Integrand size = 41, antiderivative size = 205 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d} \] Output:
-1/2*(2*A*b^3-B*a^3-2*B*a*b^2+a^2*b*(A+2*C))*x/a^4+2*b^2*(A*b^2-a*(B*b-C*a ))*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(1/2)/(a+ b)^(1/2)/d+1/3*(3*A*b^2-3*B*a*b+a^2*(2*A+3*C))*sin(d*x+c)/a^3/d-1/2*(A*b-B *a)*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
Time = 1.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {6 \left (-2 A b^3+a^3 B+2 a b^2 B-a^2 b (A+2 C)\right ) (c+d x)-\frac {24 b^2 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+3 a \left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sin (c+d x)+3 a^2 (-A b+a B) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^4 d} \] Input:
Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b* Sec[c + d*x]),x]
Output:
(6*(-2*A*b^3 + a^3*B + 2*a*b^2*B - a^2*b*(A + 2*C))*(c + d*x) - (24*b^2*(A *b^2 + a*(-(b*B) + a*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^ 2]])/Sqrt[a^2 - b^2] + 3*a*(4*A*b^2 - 4*a*b*B + a^2*(3*A + 4*C))*Sin[c + d *x] + 3*a^2*(-(A*b) + a*B)*Sin[2*(c + d*x)] + a^3*A*Sin[3*(c + d*x)])/(12* a^4*d)
Time = 1.50 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-2 A b \sec ^2(c+d x)-a (2 A+3 C) \sec (c+d x)+3 (A b-a B)\right )}{a+b \sec (c+d x)}dx}{3 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-a (2 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+3 (A b-a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-3 b (A b-a B) \sec ^2(c+d x)+a (A b+3 a B) \sec (c+d x)+2 \left (\frac {1}{2} (4 A+6 C) a^2-3 b B a+3 A b^2\right )\right )}{a+b \sec (c+d x)}dx}{2 a}}{3 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {-3 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (A b+3 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+2 \left (\frac {1}{2} (4 A+6 C) a^2-3 b B a+3 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {\int \frac {3 \left (-B a^3+b (A+2 C) a^2-2 b^2 B a+b (A b-a B) \sec (c+d x) a+2 A b^3\right )}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \int \frac {-B a^3+b (A+2 C) a^2-2 b^2 B a+b (A b-a B) \sec (c+d x) a+2 A b^3}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \int \frac {-B a^3+b (A+2 C) a^2-2 b^2 B a+b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^3}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 4407 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {x \left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right )}{a}-\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {x \left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right )}{a}-\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {x \left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right )}{a}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {x \left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right )}{a}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {x \left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right )}{a}-\frac {4 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 \sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {x \left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right )}{a}-\frac {4 b^2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
Input:
Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
Output:
(A*Cos[c + d*x]^2*Sin[c + d*x])/(3*a*d) - ((3*(A*b - a*B)*Cos[c + d*x]*Sin [c + d*x])/(2*a*d) - ((-3*(((2*A*b^3 - a^3*B - 2*a*b^2*B + a^2*b*(A + 2*C) )*x)/a - (4*b^2*(A*b^2 - a*(b*B - a*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x) /2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d)))/a + (2*(3*A*b^2 - 3*a*b *B + a^2*(2*A + 3*C))*Sin[c + d*x])/(a*d))/(2*a))/(3*a)
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]
Time = 0.63 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (\frac {\left (-a^{3} A -\frac {1}{2} A \,a^{2} b -a A \,b^{2}+\frac {1}{2} B \,a^{3}+B \,a^{2} b -a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3} A -2 a A \,b^{2}+2 B \,a^{2} b -2 a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3} A -a A \,b^{2}+B \,a^{2} b -a^{3} C +\frac {1}{2} A \,a^{2} b -\frac {1}{2} B \,a^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-B \,a^{3}-2 B a \,b^{2}+2 a^{2} b C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}+\frac {2 b^{2} \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(275\) |
default | \(\frac {-\frac {2 \left (\frac {\left (-a^{3} A -\frac {1}{2} A \,a^{2} b -a A \,b^{2}+\frac {1}{2} B \,a^{3}+B \,a^{2} b -a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3} A -2 a A \,b^{2}+2 B \,a^{2} b -2 a^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3} A -a A \,b^{2}+B \,a^{2} b -a^{3} C +\frac {1}{2} A \,a^{2} b -\frac {1}{2} B \,a^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-B \,a^{3}-2 B a \,b^{2}+2 a^{2} b C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}+\frac {2 b^{2} \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(275\) |
Input:
int(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x,method =_RETURNVERBOSE)
Output:
1/d*(-2/a^4*(((-a^3*A-1/2*A*a^2*b-a*A*b^2+1/2*B*a^3+B*a^2*b-a^3*C)*tan(1/2 *d*x+1/2*c)^5+(-2/3*a^3*A-2*a*A*b^2+2*B*a^2*b-2*a^3*C)*tan(1/2*d*x+1/2*c)^ 3+(-a^3*A-a*A*b^2+B*a^2*b-a^3*C+1/2*A*a^2*b-1/2*B*a^3)*tan(1/2*d*x+1/2*c)) /(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*(A*a^2*b+2*A*b^3-B*a^3-2*B*a*b^2+2*C*a^2*b )*arctan(tan(1/2*d*x+1/2*c)))+2*b^2*(A*b^2-B*a*b+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)))
