Integrand size = 41, antiderivative size = 276 \[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x}{8 a^5}-\frac {2 b^3 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 \sqrt {a-b} \sqrt {a+b} d}-\frac {\left (3 A b^3-2 a^3 B-3 a b^2 B+a^2 b (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 d}+\frac {\left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {(A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 a d} \] Output:
1/8*(8*A*b^4-4*B*a^3*b-8*B*a*b^3+4*a^2*b^2*(A+2*C)+a^4*(3*A+4*C))*x/a^5-2* b^3*(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^5/(a-b)^(1/2)/(a+b)^(1/2)/d-1/3*(3*A*b^3-2*B*a^3-3*B*a*b^2+a^2*b*(2*A+ 3*C))*sin(d*x+c)/a^4/d+1/8*(4*A*b^2-4*B*a*b+a^2*(3*A+4*C))*cos(d*x+c)*sin( d*x+c)/a^3/d-1/3*(A*b-B*a)*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
Time = 1.49 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {12 \left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) (c+d x)+\frac {192 b^3 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+24 a \left (-4 A b^3+3 a^3 B+4 a b^2 B-a^2 b (3 A+4 C)\right ) \sin (c+d x)+24 a^2 \left (A b^2-a b B+a^2 (A+C)\right ) \sin (2 (c+d x))+8 a^3 (-A b+a B) \sin (3 (c+d x))+3 a^4 A \sin (4 (c+d x))}{96 a^5 d} \] Input:
Integrate[(Cos[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b* Sec[c + d*x]),x]
Output:
(12*(8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4* C))*(c + d*x) + (192*b^3*(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] + 24*a*(-4*A*b^3 + 3*a^3*B + 4*a*b^2*B - a^2*b*(3*A + 4*C))*Sin[c + d*x] + 24*a^2*(A*b^2 - a*b*B + a ^2*(A + C))*Sin[2*(c + d*x)] + 8*a^3*(-(A*b) + a*B)*Sin[3*(c + d*x)] + 3*a ^4*A*Sin[4*(c + d*x)])/(96*a^5*d)
Time = 2.16 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.09, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.390, Rules used = {3042, 4592, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\int \frac {\cos ^3(c+d x) \left (-3 A b \sec ^2(c+d x)-a (3 A+4 C) \sec (c+d x)+4 (A b-a B)\right )}{a+b \sec (c+d x)}dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\int \frac {-3 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-a (3 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 (A b-a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{4 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-8 b (A b-a B) \sec ^2(c+d x)+a (A b+8 a B) \sec (c+d x)+3 \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right )\right )}{a+b \sec (c+d x)}dx}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {-8 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (A b+8 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-3 b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \sec ^2(c+d x)+a \left (-3 (3 A+4 C) a^2-4 b B a+4 A b^2\right ) \sec (c+d x)+8 \left (-2 B a^3+b (2 A+3 C) a^2-3 b^2 B a+3 A b^3\right )\right )}{a+b \sec (c+d x)}dx}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\int \frac {-3 b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (-3 (3 A+4 C) a^2-4 b B a+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+8 \left (-2 B a^3+b (2 A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {\int \frac {3 \left ((3 A+4 C) a^4-4 b B a^3+4 b^2 (A+2 C) a^2-8 b^3 B a+b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \sec (c+d x) a+8 A b^4\right )}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \int \frac {(3 A+4 C) a^4-4 b B a^3+4 b^2 (A+2 C) a^2-8 b^3 B a+b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \sec (c+d x) a+8 A b^4}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \int \frac {(3 A+4 C) a^4-4 b B a^3+4 b^2 (A+2 C) a^2-8 b^3 B a+b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^4}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^3 \left (A b^2-a (b B-a C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^3 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right )}{a}-\frac {16 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \sin (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {x \left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right )}{a}-\frac {16 b^3 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
