Integrand size = 35, antiderivative size = 151 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {4 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 C \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \]
-4*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))/a^2/d+1/3*(A-5*C)*(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))/a^2/d+4*C*sin(d*x+c)/a^2/d/ cos(d*x+c)^(1/2)+1/3*(A-5*C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))/cos(d*x+c)^(1 /2)-1/3*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 11.41 (sec) , antiderivative size = 836, normalized size of antiderivative = 5.54 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {4 A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (A+2 C+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^2}+\frac {20 C \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (A+2 C+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {16 C \cot \left (\frac {c}{2}\right ) \sec (c)}{d}+\frac {16 C \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {16 C \sec (c) \sec (c+d x) \sin (d x)}{d}+\frac {4 (A+C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {8 C \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2} \]
(-4*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcT an[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]* Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3 *d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^ 2) + (20*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/ 4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c ]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]] ])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d *x])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*(A + C*Sec[c + d*x]^2)* ((16*C*Cot[c/2]*Sec[c])/d + (16*C*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2] )/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/ (3*d) + (16*C*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (4*(A + C)*Sec[c/2 + (d*x) /2]^2*Tan[c/2])/(3*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x ])^2) + (8*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2) *((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin [d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])...
Time = 0.96 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4602, 3042, 3521, 27, 3042, 3457, 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.
\(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sec (c+d x)^2}{\cos (c+d x)^{3/2} (a \sec (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 4602 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+C}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3521 |
\(\displaystyle \frac {\int \frac {a (A+7 C)+3 a (A-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (A+7 C)+3 a (A-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (A+7 C)+3 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {12 C a^2+(A-5 C) \cos (c+d x) a^2}{\cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {12 C a^2+(A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+12 a^2 C \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
-1/3*((A + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2) + ((2*(A - 5*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])) + ((2*a^2*(A - 5*C)*EllipticF[(c + d*x)/2, 2])/d + 12*a^2*C*((-2*EllipticE[( c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/a^2)/(6*a^2)
3.12.18.3.1 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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[d^( m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A*Cos [e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(191)=382\).
Time = 1.88 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.98
method | result | size |
default | \(-\frac {-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 C \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +43 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +37 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(450\) |
-1/6*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(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)^(1/2)*(A*EllipticF(cos(1/2*d*x +1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*EllipticE( cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*(-2 *sin(1/2*d*x+1/2*c)^4+sin(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)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1 /2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*EllipticE(cos(1/2*d*x+ 1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-48*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2* d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/ 2*d*x+1/2*c)^2)^(1/2)*(A+43*C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c) ^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+37*C)*sin(1/2*d*x+1/2*c)^2)/a^2/cos(1/2* d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2* d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.43 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (12 \, C \cos \left (d x + c\right )^{2} + {\left (A + 19 \, C\right )} \cos \left (d x + c\right ) + 6 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (i \, \sqrt {2} C \cos \left (d x + c\right )^{3} + 2 i \, \sqrt {2} C \cos \left (d x + c\right )^{2} + i \, \sqrt {2} C \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (-i \, \sqrt {2} C \cos \left (d x + c\right )^{3} - 2 i \, \sqrt {2} C \cos \left (d x + c\right )^{2} - i \, \sqrt {2} C \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]
1/6*(2*(12*C*cos(d*x + c)^2 + (A + 19*C)*cos(d*x + c) + 6*C)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-I*A + 5*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(I *A - 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A + 5*I*C)*cos(d*x + c))*weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - 5*I*C) *cos(d*x + c)^3 - 2*sqrt(2)*(-I*A + 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A - 5*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 12*(I*sqrt(2)*C*cos(d*x + c)^3 + 2*I*sqrt(2)*C*cos(d*x + c)^2 + I* sqrt(2)*C*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*(-I*sqrt(2)*C*cos(d*x + c)^3 - 2*I*sq rt(2)*C*cos(d*x + c)^2 - I*sqrt(2)*C*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]