Integrand size = 31, antiderivative size = 216 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {1}{8} \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) x+\frac {b^4 B \text {arctanh}(\sin (c+d x))}{d}+\frac {a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d} \] Output:
1/8*(3*A*a^4+24*A*a^2*b^2+8*A*b^4+16*B*a^3*b+32*B*a*b^3)*x+b^4*B*arctanh(s in(d*x+c))/d+1/6*a*(16*A*a^2*b+19*A*b^3+4*B*a^3+34*B*a*b^2)*sin(d*x+c)/d+1 /24*a^2*(9*A*a^2+26*A*b^2+32*B*a*b)*cos(d*x+c)*sin(d*x+c)/d+1/12*a*(7*A*b+ 4*B*a)*cos(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/4*a*A*cos(d*x+c)^3*( a+b*sec(d*x+c))^3*sin(d*x+c)/d
Time = 1.83 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.97 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {12 \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) (c+d x)-96 b^4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 b^4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+24 a \left (12 a^2 A b+16 A b^3+3 a^3 B+24 a b^2 B\right ) \sin (c+d x)+24 a^2 \left (a^2 A+6 A b^2+4 a b B\right ) \sin (2 (c+d x))+8 a^3 (4 A b+a B) \sin (3 (c+d x))+3 a^4 A \sin (4 (c+d x))}{96 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
(12*(3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*(c + d*x) - 96*b^4*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 96*b^4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 24*a*(12*a^2*A*b + 16*A*b^3 + 3*a^3*B + 2 4*a*b^2*B)*Sin[c + d*x] + 24*a^2*(a^2*A + 6*A*b^2 + 4*a*b*B)*Sin[2*(c + d* x)] + 8*a^3*(4*A*b + a*B)*Sin[3*(c + d*x)] + 3*a^4*A*Sin[4*(c + d*x)])/(96 *d)
Time = 1.48 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4513, 25, 3042, 4582, 3042, 4562, 25, 3042, 4535, 24, 3042, 4533, 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.
\(\displaystyle \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4513 |
\(\displaystyle \frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (4 b^2 B \sec ^2(c+d x)+\left (3 A a^2+8 b B a+4 A b^2\right ) \sec (c+d x)+a (7 A b+4 a B)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (4 b^2 B \sec ^2(c+d x)+\left (3 A a^2+8 b B a+4 A b^2\right ) \sec (c+d x)+a (7 A b+4 a B)\right )dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (4 b^2 B \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 A a^2+8 b B a+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (7 A b+4 a B)\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 4582 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (12 B \sec ^2(c+d x) b^3+a \left (9 A a^2+32 b B a+26 A b^2\right )+\left (8 B a^3+23 A b a^2+36 b^2 B a+12 A b^3\right ) \sec (c+d x)\right )dx+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (12 B \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+a \left (9 A a^2+32 b B a+26 A b^2\right )+\left (8 B a^3+23 A b a^2+36 b^2 B a+12 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 4562 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} \int -\cos (c+d x) \left (24 B \sec ^2(c+d x) b^4+4 a \left (4 B a^3+16 A b a^2+34 b^2 B a+19 A b^3\right )+3 \left (3 A a^4+16 b B a^3+24 A b^2 a^2+32 b^3 B a+8 A b^4\right ) \sec (c+d x)\right )dx\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \cos (c+d x) \left (24 B \sec ^2(c+d x) b^4+4 a \left (4 B a^3+16 A b a^2+34 b^2 B a+19 A b^3\right )+3 \left (3 A a^4+16 b B a^3+24 A b^2 a^2+32 b^3 B a+8 A b^4\right ) \sec (c+d x)\right )dx+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {24 B \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+4 a \left (4 B a^3+16 A b a^2+34 b^2 B a+19 A b^3\right )+3 \left (3 A a^4+16 b B a^3+24 A b^2 a^2+32 b^3 B a+8 A b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (24 B \sec ^2(c+d x) b^4+4 a \left (4 B a^3+16 A b a^2+34 b^2 B a+19 A b^3\right )\right )dx+3 \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \int 1dx\right )+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (24 B \sec ^2(c+d x) b^4+4 a \left (4 B a^3+16 A b a^2+34 b^2 B a+19 A b^3\right )\right )dx+3 x \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right )\right )+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int \frac {24 B \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+4 a \left (4 B a^3+16 A b a^2+34 b^2 B a+19 A