Integrand size = 33, antiderivative size = 232 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\left (8 A b^4+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x}{8 a^5}-\frac {2 b^3 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 \sqrt {a-b} \sqrt {a+b} d}-\frac {b \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 d}+\frac {\left (4 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 a d} \]
1/8*(8*A*b^4+4*a^2*b^2*(A+2*C)+a^4*(3*A+4*C))*x/a^5-1/3*b*(3*A*b^2+a^2*(2* A+3*C))*sin(d*x+c)/a^4/d+1/8*(4*A*b^2+a^2*(3*A+4*C))*cos(d*x+c)*sin(d*x+c) /a^3/d-1/3*A*b*cos(d*x+c)^2*sin(d*x+c)/a^2/d+1/4*A*cos(d*x+c)^3*sin(d*x+c) /a/d-2*b^3*(A*b^2+C*a^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2 ))/a^5/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.30 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {12 \left (8 A b^4+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) (c+d x)+\frac {192 b^3 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-24 a b \left (4 A b^2+a^2 (3 A+4 C)\right ) \sin (c+d x)+24 a^2 \left (A b^2+a^2 (A+C)\right ) \sin (2 (c+d x))-8 a^3 A b \sin (3 (c+d x))+3 a^4 A \sin (4 (c+d x))}{96 a^5 d} \]
(12*(8*A*b^4 + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*(c + d*x) + (192*b^3 *(A*b^2 + a^2*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqr t[a^2 - b^2] - 24*a*b*(4*A*b^2 + a^2*(3*A + 4*C))*Sin[c + d*x] + 24*a^2*(A *b^2 + a^2*(A + C))*Sin[2*(c + d*x)] - 8*a^3*A*b*Sin[3*(c + d*x)] + 3*a^4* A*Sin[4*(c + d*x)])/(96*a^5*d)
Time = 1.87 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.11, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4593, 3042, 4592, 3042, 4592, 3042, 4592, 27, 3042, 4407, 3042, 4318, 3042, 3138, 221}
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.
\(\displaystyle \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4593 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\int \frac {\cos ^3(c+d x) \left (-3 A b \sec ^2(c+d x)-a (3 A+4 C) \sec (c+d x)+4 A b\right )}{a+b \sec (c+d x)}dx}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\int \frac {-3 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-a (3 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{4 a}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-8 A b^2 \sec ^2(c+d x)+a A b \sec (c+d x)+3 \left ((3 A+4 C) a^2+4 A b^2\right )\right )}{a+b \sec (c+d x)}dx}{3 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {-8 A b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a A b \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+4 C) a^2+4 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{4 a}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-3 b \left ((3 A+4 C) a^2+4 A b^2\right ) \sec ^2(c+d x)+a \left (4 A b^2-3 a^2 (3 A+4 C)\right ) \sec (c+d x)+8 b \left ((2 A+3 C) a^2+3 A b^2\right )\right )}{a+b \sec (c+d x)}dx}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {-3 b \left ((3 A+4 C) a^2+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (4 A b^2-3 a^2 (3 A+4 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+8 b \left ((2 A+3 C) a^2+3 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {\int \frac {3 \left ((3 A+4 C) a^4+4 b^2 (A+2 C) a^2+b \left ((3 A+4 C) a^2+4 A b^2\right ) \sec (c+d x) a+8 A b^4\right )}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \int \frac {(3 A+4 C) a^4+4 b^2 (A+2 C) a^2+b \left ((3 A+4 C) a^2+4 A b^2\right ) \sec (c+d x) a+8 A b^4}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \int \frac {(3 A+4 C) a^4+4 b^2 (A+2 C) a^2+b \left ((3 A+4 C) a^2+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^4}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 4407 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)+4 a^2 b^2 (A+2 