Integrand size = 31, antiderivative size = 123 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]
-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(s in(1/2*d*x+1/2*c),2^(1/2))/d+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))/d+2/5*C*sin(d*x+c)/d/cos(d* x+c)^(5/2)+2/3*B*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(5*A+3*C)*sin(d*x+c)/d/ cos(d*x+c)^(1/2)
Time = 0.57 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-6 (5 A+3 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 B \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+10 B \sin (c+d x)+15 A \sin (2 (c+d x))+9 C \sin (2 (c+d x))+6 C \tan (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
(-6*(5*A + 3*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 10*B*Sin[c + d*x] + 15*A*Sin[2*(c + d*x)] + 9*C*Sin[2*(c + d*x)] + 6*C*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/ 2))
Time = 0.58 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4552, 3042, 3500, 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.
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4552 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle \frac {2}{5} \int \frac {5 B+(5 A+3 C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {5 B+(5 A+3 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {5 B+(5 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{5} \left ((5 A+3 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 B \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx\right )+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left ((5 A+3 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 B \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\right )+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {1}{5} \left ((5 A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )+5 B \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\right )+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left ((5 A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+5 B \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\right )+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (5 B \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+(5 A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left ((5 A+3 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+5 B \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )\right )+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*C*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (5*B*((2*EllipticF[(c + d*x) /2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) + (5*A + 3*C)*( (-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) ))/5
3.12.85.3.1 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_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*( x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b^2 Int[(b*Cos [e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ [{b, e, f, A, B, C, m}, x] && !IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(798\) vs. \(2(159)=318\).
Time = 3.60 (sec) , antiderivative size = 799, normalized size of antiderivative = 6.50
-2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(8*sin(1/2 *d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d* x+1/2*c)^3*(120*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-60*A*(2*sin(1/2* d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1 /2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli pticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2 *d*x+1/2*c)^4+72*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-36*C*(2*sin(1/2 *d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+ 1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-120*A*sin(1/2*d*x+1/2*c)^4*cos(1/2*d* x+1/2*c)+60*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c ),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-20*B*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x +1/2*c)^2-72*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+36*C*(2*sin(1/2*d*x +1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2* c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+30*A*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2 *c)-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell ipticE(cos(1/2*d*x+1/2*c),2^(1/2))+10*B*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1 /2*c)-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+24*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.67 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-5 i \, \sqrt {2} B \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (5 i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, {\left (5 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \]
1/15*(-5*I*sqrt(2)*B*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c ) + I*sin(d*x + c)) + 5*I*sqrt(2)*B*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(5*I*A + 3*I*C)*cos(d*x + c )^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin (d*x + c))) - 3*sqrt(2)*(-5*I*A - 3*I*C)*cos(d*x + c)^3*weierstrassZeta(-4 , 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(5* A + 3*C)*cos(d*x + c)^2 + 5*B*cos(d*x + c) + 3*C)*sqrt(cos(d*x + c))*sin(d *x + c))/(d*cos(d*x + c)^3)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Time = 18.51 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.88 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {6\,C\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+10\,B\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]