Integrand size = 33, antiderivative size = 150 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^3 d}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \] Output:
2/5*(3*A+5*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(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c ,2^(1/2))*(b*sec(d*x+c))^(1/2)/b^3/d+2/3*B*sin(d*x+c)/b^2/d/(b*sec(d*x+c)) ^(1/2)+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 = 1.92 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {e^{-i d x} \sqrt {b \sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (10 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-2 i (3 A+5 C) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\cos (c+d x) (6 i (3 A+5 C)+10 B \sin (c+d x)+3 A \sin (2 (c+d x)))\right )}{15 b^3 d} \] Input:
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x ]
Output:
(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 = 0.87 (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 |
| \(\displaystyle \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b}\) |
Input:
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x]
Output:
(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))
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 = 3.44 (sec) , antiderivative size = 455, normalized size of antiderivative = 3.03
| method | result | size |
| default | \(\frac {\frac {2 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (9 \cos \left (d x +c \right )+18+9 \sec \left (d x +c \right )\right )}{15}+\frac {2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (15 \cos \left (d x +c \right )+30+15 \sec \left (d x +c \right )\right )}{15}+\frac {2 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-9 \cos \left (d x +c \right )-18-9 \sec \left (d x +c \right )\right )}{15}+\frac {2 i B \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-5 \cos \left (d x +c \right )-10-5 \sec \left (d x +c \right )\right )}{15}+\frac {2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-15 \cos \left (d x +c \right )-30-15 \sec \left (d x +c \right )\right )}{15}+\frac {2 \sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+9\right ) A}{15}+\frac {2 \sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+5\right ) B}{15}+2 C \sin \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{2}}\) | \(455\) |
| parts | \(\frac {2 A \left (\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )-3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{2}}+\frac {B \left (\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-1-\sec \left (d x +c \right )\right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {b \sec \left (d x +c \right )}\, b^{2}}+\frac {2 C \left (i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{b^{2} d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}\) | \(463\) |
Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNV ERBOSE)
Output:
2/15/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)/b^2*(I*A*(1/(cos(d*x+c)+1))^(1/ 2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I )*(9*cos(d*x+c)+18+9*sec(d*x+c))+I*C*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(15*cos(d*x+c )+30+15*sec(d*x+c))+I*A*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(-9*cos(d*x+c)-18-9*sec(d* x+c))+I*B*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*Ellip ticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(-5*cos(d*x+c)-10-5*sec(d*x+c))+I*C*(1/( cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d* x+c)-cot(d*x+c)),I)*(-15*cos(d*x+c)-30-15*sec(d*x+c))+sin(d*x+c)*(3*cos(d* x+c)^2+3*cos(d*x+c)+9)*A+sin(d*x+c)*(5*cos(d*x+c)+5)*B+15*C*sin(d*x+c))
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.17 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {-5 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, A \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, b^{3} d} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorith m="fricas")
Output:
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)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(5/2),x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/(b*sec(c + d*x))**(5/2), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorith m="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(b*sec(d*x + c))^(5/2), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorith m="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(b*sec(d*x + c))^(5/2), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(5/2),x)
Output:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(5/2), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) c \right )}{b^{3}} \] Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x)
Output:
(sqrt(b)*(int(sqrt(sec(c + d*x))/sec(c + d*x)**3,x)*a + int(sqrt(sec(c + d *x))/sec(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x))/sec(c + d*x),x)*c))/b** 3