Integrand size = 45, antiderivative size = 201 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^{3/2} (24 A+14 B+11 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a^2 (24 A+30 B+19 C) \sin (c+d x)}{24 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (2 B+C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]
1/3*C*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+1/8*a^(3/2)*(24 *A+14*B+11*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c )^(1/2)*sec(d*x+c)^(1/2)/d+1/24*a^2*(24*A+30*B+19*C)*sin(d*x+c)/d/cos(d*x+ c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/4*a*(2*B+C)*sin(d*x+c)*(a+a*sec(d*x+c))^ (1/2)/d/cos(d*x+c)^(3/2)
Time = 7.49 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a^2 \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left ((42 B+33 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-72 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+12 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+22 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+24 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+42 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+33 C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Sqrt[Cos[c + d*x]],x]
(a^2*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*((42*B + 33*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 72*A*ArcSin[Sqrt[Sec[c + d*x]]] + 12*B*Sqrt[1 - Sec[c + d *x]]*Sec[c + d*x]^(3/2) + 22*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 8*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 24*A*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 42*B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 3 3*C*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Sin[c + d*x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.30 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4506, 27, 3042, 4504, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\sqrt {\cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{2} \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2} (a (6 A+C)+3 a (2 B+C) \sec (c+d x))dx}{3 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2} (a (6 A+C)+3 a (2 B+C) \sec (c+d x))dx}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (6 A+C)+3 a (2 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {1}{2} \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left ((24 A+6 B+7 C) a^2+(24 A+30 B+19 C) \sec (c+d x) a^2\right )dx+\frac {3 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left ((24 A+6 B+7 C) a^2+(24 A+30 B+19 C) \sec (c+d x) a^2\right )dx+\frac {3 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((24 A+6 B+7 C) a^2+(24 A+30 B+19 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {3 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {3}{2} a^2 (24 A+14 B+11 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (24 A+30 B+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {3}{2} a^2 (24 A+14 B+11 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (24 A+30 B+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \left (\frac {a^3 (24 A+30 B+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 a^2 (24 A+14 B+11 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{5/2} (24 A+14 B+11 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (24 A+30 B+19 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\right )\) |
Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/S qrt[Cos[c + d*x]],x]
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a^2*(2*B + C)*Sec[c + d*x]^(3/2)*S qrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + ((3*a^(5/2)*(24*A + 14*B + 1 1*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^3*(2 4*A + 30*B + 19*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/4)/(6*a))
3.13.59.3.1 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[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(543\) vs. \(2(171)=342\).
Time = 1.00 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.71
| method | result | size |
| default | \(\frac {a \left (48 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-72 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-72 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+84 B \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-42 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-42 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+66 C \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-33 C \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-33 C \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+24 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+44 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+16 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(544\) |
int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2 ),x,method=_RETURNVERBOSE)
1/48*a/d*(48*A*sin(d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c)))^(1/2)-72*A*cos( d*x+c)^3*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(1+cos(d*x+c))/(-1/(1+cos(d *x+c)))^(1/2))-72*A*cos(d*x+c)^3*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+c os(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+84*B*cos(d*x+c)^2*sin(d*x+c)*(-1/(1+ cos(d*x+c)))^(1/2)-42*B*cos(d*x+c)^3*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1) /(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-42*B*cos(d*x+c)^3*arctan(1/2*(c os(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+66*C*sin (d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c)))^(1/2)-33*C*cos(d*x+c)^3*arctan(1/ 2*(-cos(d*x+c)+sin(d*x+c)-1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-33* C*cos(d*x+c)^3*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+ cos(d*x+c)))^(1/2))+24*B*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+4 4*C*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+16*C*sin(d*x+c)*(-1/(1 +cos(d*x+c)))^(1/2))*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d* x+c)))^(1/2)/cos(d*x+c)^(5/2)
Time = 0.42 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.32 \[ \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 {\cos (c+d x)}} \, dx=\left [\frac {4 \, {\left (3 \, {\left (8 \, A + 14 \, B + 11 \, C\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 11 \, C\right )} a \cos \left (d x + c\right ) + 8 \, C a\right )} \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 ) + 3 \, {\left ({\left (24 \, A + 14 \, B + 11 \, C\right )} a \cos \left (d x + c\right )^{4} + {\left (24 \, A + 14 \, B + 11 \, C\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, {\left (3 \, {\left (8 \, A + 14 \, B + 11 \, C\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 11 \, C\right )} a \cos \left (d x + c\right ) + 8 \, C a\right )} \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 ) + 3 \, {\left ({\left (24 \, A + 14 \, B + 11 \, C\right )} a \cos \left (d x + c\right )^{4} + {\left (24 \, A + 14 \, B + 11 \, C\right )} a \cos \left (d x + c\right )^{3}\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 )}{48 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(1/2),x, algorithm="fricas")
[1/96*(4*(3*(8*A + 14*B + 11*C)*a*cos(d*x + c)^2 + 2*(6*B + 11*C)*a*cos(d* x + c) + 8*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c)) *sin(d*x + c) + 3*((24*A + 14*B + 11*C)*a*cos(d*x + c)^4 + (24*A + 14*B + 11*C)*a*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a* cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin( d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/( d*cos(d*x + c)^4 + d*cos(d*x + c)^3), 1/48*(2*(3*(8*A + 14*B + 11*C)*a*cos (d*x + c)^2 + 2*(6*B + 11*C)*a*cos(d*x + c) + 8*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((24*A + 14*B + 11* C)*a*cos(d*x + c)^4 + (24*A + 14*B + 11*C)*a*cos(d*x + c)^3)*sqrt(-a)*arct an(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*s in(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 5748 vs. \(2 (171) = 342\).
Time = 0.75 (sec) , antiderivative size = 5748, normalized size of antiderivative = 28.60 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display} \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(1/2),x, algorithm="maxima")
1/96*(24*(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*lo g(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*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*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))*sin(2*d*x + 2*c)^2 + 4*sqrt (2)*a*sin(3/2*d*x + 3/2*c) - 4*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 2*(2*sqrt( 2)*a*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 3*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) - 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*...
\[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(1/2),x, algorithm="giac")
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\cos \left (c+d\,x\right )}} \,d x \]
int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c os(c + d*x)^(1/2),x)