C)+8 A b^4\right )}{a}-\frac {8 b^3 \left (a^2 C+A b^2\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)+4 a^2 b^2 (A+2 C)+8 A b^4\right )}{a}-\frac {8 b^3 \left (a^2 C+A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)+4 a^2 b^2 (A+2 C)+8 A b^4\right )}{a}-\frac {8 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)+4 a^2 b^2 (A+2 C)+8 A b^4\right )}{a}-\frac {8 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)+4 a^2 b^2 (A+2 C)+8 A b^4\right )}{a}-\frac {16 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 A b \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)+4 a^2 b^2 (A+2 C)+8 A b^4\right )}{a}-\frac {16 b^3 \left (a^2 C+A b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
(A*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d) - ((4*A*b*Cos[c + d*x]^2*Sin[c + d *x])/(3*a*d) - ((3*(4*A*b^2 + a^2*(3*A + 4*C))*Cos[c + d*x]*Sin[c + d*x])/ (2*a*d) - ((-3*(((8*A*b^4 + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*x)/a - (16*b^3*(A*b^2 + a^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b] ])/(a*Sqrt[a - b]*Sqrt[a + b]*d)))/a + (8*b*(3*A*b^2 + a^2*(2*A + 3*C))*Si n[c + d*x])/(a*d))/(2*a))/(3*a))/(4*a)
3.7.83.3.1 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_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
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 + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[( -A)*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && NeQ[a^2 - b^2 , 0] && LeQ[n, -1]
Time = 0.62 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.52
method | result | size |
derivativedivides | \(\frac {-\frac {2 b^{3} \left (A \,b^{2}+C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a^{4} A -A \,a^{3} b -\frac {1}{2} A \,a^{2} b^{2}-a A \,b^{3}-\frac {1}{2} a^{4} C -a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (\frac {3}{8} a^{4} A -\frac {5}{3} A \,a^{3} b -3 a A \,b^{3}-3 a^{3} b C -\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} a^{4} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {3}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}+\frac {1}{2} a^{4} C -\frac {5}{3} A \,a^{3} b -3 a A \,b^{3}-3 a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {5}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}+\frac {1}{2} a^{4} C -A \,a^{3} b -a A \,b^{3}-a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4} A +4 A \,a^{2} b^{2}+8 A \,b^{4}+4 a^{4} C +8 C \,a^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{5}}}{d}\) | \(353\) |
default | \(\frac {-\frac {2 b^{3} \left (A \,b^{2}+C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a^{4} A -A \,a^{3} b -\frac {1}{2} A \,a^{2} b^{2}-a A \,b^{3}-\frac {1}{2} a^{4} C -a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (\frac {3}{8} a^{4} A -\frac {5}{3} A \,a^{3} b -3 a A \,b^{3}-3 a^{3} b C -\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} a^{4} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {3}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}+\frac {1}{2} a^{4} C -\frac {5}{3} A \,a^{3} b -3 a A \,b^{3}-3 a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {5}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}+\frac {1}{2} a^{4} C -A \,a^{3} b -a A \,b^{3}-a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4} A +4 A \,a^{2} b^{2}+8 A \,b^{4}+4 a^{4} C +8 C \,a^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{5}}}{d}\) | \(353\) |
