Integrand size = 45, antiderivative size = 183 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^{3/2} (8 A+12 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^2 (8 A-4 B-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a (4 B+3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \]
1/4*a^(3/2)*(8*A+12*B+7*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/ 2))/d+1/2*C*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+1/4*a^2*( 8*A-4*B-5*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a*(4 *B+3*C)*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d
Time = 2.44 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.78 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 \left ((8 A+7 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-12 B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (8 A \sin (c+d x)+(4 B+7 C+2 C \sec (c+d x)) \tan (c+d x))\right )}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Sqrt[Sec[c + d*x]],x]
(a^2*((8*A + 7*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 12*B*ArcSi n[Sqrt[Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d* x])]*(8*A*Sin[c + d*x] + (4*B + 7*C + 2*C*Sec[c + d*x])*Tan[c + d*x])))/(4 *d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.08 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4576, 27, 3042, 4506, 27, 3042, 4503, 3042, 4288, 222}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{3/2} (a (4 A-C)+a (4 B+3 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{2 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^{3/2} (a (4 A-C)+a (4 B+3 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (4 A-C)+a (4 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{2} \left (a^2 (8 A+12 B+7 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (a^2 (8 A+12 B+7 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a^2 (8 A+12 B+7 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {2 a^{5/2} (8 A+12 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/S qrt[Sec[c + d*x]],x]
(C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d) + ((a ^2*(4*B + 3*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + ((2*a^(5/2)*(8*A + 12*B + 7*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(8*A - 4*B - 5*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2)/(4*a)
3.6.89.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(536\) vs. \(2(157)=314\).
Time = 1.82 (sec) , antiderivative size = 537, normalized size of antiderivative = 2.93
| method | result | size |
| default | \(\frac {a \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (8 A \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+8 A \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+12 B \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+12 B \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+7 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+7 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}-16 A \cos \left (d x +c \right )^{3} \cot \left (d x +c \right )+16 A \cos \left (d x +c \right )^{2} \cot \left (d x +c \right )-8 B \cos \left (d x +c \right )^{2} \cot \left (d x +c \right )-14 C \cos \left (d x +c \right )^{2} \cot \left (d x +c \right )+8 B \cos \left (d x +c \right ) \cot \left (d x +c \right )+10 C \cos \left (d x +c \right ) \cot \left (d x +c \right )+4 C \cot \left (d x +c \right )\right )}{8 d}\) | \(537\) |
| parts | \(-\frac {A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-\arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-\sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}-\frac {B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (3 \cos \left (d x +c \right ) \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-2 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {C a \left (7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-14 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-4 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(542\) |
int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2 ),x,method=_RETURNVERBOSE)
1/8*a/d*sec(d*x+c)^(5/2)*(a*(1+sec(d*x+c)))^(1/2)*(8*A*(-1/(cos(d*x+c)+1)) ^(1/2)*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c )+1))^(1/2))*cos(d*x+c)^3+8*A*(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cos(d* x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3+ 12*B*(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos( d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3+12*B*(-1/(cos(d*x+c)+1)) ^(1/2)*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c) +1))^(1/2))*cos(d*x+c)^3+7*C*(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cos(d* x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3+ 7*C*(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d* x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3-16*A*cos(d*x+c)^3*cot(d*x+ c)+16*A*cos(d*x+c)^2*cot(d*x+c)-8*B*cos(d*x+c)^2*cot(d*x+c)-14*C*cos(d*x+c )^2*cot(d*x+c)+8*B*cos(d*x+c)*cot(d*x+c)+10*C*cos(d*x+c)*cot(d*x+c)+4*C*co t(d*x+c))
Time = 0.47 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.44 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\left [\frac {{\left ({\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (8 \, A a \cos \left (d x + c\right )^{2} + {\left (4 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (8 \, A a \cos \left (d x + c\right )^{2} + {\left (4 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(1/2),x, algorithm="fricas")
[1/16*(((8*A + 12*B + 7*C)*a*cos(d*x + c)^2 + (8*A + 12*B + 7*C)*a*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c )^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin( d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4* (8*A*a*cos(d*x + c)^2 + (4*B + 7*C)*a*cos(d*x + c) + 2*C*a)*sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) ^2 + d*cos(d*x + c)), 1/8*(((8*A + 12*B + 7*C)*a*cos(d*x + c)^2 + (8*A + 1 2*B + 7*C)*a*cos(d*x + c))*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a* cos(d*x + c) - 2*a)) + 2*(8*A*a*cos(d*x + c)^2 + (4*B + 7*C)*a*cos(d*x + c ) + 2*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d *x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3661 vs. \(2 (157) = 314\).
Time = 0.60 (sec) , antiderivative size = 3661, normalized size of antiderivative = 20.01 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display} \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(1/2),x, algorithm="maxima")
1/16*(4*sqrt(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + s qrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt( 2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a* log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/ 2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 8*a*sin(1/2*d*x + 1 /2*c))*A*sqrt(a) + 4*(3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt (2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*c os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2* c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt( 2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2 *sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*si n(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2* d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/...
\[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(1/2),x, algorithm="giac")
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 1/cos(c + d*x))^(1/2),x)