3.5.32 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [432]

Optimal. Leaf size=171 \[ \frac {1}{2} a^3 (7 A+6 B+2 C) x+\frac {a^3 (2 A+6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(A-2 B-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.30, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4171, 4103, 4081, 3855} \begin {gather*} \frac {a^3 (2 A+6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(A-2 B-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+6 B+2 C)+\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}-\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(7*A + 6*B + 2*C)*x)/2 + (a^3*(2*A + 6*B + 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*Si
n[c + d*x])/(2*a*d) - ((A - 2*B - 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 4171

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\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 (a (3 A+2 B)-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+2 B-C)-2 a^2 (A-2 B-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-2 B-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+6 B+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-2 B-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+6 B+2 C)-2 a^4 (2 A+6 B+7 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac {1}{2} a^3 (7 A+6 B+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-2 B-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+6 B+7 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^3 (7 A+6 B+2 C) x+\frac {a^3 (2 A+6 B+7 C) \tanh ^{-1}(\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-2 B-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(406\) vs. \(2(171)=342\).
time = 5.91, size = 406, normalized size = 2.37 \begin {gather*} \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+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 (7 A+6 B+2 C) x-\frac {2 (2 A+6 B+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+6 B+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 {4 (3 A+B) \cos (d x) \sin (c)}{d}+\frac {A \cos (2 d x) \sin (2 c)}{d}+\frac {4 (3 A+B) \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 {4 (B+3 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 {4 (B+3 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+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(7*A +
6*B + 2*C)*x - (2*(2*A + 6*B + 7*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(2*A + 6*B + 7*C)*Log[Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(3*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (4*(3*A +
B)*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) + (4*(B + 3
*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] + S
in[(c + d*x)/2])^2) + (4*(B + 3*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 + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

________________________________________________________________________________________

Maple [A]
time = 0.84, size = 206, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} B \tan \left (d x +c \right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{3} \left (d x +c \right )+3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} C \tan \left (d x +c \right )+3 A \,a^{3} \sin \left (d x +c \right )+3 a^{3} B \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} B \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )}{d}\)	\(206\)
default	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} B \tan \left (d x +c \right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{3} \left (d x +c \right )+3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} C \tan \left (d x +c \right )+3 A \,a^{3} \sin \left (d x +c \right )+3 a^{3} B \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} B \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )}{d}\)	\(206\)
risch	\(\frac {7 a^{3} A x}{2}+3 a^{3} B x +a^{3} x C -\frac {i a^{3} B \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {3 i A \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i A \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i a^{3} B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{3} \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B -6 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 i A \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(343\)
norman	\(\frac {\left (\frac {7}{2} A \,a^{3}+a^{3} C +3 a^{3} B \right ) x +\left (-\frac {7}{2} A \,a^{3}-a^{3} C -3 a^{3} B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {7}{2} A \,a^{3}-a^{3} C -3 a^{3} B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {7}{2} A \,a^{3}+a^{3} C +3 a^{3} B \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-7 A \,a^{3}-6 a^{3} B -2 a^{3} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-7 A \,a^{3}-6 a^{3} B -2 a^{3} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (14 A \,a^{3}+12 a^{3} B +4 a^{3} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{3} \left (7 A +4 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {5 a^{3} \left (A -C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (11 A -7 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (A +4 B +5 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (13 A -7 C +4 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (23 A +8 B +5 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a^{3} \left (2 A +6 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (2 A +6 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(466\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.32, size = 237, normalized size = 1.39 \begin {gather*} \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} + 12 \, {\left (d x + c\right )} B 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 \, B 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 ) + 4 \, B a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [A]
time = 2.45, size = 165, normalized size = 0.96 \begin {gather*} \frac {2 \, {\left (7 \, A + 6 \, B + 2 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + 6 \, B + 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 + 6 \, B + 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} + 2 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 3 \, C\right )} 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 0.53, size = 280, normalized size = 1.64 \begin {gather*} \frac {{\left (7 \, A a^{3} + 6 \, B a^{3} + 2 \, C a^{3}\right )} {\left (d x + c\right )} + {\left (2 \, A a^{3} + 6 \, B 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} + 6 \, B 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} - 4 \, B 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 ) + 4 \, B 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((7*A*a^3 + 6*B*a^3 + 2*C*a^3)*(d*x + c) + (2*A*a^3 + 6*B*a^3 + 7*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)
) - (2*A*a^3 + 6*B*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 - 4*B*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) + 4*B*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]
time = 4.10, size = 329, normalized size = 1.92 \begin {gather*} \frac {2\,\left (\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 )}{2}-A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}+3\,B\,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 )-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}+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\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,7{}\mathrm {i}}{2}\right )}{d}+\frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {3\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*((7*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - A*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*
x)/2))*1i + 3*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 +
(d*x)/2))*3i + C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (C*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/
2 + (d*x)/2))*7i)/2))/d + ((A*a^3*sin(2*c + 2*d*x))/8 + (3*A*a^3*sin(3*c + 3*d*x))/4 + (A*a^3*sin(4*c + 4*d*x)
)/16 + (B*a^3*sin(2*c + 2*d*x))/2 + (B*a^3*sin(3*c + 3*d*x))/4 + (3*C*a^3*sin(2*c + 2*d*x))/2 + (3*A*a^3*sin(c
+ d*x))/4 + (B*a^3*sin(c + d*x))/4 + (C*a^3*sin(c + d*x))/2)/(d*(cos(2*c + 2*d*x)/2 + 1/2))

________________________________________________________________________________________