Integrand size = 33, antiderivative size = 175 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {b \left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {2 b^2 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {A b \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d} \]
-1/2*b*(2*A*b^2+a^2*(A+2*C))*x/a^4+1/3*(3*A*b^2+a^2*(2*A+3*C))*sin(d*x+c)/ a^3/d-1/2*A*b*cos(d*x+c)*sin(d*x+c)/a^2/d+1/3*A*cos(d*x+c)^2*sin(d*x+c)/a/ d+2*b^2*(A*b^2+C*a^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/ a^4/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {-6 b \left (2 A b^2+a^2 (A+2 C)\right ) (c+d x)-\frac {24 b^2 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+3 a \left (4 A b^2+a^2 (3 A+4 C)\right ) \sin (c+d x)-3 a^2 A b \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^4 d} \]
(-6*b*(2*A*b^2 + a^2*(A + 2*C))*(c + d*x) - (24*b^2*(A*b^2 + a^2*C)*ArcTan h[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 3*a*(4*A *b^2 + a^2*(3*A + 4*C))*Sin[c + d*x] - 3*a^2*A*b*Sin[2*(c + d*x)] + a^3*A* Sin[3*(c + d*x)])/(12*a^4*d)
Time = 1.31 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4593, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4593 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-2 A b \sec ^2(c+d x)-a (2 A+3 C) \sec (c+d x)+3 A b\right )}{a+b \sec (c+d x)}dx}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-a (2 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+3 A b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-3 A b^2 \sec ^2(c+d x)+a A b \sec (c+d x)+2 \left (\frac {1}{2} (4 A+6 C) a^2+3 A b^2\right )\right )}{a+b \sec (c+d x)}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {-3 A b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a A b \csc \left (c+d x+\frac {\pi }{2}\right )+2 \left (\frac {1}{2} (4 A+6 C) a^2+3 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {\int \frac {3 \left (a A \sec (c+d x) b^2+\left ((A+2 C) a^2+2 A b^2\right ) b\right )}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \int \frac {a A \sec (c+d x) b^2+\left ((A+2 C) a^2+2 A b^2\right ) b}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \int \frac {a A \csc \left (c+d x+\frac {\pi }{2}\right ) b^2+\left ((A+2 C) a^2+2 A b^2\right ) b}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2 (A+2 C)+2 A b^2\right )}{a}-\frac {2 b^2 \left (a^2 C+A b^2\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2 (A+2 C)+2 A b^2\right )}{a}-\frac {2 b^2 \left (a^2 C+A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2 (A+2 C)+2 A b^2\right )}{a}-\frac {2 b \left (a^2 C+A b^2\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2 (A+2 C)+2 A b^2\right )}{a}-\frac {2 b \left (a^2 C+A b^2\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2 (A+2 C)+2 A b^2\right )}{a}-\frac {4 b \left (a^2 C+A b^2\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{a d}-\frac {3 \left (\frac {b x \left (a^2 (A+2 C)+2 A b^2\right )}{a}-\frac {4 b^2 \left (a^2 C+A b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
(A*Cos[c + d*x]^2*Sin[c + d*x])/(3*a*d) - ((3*A*b*Cos[c + d*x]*Sin[c + d*x ])/(2*a*d) - ((-3*((b*(2*A*b^2 + a^2*(A + 2*C))*x)/a - (4*b^2*(A*b^2 + a^2 *C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sq rt[a + b]*d)))/a + (2*(3*A*b^2 + a^2*(2*A + 3*C))*Sin[c + d*x])/(a*d))/(2* a))/(3*a)
3.7.82.3.1 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 = 0.48 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.29
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b^{2} \left (A \,b^{2}+C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (\frac {\left (-a^{3} A -\frac {1}{2} A \,a^{2} b -a A \,b^{2}-a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3} A -2 a A \,b^{2}-2 a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3} A -a A \,b^{2}-a^{3} C +\frac {1}{2} A \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {b \left (a^{2} A +2 A \,b^{2}+2 C \,a^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}}{d}\) | \(225\) |
| default | \(\frac {\frac {2 b^{2} \left (A \,b^{2}+C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (\frac {\left (-a^{3} A -\frac {1}{2} A \,a^{2} b -a A \,b^{2}-a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3} A -2 a A \,b^{2}-2 a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3} A -a A \,b^{2}-a^{3} C +\frac {1}{2} A \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {b \left (a^{2} A +2 A \,b^{2}+2 C \,a^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}}{d}\) | \(225\) |
