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

Optimal. Leaf size=167 \[ \frac {a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \sin (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{4 d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

[Out]

1/8*b*(12*a^2*(2*A+C)+b^2*(4*A+3*C))*x+a^3*A*arctanh(sin(d*x+c))/d+1/2*a*(6*A*b^2+(a^2+4*b^2)*C)*sin(d*x+c)/d+
1/8*b*(2*a^2*C+b^2*(4*A+3*C))*cos(d*x+c)*sin(d*x+c)/d+1/4*a*C*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/4*C*(a+b*cos(d
*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.54, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3050, 3049, 3033, 3023, 2735, 3770} \[ \frac {a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \sin (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+\frac {a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{4 d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*x)/8 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + (a*(6*A*b^2 + (a^2 + 4*b^2)*C
)*Sin[c + d*x])/(2*d) + (b*(2*a^2*C + b^2*(4*A + 3*C))*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*C*(a + b*Cos[c +
d*x])^2*Sin[c + d*x])/(4*d) + (C*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*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 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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 (c+d x) \, dx &=\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x))^2 \left (4 a A+b (4 A+3 C) \cos (c+d x)+3 a C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {a C (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{12} \int (a+b \cos (c+d x)) \left (12 a^2 A+3 a b (8 A+5 C) \cos (c+d x)+3 \left (2 a^2 C+b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a C (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{24} \int \left (24 a^3 A+3 b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \cos (c+d x)+12 a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sin (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a C (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{24} \int \left (24 a^3 A+3 b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{8} b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) x+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sin (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a C (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) x+\frac {a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sin (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a C (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 180, normalized size = 1.08 \[ \frac {-32 a^3 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 a^3 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 b (c+d x) \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+8 a \left (4 a^2 C+12 A b^2+9 b^2 C\right ) \sin (c+d x)+8 b \left (C \left (3 a^2+b^2\right )+A b^2\right ) \sin (2 (c+d x))+8 a b^2 C \sin (3 (c+d x))+b^3 C \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*(c + d*x) - 32*a^3*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*a
^3*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*a*(12*A*b^2 + 4*a^2*C + 9*b^2*C)*Sin[c + d*x] + 8*b*(A*b^2 +
(3*a^2 + b^2)*C)*Sin[2*(c + d*x)] + 8*a*b^2*C*Sin[3*(c + d*x)] + b^3*C*Sin[4*(c + d*x)])/(32*d)

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fricas [A]  time = 0.97, size = 146, normalized size = 0.87 \[ \frac {4 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} d x + {\left (2 \, C b^{3} \cos \left (d x + c\right )^{3} + 8 \, C a b^{2} \cos \left (d x + c\right )^{2} + 8 \, C a^{3} + 8 \, {\left (3 \, A + 2 \, C\right )} a b^{2} + {\left (12 \, C a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

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),x, algorithm="fricas")

[Out]

1/8*(4*A*a^3*log(sin(d*x + c) + 1) - 4*A*a^3*log(-sin(d*x + c) + 1) + (12*(2*A + C)*a^2*b + (4*A + 3*C)*b^3)*d
*x + (2*C*b^3*cos(d*x + c)^3 + 8*C*a*b^2*cos(d*x + c)^2 + 8*C*a^3 + 8*(3*A + 2*C)*a*b^2 + (12*C*a^2*b + (4*A +
3*C)*b^3)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 1.89, size = 503, normalized size = 3.01 \[ \frac {8 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, C b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, 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 )}^{4}}}{8 \, 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),x, algorithm="giac")

[Out]

1/8*(8*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (24*A*a^2*b + 1
2*C*a^2*b + 4*A*b^3 + 3*C*b^3)*(d*x + c) + 2*(8*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)
^7 + 24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 4*A*b^3*tan(1/2*d*x + 1/2*c)^7 -
5*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 72*A*a*
b^2*tan(1/2*d*x + 1/2*c)^5 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*C*b^3*tan(
1/2*d*x + 1/2*c)^5 + 24*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*A*a*b^2*tan(1/2*
d*x + 1/2*c)^3 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/
2*c)^3 + 8*C*a^3*tan(1/2*d*x + 1/2*c) + 12*C*a^2*b*tan(1/2*d*x + 1/2*c) + 24*A*a*b^2*tan(1/2*d*x + 1/2*c) + 24
*C*a*b^2*tan(1/2*d*x + 1/2*c) + 4*A*b^3*tan(1/2*d*x + 1/2*c) + 5*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/
2*c)^2 + 1)^4)/d

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maple [A]  time = 0.27, size = 252, normalized size = 1.51 \[ \frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3} C \sin \left (d x +c \right )}{d}+3 A x \,a^{2} b +\frac {3 A \,a^{2} b c}{d}+\frac {3 C \,a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 C \,a^{2} b x}{2}+\frac {3 C \,a^{2} b c}{2 d}+\frac {3 A a \,b^{2} \sin \left (d x +c \right )}{d}+\frac {C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a \,b^{2}}{d}+\frac {2 C a \,b^{2} \sin \left (d x +c \right )}{d}+\frac {A \,b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {A x \,b^{3}}{2}+\frac {A \,b^{3} c}{2 d}+\frac {b^{3} C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 b^{3} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 b^{3} C x}{8}+\frac {3 b^{3} C c}{8 d} \]

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),x)

[Out]

