + a*Cos[c + d*x])^4) + (a^4*A*
Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*Sin[2*(c + d*x)])/(4*d*(b + a*Cos[c + d*x])^4)

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fricas [A]  time = 0.52, size = 208, normalized size = 0.95 \[ \frac {3 \, {\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} d x \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, C a^{3} b + {\left (2 \, A + C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (2 \, C a^{3} b + {\left (2 \, A + C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} b \cos \left (d x + c\right )^{3} + 12 \, C a b^{3} \cos \left (d x + c\right ) + 2 \, C b^{4} + 2 \, {\left (18 \, C a^{2} b^{2} + {\left (3 \, A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*((A + 2*C)*a^4 + 12*A*a^2*b^2)*d*x*cos(d*x + c)^3 + 6*(2*C*a^3*b + (2*A + C)*a*b^3)*cos(d*x + c)^3*log(
sin(d*x + c) + 1) - 6*(2*C*a^3*b + (2*A + C)*a*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (3*A*a^4*cos(d*x +
c)^4 + 24*A*a^3*b*cos(d*x + c)^3 + 12*C*a*b^3*cos(d*x + c) + 2*C*b^4 + 2*(18*C*a^2*b^2 + (3*A + 2*C)*b^4)*cos
(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.35, size = 397, normalized size = 1.81 \[ \frac {3 \, {\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )} {\left (d x + c\right )} + 12 \, {\left (2 \, C a^{3} b + 2 \, A a b^{3} + C a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (2 \, C a^{3} b + 2 \, A a b^{3} + C a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {4 \, {\left (18 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(A*a^4 + 2*C*a^4 + 12*A*a^2*b^2)*(d*x + c) + 12*(2*C*a^3*b + 2*A*a*b^3 + C*a*b^3)*log(abs(tan(1/2*d*x +
1/2*c) + 1)) - 12*(2*C*a^3*b + 2*A*a*b^3 + C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(A*a^4*tan(1/2*d*x
+ 1/2*c)^3 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - A*a^4*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b*tan(1/2*d*x + 1/2*c))/
(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 4*(18*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3
*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*b^4*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*A*b^4
*tan(1/2*d*x + 1/2*c)^3 - 2*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 6*C*a*b^3*tan(1
/2*d*x + 1/2*c) + 3*A*b^4*tan(1/2*d*x + 1/2*c) + 3*C*b^4*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 = 1.13, size = 258, normalized size = 1.18 \[ \frac {A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {A \,a^{4} x}{2}+\frac {A \,a^{4} c}{2 d}+a^{4} C x +\frac {C \,a^{4} c}{d}+\frac {4 A \,a^{3} b \sin \left (d x +c \right )}{d}+\frac {4 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+6 A x \,a^{2} b^{2}+\frac {6 A \,a^{2} b^{2} c}{d}+\frac {6 C \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {4 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 C a \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {2 C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{4} \tan \left (d x +c \right )}{d}+\frac {2 C \,b^{4} \tan \left (d x +c \right )}{3 d}+\frac {C \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 221, normalized size = 1.01 \[ \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 12 \, {\left (d x + c\right )} C a^{4} + 72 \, {\left (d x + c\right )} A a^{2} b^{2} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{4} - 12 \, C a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{3} b \sin \left (d x + c\right ) + 72 \, C a^{2} b^{2} \tan \left (d x + c\right ) + 12 \, A b^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 12*(d*x + c)*C*a^4 + 72*(d*x + c)*A*a^2*b^2 + 4*(tan(d*x + c)
^3 + 3*tan(d*x + c))*C*b^4 - 12*C*a*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)) + 24*C*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*a*b^3*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 48*A*a^3*b*sin(d*x + c) + 72*C*a^2*b^2*tan(d*x + c) + 12*A*b^4*tan(d*x + c))/d

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mupad [B]  time = 5.90, size = 2664, normalized size = 12.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*atan(-((a^2*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^