risch | \(\frac {3 A x}{8 a}+\frac {x A \,b^{2}}{2 a^{3}}+\frac {x A \,b^{4}}{a^{5}}+\frac {x C}{2 a}+\frac {x C \,b^{2}}{a^{3}}+\frac {i b \,{\mathrm e}^{i \left (d x +c \right )} C}{2 d \,a^{2}}-\frac {3 i b \,{\mathrm e}^{-i \left (d x +c \right )} A}{8 d \,a^{2}}-\frac {i b^{3} {\mathrm e}^{-i \left (d x +c \right )} A}{2 d \,a^{4}}+\frac {3 i A b \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )} C}{2 d \,a^{2}}+\frac {i b^{3} {\mathrm e}^{i \left (d x +c \right )} A}{2 d \,a^{4}}+\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{5}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{5}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {A \sin \left (4 d x +4 c \right )}{32 a d}-\frac {A b \sin \left (3 d x +3 c \right )}{12 a^{2} d}+\frac {A \sin \left (2 d x +2 c \right )}{4 a d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{2}}{4 d \,a^{3}}+\frac {\sin \left (2 d x +2 c \right ) C}{4 a d}\) | \(560\) |
1/d*(-2*b^3*(A*b^2+C*a^2)/a^5/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d* x+1/2*c)/((a+b)*(a-b))^(1/2))+2/a^5*(((-5/8*a^4*A-A*a^3*b-1/2*A*a^2*b^2-a* A*b^3-1/2*a^4*C-a^3*b*C)*tan(1/2*d*x+1/2*c)^7+(3/8*a^4*A-5/3*A*a^3*b-3*a*A *b^3-3*a^3*b*C-1/2*A*a^2*b^2-1/2*a^4*C)*tan(1/2*d*x+1/2*c)^5+(-3/8*a^4*A+1 /2*A*a^2*b^2+1/2*a^4*C-5/3*A*a^3*b-3*a*A*b^3-3*a^3*b*C)*tan(1/2*d*x+1/2*c) ^3+(5/8*a^4*A+1/2*A*a^2*b^2+1/2*a^4*C-A*a^3*b-a*A*b^3-a^3*b*C)*tan(1/2*d*x +1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^4+1/8*(3*A*a^4+4*A*a^2*b^2+8*A*b^4+4*C*a ^4+8*C*a^2*b^2)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.30 (sec) , antiderivative size = 599, normalized size of antiderivative = 2.58 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [\frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{6} + {\left (A + 4 \, C\right )} a^{4} b^{2} + 4 \, {\left (A - 2 \, C\right )} a^{2} b^{4} - 8 \, A b^{6}\right )} d x + 12 \, {\left (C a^{2} b^{3} + A b^{5}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (8 \, {\left (2 \, A + 3 \, C\right )} a^{5} b + 8 \, {\left (A - 3 \, C\right )} a^{3} b^{3} - 24 \, A a b^{5} - 6 \, {\left (A a^{6} - A a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (A a^{5} b - A a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{6} + {\left (A - 4 \, C\right )} a^{4} b^{2} - 4 \, A a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{7} - a^{5} b^{2}\right )} d}, \frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{6} + {\left (A + 4 \, C\right )} a^{4} b^{2} + 4 \, {\left (A - 2 \, C\right )} a^{2} b^{4} - 8 \, A b^{6}\right )} d x - 24 \, {\left (C a^{2} b^{3} + A b^{5}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (8 \, {\left (2 \, A + 3 \, C\right )} a^{5} b + 8 \, {\left (A - 3 \, C\right )} a^{3} b^{3} - 24 \, A a b^{5} - 6 \, {\left (A a^{6} - A a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (A a^{5} b - A a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{6} + {\left (A - 4 \, C\right )} a^{4} b^{2} - 4 \, A a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{7} - a^{5} b^{2}\right )} d}\right ] \]
[1/24*(3*((3*A + 4*C)*a^6 + (A + 4*C)*a^4*b^2 + 4*(A - 2*C)*a^2*b^4 - 8*A* b^6)*d*x + 12*(C*a^2*b^3 + A*b^5)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*si n(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - (8*(2*A + 3*C)*a^5*b + 8*(A - 3*C)*a^3*b^3 - 24*A*a*b^5 - 6*(A*a^6 - A* a^4*b^2)*cos(d*x + c)^3 + 8*(A*a^5*b - A*a^3*b^3)*cos(d*x + c)^2 - 3*((3*A + 4*C)*a^6 + (A - 4*C)*a^4*b^2 - 4*A*a^2*b^4)*cos(d*x + c))*sin(d*x + c)) /((a^7 - a^5*b^2)*d), 1/24*(3*((3*A + 4*C)*a^6 + (A + 4*C)*a^4*b^2 + 4*(A - 2*C)*a^2*b^4 - 8*A*b^6)*d*x - 24*(C*a^2*b^3 + A*b^5)*sqrt(-a^2 + b^2)*ar ctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (8*(2*A + 3*C)*a^5*b + 8*(A - 3*C)*a^3*b^3 - 24*A*a*b^5 - 6*(A*a^6 - A*a^4 *b^2)*cos(d*x + c)^3 + 8*(A*a^5*b - A*a^3*b^3)*cos(d*x + c)^2 - 3*((3*A + 4*C)*a^6 + (A - 4*C)*a^4*b^2 - 4*A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/(( a^7 - a^5*b^2)*d)]
\[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 574 vs. \(2 (213) = 426\).
