Integrand size = 33, antiderivative size = 184 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 b^2 \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}-\frac {b \left (2 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}+\frac {\left (3 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 a^3 d}-\frac {A b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d} \] Output:
2*b^2*(A*b^2+C*a^2)*arctan((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-1/2*b*(2*A*b^2+a^2*(A+2*C))*arctanh(sin(d*x+c)) /a^4/d+1/3*(3*A*b^2+a^2*(2*A+3*C))*tan(d*x+c)/a^3/d-1/2*A*b*sec(d*x+c)*tan (d*x+c)/a^2/d+1/3*A*sec(d*x+c)^2*tan(d*x+c)/a/d
Leaf count is larger than twice the leaf count of optimal. \(413\) vs. \(2(184)=368\).
Time = 3.86 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.24 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\frac {24 b^2 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+6 b \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 b \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 A (a-3 b)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 a \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {a^2 A (a-3 b)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{12 a^4 d} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d*x]),x]
Output:
((-24*b^2*(A*b^2 + a^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b ^2]])/Sqrt[-a^2 + b^2] + 6*b*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2 ] - Sin[(c + d*x)/2]] - 6*b*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*A*(a - 3*b))/(Cos[(c + d*x)/2] - Sin[(c + d*x) /2])^2 + (2*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^ 3 + (4*a*(3*A*b^2 + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (a^2*A*(a - 3*b))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a*(3*A*b^2 + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + S in[(c + d*x)/2]))/(12*a^4*d)
Time = 1.36 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.09, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3535, 25, 3042, 3534, 25, 3042, 3534, 27, 3042, 3480, 3042, 3138, 218, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-2 A b \cos ^2(c+d x)-a (2 A+3 C) \cos (c+d x)+3 A b\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {\left (-2 A b \cos ^2(c+d x)-a (2 A+3 C) \cos (c+d x)+3 A b\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int -\frac {\left (-3 A b^2 \cos ^2(c+d x)+a A b \cos (c+d x)+2 \left ((2 A+3 C) a^2+3 A b^2\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}}{3 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-3 A b^2 \cos ^2(c+d x)+a A b \cos (c+d x)+2 \left (\frac {1}{2} (4 A+6 C) a^2+3 A b^2\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-3 A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2+a A b \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (\frac {1}{2} (4 A+6 C) a^2+3 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {3 \left (a A \cos (c+d x) b^2+\left ((A+2 C) a^2+2 A b^2\right ) b\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {\left (a A \cos (c+d x) b^2+\left ((A+2 C) a^2+2 A b^2\right ) b\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {a A \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left ((A+2 C) a^2+2 A b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2 (A+2 C)+2 A b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2 (A+2 C)+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2 (A+2 C)+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2 (A+2 C)+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^2 (A+2 C)+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b^2 \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d*x]),x]
Output:
(A*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d) - ((3*A*b*Sec[c + d*x]*Tan[c + d*x ])/(2*a*d) - ((-3*((-4*b^2*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d* x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (b*(2*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a*d)))/a + (2*(3*A*b^2 + a^2*(2*A + 3*C))* Tan[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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.81 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.62
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}+2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A \left (a +b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b \left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {2 b^{2} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}+2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A \left (a +b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}}{d}\) | \(299\) |
| default | \(\frac {-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}+2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A \left (a +b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b \left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {2 b^{2} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}+2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A \left (a +b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}}{d}\) | \(299\) |
| risch | \(\frac {i \left (3 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a A b \,{\mathrm e}^{i \left (d x +c \right )}+4 A \,a^{2}+6 A \,b^{2}+6 a^{2} C \right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d \,a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{4} d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{2} d}-\frac {A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d \,a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{4} d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{2} d}\) | \(588\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x,method=_RETURNVERBO SE)
