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

Optimal. Leaf size=168 \[ \frac {b \left (C \left (6 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} a x \left (a^2 (A+2 C)+6 A b^2\right )-\frac {3 a b^2 (3 A-2 C) \tan (c+d x)}{2 d}+\frac {3 A b \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}-\frac {b^3 (4 A-C) \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4095, 4094, 4048, 3770, 3767, 8} \[ \frac {b \left (C \left (6 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} a x \left (a^2 (A+2 C)+6 A b^2\right )-\frac {3 a b^2 (3 A-2 C) \tan (c+d x)}{2 d}+\frac {3 A b \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}-\frac {b^3 (4 A-C) \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*A*b^2 + a^2*(A + 2*C))*x)/2 + (b*(2*A*b^2 + (6*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (3*A*b*(a +
b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (A*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) - (3*a*b^2*
(3*A - 2*C)*Tan[c + d*x])/(2*d) - (b^3*(4*A - C)*Sec[c + d*x]*Tan[c + d*x])/(2*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 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))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (A+2 C) \sec (c+d x)-2 b (A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {3 A b (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac {1}{2} \int (a+b \sec (c+d x)) \left (6 A b^2+a^2 (A+2 C)-a b (A-4 C) \sec (c+d x)-2 b^2 (4 A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {3 A b (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b^3 (4 A-C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a \left (6 A b^2+a^2 (A+2 C)\right )+2 b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \sec (c+d x)-6 a b^2 (3 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {1}{2} a \left (6 A b^2+a^2 (A+2 C)\right ) x+\frac {3 A b (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b^3 (4 A-C) \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \left (3 a b^2 (3 A-2 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a \left (6 A b^2+a^2 (A+2 C)\right ) x+\frac {b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {3 A b (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b^3 (4 A-C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\left (3 a b^2 (3 A-2 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=\frac {1}{2} a \left (6 A b^2+a^2 (A+2 C)\right ) x+\frac {b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {3 A b (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {3 a b^2 (3 A-2 C) \tan (c+d x)}{2 d}-\frac {b^3 (4 A-C) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.91, size = 287, normalized size = 1.71 \[ \frac {a^3 A \sin (2 (c+d x))+2 a (c+d x) \left (a^2 (A+2 C)+6 A b^2\right )-2 b \left (C \left (6 a^2+b^2\right )+2 A b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b \left (C \left (6 a^2+b^2\right )+2 A b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 A b \sin (c+d x)+\frac {12 a b^2 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 a b^2 C \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+\frac {b^3 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^3 C}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 171, normalized size = 1.02 \[ \frac {2 \, {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, A a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + {\left (6 \, C a^{2} b + {\left (2 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, C a^{2} b + {\left (2 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} \cos \left (d x + c\right )^{3} + 6 \, A a^{2} b \cos \left (d x + c\right )^{2} + 6 \, C a b^{2} \cos \left (d x + c\right ) + C b^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.32, size = 387, normalized size = 2.30 \[ \frac {{\left (A a^{3} + 2 \, C a^{3} + 6 \, A a b^{2}\right )} {\left (d x + c\right )} + {\left (6 \, C a^{2} b + 2 \, A b^{3} + C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (6 \, C a^{2} b + 2 \, A b^{3} + C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((A*a^3 + 2*C*a^3 + 6*A*a*b^2)*(d*x + c) + (6*C*a^2*b + 2*A*b^3 + C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)
) - (6*C*a^2*b + 2*A*b^3 + C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(A*a^3*tan(1/2*d*x + 1/2*c)^7 - 6*A*a
^2*b*tan(1/2*d*x + 1/2*c)^7 + 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - C*b^3*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^3*tan(1/
2*d*x + 1/2*c)^5 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*C*b^3*tan(1/2*d*x +
1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)
^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 - A*a^3*tan(1/2*d*x + 1/2*c) - 6*A*a^2*b*tan(1/2*d*x + 1/2*c) - 6*C*a*b^2*
tan(1/2*d*x + 1/2*c) - C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d

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maple [A]  time = 0.98, size = 196, normalized size = 1.17 \[ \frac {A \,a^{3} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {a^{3} A x}{2}+\frac {A \,a^{3} c}{2 d}+a^{3} C x +\frac {C \,a^{3} c}{d}+\frac {3 A \,a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 C \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+3 A x a \,b^{2}+\frac {3 A a \,b^{2} c}{d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{3} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {b^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 179, normalized size = 1.07 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 4 \, {\left (d x + c\right )} C a^{3} + 12 \, {\left (d x + c\right )} A a b^{2} - C b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{2} b \sin \left (d x + c\right ) + 12 \, C a b^{2} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.47, size = 282, normalized size = 1.68 \[ \frac {2\,\left (\frac {A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+A\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+3\,A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+3\,C\,a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,A\,a^2\,b\,\sin \left (c+d\,x\right )}{4}+\frac {3\,A\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

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))^3,x)

[Out]

(2*((A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
+ C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (C*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2
+ 3*A*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 3*C*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
))))/d + ((A*a^3*sin(2*c + 2*d*x))/8 + (A*a^3*sin(4*c + 4*d*x))/16 + (C*b^3*sin(c + d*x))/2 + (3*A*a^2*b*sin(c
+ d*x))/4 + (3*A*a^2*b*sin(3*c + 3*d*x))/4 + (3*C*a*b^2*sin(2*c + 2*d*x))/2)/(d*(cos(2*c + 2*d*x)/2 + 1/2))

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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))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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