| risch | \(-\frac {A b x}{2 a^{2}}-\frac {x \,b^{3} A}{a^{4}}-\frac {x b C}{a^{2}}-\frac {3 i A \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{2}}{2 d \,a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}+\frac {3 i A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{2}}{2 d \,a^{3}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {A \sin \left (3 d x +3 c \right )}{12 a d}-\frac {A b \sin \left (2 d x +2 c \right )}{4 a^{2} d}\) | \(483\) |
1/d*(2*b^2*(A*b^2+C*a^2)/a^4/((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^4*(((-a^3*A-1/2*A*a^2*b-a*A*b^2-a^3*C)*ta n(1/2*d*x+1/2*c)^5+(-2/3*a^3*A-2*a*A*b^2-2*a^3*C)*tan(1/2*d*x+1/2*c)^3+(-a ^3*A-a*A*b^2-a^3*C+1/2*A*a^2*b)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^ 2)^3+1/2*b*(A*a^2+2*A*b^2+2*C*a^2)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.29 (sec) , antiderivative size = 483, normalized size of antiderivative = 2.76 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [-\frac {3 \, {\left ({\left (A + 2 \, C\right )} a^{4} b + {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, A b^{5}\right )} d x - 3 \, {\left (C a^{2} b^{2} + A b^{4}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (A a^{4} b - A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, -\frac {3 \, {\left ({\left (A + 2 \, C\right )} a^{4} b + {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, A b^{5}\right )} d x - 6 \, {\left (C a^{2} b^{2} + A b^{4}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (A a^{4} b - A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \]
[-1/6*(3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*d*x - 3*(C*a^2*b^ 2 + A*b^4)*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)*sin(d*x + c) + 2*a^2 - b^ 2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - (2*(2*A + 3*C)*a^5 + 2*(A - 3*C)*a^3*b^2 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 - 3*(A*a^4*b - A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d), - 1/6*(3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*d*x - 6*(C*a^2*b^2 + A*b^4)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/(( a^2 - b^2)*sin(d*x + c))) - (2*(2*A + 3*C)*a^5 + 2*(A - 3*C)*a^3*b^2 - 6*A *a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 - 3*(A*a^4*b - A*a^2*b^3)*co s(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d)]
\[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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. 326 vs. \(2 (158) = 316\).
Time = 0.34 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {\frac {3 \, {\left (A a^{2} b + 2 \, C a^{2} b + 2 \, A b^{3}\right )} {\left (d x + c\right )}}{a^{4}} - \frac {12 \, {\left (C a^{2} b^{2} + A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{4}} - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]
-1/6*(3*(A*a^2*b + 2*C*a^2*b + 2*A*b^3)*(d*x + c)/a^4 - 12*(C*a^2*b^2 + A* b^4)*(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^4) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c )^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4* A*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2* tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1/2*d* x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c))/ ((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d
Time = 21.10 (sec) , antiderivative size = 3942, normalized size of antiderivative = 22.53 \[ \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)^5*(2*A*a^2 + 2*A*b^2 + 2*C*a^2 + A*a*b))/a^3 + (4*tan (c/2 + (d*x)/2)^3*(A*a^2 + 3*A*b^2 + 3*C*a^2))/(3*a^3) + (tan(c/2 + (d*x)/ 2)*(2*A*a^2 + 2*A*b^2 + 2*C*a^2 - A*a*b))/a^3)/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) - (atan(((((8*tan(c/ 2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*C^2*a^4* b^5 - 16*C^2*a^5*b^4 + 12*C^2*a^6*b^3 - 4*C^2*a^7*b^2 + 16*A*C*a^2*b^7 - 3 2*A*C*a^3*b^6 + 28*A*C*a^4*b^5 - 20*A*C*a^5*b^4 + 12*A*C*a^6*b^3 - 4*A*C*a ^7*b^2))/a^6 + (((8*(4*A*a^8*b^5 - 6*A*a^9*b^4 + 2*A*a^10*b^3 - 2*A*a^11*b ^2 + 4*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 + (8*tan( c/2 + (d*x)/2)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/a^10)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2))/a^4)*(A*b^3*1i + (a^2*b*(A + 2*C)*1i)/2)*1i)/a^4 + (((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - 16* A^2*a*b^8 + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b ^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*C^2*a^4*b^5 - 16*C^2*a^5*b^4 + 12*C^2 *a^6*b^3 - 4*C^2*a^7*b^2 + 16*A*C*a^2*b^7 - 32*A*C*a^3*b^6 + 28*A*C*a^4*b^ 5 - 20*A*C*a^5*b^4 + 12*A*C*a^6*b^3 - 4*A*C*a^7*b^2))/a^6 - (((8*(4*A*a^8* b^5 - 6*A*a^9*b^4 + 2*A*a^10*b^3 - 2*A*a^11*b^2 + 4*C*a^10*b^3 - 8*C*a^11* b^2 + 2*A*a^12*b + 4*C*a^12*b))/a^9 - (8*tan(c/2 + (d*x)/2)*(A*b^3*1i + (a ^2*b*(A + 2*C)*1i)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/a^10)*(A*b^3...