Input:
Int[(Cos[c + d*x]^4*(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]^3*Sin[c + d*x])/(4*a*d) - ((4*(A*b - a*B)*Cos[c + d*x]^2*S in[c + d*x])/(3*a*d) - ((3*(4*A*b^2 - 4*a*b*B + a^2*(3*A + 4*C))*Cos[c + d *x]*Sin[c + d*x])/(2*a*d) - ((-3*(((8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^ 2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*x)/a - (16*b^3*(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 + (8*(3*A*b^3 - 2*a^3*B - 3*a*b^2*B + a^2*b*(2*A + 3*C))*Sin[ c + d*x])/(a*d))/(2*a))/(3*a))/(4*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 = 1.02 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.65
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 b^{3} \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a^{4} A -A \,a^{3} b -\frac {1}{2} A \,a^{2} b^{2}-a A \,b^{3}+B \,a^{4}+\frac {1}{2} B \,a^{3} b +B \,a^{2} b^{2}-\frac {1}{2} a^{4} C -a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (-3 a^{3} b C +\frac {3}{8} a^{4} A -\frac {1}{2} a^{4} C +\frac {1}{2} B \,a^{3} b -3 a A \,b^{3}+3 B \,a^{2} b^{2}-\frac {1}{2} A \,a^{2} b^{2}+\frac {5}{3} B \,a^{4}-\frac {5}{3} A \,a^{3} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {3}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} B \,a^{3} b +\frac {1}{2} a^{4} C -3 a^{3} b C -3 a A \,b^{3}+3 B \,a^{2} b^{2}+\frac {5}{3} B \,a^{4}-\frac {5}{3} A \,a^{3} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {5}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} B \,a^{3} b +\frac {1}{2} a^{4} C -A \,a^{3} b -a A \,b^{3}+B \,a^{4}+B \,a^{2} b^{2}-a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4} A +4 A \,a^{2} b^{2}+8 A \,b^{4}-4 B \,a^{3} b -8 B a \,b^{3}+4 a^{4} C +8 C \,a^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{5}}}{d}\) | \(456\) |
| default | \(\frac {-\frac {2 b^{3} \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a^{4} A -A \,a^{3} b -\frac {1}{2} A \,a^{2} b^{2}-a A \,b^{3}+B \,a^{4}+\frac {1}{2} B \,a^{3} b +B \,a^{2} b^{2}-\frac {1}{2} a^{4} C -a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (-3 a^{3} b C +\frac {3}{8} a^{4} A -\frac {1}{2} a^{4} C +\frac {1}{2} B \,a^{3} b -3 a A \,b^{3}+3 B \,a^{2} b^{2}-\frac {1}{2} A \,a^{2} b^{2}+\frac {5}{3} B \,a^{4}-\frac {5}{3} A \,a^{3} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {3}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} B \,a^{3} b +\frac {1}{2} a^{4} C -3 a^{3} b C -3 a A \,b^{3}+3 B \,a^{2} b^{2}+\frac {5}{3} B \,a^{4}-\frac {5}{3} A \,a^{3} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {5}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} B \,a^{3} b +\frac {1}{2} a^{4} C -A \,a^{3} b -a A \,b^{3}+B \,a^{4}+B \,a^{2} b^{2}-a^{3} b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4} A +4 A \,a^{2} b^{2}+8 A \,b^{4}-4 B \,a^{3} b -8 B a \,b^{3}+4 a^{4} C +8 C \,a^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{5}}}{d}\) | \(456\) |