b^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 x \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right )\right )+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (24 b^4 B \int \sec (c+d x)dx+\frac {4 a \left (4 a^3 B+16 a^2 A b+34 a b^2 B+19 A b^3\right ) \sin (c+d x)}{d}+3 x \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right )\right )+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (24 b^4 B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {4 a \left (4 a^3 B+16 a^2 A b+34 a b^2 B+19 A b^3\right ) \sin (c+d x)}{d}+3 x \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right )\right )+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (\frac {4 a \left (4 a^3 B+16 a^2 A b+34 a b^2 B+19 A b^3\right ) \sin (c+d x)}{d}+3 x \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right )+\frac {24 b^4 B \text {arctanh}(\sin (c+d x))}{d}\right )\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}\) |
Input:
Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
(a*A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + ((a*(7*A* b + 4*a*B)*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + ((a ^2*(9*a^2*A + 26*A*b^2 + 32*a*b*B)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (3*( 3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*x + (24*b^4*B* ArcTanh[Sin[c + d*x]])/d + (4*a*(16*a^2*A*b + 19*A*b^3 + 4*a^3*B + 34*a*b^ 2*B)*Sin[c + d*x])/d)/2)/3)/4
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] + Sim p[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[ a*(a*B*n - A*b*(m - n - 1)) + (2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] & & LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Time = 1.08 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {-96 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}+96 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}+24 a^{2} \left (A \,a^{2}+6 A \,b^{2}+4 B a b \right ) \sin \left (2 d x +2 c \right )+\left (32 A \,a^{3} b +8 B \,a^{4}\right ) \sin \left (3 d x +3 c \right )+3 a^{4} A \sin \left (4 d x +4 c \right )+\left (288 A \,a^{3} b +384 A a \,b^{3}+72 B \,a^{4}+576 B \,a^{2} b^{2}\right ) \sin \left (d x +c \right )+36 x \left (a^{4} A +8 A \,a^{2} b^{2}+\frac {8}{3} A \,b^{4}+\frac {16}{3} a^{3} b B +\frac {32}{3} a \,b^{3} B \right ) d}{96 d}\) | \(189\) |
derivativedivides | \(\frac {a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {4 A \,a^{3} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{3} b B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 A \,a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{2} b^{2} \sin \left (d x +c \right )+4 A a \,b^{3} \sin \left (d x +c \right )+4 a \,b^{3} B \left (d x +c \right )+A \,b^{4} \left (d x +c \right )+B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(218\) |
default | \(\frac {a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {4 A \,a^{3} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{3} b B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 A \,a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{2} b^{2} \sin \left (d x +c \right )+4 A a \,b^{3} \sin \left (d x +c \right )+4 a \,b^{3} B \left (d x +c \right )+A \,b^{4} \left (d x +c \right )+B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(218\) |
risch | \(\frac {3 a^{4} A x}{8}+3 A \,a^{2} b^{2} x +x A \,b^{4}+2 B \,a^{3} b x +4 x a \,b^{3} B -\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3} b}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2} b^{2}}{d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} A a \,b^{3}}{d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} A a \,b^{3}}{d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3} b}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2} b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{4}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{4}}{d}+\frac {a^{4} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{3} b}{3 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{4}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a^{4} A}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) A \,a^{2} b^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} b B}{d}\) | \(358\) |