Time = 0.12 (sec) , antiderivative size = 599, normalized size of antiderivative = 2.92 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [\frac {3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x + 3 \, {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} - 6 \, B a^{4} b + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, \frac {3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x + 6 \, {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (2 \, {\left (2 \, A + 3 \, C\right )} a^{5} - 6 \, B a^{4} b + 2 \, {\left (A - 3 \, C\right )} a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
[1/6*(3*(B*a^5 - (A + 2*C)*a^4*b + B*a^3*b^2 - (A - 2*C)*a^2*b^3 - 2*B*a*b ^4 + 2*A*b^5)*d*x + 3*(C*a^2*b^2 - B*a*b^3 + 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*co s(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*co s(d*x + c) + b^2)) + (2*(2*A + 3*C)*a^5 - 6*B*a^4*b + 2*(A - 3*C)*a^3*b^2 + 6*B*a^2*b^3 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 + 3*(B*a^ 5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a ^4*b^2)*d), 1/6*(3*(B*a^5 - (A + 2*C)*a^4*b + B*a^3*b^2 - (A - 2*C)*a^2*b^ 3 - 2*B*a*b^4 + 2*A*b^5)*d*x + 6*(C*a^2*b^2 - B*a*b^3 + 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 - 6*B*a^4*b + 2*(A - 3*C)*a^3*b^2 + 6*B*a^2*b^3 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 + 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d)]
\[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)), x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**3/(a + b*s ec(c + d*x)), x)
Exception generated. \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
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. 424 vs. \(2 (187) = 374\).
Time = 0.32 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {3 \, {\left (B a^{3} - A a^{2} b - 2 \, C a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} {\left (d x + c\right )}}{a^{4}} + \frac {12 \, {\left (C a^{2} b^{2} - B a b^{3} + 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} - 3 \, B 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 \, B 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 \, B a b \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 ) + 3 \, B 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 \, B 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} \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
1/6*(3*(B*a^3 - A*a^2*b - 2*C*a^2*b + 2*B*a*b^2 - 2*A*b^3)*(d*x + c)/a^4 + 12*(C*a^2*b^2 - B*a*b^3 + 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 - 3*B*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*B*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*B*a*b*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) + 3*B*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*B*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 = 20.50 (sec) , antiderivative size = 7110, normalized size of antiderivative = 34.68 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:
int((cos(c + d*x)^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)),x)
Output:
((4*tan(c/2 + (d*x)/2)^3*(A*a^2 + 3*A*b^2 + 3*C*a^2 - 3*B*a*b))/(3*a^3) + (tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + B*a^2 + 2*C*a^2 - A*a*b - 2*B*a*b ))/a^3 + (tan(c/2 + (d*x)/2)^5*(2*A*a^2 + 2*A*b^2 - B*a^2 + 2*C*a^2 + A*a* b - 2*B*a*b))/a^3)/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + t an(c/2 + (d*x)/2)^6 + 1)) + (atan(-((((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B ^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 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*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B ^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*B*a*b^8 + 2*A*B*a^8*b + 4*B*C*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^ 3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*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 - 16*B*C*a^3*b^6 + 32*B*C*a^4*b^5 - 28*B*C*a^5*b ^4 + 20*B*C*a^6*b^3 - 12*B*C*a^7*b^2))/a^6 + (((8*(4*A*a^8*b^5 - 2*B*a^13 - 6*A*a^9*b^4 + 2*A*a^10*b^3 - 2*A*a^11*b^2 - 4*B*a^9*b^4 + 6*B*a^10*b^3 - 2*B*a^11*b^2 + 4*C*a^10*b^3 - 8*C*a^11*b^2 + 2*A*a^12*b + 2*B*a^12*b + 4* C*a^12*b))/a^9 + (8*tan(c/2 + (d*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2) *(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i))/a^10) *(A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i))/a^4)* (A*b^3*1i - (B*a^3*1i)/2 + a^2*((A*b*1i)/2 + C*b*1i) - B*a*b^2*1i)*1i)/...
Time = 0.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {6 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{2} c -\sin \left (d x +c \right )^{3} a^{4}+\sin \left (d x +c \right )^{3} a^{2} b^{2}+3 \sin \left (d x +c \right ) a^{4}+3 \sin \left (d x +c \right ) a^{3} c -3 \sin \left (d x +c \right ) a^{2} b^{2}-3 \sin \left (d x +c \right ) a \,b^{2} c -3 a^{2} b \,c^{2}-3 a^{2} b c d x +3 b^{3} c^{2}+3 b^{3} c d x}{3 a^{2} d \left (a^{2}-b^{2}\right )} \] Input:
int(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)
Output:
(6*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr t( - a**2 + b**2))*b**2*c - sin(c + d*x)**3*a**4 + sin(c + d*x)**3*a**2*b* *2 + 3*sin(c + d*x)*a**4 + 3*sin(c + d*x)*a**3*c - 3*sin(c + d*x)*a**2*b** 2 - 3*sin(c + d*x)*a*b**2*c - 3*a**2*b*c**2 - 3*a**2*b*c*d*x + 3*b**3*c**2 + 3*b**3*c*d*x)/(3*a**2*d*(a**2 - b**2))