Integrand size = 33, antiderivative size = 194 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {4 a^2 (4 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (7 A+6 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+3 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]
-4/5*a^2*(4*A+3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/21*a^2*(7*A+6*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+2/35*a^2 *(7*A+9*B)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+4/21*a^2*(7*A+6*B)*sin(d*x+c)/d/c os(d*x+c)^(3/2)+2/7*B*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(7/2)+4 /5*a^2*(4*A+3*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.82 (sec) , antiderivative size = 1067, normalized size of antiderivative = 5.50 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\cos ^{\frac {7}{2}}(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \left (\frac {(4 A+3 B) \csc (c) \sec (c)}{5 d}+\frac {B \sec (c) \sec ^4(c+d x) \sin (d x)}{14 d}+\frac {\sec (c) \sec ^3(c+d x) (5 B \sin (c)+7 A \sin (d x)+14 B \sin (d x))}{70 d}+\frac {\sec (c) \sec ^2(c+d x) (21 A \sin (c)+42 B \sin (c)+70 A \sin (d x)+60 B \sin (d x))}{210 d}+\frac {\sec (c) \sec (c+d x) (35 A \sin (c)+30 B \sin (c)+84 A \sin (d x)+63 B \sin (d x))}{105 d}\right )}{B+A \cos (c+d x)}-\frac {A \cos ^3(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \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 (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)}}-\frac {2 B \cos ^3(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \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)))}}{7 d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)}}+\frac {2 A \cos ^3(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \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 )}{5 d (B+A \cos (c+d x))}+\frac {3 B \cos ^3(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \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 )}{10 d (B+A \cos (c+d x))} \]
(Cos[c + d*x]^(7/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec [c + d*x])*(((4*A + 3*B)*Csc[c]*Sec[c])/(5*d) + (B*Sec[c]*Sec[c + d*x]^4*S in[d*x])/(14*d) + (Sec[c]*Sec[c + d*x]^3*(5*B*Sin[c] + 7*A*Sin[d*x] + 14*B *Sin[d*x]))/(70*d) + (Sec[c]*Sec[c + d*x]^2*(21*A*Sin[c] + 42*B*Sin[c] + 7 0*A*Sin[d*x] + 60*B*Sin[d*x]))/(210*d) + (Sec[c]*Sec[c + d*x]*(35*A*Sin[c] + 30*B*Sin[c] + 84*A*Sin[d*x] + 63*B*Sin[d*x]))/(105*d)))/(B + A*Cos[c + d*x]) - (A*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[ d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*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*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Se c[c + d*x])*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]]]])/(7*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (2 *A*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*((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[T an[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcT...
Time = 1.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.96, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3433, 3042, 3454, 27, 3042, 3447, 3042, 3500, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^2 (A \cos (c+d x)+B)}{\cos ^{\frac {9}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {2}{7} \int \frac {(\cos (c+d x) a+a) (a (7 A+9 B)+a (7 A+3 B) \cos (c+d x))}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int \frac {(\cos (c+d x) a+a) (a (7 A+9 B)+a (7 A+3 B) \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (7 A+9 B)+a (7 A+3 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{7} \int \frac {(7 A+3 B) \cos ^2(c+d x) a^2+(7 A+9 B) a^2+\left ((7 A+3 B) a^2+(7 A+9 B) a^2\right ) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \frac {(7 A+3 B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(7 A+9 B) a^2+\left ((7 A+3 B) a^2+(7 A+9 B) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {5 (7 A+6 B) a^2+7 (4 A+3 B) \cos (c+d x) a^2}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {5 (7 A+6 B) a^2+7 (4 A+3 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (5 a^2 (7 A+6 B) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx+7 a^2 (4 A+3 B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (5 a^2 (7 A+6 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+7 a^2 (4 A+3 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (5 a^2 (7 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 a^2 (4 A+3 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (5 a^2 (7 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 a^2 (4 A+3 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (5 a^2 (7 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 a^2 (4 A+3 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \left (5 a^2 (7 A+6 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 a^2 (4 A+3 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
(2*B*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2 *a^2*(7*A + 9*B)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(5*a^2*(7*A + 6*B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) + 7*a^2*(4*A + 3*B)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*S in[c + d*x])/(d*Sqrt[Cos[c + d*x]]))))/5)/7
3.5.95.3.1 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_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(823\) vs. \(2(226)=452\).
Time = 23.11 (sec) , antiderivative size = 824, normalized size of antiderivative = 4.25
-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/4*A/si n(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin (1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(sin(1 /2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d* x+1/2*c)^2-1)^(1/2))+1/4*B*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c )^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2* d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d *x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c )^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2)))+(1/2*A+1/4*B)*(-1/6*cos(1/2*d*x+1/2*c)*(-2* sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2 )^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2 *sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/ 2*c),2^(1/2)))+1/5*(1/4*A+1/2*B)/sin(1/2*d*x+1/2*c)^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)*(24*sin(1/2*d*x+1/2* c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(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- 24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2) *EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*si n(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.36 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (42 \, {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 21 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, B a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d \cos \left (d x + c\right )^{4}} \]
-2/105*(5*I*sqrt(2)*(7*A + 6*B)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(7*A + 6*B)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I*sqr t(2)*(4*A + 3*B)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*(4*A + 3*B)*a^2 *cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (42*(4*A + 3*B)*a^2*cos(d*x + c)^3 + 10*(7*A + 6 *B)*a^2*cos(d*x + c)^2 + 21*(A + 2*B)*a^2*cos(d*x + c) + 15*B*a^2)*sqrt(co s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=a^{2} \left (\int \frac {A}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A \sec {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 B \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx\right ) \]
a**2*(Integral(A/cos(c + d*x)**(3/2), x) + Integral(2*A*sec(c + d*x)/cos(c + d*x)**(3/2), x) + Integral(A*sec(c + d*x)**2/cos(c + d*x)**(3/2), x) + Integral(B*sec(c + d*x)/cos(c + d*x)**(3/2), x) + Integral(2*B*sec(c + d*x )**2/cos(c + d*x)**(3/2), x) + Integral(B*sec(c + d*x)**3/cos(c + d*x)**(3 /2), x))
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Time = 16.88 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {30\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,B\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,B\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]
(6*A*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 20*A* a^2*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 30*A*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (30*B*a^2 *sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2) + 84*B*a^2*cos( c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 70*B* a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2 ))/(105*d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))