Integrand size = 25, antiderivative size = 77 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \]
2/5*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(si n(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/5*b^2* C*tan(d*x+c)/d/(b*sec(d*x+c))^(5/2)
Time = 0.57 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {\frac {4 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+2 C \sin (2 (c+d x))}{10 d \sqrt {b \sec (c+d x)}} \]
((4*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*C*Sin[2* (c + d*x)])/(10*d*Sqrt[b*Sec[c + d*x]])
Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3717, 3042, 4533, 3042, 4258, 3042, 3119}
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.
\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3717 |
\(\displaystyle b^2 \int \frac {A \sec ^2(c+d x)+C}{(b \sec (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )^2+C}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{5 b^2}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \sqrt {\cos (c+d x)}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {(5 A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle b^2 \left (\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\right )\) |
b^2*((2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]] *Sqrt[b*Sec[c + d*x]]) + (2*C*Tan[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/2)))
3.1.31.3.1 Defintions of rubi rules used
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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p) Int[(d*Csc[e + f*x])^(m - n*p )*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && !IntegerQ[m] && IntegersQ[n, p]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Result contains complex when optimal does not.
Time = 16.52 (sec) , antiderivative size = 795, normalized size of antiderivative = 10.32
method | result | size |
default | \(\text {Expression too large to display}\) | \(795\) |
parts | \(\text {Expression too large to display}\) | \(806\) |
2/5/d/(1+cos(d*x+c))/(b*sec(d*x+c))^(1/2)*(3*I*sec(d*x+c)*C*(1/(1+cos(d*x+ c)))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c )))^(1/2)+3*I*cos(d*x+c)*C*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x +c)))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-3*I*cos(d*x+c)*C*(1/(1+ cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(csc(d*x+ c)-cot(d*x+c)),I)+5*I*sec(d*x+c)*A*(1/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(c sc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-10*I*A*(1/(1+co s(d*x+c)))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(1+cos (d*x+c)))^(1/2)+10*I*A*(1/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(csc(d*x+c)-co t(d*x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-6*I*C*(1/(1+cos(d*x+c)))^(1 /2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 2)-3*I*sec(d*x+c)*C*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d *x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+6*I*C*(1/(1+cos(d*x+c)))^(1/2) *EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+ 5*I*cos(d*x+c)*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 )*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-5*I*sec(d*x+c)*A*(1/(1+cos(d*x+c) ))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c)) )^(1/2)-5*I*cos(d*x+c)*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c )))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)+C*cos(d*x+c)^2*sin(d*x+c) +C*cos(d*x+c)*sin(d*x+c)+5*A*sin(d*x+c)+3*sin(d*x+c)*C)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {2 \, C \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 3 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, b d} \]
1/5*(2*C*sqrt(b/cos(d*x + c))*cos(d*x + c)^2*sin(d*x + c) + sqrt(2)*(5*I*A + 3*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) + I*sin(d*x + c))) + sqrt(2)*(-5*I*A - 3*I*C)*sqrt(b)*weierstrassZe ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b*d )
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\sqrt {b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\sqrt {b \sec \left (d x + c\right )}} \,d x } \]
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\sqrt {b \sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]