3.960 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=196 \[ \frac{a \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+3 A b^2\right )}{3 d}+\frac{\left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \sin (c+d x) (3 a B+5 A b-6 b C)}{6 d}+\frac{(a B+A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+b^2 x (3 a C+b B) \]

[Out]

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

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Rubi [A]  time = 0.641592, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3047, 3031, 3023, 2735, 3770} \[ \frac{a \tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+3 A b^2\right )}{3 d}+\frac{\left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \sin (c+d x) (3 a B+5 A b-6 b C)}{6 d}+\frac{(a B+A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+b^2 x (3 a C+b B) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

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

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3770

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

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x))^2 \left (3 (A b+a B)+(2 a A+3 b B+3 a C) \cos (c+d x)-b (A-3 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{(A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (a+b \cos (c+d x)) \left (2 \left (3 A b^2+6 a b B+\frac{1}{2} a^2 (4 A+6 C)\right )+\left (3 a^2 B+6 b^2 B+a b (5 A+12 C)\right ) \cos (c+d x)-b (5 A b+3 a B-6 b C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{(A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 \left (2 A b^3+a^3 B+6 a b^2 B+3 a^2 b (A+2 C)\right )-6 b^2 (b B+3 a C) \cos (c+d x)+b^2 (5 A b+3 a B-6 b C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 (5 A b+3 a B-6 b C) \sin (c+d x)}{6 d}+\frac{a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{(A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 \left (2 A b^3+a^3 B+6 a b^2 B+3 a^2 b (A+2 C)\right )-6 b^2 (b B+3 a C) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^2 (b B+3 a C) x-\frac{b^2 (5 A b+3 a B-6 b C) \sin (c+d x)}{6 d}+\frac{a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{(A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{2} \left (-2 A b^3-a^3 B-6 a b^2 B-3 a^2 b (A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=b^2 (b B+3 a C) x+\frac{\left (2 A b^3+a^3 B+6 a b^2 B+3 a^2 b (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 (5 A b+3 a B-6 b C) \sin (c+d x)}{6 d}+\frac{a \left (3 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{(A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 5.74256, size = 429, normalized size = 2.19 \[ \frac{\frac{4 a \sin \left (\frac{1}{2} (c+d x)\right ) \left (a^2 (2 A+3 C)+9 a b B+9 A b^2\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a \sin \left (\frac{1}{2} (c+d x)\right ) \left (a^2 (2 A+3 C)+9 a b B+9 A b^2\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-6 \left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 \left (3 a^2 b (A+2 C)+a^3 B+6 a b^2 B+2 A b^3\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^2 (a (A+3 B)+9 A b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2 (a (A+3 B)+9 A b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+12 b^2 (c+d x) (3 a C+b B)+12 b^3 C \sin (c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

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

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Maple [A]  time = 0.075, size = 294, normalized size = 1.5 \begin{align*}{\frac{A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{3}Bx+{\frac{B{b}^{3}c}{d}}+{\frac{C{b}^{3}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{aA{b}^{2}\tan \left ( dx+c \right ) }{d}}+3\,{\frac{a{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,a{b}^{2}Cx+3\,{\frac{Ca{b}^{2}c}{d}}+{\frac{3\,A{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{2}bB\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,A{a}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

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

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Maxima [A]  time = 1.02746, size = 378, normalized size = 1.93 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 36 \,{\left (d x + c\right )} C a b^{2} + 12 \,{\left (d x + c\right )} B b^{3} - 3 \, B 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 )} - 9 \, A a^{2} b{\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 )} + 18 \, 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 )} + 18 \, B 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 )} + 6 \, 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 \, C b^{3} \sin \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right ) + 36 \, B a^{2} b \tan \left (d x + c\right ) + 36 \, A a b^{2} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.06222, size = 543, normalized size = 2.77 \begin{align*} \frac{12 \,{\left (3 \, C a b^{2} + B b^{3}\right )} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{3} + 3 \,{\left (A + 2 \, C\right )} a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (B a^{3} + 3 \,{\left (A + 2 \, C\right )} a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, C b^{3} \cos \left (d x + c\right )^{3} + 2 \, A a^{3} + 2 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{3} + 9 \, B a^{2} b + 9 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.25153, size = 591, normalized size = 3.02 \begin{align*} \frac{\frac{12 \, C b^{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 (3 \, C a b^{2} + B b^{3}\right )}{\left (d x + c\right )} + 3 \,{\left (B a^{3} + 3 \, A a^{2} b + 6 \, C a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (B a^{3} + 3 \, A a^{2} b + 6 \, C a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, A a b^{2} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(12*C*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(3*C*a*b^2 + B*b^3)*(d*x + c) + 3*(B*a^3 +
3*A*a^2*b + 6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a^3 + 3*A*a^2*b + 6*C*
a^2*b + 6*B*a*b^2 + 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^3*
tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*a^2*b*tan(1/
2*d*x + 1/2*c)^5 + 18*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^3*tan(1/2*d*x +
1/2*c)^3 - 36*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1/2*d*x + 1/2*
c) + 3*B*a^3*tan(1/2*d*x + 1/2*c) + 6*C*a^3*tan(1/2*d*x + 1/2*c) + 9*A*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a^2*b
*tan(1/2*d*x + 1/2*c) + 18*A*a*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d