| risch | \(\frac {A \sin \left (2 d x +2 c \right )}{4 a d}+\frac {3 A x}{8 a}+\frac {A \sin \left (4 d x +4 c \right )}{32 a d}+\frac {x C}{2 a}+\frac {\sin \left (2 d x +2 c \right ) C}{4 a d}-\frac {3 i B \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B}{8 a d}-\frac {x B \,b^{3}}{a^{4}}+\frac {x C \,b^{2}}{a^{3}}+\frac {\sin \left (3 d x +3 c \right ) B}{12 a d}+\frac {3 i A b \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A b}{8 a^{2} d}-\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 {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{3}}{2 a^{4} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,b^{2}}{2 a^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} b C}{2 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{3}}{2 a^{4} d}+\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^{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 ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}+\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 {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{2}}{2 a^{3} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} b C}{2 a^{2} d}-\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^{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 ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}-\frac {\sin \left (3 d x +3 c \right ) A b}{12 a^{2} d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{2}}{4 a^{3} d}-\frac {\sin \left (2 d x +2 c \right ) B b}{4 a^{2} d}+\frac {x A \,b^{4}}{a^{5}}+\frac {x A \,b^{2}}{2 a^{3}}-\frac {x B b}{2 a^{2}}\) | \(846\) |
Input:
int(cos(d*x+c)^4*(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*b^3*(A*b^2-B*a*b+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+B*a^4+1/2*B*a^3*b+B*a^2*b^2-1/2*a^4*C-a^3*b*C)*tan(1/2*d*x+1/2 *c)^7+(-3*a^3*b*C+3/8*a^4*A-1/2*a^4*C+1/2*B*a^3*b-3*a*A*b^3+3*B*a^2*b^2-1/ 2*A*a^2*b^2+5/3*B*a^4-5/3*A*a^3*b)*tan(1/2*d*x+1/2*c)^5+(-3/8*a^4*A+1/2*A* a^2*b^2-1/2*B*a^3*b+1/2*a^4*C-3*a^3*b*C-3*a*A*b^3+3*B*a^2*b^2+5/3*B*a^4-5/ 3*A*a^3*b)*tan(1/2*d*x+1/2*c)^3+(5/8*a^4*A+1/2*A*a^2*b^2-1/2*B*a^3*b+1/2*a ^4*C-A*a^3*b-a*A*b^3+B*a^4+B*a^2*b^2-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*B*a^3*b-8*B*a*b^3+4* C*a^4+8*C*a^2*b^2)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.14 (sec) , antiderivative size = 765, normalized size of antiderivative = 2.77 \[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
[1/24*(3*((3*A + 4*C)*a^6 - 4*B*a^5*b + (A + 4*C)*a^4*b^2 - 4*B*a^3*b^3 + 4*(A - 2*C)*a^2*b^4 + 8*B*a*b^5 - 8*A*b^6)*d*x + 12*(C*a^2*b^3 - B*a*b^4 + 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)*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)) + (16*B*a^6 - 8*(2*A + 3* C)*a^5*b + 8*B*a^4*b^2 - 8*(A - 3*C)*a^3*b^3 - 24*B*a^2*b^4 + 24*A*a*b^5 + 6*(A*a^6 - A*a^4*b^2)*cos(d*x + c)^3 + 8*(B*a^6 - A*a^5*b - B*a^4*b^2 + A *a^3*b^3)*cos(d*x + c)^2 + 3*((3*A + 4*C)*a^6 - 4*B*a^5*b + (A - 4*C)*a^4* b^2 + 4*B*a^3*b^3 - 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 - 4*B*a^5*b + (A + 4*C)*a^4*b^2 - 4*B*a^3 *b^3 + 4*(A - 2*C)*a^2*b^4 + 8*B*a*b^5 - 8*A*b^6)*d*x - 24*(C*a^2*b^3 - B* a*b^4 + A*b^5)*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))) + (16*B*a^6 - 8*(2*A + 3*C)*a^5*b + 8*B*a^ 4*b^2 - 8*(A - 3*C)*a^3*b^3 - 24*B*a^2*b^4 + 24*A*a*b^5 + 6*(A*a^6 - A*a^4 *b^2)*cos(d*x + c)^3 + 8*(B*a^6 - A*a^5*b - B*a^4*b^2 + A*a^3*b^3)*cos(d*x + c)^2 + 3*((3*A + 4*C)*a^6 - 4*B*a^5*b + (A - 4*C)*a^4*b^2 + 4*B*a^3*b^3 - 4*A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5*b^2)*d)]
\[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**4*(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)**4/(a + b*s ec(c + d*x)), x)
Exception generated. \[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^4*(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. 801 vs. \(2 (257) = 514\).
