Integrand size = 33, antiderivative size = 156 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=3 a^3 C x+\frac {a^3 (5 A+6 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {(5 A+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
3*a^3*C*x+1/2*a^3*(5*A+6*C)*arctanh(sin(d*x+c))/d-5/2*a^3*A*sin(d*x+c)/d+1 /3*(5*A+3*C)*(a^3+a^3*cos(d*x+c))*tan(d*x+c)/d+1/2*A*(a^2+a^2*cos(d*x+c))^ 2*sec(d*x+c)*tan(d*x+c)/a/d+1/3*A*(a+a*cos(d*x+c))^3*sec(d*x+c)^2*tan(d*x+ c)/d
Leaf count is larger than twice the leaf count of optimal. \(475\) vs. \(2(156)=312\).
Time = 8.24 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.04 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {1}{48} a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (18 C x-\frac {3 (5 A+6 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {3 (5 A+6 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {6 C \cos (d x) \sin (c)}{d}+\frac {6 C \cos (c) \sin (d x)}{d}+\frac {A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {A \left (5 \cos \left (\frac {c}{2}\right )-4 \sin \left (\frac {c}{2}\right )\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 (11 A+3 C) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {A \left (5 \cos \left (\frac {c}{2}\right )+4 \sin \left (\frac {c}{2}\right )\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 (11 A+3 C) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \]
(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(18*C*x - (3*(5*A + 6*C)*Log[ Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (3*(5*A + 6*C)*Log[Cos[(c + d*x) /2] + Sin[(c + d*x)/2]])/d + (6*C*Cos[d*x]*Sin[c])/d + (6*C*Cos[c]*Sin[d*x ])/d + (A*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[( c + d*x)/2])^3) + (A*(5*Cos[c/2] - 4*Sin[c/2]))/(d*(Cos[c/2] - Sin[c/2])*( Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (2*(11*A + 3*C)*Sin[(d*x)/2])/(d *(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (A*Sin[(d* x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - (A*(5*Cos[c/2] + 4*Sin[c/2]))/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (2*(11*A + 3*C)*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[ c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/48
Time = 1.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3523, 3042, 3454, 3042, 3454, 27, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 3523 |
\(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 (3 a A-a (A-3 C) \cos (c+d x)) \sec ^3(c+d x)dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a A-a (A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \frac {\frac {1}{2} \int (\cos (c+d x) a+a)^2 \left (2 a^2 (5 A+3 C)-a^2 (5 A-6 C) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a^2 (5 A+3 C)-a^2 (5 A-6 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \frac {\frac {1}{2} \left (\int 3 (\cos (c+d x) a+a) \left (a^3 (5 A+6 C)-5 a^3 A \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int (\cos (c+d x) a+a) \left (a^3 (5 A+6 C)-5 a^3 A \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^3 (5 A+6 C)-5 a^3 A \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int \left (-5 A \cos ^2(c+d x) a^4+(5 A+6 C) a^4+\left (a^4 (5 A+6 C)-5 a^4 A\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int \frac {-5 A \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+(5 A+6 C) a^4+\left (a^4 (5 A+6 C)-5 a^4 A\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (\int \left ((5 A+6 C) a^4+6 C \cos (c+d x) a^4\right ) \sec (c+d x)dx-\frac {5 a^4 A \sin (c+d x)}{d}\right )+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (\int \frac {(5 A+6 C) a^4+6 C \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 a^4 A \sin (c+d x)}{d}\right )+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (a^4 (5 A+6 C) \int \sec (c+d x)dx-\frac {5 a^4 A \sin (c+d x)}{d}+6 a^4 C x\right )+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (a^4 (5 A+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^4 A \sin (c+d x)}{d}+6 a^4 C x\right )+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (\frac {a^4 (5 A+6 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^4 A \sin (c+d x)}{d}+6 a^4 C x\right )+\frac {2 (5 A+3 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{d}\right )+\frac {3 A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}\) |
(A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*A*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (3*(6*a^4*C*x + ( a^4*(5*A + 6*C)*ArcTanh[Sin[c + d*x]])/d - (5*a^4*A*Sin[c + d*x])/d) + (2* (5*A + 3*C)*(a^4 + a^4*Cos[c + d*x])*Tan[c + d*x])/d)/2)/(3*a)
3.1.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(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)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
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^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] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 8.36 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96
method | result | size |
parts | \(-\frac {A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{3}+3 C \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 A \,a^{3}+C \,a^{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{3} C \sin \left (d x +c \right )}{d}+\frac {3 A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 C \,a^{3} \left (d x +c \right )}{d}\) | \(149\) |
derivativedivides | \(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \sin \left (d x +c \right )+3 A \,a^{3} \tan \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+3 A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \,a^{3} \tan \left (d x +c \right )}{d}\) | \(151\) |