Time = 0.33 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.47 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {3 \, {\left (3 \, A a^{4} + 4 \, C a^{4} + 4 \, A a^{2} b^{2} + 8 \, C a^{2} b^{2} + 8 \, A b^{4}\right )} {\left (d x + c\right )}}{a^{5}} - \frac {48 \, {\left (C a^{2} b^{3} + A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{5}} - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, 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} + 40 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A 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} a^{4}}}{24 \, d} \]
1/24*(3*(3*A*a^4 + 4*C*a^4 + 4*A*a^2*b^2 + 8*C*a^2*b^2 + 8*A*b^4)*(d*x + c )/a^5 - 48*(C*a^2*b^3 + A*b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a ^2 + b^2)))/(sqrt(-a^2 + b^2)*a^5) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 24*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*C *a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*A*b ^3*tan(1/2*d*x + 1/2*c)^7 - 9*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^3*tan( 1/2*d*x + 1/2*c)^5 + 40*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 72*C*a^2*b*tan(1/ 2*d*x + 1/2*c)^5 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*A*b^3*tan(1/2*d* x + 1/2*c)^5 + 9*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 + 40*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*C*a^2*b*tan(1/2*d*x + 1/2*c )^3 - 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 72*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^3*tan(1/2*d*x + 1/2*c) - 12*C*a^3*tan(1/2*d*x + 1/2*c) + 24*A*a^2 *b*tan(1/2*d*x + 1/2*c) + 24*C*a^2*b*tan(1/2*d*x + 1/2*c) - 12*A*a*b^2*tan (1/2*d*x + 1/2*c) + 24*A*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^ 2 + 1)^4*a^4))/d
Time = 23.45 (sec) , antiderivative size = 5828, normalized size of antiderivative = 25.12 \[ \int \frac {\cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)^7*(5*A*a^3 + 8*A*b^3 + 4*C*a^3 + 4*A*a*b^2 + 8*A*a^ 2*b + 8*C*a^2*b))/(4*a^4) + (tan(c/2 + (d*x)/2)^3*(9*A*a^3 + 72*A*b^3 - 12 *C*a^3 - 12*A*a*b^2 + 40*A*a^2*b + 72*C*a^2*b))/(12*a^4) + (tan(c/2 + (d*x )/2)^5*(72*A*b^3 - 9*A*a^3 + 12*C*a^3 + 12*A*a*b^2 + 40*A*a^2*b + 72*C*a^2 *b))/(12*a^4) - (tan(c/2 + (d*x)/2)*(5*A*a^3 - 8*A*b^3 + 4*C*a^3 + 4*A*a*b ^2 - 8*A*a^2*b - 8*C*a^2*b))/(4*a^4))/(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)) - (a tan(((((((12*A*a^16 + 16*C*a^16 + 32*A*a^10*b^6 - 48*A*a^11*b^5 + 16*A*a^1 2*b^4 - 4*A*a^13*b^3 + 4*A*a^14*b^2 + 32*C*a^12*b^4 - 48*C*a^13*b^3 + 16*C *a^14*b^2 - 12*A*a^15*b - 16*C*a^15*b)/a^12 - (tan(c/2 + (d*x)/2)*(128*a^1 2*b + 128*a^10*b^3 - 256*a^11*b^2)*(a^2*((A*b^2*1i)/2 + C*b^2*1i) + A*b^4* 1i + a^4*((A*3i)/8 + (C*1i)/2)))/(2*a^13))*(a^2*((A*b^2*1i)/2 + C*b^2*1i) + A*b^4*1i + a^4*((A*3i)/8 + (C*1i)/2)))/a^5 + (tan(c/2 + (d*x)/2)*(9*A^2* a^11 - 128*A^2*b^11 + 16*C^2*a^11 + 256*A^2*a*b^10 - 27*A^2*a^10*b - 48*C^ 2*a^10*b - 256*A^2*a^2*b^9 + 256*A^2*a^3*b^8 - 256*A^2*a^4*b^7 + 256*A^2*a ^5*b^6 - 216*A^2*a^6*b^5 + 136*A^2*a^7*b^4 - 81*A^2*a^8*b^3 + 51*A^2*a^9*b ^2 - 128*C^2*a^4*b^7 + 256*C^2*a^5*b^6 - 256*C^2*a^6*b^5 + 256*C^2*a^7*b^4 - 208*C^2*a^8*b^3 + 112*C^2*a^9*b^2 + 24*A*C*a^11 - 72*A*C*a^10*b - 256*A *C*a^2*b^9 + 512*A*C*a^3*b^8 - 512*A*C*a^4*b^7 + 512*A*C*a^5*b^6 - 464*A*C *a^6*b^5 + 368*A*C*a^7*b^4 - 264*A*C*a^8*b^3 + 152*A*C*a^9*b^2))/(2*a^8...