Time = 0.29 (sec) , antiderivative size = 801, normalized size of antiderivative = 2.90 \[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
1/24*(3*(3*A*a^4 + 4*C*a^4 - 4*B*a^3*b + 4*A*a^2*b^2 + 8*C*a^2*b^2 - 8*B*a *b^3 + 8*A*b^4)*(d*x + c)/a^5 - 48*(C*a^2*b^3 - B*a*b^4 + 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 - 24*B*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 - 12*B*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*t an(1/2*d*x + 1/2*c)^7 - 24*B*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 - 40*B*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 - 12*B*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*B*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 - 40 *B*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 + 12*B*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*B*a*b^2*t an(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) - 24*B*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) + 12*B*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*...
Time = 21.86 (sec) , antiderivative size = 9648, normalized size of antiderivative = 34.96 \[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:
int((cos(c + d*x)^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)),x)
Output:
- ((tan(c/2 + (d*x)/2)^7*(5*A*a^3 + 8*A*b^3 - 8*B*a^3 + 4*C*a^3 + 4*A*a*b^ 2 + 8*A*a^2*b - 8*B*a*b^2 - 4*B*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 - 40*B*a^3 - 12*C*a^3 - 12*A*a*b^2 + 40*A*a^ 2*b - 72*B*a*b^2 + 12*B*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 - 40*B*a^3 + 12*C*a^3 + 12*A*a*b^2 + 40*A*a^2*b - 72*B*a*b^2 - 12*B*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 + 8*B*a^3 + 4*C*a^3 + 4*A*a*b^2 - 8*A*a^2*b + 8*B*a*b^2 - 4 *B*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)) - (atan((( ((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 - 8 1*A^2*a^8*b^3 + 51*A^2*a^9*b^2 - 128*B^2*a^2*b^9 + 256*B^2*a^3*b^8 - 256*B ^2*a^4*b^7 + 256*B^2*a^5*b^6 - 208*B^2*a^6*b^5 + 112*B^2*a^7*b^4 - 48*B^2* a^8*b^3 + 16*B^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 + 256*A*B*a*b^10 - 24*A*B*a^10*b - 72*A*C*a^10*b - 32*B*C*a^10*b - 512*A*B* a^2*b^9 + 512*A*B*a^3*b^8 - 512*A*B*a^4*b^7 + 464*A*B*a^5*b^6 - 368*A*B*a^ 6*b^5 + 264*A*B*a^7*b^4 - 152*A*B*a^8*b^3 + 72*A*B*a^9*b^2 - 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*...
Time = 0.16 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {-16 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{3} c -2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{5}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b^{2}+5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{5}+4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} c -5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}-4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2} c -8 \sin \left (d x +c \right ) a^{3} b c +8 \sin \left (d x +c \right ) a \,b^{3} c +3 a^{5} c +3 a^{5} d x +4 a^{4} c^{2}+4 a^{4} c d x -3 a^{3} b^{2} c -3 a^{3} b^{2} d x +4 a^{2} b^{2} c^{2}+4 a^{2} b^{2} c d x -8 b^{4} c^{2}-8 b^{4} c d x}{8 a^{3} d \left (a^{2}-b^{2}\right )} \] Input:
int(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)
Output:
( - 16*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b) /sqrt( - a**2 + b**2))*b**3*c - 2*cos(c + d*x)*sin(c + d*x)**3*a**5 + 2*co s(c + d*x)*sin(c + d*x)**3*a**3*b**2 + 5*cos(c + d*x)*sin(c + d*x)*a**5 + 4*cos(c + d*x)*sin(c + d*x)*a**4*c - 5*cos(c + d*x)*sin(c + d*x)*a**3*b**2 - 4*cos(c + d*x)*sin(c + d*x)*a**2*b**2*c - 8*sin(c + d*x)*a**3*b*c + 8*s in(c + d*x)*a*b**3*c + 3*a**5*c + 3*a**5*d*x + 4*a**4*c**2 + 4*a**4*c*d*x - 3*a**3*b**2*c - 3*a**3*b**2*d*x + 4*a**2*b**2*c**2 + 4*a**2*b**2*c*d*x - 8*b**4*c**2 - 8*b**4*c*d*x)/(8*a**3*d*(a**2 - b**2))