Integrand size = 25, antiderivative size = 115 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac {2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]
2/45*(9*A+7*C)*sin(d*x+c)/b/d/(b*sec(d*x+c))^(3/2)+2/15*(9*A+7*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))/b^2/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/9*b^2*C*tan(d*x+c)/d/(b* sec(d*x+c))^(9/2)
Time = 1.00 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {\frac {48 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+4 (18 A+19 C+5 C \cos (2 (c+d x))) \sin (2 (c+d x))}{360 b^2 d \sqrt {b \sec (c+d x)}} \]
((48*(9*A + 7*C)*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 4*(18*A + 19*C + 5*C*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(360*b^2*d*Sqrt[b*Sec[c + d*x]])
Time = 0.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3717, 3042, 4533, 3042, 4256, 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.
\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3717 |
\(\displaystyle b^2 \int \frac {A \sec ^2(c+d x)+C}{(b \sec (c+d x))^{9/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )^2+C}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}dx\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle b^2 \left (\frac {(9 A+7 C) \int \frac {1}{(b \sec (c+d x))^{5/2}}dx}{9 b^2}+\frac {2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {(9 A+7 C) \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 b^2}+\frac {2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle b^2 \left (\frac {(9 A+7 C) \left (\frac {3 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{9 b^2}+\frac {2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {(9 A+7 C) \left (\frac {3 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{9 b^2}+\frac {2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle b^2 \left (\frac {(9 A+7 C) \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{9 b^2}+\frac {2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {(9 A+7 C) \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{9 b^2}+\frac {2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle b^2 \left (\frac {(9 A+7 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{9 b^2}+\frac {2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\right )\) |
b^2*(((9*A + 7*C)*((6*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x ]]*Sqrt[b*Sec[c + d*x]]) + (2*Sin[c + d*x])/(5*b*d*(b*Sec[c + d*x])^(3/2)) ))/(9*b^2) + (2*C*Tan[c + d*x])/(9*d*(b*Sec[c + d*x])^(9/2)))
3.1.33.3.1 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[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* n]
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 = 23.00 (sec) , antiderivative size = 866, normalized size of antiderivative = 7.53
method | result | size |
default | \(\text {Expression too large to display}\) | \(866\) |
parts | \(\text {Expression too large to display}\) | \(876\) |
-2/45/b^2/d/(1+cos(d*x+c))/(b*sec(d*x+c))^(1/2)*(21*I*cos(d*x+c)*C*Ellipti cF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+co s(d*x+c)))^(1/2)-21*I*sec(d*x+c)*C*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)* (1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-27*I*sec(d*x+c) *A*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d* x+c)/(1+cos(d*x+c)))^(1/2)+54*I*A*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*( 1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-5*C*sin(d*x+c)*c os(d*x+c)^4+27*I*sec(d*x+c)*A*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1 +cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+42*I*C*EllipticF(I*( csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+ c)))^(1/2)+21*I*sec(d*x+c)*C*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+ cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-21*I*cos(d*x+c)*C*Ell ipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/( 1+cos(d*x+c)))^(1/2)-5*C*cos(d*x+c)^3*sin(d*x+c)-42*I*C*EllipticE(I*(csc(d *x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^ (1/2)+27*I*cos(d*x+c)*A*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d *x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-9*A*sin(d*x+c)*cos(d*x+c)^ 2-27*I*cos(d*x+c)*A*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c )))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-54*I*A*EllipticE(I*(csc(d*x+c) -cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-9 i \, A - 7 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (9 i \, A + 7 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, C \cos \left (d x + c\right )^{4} + {\left (9 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45 \, b^{3} d} \]
-1/45*(3*sqrt(2)*(-9*I*A - 7*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstra ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(9*I*A + 7*I *C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(5*C*cos(d*x + c)^4 + (9*A + 7*C)*cos(d*x + c)^2)* sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^3*d)
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]