Integrand size = 45, antiderivative size = 197 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (2 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {a^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \]
2/3*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)*cos(d*x+c)^(1/2)/d+a^(3/2)*(2*B+3* 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+1/3*a^2*(8*A+6*B-3*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*s ec(d*x+c))^(1/2)-1/3*a*(2*A-3*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d *x+c)^(1/2)
Time = 7.37 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.94 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sqrt {\cos (c+d x)} \sqrt {a (1+\sec (c+d x))} \left (6 B \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-9 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+\sqrt {1-\sec (c+d x)} (2 A \sin (c+d x)+(10 A+6 B+3 C \sec (c+d x)) \tan (c+d x))\right )}{3 d (1+\cos (c+d x)) \sqrt {\sec (c+d x)} \sqrt {(-1+\cos (c+d x)) \sec ^2(c+d x)}} \]
Integrate[Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x ] + C*Sec[c + d*x]^2),x]
(a*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])]*(6*B*ArcSin[Sqrt[1 - Sec[ c + d*x]]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] - 9*C*ArcSin[Sqrt[Sec[c + d*x]] ]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(2*A*Sin[c + d* x] + (10*A + 6*B + 3*C*Sec[c + d*x])*Tan[c + d*x])))/(3*d*(1 + Cos[c + d*x ])*Sqrt[Sec[c + d*x]]*Sqrt[(-1 + Cos[c + d*x])*Sec[c + d*x]^2])
Time = 1.30 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.07, 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, 4574, 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.
| \(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^{3/2} (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{3/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\sec (c+d x) a+a)^{3/2} (3 a (A+B)-a (2 A-3 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\sec (c+d x) a+a)^{3/2} (3 a (A+B)-a (2 A-3 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a (A+B)-a (2 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^2 (2 B+3 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{2} \left (\frac {6 a^{5/2} (2 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Sec[c + d*x])^(3/2)*Sin [c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (-((a^2*(2*A - 3*C)*Sqrt[Sec[c + d*x ]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d) + ((6*a^(5/2)*(2*B + 3*C)*Arc Sinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(8*A + 6 *B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2 )/(3*a))
3.13.57.3.1 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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[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. \(367\) vs. \(2(169)=338\).
Time = 1.08 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.87
| method | result | size |
| default | \(\frac {a \left (4 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+20 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+12 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+6 B \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )-6 B \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )+9 C \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )-9 C \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+6 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{6 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\cos \left (d x +c \right )}}\) | \(368\) |
int(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x,method=_RETURNVERBOSE)
1/6*a/d*(4*A*sin(d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c)))^(1/2)+20*A*cos(d* x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+12*B*cos(d*x+c)*sin(d*x+c)*(-1/( 1+cos(d*x+c)))^(1/2)+6*B*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))*cos(d*x+c)-6*B*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))*cos(d*x+c)+9*C*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))*cos( d*x+c)-9*C*cos(d*x+c)*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))+6*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)^(1 /2)
Time = 0.33 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.18 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {4 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]
integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="fricas")
[1/12*(4*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 3*C*a)*sqr t((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*( (2*B + 3*C)*a*cos(d*x + c)^2 + (2*B + 3*C)*a*cos(d*x + c))*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)/(c os(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), 1/6 *(2*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 3*C*a)*sqrt((a* cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((2*B + 3*C)*a*cos(d*x + c)^2 + (2*B + 3*C)*a*cos(d*x + c))*sqrt(-a)*arctan(2*sq rt(-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)))/(d*cos(d*x + c)^2 + d*cos (d*x + c))]
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1895 vs. \(2 (169) = 338\).
Time = 0.51 (sec) , antiderivative size = 1895, normalized size of antiderivative = 9.62 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="maxima")
1/60*(20*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c ))*A*sqrt(a) + 6*(2*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 40*sqrt(2)*a*sin(1/2* d*x + 1/2*c) - 2*sqrt(2)*a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d *x + 3/2*c))) - 20*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2 *d*x + 3/2*c))) + 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3 /2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3 /2*c))) + 2) - 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2 *c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* c))) + 2) + 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) )^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) )) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) ) + 2) - 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2 *c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))...
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(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. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]