default | \(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \sin \left (d x +c \right )+3 A \,a^{3} \tan \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+3 A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \,a^{3} \tan \left (d x +c \right )}{d}\) | \(151\) |
parallelrisch | \(\frac {3 \left (-\frac {5 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {6 C}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {5 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {6 C}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+d x C \cos \left (3 d x +3 c \right )+\left (A +\frac {C}{3}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {11 A}{9}+\frac {C}{3}\right ) \sin \left (3 d x +3 c \right )+\frac {\sin \left (4 d x +4 c \right ) C}{6}+3 d x C \cos \left (d x +c \right )+\frac {5 \left (A +\frac {C}{5}\right ) \sin \left (d x +c \right )}{3}\right ) a^{3}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(185\) |
risch | \(3 a^{3} C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{2 d}-\frac {i a^{3} \left (9 A \,{\mathrm e}^{5 i \left (d x +c \right )}-18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{4 i \left (d x +c \right )}-48 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A \,{\mathrm e}^{i \left (d x +c \right )}-22 A -6 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {5 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {5 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(236\) |
norman | \(\frac {-3 a^{3} C x -\frac {5 A \,a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-6 a^{3} C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a^{3} C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 a^{3} C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 a^{3} C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 a^{3} C x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a^{3} C x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{3} C x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a^{3} \left (5 A -48 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (11 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} \left (17 A -24 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (35 A +12 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{3} \left (85 A +12 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (125 A +24 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (145 A -12 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a^{3} \left (5 A +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (5 A +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(421\) |
-A*a^3/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(A*a^3+3*C*a^3)/d*ln(sec(d*x+c )+tan(d*x+c))+(3*A*a^3+C*a^3)/d*tan(d*x+c)+a^3*C*sin(d*x+c)/d+3*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)))+3*C*a^3/d*(d*x+c)
Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {36 \, C a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, C a^{3} \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 9 \, A a^{3} \cos \left (d x + c\right ) + 2 \, A a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
1/12*(36*C*a^3*d*x*cos(d*x + c)^3 + 3*(5*A + 6*C)*a^3*cos(d*x + c)^3*log(s in(d*x + c) + 1) - 3*(5*A + 6*C)*a^3*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*C*a^3*cos(d*x + c)^3 + 2*(11*A + 3*C)*a^3*cos(d*x + c)^2 + 9*A*a^3 *cos(d*x + c) + 2*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.13 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 36 \, {\left (d x + c\right )} C a^{3} - 9 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{3} \sin \left (d x + c\right ) + 36 \, A a^{3} \tan \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \]
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 36*(d*x + c)*C*a^3 - 9*A *a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(si n(d*x + c) - 1)) + 6*A*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a^3*sin (d*x + c) + 36*A*a^3*tan(d*x + c) + 12*C*a^3*tan(d*x + c))/d
Time = 0.32 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.40 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {18 \, {\left (d x + c\right )} C a^{3} + \frac {12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 3 \, {\left (5 \, A a^{3} + 6 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a^{3} + 6 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
1/6*(18*(d*x + c)*C*a^3 + 12*C*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2 *c)^2 + 1) + 3*(5*A*a^3 + 6*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3* (5*A*a^3 + 6*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*A*a^3*tan(1 /2*d*x + 1/2*c)^5 + 6*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*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 + 33*A*a^3*tan(1/2*d*x + 1/2* c) + 6*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 1.24 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.28 \[ \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {11\,A\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]
(C*a^3*sin(c + d*x))/d + (5*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) /2)))/d + (6*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*C*a ^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (11*A*a^3*sin(c + d*x ))/(3*d*cos(c + d*x)) + (3*A*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (A*a ^3*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (C*a^3*sin(c + d*x))/(d*cos(c + d* x))