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

Optimal. Leaf size=167 \[ -\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}+\frac {1}{2} a x \left (2 a^2 C+6 A b^2+3 b^2 C\right )+\frac {3 a^2 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{6 d}-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]

[Out]

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

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Rubi [A]  time = 0.50, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3048, 3049, 3033, 3023, 2735, 3770} \[ -\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}+\frac {1}{2} a x \left (2 a^2 C+6 A b^2+3 b^2 C\right )+\frac {3 a^2 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{6 d}-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*A*b^2 + 2*a^2*C + 3*b^2*C)*x)/2 + (3*a^2*A*b*ArcTanh[Sin[c + d*x]])/d - (b*(a^2*(6*A - 8*C) - b^2*(3*A +
2*C))*Sin[c + d*x])/(3*d) - (a*b^2*(6*A - 5*C)*Cos[c + d*x]*Sin[c + d*x])/(6*d) - (b*(3*A - C)*(a + b*Cos[c +
d*x])^2*Sin[c + d*x])/(3*d) + (A*(a + b*Cos[c + d*x])^3*Tan[c + d*x])/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 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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^2 \left (3 A b+a C \cos (c+d x)-b (3 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{3} \int (a+b \cos (c+d x)) \left (9 a A b+\left (3 A b^2+3 a^2 C+2 b^2 C\right ) \cos (c+d x)-a b (6 A-5 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{6} \int \left (18 a^2 A b+3 a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \cos (c+d x)-2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac {a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{6} \int \left (18 a^2 A b+3 a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) x-\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac {a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\left (3 a^2 A b\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) x+\frac {3 a^2 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac {a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.89, size = 185, normalized size = 1.11 \[ \frac {12 a^3 A \tan (c+d x)+12 a^3 c C+12 a^3 C d x+3 b \left (3 C \left (4 a^2+b^2\right )+4 A b^2\right ) \sin (c+d x)-36 a^2 A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+36 a^2 A b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+36 a A b^2 c+36 a A b^2 d x+9 a b^2 C \sin (2 (c+d x))+18 a b^2 c C+18 a b^2 C d x+b^3 C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a*A*b^2*c + 12*a^3*c*C + 18*a*b^2*c*C + 36*a*A*b^2*d*x + 12*a^3*C*d*x + 18*a*b^2*C*d*x - 36*a^2*A*b*Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] + 36*a^2*A*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*b*(4*A*b^2 + 3*(4
*a^2 + b^2)*C)*Sin[c + d*x] + 9*a*b^2*C*Sin[2*(c + d*x)] + b^3*C*Sin[3*(c + d*x)] + 12*a^3*A*Tan[c + d*x])/(12
*d)

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fricas [A]  time = 1.06, size = 158, normalized size = 0.95 \[ \frac {9 \, A a^{2} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, A a^{2} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, C a^{3} + 3 \, {\left (2 \, A + C\right )} a b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (2 \, C b^{3} \cos \left (d x + c\right )^{3} + 9 \, C a b^{2} \cos \left (d x + c\right )^{2} + 6 \, A a^{3} + 2 \, {\left (9 \, C a^{2} b + {\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.14, size = 306, normalized size = 1.83 \[ \frac {18 \, A a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, A a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (2 \, C a^{3} + 6 \, A a b^{2} + 3 \, C a b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(18*A*a^2*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*A*a^2*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 12*A*a^3*
tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 3*(2*C*a^3 + 6*A*a*b^2 + 3*C*a*b^2)*(d*x + c) + 2*(18*C*a^
2*b*tan(1/2*d*x + 1/2*c)^5 - 9*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^3*tan(1
/2*d*x + 1/2*c)^5 + 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^3*tan(1/2*d*x
+ 1/2*c)^3 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c) + 9*C*a*b^2*tan(1/2*d*x + 1/2*c) + 6*A*b^3*tan(1/2*d*x + 1/2*c) +
6*C*b^3*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 = 0.27, size = 183, normalized size = 1.10 \[ \frac {A \,a^{3} \tan \left (d x +c \right )}{d}+a^{3} C x +\frac {C \,a^{3} c}{d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 C \,a^{2} b \sin \left (d x +c \right )}{d}+3 A x a \,b^{2}+\frac {3 A a \,b^{2} c}{d}+\frac {3 C a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a \,b^{2} C x}{2}+\frac {3 C a \,b^{2} c}{2 d}+\frac {A \,b^{3} \sin \left (d x +c \right )}{d}+\frac {C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) b^{3}}{3 d}+\frac {2 b^{3} C \sin \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 141, normalized size = 0.84 \[ \frac {12 \, {\left (d x + c\right )} C a^{3} + 36 \, {\left (d x + c\right )} A a b^{2} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 18 \, A a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, C a^{2} b \sin \left (d x + c\right ) + 12 \, A b^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.28, size = 238, normalized size = 1.43 \[ \frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+3\,C\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {5\,C\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{12}+\frac {C\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{24}+A\,a^3\,\sin \left (c+d\,x\right )+\frac {3\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{8}+\frac {3\,C\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,C\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{8}}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 6*A*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) -
A*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*6i + 3*C*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/d + ((A*b^3*sin(2*c + 2*d*x))/2 + (5*C*b^3*sin(2*c + 2*d*x))/12 + (C*b^3*sin(4*c + 4*d*x))/24 + A*a^3*
sin(c + d*x) + (3*C*a*b^2*sin(c + d*x))/8 + (3*C*a^2*b*sin(2*c + 2*d*x))/2 + (3*C*a*b^2*sin(3*c + 3*d*x))/8)/(
d*cos(c + d*x))

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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+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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