\(\int (a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [544]

Optimal result

Integrand size = 33, antiderivative size = 168 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {1}{2} b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) x+\frac {a \left (6 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {3 a b^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac {b^3 (4 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 A b (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3127, 3126, 3112, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a \left (a^2 (A+2 C)+6 A b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {1}{2} b x \left (C \left (6 a^2+b^2\right )+2 A b^2\right )-\frac {3 a b^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {3 A b \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}-\frac {b^3 (4 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]

[In]

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

[Out]

(b*(2*A*b^2 + (6*a^2 + b^2)*C)*x)/2 + (a*(6*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a*b^2*(3*
A - 2*C)*Sin[c + d*x])/(2*d) - (b^3*(4*A - C)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (3*A*b*(a + b*Cos[c + d*x])^2
*Tan[c + d*x])/(2*d) + (A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 2814

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 3102

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 3112

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 3126

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 3127

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*Si
n[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 3855

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

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

Mathematica [A] (verified)

Time = 4.06 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.70 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) (c+d x)-2 a \left (6 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a \left (6 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^3 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 a^2 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^3 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 a^2 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+12 a b^2 C \sin (c+d x)+b^3 C \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.44 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93

method	result	size
parts	\(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,b^{3}+3 C \,a^{2} b \right ) \left (d x +c \right )}{d}+\frac {\left (3 A a \,b^{2}+C \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {3 \sin \left (d x +c \right ) C a \,b^{2}}{d}\)	\(156\)
derivativedivides	\(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \tan \left (d x +c \right )+3 C \,a^{2} b \left (d x +c \right )+3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (d x +c \right )+C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(157\)
default	\(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \tan \left (d x +c \right )+3 C \,a^{2} b \left (d x +c \right )+3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (d x +c \right )+C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(157\)
parallelrisch	\(\frac {-4 \left (6 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \left (6 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 x b d \left (3 a^{2} C +\left (A +\frac {C}{2}\right ) b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (24 A \,a^{2} b +2 C \,b^{3}\right ) \sin \left (2 d x +2 c \right )+12 C \sin \left (3 d x +3 c \right ) a \,b^{2}+C \sin \left (4 d x +4 c \right ) b^{3}+\left (8 A \,a^{3}+12 C a \,b^{2}\right ) \sin \left (d x +c \right )+8 x b d \left (3 a^{2} C +\left (A +\frac {C}{2}\right ) b^{2}\right )}{8 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(230\)
risch	\(A \,b^{3} x +3 C \,a^{2} b x +\frac {b^{3} C x}{2}-\frac {i C \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C a \,b^{2}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C a \,b^{2}}{2 d}+\frac {i C \,b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i A \,a^{2} \left ({\mathrm e}^{3 i \left (d x +c \right )} a -6 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}-6 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(296\)
norman	\(\frac {\left (A \,b^{3}+3 C \,a^{2} b +\frac {1}{2} C \,b^{3}\right ) x +\left (-5 A \,b^{3}-15 C \,a^{2} b -\frac {5}{2} C \,b^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-5 A \,b^{3}-15 C \,a^{2} b -\frac {5}{2} C \,b^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,b^{3}+3 C \,a^{2} b +\frac {1}{2} C \,b^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,b^{3}+3 C \,a^{2} b +\frac {1}{2} C \,b^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,b^{3}+3 C \,a^{2} b +\frac {1}{2} C \,b^{3}\right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,b^{3}+9 C \,a^{2} b +\frac {3}{2} C \,b^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,b^{3}+9 C \,a^{2} b +\frac {3}{2} C \,b^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (A \,a^{3}-6 A \,a^{2} b +6 C a \,b^{2}-C \,b^{3}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (A \,a^{3}+6 A \,a^{2} b +6 C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {3 \left (5 A \,a^{3}-10 A \,a^{2} b -2 C a \,b^{2}+C \,b^{3}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (5 A \,a^{3}+10 A \,a^{2} b -2 C a \,b^{2}-C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (5 A \,a^{2}-6 b^{2} C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a \left (A \,a^{2}-4 A a b +2 b^{2} C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a \left (A \,a^{2}+4 A a b +2 b^{2} 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 )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a \left (A \,a^{2}+6 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A \,a^{2}+6 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(617\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (6 \, C a^{2} b + {\left (2 \, A + C\right )} b^{3}\right )} d x \cos \left (d x + c\right )^{2} + {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b^{3} \cos \left (d x + c\right )^{3} + 6 \, C a b^{2} \cos \left (d x + c\right )^{2} + 6 \, A a^{2} b \cos \left (d x + c\right ) + A a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.07 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} C a^{2} b + 4 \, {\left (d x + c\right )} A b^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} - A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b^{2} \sin \left (d x + c\right ) + 12 \, A a^{2} b \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(12*(d*x + c)*C*a^2*b + 4*(d*x + c)*A*b^3 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*b^3 - A*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*C*a^3*(log(sin(d*x + c) + 1) - log(
sin(d*x + c) - 1)) + 6*A*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a*b^2*sin(d*x + c) + 12*
A*a^2*b*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (156) = 312\).

Time = 0.35 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.29 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (6 \, C a^{2} b + 2 \, A b^{3} + C b^{3}\right )} {\left (d x + c\right )} + {\left (A a^{3} + 2 \, C a^{3} + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{3} + 2 \, C a^{3} + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*((6*C*a^2*b + 2*A*b^3 + C*b^3)*(d*x + c) + (A*a^3 + 2*C*a^3 + 6*A*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)
) - (A*a^3 + 2*C*a^3 + 6*A*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A*a^3*tan(1/2*d*x + 1/2*c)^7 - 6*A*a
^2*b*tan(1/2*d*x + 1/2*c)^7 + 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - C*b^3*tan(1/2*d*x + 1/2*c)^7 + 3*A*a^3*tan(1/
2*d*x + 1/2*c)^5 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*b^3*tan(1/2*d*x +
1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)
^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 + A*a^3*tan(1/2*d*x + 1/2*c) + 6*A*a^2*b*tan(1/2*d*x + 1/2*c) + 6*C*a*b^2*
tan(1/2*d*x + 1/2*c) + C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 3.69 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.68 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2\,\left (\frac {A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+A\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+3\,A\,a\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+3\,C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {C\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {C\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,C\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

[In]

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

[Out]

(2*((A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + A*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
+ C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (C*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2
+ 3*A*a*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 3*C*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
))))/d + ((C*b^3*sin(2*c + 2*d*x))/8 + (C*b^3*sin(4*c + 4*d*x))/16 + (A*a^3*sin(c + d*x))/2 + (3*C*a*b^2*sin(c
+ d*x))/4 + (3*A*a^2*b*sin(2*c + 2*d*x))/2 + (3*C*a*b^2*sin(3*c + 3*d*x))/4)/(d*(cos(2*c + 2*d*x)/2 + 1/2))