Integrand size = 33, antiderivative size = 150 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^3 d}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \]
2/3*B*sin(d*x+c)/b^2/d/(b*sec(d*x+c))^(1/2)+2/5*(3*A+5*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)+2/3*B*(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))*cos(d*x+c)^(1/2) *(b*sec(d*x+c))^(1/2)/b^3/d+2/5*A*tan(d*x+c)/d/(b*sec(d*x+c))^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {e^{-i d x} \sqrt {b \sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (10 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-2 i (3 A+5 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\cos (c+d x) (6 i (3 A+5 C)+10 B \sin (c+d x)+3 A \sin (2 (c+d x)))\right )}{15 b^3 d} \]
(Sqrt[b*Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(10*B*Sqrt[Cos[c + d*x]]*Ell ipticF[(c + d*x)/2, 2] - (2*I)*(3*A + 5*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2* I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + Co s[c + d*x]*((6*I)*(3*A + 5*C) + 10*B*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)])) )/(15*b^3*d*E^(I*d*x))
Time = 0.84 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4535, 3042, 4256, 3042, 4258, 3042, 3120, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \int \frac {C \sec ^2(c+d x)+A}{(b \sec (c+d x))^{5/2}}dx+\frac {B \int \frac {1}{(b \sec (c+d x))^{3/2}}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx+\frac {B \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{b}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx+\frac {B \left (\frac {\int \sqrt {b \sec (c+d x)}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx+\frac {B \left (\frac {\int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx+\frac {B \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx+\frac {B \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {(3 A+5 C) \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{5 b^2}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(3 A+5 C) \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {(3 A+5 C) \int \sqrt {\cos (c+d x)}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(3 A+5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
(2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt [b*Sec[c + d*x]]) + (B*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sq rt[b*Sec[c + d*x]])/(3*b^2*d) + (2*Sin[c + d*x])/(3*b*d*Sqrt[b*Sec[c + d*x ]])))/b + (2*A*Tan[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/2))
3.1.69.3.1 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[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[(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]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Result contains complex when optimal does not.
Time = 2.69 (sec) , antiderivative size = 956, normalized size of antiderivative = 6.37
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(956\) |
| default | \(\text {Expression too large to display}\) | \(1004\) |
2/5*A/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)/b^2*(3*I*(1/(cos(d*x+c)+1))^(1 /2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-cot(d*x+c)+csc(d*x+c)) ,I)*cos(d*x+c)-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1 /2)*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*cos(d*x+c)+6*I*(1/(cos(d*x+c)+ 1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-cot(d*x+c)+csc(d *x+c)),I)-6*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*E llipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)+3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d *x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-cot(d*x+c)+csc(d*x+c)),I)*sec(d* x+c)-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipt icF(I*(-cot(d*x+c)+csc(d*x+c)),I)*sec(d*x+c)+cos(d*x+c)^2*sin(d*x+c)+sin(d *x+c)*cos(d*x+c)+3*sin(d*x+c))-2/3*B/d/(b*sec(d*x+c))^(1/2)/b^2*(I*(1/(cos (d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-cot(d*x+ c)+csc(d*x+c)),I)+I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^( 1/2)*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*sec(d*x+c)-sin(d*x+c))+2*C/b^ 2/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)*(I*EllipticE(I*(-cot(d*x+c)+csc(d* x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d* x+c)-I*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+2*I*(1/(cos(d*x+c)+1))^(1/2)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-cot(d*x+c)+csc(d*x+c)),I)-2* I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.17 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {-5 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, A \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, b^{3} d} \]
1/15*(-5*I*sqrt(2)*B*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s in(d*x + c)) + 5*I*sqrt(2)*B*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(-3*I*A - 5*I*C)*sqrt(b)*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt (2)*(3*I*A + 5*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*A*cos(d*x + c)^2 + 5*B*cos(d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^3*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]