norman | \(\frac {\left (\frac {3}{8} a^{4} A +3 A \,a^{2} b^{2}+A \,b^{4}+2 a^{3} b B +4 a \,b^{3} B \right ) x +\left (-\frac {3}{2} a^{4} A -12 A \,a^{2} b^{2}-4 A \,b^{4}-8 a^{3} b B -16 a \,b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{2} a^{4} A -12 A \,a^{2} b^{2}-4 A \,b^{4}-8 a^{3} b B -16 a \,b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {3}{8} a^{4} A +3 A \,a^{2} b^{2}+A \,b^{4}+2 a^{3} b B +4 a \,b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (\frac {9}{4} a^{4} A +18 A \,a^{2} b^{2}+6 A \,b^{4}+12 a^{3} b B +24 a \,b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {a \left (5 a^{3} A -32 A \,a^{2} b +24 A a \,b^{2}-32 A \,b^{3}-8 B \,a^{3}+16 B \,a^{2} b -48 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}+\frac {a \left (5 a^{3} A +32 A \,a^{2} b +24 A a \,b^{2}+32 A \,b^{3}+8 B \,a^{3}+16 B \,a^{2} b +48 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a \left (45 a^{3} A -32 A \,a^{2} b +24 A a \,b^{2}+96 A \,b^{3}-8 B \,a^{3}+16 B \,a^{2} b +144 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {a \left (45 a^{3} A +32 A \,a^{2} b +24 A a \,b^{2}-96 A \,b^{3}+8 B \,a^{3}+16 B \,a^{2} b -144 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {a \left (69 a^{3} A -224 A \,a^{2} b +216 A a \,b^{2}-96 A \,b^{3}-56 B \,a^{3}+144 B \,a^{2} b -144 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}-\frac {a \left (69 a^{3} A +224 A \,a^{2} b +216 A a \,b^{2}+96 A \,b^{3}+56 B \,a^{3}+144 B \,a^{2} b +144 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}-\frac {a \left (165 a^{3} A -32 A \,a^{2} b -360 A a \,b^{2}-288 A \,b^{3}-8 B \,a^{3}-240 B \,a^{2} b -432 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 d}+\frac {a \left (165 a^{3} A +32 A \,a^{2} b -360 A a \,b^{2}+288 A \,b^{3}+8 B \,a^{3}-240 B \,a^{2} b +432 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {B \,b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \,b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(817\) |
Input:
int(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBO SE)
Output:
1/96*(-96*B*ln(tan(1/2*d*x+1/2*c)-1)*b^4+96*B*ln(tan(1/2*d*x+1/2*c)+1)*b^4 +24*a^2*(A*a^2+6*A*b^2+4*B*a*b)*sin(2*d*x+2*c)+(32*A*a^3*b+8*B*a^4)*sin(3* d*x+3*c)+3*a^4*A*sin(4*d*x+4*c)+(288*A*a^3*b+384*A*a*b^3+72*B*a^4+576*B*a^ 2*b^2)*sin(d*x+c)+36*x*(a^4*A+8*A*a^2*b^2+8/3*A*b^4+16/3*a^3*b*B+32/3*a*b^ 3*B)*d)/d
Time = 0.11 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.85 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {12 \, B b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, B b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} d x + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 16 \, B a^{4} + 64 \, A a^{3} b + 144 \, B a^{2} b^{2} + 96 \, A a b^{3} + 8 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="f ricas")
Output:
1/24*(12*B*b^4*log(sin(d*x + c) + 1) - 12*B*b^4*log(-sin(d*x + c) + 1) + 3 *(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*d*x + (6*A*a ^4*cos(d*x + c)^3 + 16*B*a^4 + 64*A*a^3*b + 144*B*a^2*b^2 + 96*A*a*b^3 + 8 *(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^2 + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b ^2)*cos(d*x + c))*sin(d*x + c))/d
Timed out. \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} b + 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} b + 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} + 384 \, {\left (d x + c\right )} B a b^{3} + 96 \, {\left (d x + c\right )} A b^{4} + 48 \, B b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 576 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 384 \, A a b^{3} \sin \left (d x + c\right )}{96 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="m axima")
Output:
1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 32 *(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 - 128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3*b + 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3*b + 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2*b^2 + 384*(d*x + c)*B*a*b^3 + 96*(d*x + c) *A*b^4 + 48*B*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 576*B* a^2*b^2*sin(d*x + c) + 384*A*a*b^3*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 603 vs. \(2 (206) = 412\).