Output:
1/d*(-1/3*A/a/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(2*A*a^2+A*a*b+2*A*b^2+2*C*a^2) /a^3/(tan(1/2*d*x+1/2*c)-1)-1/2*A*(a+b)/a^2/(tan(1/2*d*x+1/2*c)-1)^2+1/2*b *(A*a^2+2*A*b^2+2*C*a^2)/a^4*ln(tan(1/2*d*x+1/2*c)-1)+2*b^2*(A*b^2+C*a^2)/ a^4/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2 ))-1/3*A/a/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(2*A*a^2+A*a*b+2*A*b^2+2*C*a^2)/a^ 3/(tan(1/2*d*x+1/2*c)+1)+1/2*A*(a+b)/a^2/(tan(1/2*d*x+1/2*c)+1)^2-1/2*b*(A *a^2+2*A*b^2+2*C*a^2)/a^4*ln(tan(1/2*d*x+1/2*c)+1))
Time = 1.47 (sec) , antiderivative size = 633, normalized size of antiderivative = 3.44 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="f ricas")
Output:
[-1/12*(6*(C*a^2*b^2 + A*b^4)*sqrt(-a^2 + b^2)*cos(d*x + c)^3*log((2*a*b*c os(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2* A*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) - 2*(2*A*a^5 - 2*A*a^3*b^2 + 2*((2*A + 3*C)*a^5 + (A - 3*C)*a^3*b^2 - 3*A*a*b^4)*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*cos(d*x + c)^3), 1/12*(12*(C*a^2*b^2 + A*b^4)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^3 - 3*((A + 2*C)*a^4 *b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3 *((A + 2*C)*a^4*b + (A - 2*C)*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^3*log(-sin(d *x + c) + 1) + 2*(2*A*a^5 - 2*A*a^3*b^2 + 2*((2*A + 3*C)*a^5 + (A - 3*C)*a ^3*b^2 - 3*A*a*b^4)*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*cos(d*x + c)^3)]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c)),x)
Output:
Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**4/(a + b*cos(c + d*x)), x)
Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="m axima")
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*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (167) = 334\).
Time = 0.18 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.02 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {3 \, {\left (A a^{2} b + 2 \, C a^{2} b + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, {\left (A a^{2} b + 2 \, C a^{2} b + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {12 \, {\left (C a^{2} b^{2} + A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}}}{6 \, d} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="g iac")
Output:
-1/6*(3*(A*a^2*b + 2*C*a^2*b + 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) /a^4 - 3*(A*a^2*b + 2*C*a^2*b + 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1) )/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 = 2.73 (sec) , antiderivative size = 3927, normalized size of antiderivative = 21.34 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x))),x)
Output:
(b^2*atan(((b^2*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((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 - 1 6*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 + (b^2*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((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*b^2*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b) )^(1/2)*(A*b^2 + C*a^2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a ^4*b^2))))/(a^6 - a^4*b^2))*1i)/(a^6 - a^4*b^2) + (b^2*(-(a + b)*(a - b))^ (1/2)*(A*b^2 + C*a^2)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - 16*A^2*a*b^8 + 1 6*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 - (b^2*(-(a + b)*(a - b))^(1/ 2)*(A*b^2 + C*a^2)*((8*(4*A*a^8*b^5 - 6*A*a^9*b^4 + 2*A*a^10*b^3 - 2*A*a^1 1*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*b ^2*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2))))/(a^6 - a^4*b^2))*1i)/(a^ 6 - a^4*b^2))/((16*(4*A^3*b^11 - 6*A^3*a*b^10 + 6*A^3*a^2*b^9 - 5*A^3*a...
Time = 0.17 (sec) , antiderivative size = 1043, normalized size of antiderivative = 5.67 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx =\text {Too large to display} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x)
Output:
(12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a*b**2*c + 12*sqrt(a**2 - b**2) *atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*b**4 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a*b**2*c - 12*sqrt( a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b* *2))*cos(c + d*x)*b**4 + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2* a**3*b*c + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b **3 - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3*c - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**5 - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**4*b - 6*cos(c + d*x)*log(tan((c + d*x)/2 ) - 1)*a**3*b*c - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**3 + 6*c os(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**3*c + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**5 - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**2*a**4*b - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a* *3*b*c - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b** 3 + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3*c + 6* cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**5 + 3*cos(c + d* x)*log(tan((c + d*x)/2) + 1)*a**4*b + 6*cos(c + d*x)*log(tan((c + d*x)/...