1/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+a^3*C*sin(d*x+c)/d+3*A*x*a^2*b+3/d*A*a^2*b*c+3/2/d*C*a^2*b*cos(d*x+c)*sin(
d*x+c)+3/2*C*a^2*b*x+3/2/d*C*a^2*b*c+3/d*A*a*b^2*sin(d*x+c)+1/d*C*sin(d*x+c)*cos(d*x+c)^2*a*b^2+2/d*C*a*b^2*si
n(d*x+c)+1/2/d*A*b^3*cos(d*x+c)*sin(d*x+c)+1/2*A*x*b^3+1/2/d*A*b^3*c+1/4/d*b^3*C*sin(d*x+c)*cos(d*x+c)^3+3/8/d
*b^3*C*cos(d*x+c)*sin(d*x+c)+3/8*b^3*C*x+3/8/d*b^3*C*c

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maxima [A]  time = 0.70, size = 167, normalized size = 1.00 \[ \frac {96 \, {\left (d x + c\right )} A a^{2} b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{2} + 8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} + 32 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 32 \, C a^{3} \sin \left (d x + c\right ) + 96 \, A a b^{2} \sin \left (d x + c\right )}{32 \, d} \]

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),x, algorithm="maxima")

[Out]

1/32*(96*(d*x + c)*A*a^2*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2*b - 32*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*C*a*b^2 + 8*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^3 + (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*C*b^3 + 32*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 32*C*a^3*sin(d*x + c) + 96*A*a*b^2*sin(d*x + c))/d

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mupad [B]  time = 3.02, size = 2008, normalized size = 12.02 \[ \text {result too large to display} \]

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),x)

[Out]

(tan(c/2 + (d*x)/2)*(A*b^3 + 2*C*a^3 + (5*C*b^3)/4 + 6*A*a*b^2 + 6*C*a*b^2 + 3*C*a^2*b) - tan(c/2 + (d*x)/2)^7
*(A*b^3 - 2*C*a^3 + (5*C*b^3)/4 - 6*A*a*b^2 - 6*C*a*b^2 + 3*C*a^2*b) + tan(c/2 + (d*x)/2)^3*(A*b^3 + 6*C*a^3 -
(3*C*b^3)/4 + 18*A*a*b^2 + 10*C*a*b^2 + 3*C*a^2*b) + tan(c/2 + (d*x)/2)^5*(6*C*a^3 - A*b^3 + (3*C*b^3)/4 + 18
*A*a*b^2 + 10*C*a*b^2 - 3*C*a^2*b))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)
^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (b*atan(((b*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96
*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*
b^2) - (b*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 + 96*A*a^2*b + 48*C*a^2*b)
*1i)/8)*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2))/8 + (b*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2
*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 120*A*C*a^2*b^4 +
288*A*C*a^4*b^2) + (b*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 + 96*A*a^2*b +
48*C*a^2*b)*1i)/8)*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2))/8)/((b*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*
b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 120*A*
C*a^2*b^4 + 288*A*C*a^4*b^2) + (b*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 +
96*A*a^2*b + 48*C*a^2*b)*1i)/8)*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*1i)/8 - (b*(tan(c/2 + (d*x)/2)*(32*A
^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12*A
*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*b^2) - (b*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*(32*A*a^3 + 16*A*b^
3 + 12*C*b^3 + 96*A*a^2*b + 48*C*a^2*b)*1i)/8)*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*1i)/8 - 192*A^3*a^8*b
+ 16*A^3*a^3*b^6 + 192*A^3*a^5*b^4 - 32*A^3*a^6*b^3 + 576*A^3*a^7*b^2 - 96*A^2*C*a^8*b + 9*A*C^2*a^3*b^6 + 72
*A*C^2*a^5*b^4 + 144*A*C^2*a^7*b^2 + 24*A^2*C*a^3*b^6 + 240*A^2*C*a^5*b^4 - 24*A^2*C*a^6*b^3 + 576*A^2*C*a^7*b
^2))*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2))/(4*d) - (A*a^3*atan((A*a^3*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8
*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 1
20*A*C*a^2*b^4 + 288*A*C*a^4*b^2) + A*a^3*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 + 96*A*a^2*b + 48*C*a^2*b))*1i + A*a
^3*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2
*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*b^2) - A*a^3*(32*A*a^3 + 16*A*b^3 + 12*C*b^
3 + 96*A*a^2*b + 48*C*a^2*b))*1i)/(A*a^3*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*
a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*b^2)
+ A*a^3*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 + 96*A*a^2*b + 48*C*a^2*b)) - 192*A^3*a^8*b - A*a^3*(tan(c/2 + (d*x)/2
)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2
+ 12*A*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*b^2) - A*a^3*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 + 96*A*a^2*b + 48*C
*a^2*b)) + 16*A^3*a^3*b^6 + 192*A^3*a^5*b^4 - 32*A^3*a^6*b^3 + 576*A^3*a^7*b^2 - 96*A^2*C*a^8*b + 9*A*C^2*a^3*
b^6 + 72*A*C^2*a^5*b^4 + 144*A*C^2*a^7*b^2 + 24*A^2*C*a^3*b^6 + 240*A^2*C*a^5*b^4 - 24*A^2*C*a^6*b^3 + 576*A^2
*C*a^7*b^2))*2i)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, dx \]

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),x)

[Out]

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

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