Integrand size = 45, antiderivative size = 203 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (8 A+12 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a^2 (8 A-4 B-5 C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a (4 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d \sqrt {\cos (c+d x)}}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt {\cos (c+d x)}} \]
1/2*C*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+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))*cos(d*x+c)^ (1/2)*sec(d*x+c)^(1/2)/d+1/4*a^2*(8*A-4*B-5*C)*sin(d*x+c)/d/cos(d*x+c)^(1/ 2)/(a+a*sec(d*x+c))^(1/2)+1/4*a*(4*B+3*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2 )/d/cos(d*x+c)^(1/2)
Time = 10.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \left (\sqrt {2} (8 A+12 B+7 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2(c+d x)+2 ((4 B+7 C) \cos (c+d x)+2 (2 A+C+2 A \cos (2 (c+d x)))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d \cos ^{\frac {3}{2}}(c+d x)} \]
Integrate[Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x ] + C*Sec[c + d*x]^2),x]
(a*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*(8*A + 12*B + 7*C) *ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^2 + 2*((4*B + 7*C)*Cos[c + d*x] + 2*(2*A + C + 2*A*Cos[2*(c + d*x)]))*Sin[(c + d*x)/2]))/(8*d*Cos[c + d*x]^(3/2))
Time = 1.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.04, 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, 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.
| \(\displaystyle \int \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{\sqrt {\sec (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{3/2} (a (4 A-C)+a (4 B+3 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{2 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{3/2} (a (4 A-C)+a (4 B+3 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (4 A-C)+a (4 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (a^2 (8 A+12 B+7 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (a^2 (8 A+12 B+7 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a^2 (8 A+12 B+7 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {2 a^{5/2} (8 A+12 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*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]]*S qrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + ((2*a^(5/2)*(8*A + 12*B + 7*C)*A rcSinh[(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))
3.13.58.3.1 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]
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. \(481\) vs. \(2(173)=346\).
Time = 0.98 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.37
| method | result | size |
| default | \(\frac {a \left (16 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+8 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-8 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+12 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-12 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+8 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+7 C \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-7 C \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+14 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+4 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) | \(482\) |
int(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x,method=_RETURNVERBOSE)
1/8*a/d*(16*A*sin(d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c)))^(1/2)+8*A*cos(d* x+c)^2*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))-8*A*cos(d*x+c)^2*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))+12*B*cos(d*x+c)^2*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))-12*B*cos(d*x+c)^2 *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))+8*B*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+7*C*cos(d*x+c)^2 *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))-7*C*cos(d*x+c)^2*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))+14*C*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c))) ^(1/2)+4*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)^(3/2)
Time = 0.43 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.19 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {4 \, {\left (8 \, A a \cos \left (d x + c\right )^{2} + {\left (4 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left (8 \, A a \cos \left (d x + c\right )^{2} + {\left (4 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
integrate(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="fricas")
[1/16*(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))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 12*B + 7*C)*a*cos(d*x + c)^3 + (8*A + 12*B + 7*C)*a*cos(d*x + c)^2)*sqr t(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)^3 + d*cos(d* x + c)^2), 1/8*(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))*sqrt(cos(d*x + c))*sin(d*x + c ) + ((8*A + 12*B + 7*C)*a*cos(d*x + c)^3 + (8*A + 12*B + 7*C)*a*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s qrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a))) /(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]
Timed out. \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3661 vs. \(2 (173) = 346\).
Time = 0.60 (sec) , antiderivative size = 3661, normalized size of antiderivative = 18.03 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^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/...
\[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cos \left (d x + c\right )} \,d x } \]
integrate(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="giac")
Timed out. \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]