6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4
+ 384*A*C*a^6*b^2) - (a^2*(A*a^2 + 12*A*b^2 + 2*C*a^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64
*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 12*A*b^2 + 2*C*a^2))/2 + (a^2*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*
a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b
^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2) + (a^2*(A*a^2 + 12*A*b^2 + 2*C*a^2)*(1
6*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 12*A*b^2 + 2*C*a^
2))/2)/(256*C^3*a^11*b - (a^2*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4
+ 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024
*A*C*a^4*b^4 + 384*A*C*a^6*b^2) - (a^2*(A*a^2 + 12*A*b^2 + 2*C*a^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128
*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 12*A*b^2 + 2*C*a^2)*1i)/2 + (a^2*(tan(c/2 + (d*x)/2)*(8*A
^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4
+ 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2) + (a^2*(A*a^2 + 12*A*b
^2 + 2*C*a^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 1
2*A*b^2 + 2*C*a^2)*1i)/2 - 6144*A^3*a^4*b^8 + 9216*A^3*a^5*b^7 - 512*A^3*a^6*b^6 + 1536*A^3*a^7*b^5 + 64*A^3*a
^9*b^3 - 256*C^3*a^6*b^6 - 1024*C^3*a^8*b^4 + 128*C^3*a^9*b^3 - 1024*C^3*a^10*b^2 + 256*A*C^2*a^11*b + 64*A^2*
C*a^11*b - 1536*A*C^2*a^4*b^8 - 7296*A*C^2*a^6*b^6 + 1536*A*C^2*a^7*b^5 - 8704*A*C^2*a^8*b^4 + 3456*A*C^2*a^9*
b^3 - 512*A*C^2*a^10*b^2 - 6144*A^2*C*a^4*b^8 + 4608*A^2*C*a^5*b^7 - 13824*A^2*C*a^6*b^6 + 13056*A^2*C*a^7*b^5
- 1024*A^2*C*a^8*b^4 + 1824*A^2*C*a^9*b^3))*(A*a^2 + 12*A*b^2 + 2*C*a^2))/d - (tan(c/2 + (d*x)/2)^5*(6*A*a^4
- 4*A*b^4 + (4*C*b^4)/3 - 24*C*a^2*b^2) - tan(c/2 + (d*x)/2)^3*(4*A*a^4 - (8*C*b^4)/3 + 16*A*a^3*b - 8*C*a*b^3
) - tan(c/2 + (d*x)/2)^7*(4*A*a^4 - (8*C*b^4)/3 - 16*A*a^3*b + 8*C*a*b^3) + tan(c/2 + (d*x)/2)*(A*a^4 + 2*A*b^
4 + 2*C*b^4 + 12*C*a^2*b^2 + 8*A*a^3*b + 4*C*a*b^3) + tan(c/2 + (d*x)/2)^9*(A*a^4 + 2*A*b^4 + 2*C*b^4 + 12*C*a
^2*b^2 - 8*A*a^3*b - 4*C*a*b^3))/(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)) - (a*b*atan((a*b*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^
8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2
+ 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2) - 2*a*b*(2*A*b^2 + 2*C*a^2 + C*b^2)*(16*
A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b))*(2*A*b^2 + 2*C*a^2 + C*b^2)*2i + a
*b*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^
2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*
b^2) + 2*a*b*(2*A*b^2 + 2*C*a^2 + C*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128
*C*a^3*b))*(2*A*b^2 + 2*C*a^2 + C*b^2)*2i)/(256*C^3*a^11*b - 6144*A^3*a^4*b^8 + 9216*A^3*a^5*b^7 - 512*A^3*a^6
*b^6 + 1536*A^3*a^7*b^5 + 64*A^3*a^9*b^3 - 256*C^3*a^6*b^6 - 1024*C^3*a^8*b^4 + 128*C^3*a^9*b^3 - 1024*C^3*a^1
0*b^2 - 2*a*b*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b
^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 3
84*A*C*a^6*b^2) - 2*a*b*(2*A*b^2 + 2*C*a^2 + C*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*
a*b^3 + 128*C*a^3*b))*(2*A*b^2 + 2*C*a^2 + C*b^2) + 2*a*b*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^
2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*
a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2) + 2*a*b*(2*A*b^2 + 2*C*a^2 + C*b^2)*(16*A*a^4 + 32
*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b))*(2*A*b^2 + 2*C*a^2 + C*b^2) + 256*A*C^2*a^11
*b + 64*A^2*C*a^11*b - 1536*A*C^2*a^4*b^8 - 7296*A*C^2*a^6*b^6 + 1536*A*C^2*a^7*b^5 - 8704*A*C^2*a^8*b^4 + 345
6*A*C^2*a^9*b^3 - 512*A*C^2*a^10*b^2 - 6144*A^2*C*a^4*b^8 + 4608*A^2*C*a^5*b^7 - 13824*A^2*C*a^6*b^6 + 13056*A
^2*C*a^7*b^5 - 1024*A^2*C*a^8*b^4 + 1824*A^2*C*a^9*b^3))*(2*A*b^2 + 2*C*a^2 + C*b^2)*4i)/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)**2*(a+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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