\(\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [106]

Optimal result

Integrand size = 33, antiderivative size = 162 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^3 (7 A+2 C) x+\frac {a^3 (2 A+7 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(A-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \]

[Out]

1/2*a^3*(7*A+2*C)*x+1/2*a^3*(2*A+7*C)*arctanh(sin(d*x+c))/d+5/2*a^3*(A-C)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+a*s
ec(d*x+c))^3*sin(d*x+c)/d-1/2*(A-C)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d-1/2*(A-4*C)*(a^3+a^3*sec(d*x+c))*sin
(d*x+c)/d

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4172, 4103, 4081, 3855} \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 (2 A+7 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}-\frac {(A-4 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac {1}{2} a^3 x (7 A+2 C)-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 a d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d} \]

[In]

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

[Out]

(a^3*(7*A + 2*C)*x)/2 + (a^3*(2*A + 7*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(A - C)*Sin[c + d*x])/(2*d) + (
A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) - ((A - C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/
(2*a*d) - ((A - 4*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(2*d)

Rule 3855

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

Rule 4081

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(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) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4103

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

Rule 4172

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^3 (3 a A-2 a (A-C) \sec (c+d x)) \, dx}{2 a} \\ & = \frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (4 A-C)-2 a^2 (A-4 C) \sec (c+d x)\right ) \, dx}{4 a} \\ & = \frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(A-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (10 a^3 (A-C)+2 a^3 (2 A+7 C) \sec (c+d x)\right ) \, dx}{4 a} \\ & = \frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(A-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac {\int \left (-2 a^4 (7 A+2 C)-2 a^4 (2 A+7 C) \sec (c+d x)\right ) \, dx}{4 a} \\ & = \frac {1}{2} a^3 (7 A+2 C) x+\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(A-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {1}{2} \left (a^3 (2 A+7 C)\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^3 (7 A+2 C) x+\frac {a^3 (2 A+7 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(A-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(364\) vs. \(2(162)=324\).

Time = 7.47 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.25 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \cos ^5(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \left (2 (7 A+2 C) x-\frac {2 (2 A+7 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (2 A+7 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {12 A \cos (d x) \sin (c)}{d}+\frac {A \cos (2 d x) \sin (2 c)}{d}+\frac {12 A \cos (c) \sin (d x)}{d}+\frac {A \cos (2 c) \sin (2 d x)}{d}+\frac {C}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {C}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{16 (A+2 C+A \cos (2 (c+d x)))} \]

[In]

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

[Out]

(a^3*Cos[c + d*x]^5*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*(2*(7*A + 2*C)*x - (2*(2*A
+ 7*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(2*A + 7*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/
d + (12*A*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (12*A*Cos[c]*Sin[d*x])/d + (A*Cos[2*c]*Sin[2*d*x])/
d + C/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (12*C*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x
)/2] - Sin[(c + d*x)/2])) - C/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (12*C*Sin[(d*x)/2])/(d*(Cos[c/2] +
Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(16*(A + 2*C + A*Cos[2*(c + d*x)]))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} C \left (d x +c \right )+3 a^{3} A \sin \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \left (d x +c \right )+3 a^{3} C \tan \left (d x +c \right )+a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(153\)
default	\(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} C \left (d x +c \right )+3 a^{3} A \sin \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \left (d x +c \right )+3 a^{3} C \tan \left (d x +c \right )+a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(153\)
parallelrisch	\(\frac {3 a^{3} \left (-\frac {2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {7 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{3}+\frac {2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {7 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{3}+\frac {7 \left (A +\frac {2 C}{7}\right ) x d \cos \left (2 d x +2 c \right )}{3}+\left (\frac {A}{6}+2 C \right ) \sin \left (2 d x +2 c \right )+A \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{12}+\left (A +\frac {2 C}{3}\right ) \sin \left (d x +c \right )+\frac {7 \left (A +\frac {2 C}{7}\right ) x d}{3}\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(162\)
risch	\(\frac {7 a^{3} A x}{2}+a^{3} x C -\frac {i a^{3} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {i a^{3} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{3} C \left ({\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-6\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\)	\(234\)
norman	\(\frac {\left (\frac {7}{2} a^{3} A +a^{3} C \right ) x +\left (-\frac {7}{2} a^{3} A -a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {7}{2} a^{3} A -a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {7}{2} a^{3} A +a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-7 a^{3} A -2 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-7 a^{3} A -2 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (14 a^{3} A +4 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {5 a^{3} \left (A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {7 a^{3} \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{3} \left (A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 a^{3} \left (11 A -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{3} \left (13 A -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {a^{3} \left (23 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a^{3} \left (2 A +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (2 A +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(403\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.91 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (7 \, A + 2 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} \cos \left (d x + c\right )^{3} + 6 \, A a^{3} \cos \left (d x + c\right )^{2} + 6 \, C a^{3} \cos \left (d x + c\right ) + C a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*(2*(7*A + 2*C)*a^3*d*x*cos(d*x + c)^2 + (2*A + 7*C)*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*A + 7*C)
*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^3*cos(d*x + c)^3 + 6*A*a^3*cos(d*x + c)^2 + 6*C*a^3*cos(d*
x + c) + C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

\[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{3} \left (\int A \cos ^{2}{\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**3*(Integral(A*cos(c + d*x)**2, x) + Integral(3*A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(3*A*cos(c + d*
x)**2*sec(c + d*x)**2, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(C*cos(c + d*x)**2*sec(c
+ d*x)**2, x) + Integral(3*C*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(3*C*cos(c + d*x)**2*sec(c + d*x)**
4, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.08 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} A a^{3} + 4 \, {\left (d x + c\right )} C a^{3} - C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right )}{4 \, d} \]

[In]

integrate(cos(d*x+c)^2*(a+a*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 + 12*(d*x + c)*A*a^3 + 4*(d*x + c)*C*a^3 - C*a^3*(2*sin(d*x + c)/(
sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*A*a^3*(log(sin(d*x + c) + 1) - log(si
n(d*x + c) - 1)) + 6*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 12*C*a^3*
tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.42 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (7 \, A a^{3} + 2 \, C a^{3}\right )} {\left (d x + c\right )} + {\left (2 \, A a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*((7*A*a^3 + 2*C*a^3)*(d*x + c) + (2*A*a^3 + 7*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*a^3 + 7*C*a
^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 5*C*a^3*tan(1/2*d*x + 1/2*c)^7 -
3*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 9*C*a^3*tan
(1/2*d*x + 1/2*c)^3 + 7*A*a^3*tan(1/2*d*x + 1/2*c) + 7*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1
)^2)/d

Mupad [B] (verification not implemented)

Time = 15.62 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.28 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

[In]

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

[Out]

(3*A*a^3*sin(c + d*x))/d + (7*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*A*a^3*atanh(sin(c/2 +
(d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*C*a^3*atanh(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*C*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (C*a^3*sin(c + d*x))/(2*d*c
os(c + d*x)^2) + (A*a^3*cos(c + d*x)*sin(c + d*x))/(2*d)