Time = 0.19 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.79 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="g iac")
Output:
1/24*(24*B*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*B*b^4*log(abs(tan(1 /2*d*x + 1/2*c) - 1)) + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^ 3 + 8*A*b^4)*(d*x + c) - 2*(15*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^4*tan (1/2*d*x + 1/2*c)^7 - 96*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 48*B*a^3*b*tan(1 /2*d*x + 1/2*c)^7 + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 144*B*a^2*b^2*ta n(1/2*d*x + 1/2*c)^7 - 96*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 9*A*a^4*tan(1/2 *d*x + 1/2*c)^5 - 40*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^3*b*tan(1/2*d* x + 1/2*c)^5 + 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 72*A*a^2*b^2*tan(1/2*d* x + 1/2*c)^5 - 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 288*A*a*b^3*tan(1/2* d*x + 1/2*c)^5 + 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^4*tan(1/2*d*x + 1 /2*c)^3 - 160*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 48*B*a^3*b*tan(1/2*d*x + 1/ 2*c)^3 - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^4*tan(1/2*d*x + 1/ 2*c) - 24*B*a^4*tan(1/2*d*x + 1/2*c) - 96*A*a^3*b*tan(1/2*d*x + 1/2*c) - 4 8*B*a^3*b*tan(1/2*d*x + 1/2*c) - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c) - 144*B *a^2*b^2*tan(1/2*d*x + 1/2*c) - 96*A*a*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2* d*x + 1/2*c)^2 + 1)^4)/d
Time = 11.71 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.71 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {3\,B\,a^4\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {2\,A\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,b^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {8\,B\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,B\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {B\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {6\,B\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {6\,A\,a^2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {4\,A\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^4*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4,x)
Output:
(3*B*a^4*sin(c + d*x))/(4*d) + (3*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (2*A*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*sin(2 *c + 2*d*x))/(4*d) + (A*a^4*sin(4*c + 4*d*x))/(32*d) + (B*a^4*sin(3*c + 3* d*x))/(12*d) + (8*B*a*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*B*a^3*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^3*b*sin( 3*c + 3*d*x))/(3*d) + (B*a^3*b*sin(2*c + 2*d*x))/d + (6*B*a^2*b^2*sin(c + d*x))/d + (6*A*a^2*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3 *A*a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (4*A*a*b^3*sin(c + d*x))/d + (3*A*a^3 *b*sin(c + d*x))/d
Time = 0.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.85 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{5}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{5}+120 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{5}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{5}-40 \sin \left (d x +c \right )^{3} a^{4} b +120 \sin \left (d x +c \right ) a^{4} b +240 \sin \left (d x +c \right ) a^{2} b^{3}+9 a^{5} c +9 a^{5} d x +120 a^{3} b^{2} c +120 a^{3} b^{2} d x +120 a \,b^{4} c +120 a \,b^{4} d x}{24 d} \] Input:
int(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)
Output:
( - 6*cos(c + d*x)*sin(c + d*x)**3*a**5 + 15*cos(c + d*x)*sin(c + d*x)*a** 5 + 120*cos(c + d*x)*sin(c + d*x)*a**3*b**2 - 24*log(tan((c + d*x)/2) - 1) *b**5 + 24*log(tan((c + d*x)/2) + 1)*b**5 - 40*sin(c + d*x)**3*a**4*b + 12 0*sin(c + d*x)*a**4*b + 240*sin(c + d*x)*a**2*b**3 + 9*a**5*c + 9*a**5*d*x + 120*a**3*b**2*c + 120*a**3*b**2*d*x + 120*a*b**4*c + 120*a*b**4*d*x)/(2 4*d)