Integrand size = 35, antiderivative size = 112 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=-\frac {(A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(A-C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}+\frac {(A+3 C) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \]
-(A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2 *d*x+1/2*c),2^(1/2))/a/d+(A-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/ 2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+(A+3*C)*sin(d*x+c)/a/d/cos( d*x+c)^(1/2)-(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.23 (sec) , antiderivative size = 1063, normalized size of antiderivative = 9.49 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx =\text {Too large to display} \]
(Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*((2*(2*C + A*Cos[c] + C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d + (8*C*Sec[c]*Sec[c + d*x]*S in[d*x])/d))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (2*A* Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, { 5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot [c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c] ]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d *x])) + (2*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPFQ[ {1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d* x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[- (Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2 ]*(A + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + Ar cTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan...
Time = 0.69 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4602, 3042, 3521, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}
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.
\(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a \sec (c+d x)+a)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sec (c+d x)^2}{\sqrt {\cos (c+d x)} (a \sec (c+d x)+a)}dx\) |
\(\Big \downarrow \) 4602 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+C}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
\(\Big \downarrow \) 3521 |
\(\displaystyle \frac {\int \frac {a (A+3 C)+a (A-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (A+3 C)+a (A-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (A+3 C)+a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {a (A+3 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+a (A-C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A+3 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+a (A-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {a (A-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {a (A-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {2 a (A-C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+a (A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}\) |
-(((A + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]))) + (( 2*a*(A - C)*EllipticF[(c + d*x)/2, 2])/d + a*(A + 3*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/(2*a^2)
3.12.11.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[d^( m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A*Cos [e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Time = 1.56 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.82
method | result | size |
default | \(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +3 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +5 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(316\) |
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*(-cos(1/2*d*x +1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+ 1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(A*EllipticF(cos(1/2*d*x+ 1/2*c),2^(1/2))+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF(cos(1/ 2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(-2*sin (1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+3*C)*sin(1/2*d*x+1/2*c)^4 +(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+5*C)*sin(1/2*d*x+ 1/2*c)^2)/sin(1/2*d*x+1/2*c)^3/(2*sin(1/2*d*x+1/2*c)^2-1)/cos(1/2*d*x+1/2* c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.57 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\frac {2 \, {\left ({\left (A + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{2 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]
1/2*(2*((A + 3*C)*cos(d*x + c) + 2*C)*sqrt(cos(d*x + c))*sin(d*x + c) + (s qrt(2)*(-I*A + I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A + I*C)*cos(d*x + c))*we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A - I*C)*cos(d*x + c))*weierstrassPInvers e(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + (sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weier strassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + (sqrt(2)*(I*A + 3* I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A + 3*I*C)*cos(d*x + c))*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*co s(d*x + c)^2 + a*d*cos(d*x + c))
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\frac {\int \frac {A}{\sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx}{a} \]
(Integral(A/(sqrt(cos(c + d*x))*sec(c + d*x) + sqrt(cos(c + d*x))), x) + I ntegral(C*sec(c + d*x)**2/(sqrt(cos(c + d*x))*sec(c + d*x) + sqrt(cos(c + d*x))), x))/a
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\sqrt {\cos \left (c+d\,x\right )}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )} \,d x \]