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

Optimal. Leaf size=145 \[ \frac {a \left (2 a^2 A+9 a b B+8 A b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 (3 a B+5 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {\left (a^3 B+3 a^2 A b+6 a b^2 B+2 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^3 B x \]

[Out]

b^3*B*x+1/2*(3*A*a^2*b+2*A*b^3+B*a^3+6*B*a*b^2)*arctanh(sin(d*x+c))/d+1/3*a*(2*A*a^2+8*A*b^2+9*B*a*b)*tan(d*x+
c)/d+1/6*a^2*(5*A*b+3*B*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*a*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.35, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2989, 3031, 3021, 2735, 3770} \[ \frac {a \left (2 a^2 A+9 a b B+8 A b^2\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^2 A b+a^3 B+6 a b^2 B+2 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 (3 a B+5 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^3 B x \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^3*B*x + ((3*a^2*A*b + 2*A*b^3 + a^3*B + 6*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*a^2*A + 8*A*b^2 + 9*
a*b*B)*Tan[c + d*x])/(3*d) + (a^2*(5*A*b + 3*a*B)*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (a*A*(a + b*Cos[c + d*x])
^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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 2989

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

Rule 3021

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

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 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 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac {a A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \cos (c+d x)) \left (a (5 A b+3 a B)+\left (2 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)+3 b^2 B \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a^2 (5 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-2 a \left (2 a^2 A+8 A b^2+9 a b B\right )-3 \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) \cos (c+d x)-6 b^3 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a \left (2 a^2 A+8 A b^2+9 a b B\right ) \tan (c+d x)}{3 d}+\frac {a^2 (5 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-3 \left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right )-6 b^3 B \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 B x+\frac {a \left (2 a^2 A+8 A b^2+9 a b B\right ) \tan (c+d x)}{3 d}+\frac {a^2 (5 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{2} \left (-3 a^2 A b-2 A b^3-a^3 B-6 a b^2 B\right ) \int \sec (c+d x) \, dx\\ &=b^3 B x+\frac {\left (3 a^2 A b+2 A b^3+a^3 B+6 a b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a \left (2 a^2 A+8 A b^2+9 a b B\right ) \tan (c+d x)}{3 d}+\frac {a^2 (5 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 108, normalized size = 0.74 \[ \frac {2 a^3 A \tan ^3(c+d x)+3 a \tan (c+d x) \left (2 a^2 A+a (a B+3 A b) \sec (c+d x)+6 a b B+6 A b^2\right )+3 \left (a^3 B+3 a^2 A b+6 a b^2 B+2 A b^3\right ) \tanh ^{-1}(\sin (c+d x))+6 b^3 B d x}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 189, normalized size = 1.30 \[ \frac {12 \, B b^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{3} + 2 \, {\left (2 \, A a^{3} + 9 \, B a^{2} b + 9 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

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))*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(12*B*b^3*d*x*cos(d*x + c)^3 + 3*(B*a^3 + 3*A*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*a^2*b + 6*B*a*b^2 + 2*A*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*A*a^3 + 2*(2
*A*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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giac [B]  time = 0.44, size = 336, normalized size = 2.32 \[ \frac {6 \, {\left (d x + c\right )} B b^{3} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a^{3} + 3 \, A a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

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))*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*B*b^3 + 3*(B*a^3 + 3*A*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*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 - 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 - 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) + 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

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maple [A]  time = 0.14, size = 223, normalized size = 1.54 \[ \frac {2 A \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 A \,a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{2} b B \tan \left (d x +c \right )}{d}+\frac {3 A \,b^{2} a \tan \left (d x +c \right )}{d}+\frac {3 B \,b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+b^{3} B x +\frac {b^{3} B c}{d} \]

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))*sec(d*x+c)^4,x)

[Out]

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(se
c(d*x+c)+tan(d*x+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*t
an(d*x+c)+3/d*A*b^2*a*tan(d*x+c)+3/d*B*b^2*a*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*b^3*ln(sec(d*x+c)+tan(d*x+c))+b^3
*B*x+1/d*b^3*B*c

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maxima [A]  time = 0.65, size = 216, normalized size = 1.49 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} B b^{3} - 3 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 9 \, A a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{2} b \tan \left (d x + c\right ) + 36 \, A a b^{2} \tan \left (d x + c\right )}{12 \, d} \]

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))*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 + 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*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)) + 36*B*a^2*b*tan(d*x + c) + 36*A*a*b^2*tan(d*x + c))/d

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mupad [B]  time = 1.95, size = 526, normalized size = 3.63 \[ \frac {\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{6}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4}-\frac {A\,b^3\,\cos \left (c+d\,x\right )\,\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}}{2}-\frac {B\,a^3\,\cos \left (c+d\,x\right )\,\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}}{4}+\frac {3\,B\,b^3\,\cos \left (c+d\,x\right )\,\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}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4}-\frac {A\,b^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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{2}-\frac {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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{4}+\frac {B\,b^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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {A\,a^2\,b\,\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 )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{4}-\frac {B\,a\,b^2\,\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 )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{2}-\frac {A\,a^2\,b\,\cos \left (c+d\,x\right )\,\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 )\,9{}\mathrm {i}}{4}-\frac {B\,a\,b^2\,\cos \left (c+d\,x\right )\,\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 )\,9{}\mathrm {i}}{2}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*a^3*sin(3*c + 3*d*x))/6 + (B*a^3*sin(2*c + 2*d*x))/4 + (A*a^3*sin(c + d*x))/2 + (3*A*a*b^2*sin(c + d*x))/4
+ (3*B*a^2*b*sin(c + d*x))/4 - (A*b^3*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/2 - (
B*a^3*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/4 + (3*B*b^3*cos(c + d*x)*atan(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (3*A*a^2*b*sin(2*c + 2*d*x))/4 + (3*A*a*b^2*sin(3*c + 3*d*x))/4 + (3*B*a^
2*b*sin(3*c + 3*d*x))/4 - (A*b^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/2 - (B*
a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/4 + (B*b^3*atan(sin(c/2 + (d*x)/2)/c
os(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 - (A*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3
*d*x)*3i)/4 - (B*a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*3i)/2 - (A*a^2*b*cos(
c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i)/4 - (B*a*b^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/
2)*1i)/cos(c/2 + (d*x)/2))*9i)/2)/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))

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

[Out]

Timed out

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