Integrand size = 43, antiderivative size = 114 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {b (A+2 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
1/2*b*(A+2*C)*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+ 1/2*A*b*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+b*B*(b*cos(d*x+ c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.73 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{3/2} \left (2 C \coth ^{-1}(\sin (c+d x)) \cos ^2(c+d x)+A \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+(A+2 B \cos (c+d x)) \sin (c+d x)\right )}{2 d \cos ^{\frac {7}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(9/2),x]
Output:
((b*Cos[c + d*x])^(3/2)*(2*C*ArcCoth[Sin[c + d*x]]*Cos[c + d*x]^2 + A*ArcT anh[Sin[c + d*x]]*Cos[c + d*x]^2 + (A + 2*B*Cos[c + d*x])*Sin[c + d*x]))/( 2*d*Cos[c + d*x]^(7/2))
Time = 0.50 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {2031, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4257}
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.
| \(\displaystyle \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^3(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \int (2 B+(A+2 C) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \int \frac {2 B+(A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \sec (c+d x)dx+2 B \int \sec ^2(c+d x)dx\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 B \int 1d(-\tan (c+d x))}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 B \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \left (\frac {(A+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 B \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]
Output:
(b*Sqrt[b*Cos[c + d*x]]*((A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((A + 2*C) *ArcTanh[Sin[c + d*x]])/d + (2*B*Tan[c + d*x])/d)/2))/Sqrt[Cos[c + d*x]]
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
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_.) + (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]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.43 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(\frac {b \left (-A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-4 C \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}+2 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+A \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(131\) |
| parts | \(\frac {A \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}\, b}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {b B \sqrt {b \cos \left (d x +c \right )}\, \sin \left (d x +c \right )}{d \cos \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 C \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}\, b}{d \sqrt {\cos \left (d x +c \right )}}\) | \(158\) |
| risch | \(-\frac {i b \sqrt {b \cos \left (d x +c \right )}\, \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-2 B \right )}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {b \sqrt {b \cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b \sqrt {b \cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(163\) |
Input:
int((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2), x,method=_RETURNVERBOSE)
Output:
1/2*b/d*(-A*ln(-cot(d*x+c)+csc(d*x+c)-1)*cos(d*x+c)^2+A*ln(-cot(d*x+c)+csc (d*x+c)+1)*cos(d*x+c)^2-4*C*arctanh(-csc(d*x+c)+cot(d*x+c))*cos(d*x+c)^2+2 *B*sin(d*x+c)*cos(d*x+c)+A*sin(d*x+c))*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/ 2)
Time = 0.12 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.11 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\left [\frac {{\left (A + 2 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (2 \, B b \cos \left (d x + c\right ) + A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{3}}, -\frac {{\left (A + 2 \, C\right )} \sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - {\left (2 \, B b \cos \left (d x + c\right ) + A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{3}}\right ] \] Input:
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (9/2),x, algorithm="fricas")
Output:
[1/4*((A + 2*C)*b^(3/2)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*c os(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/c os(d*x + c)^3) + 2*(2*B*b*cos(d*x + c) + A*b)*sqrt(b*cos(d*x + c))*sqrt(co s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3), -1/2*((A + 2*C)*sqrt(-b)*b*a rctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*c os(d*x + c)^3 - (2*B*b*cos(d*x + c) + A*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d *x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)]
Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c )**(9/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 813 vs. \(2 (98) = 196\).
Time = 0.35 (sec) , antiderivative size = 813, normalized size of antiderivative = 7.13 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (9/2),x, algorithm="maxima")
Output:
1/4*(2*(b*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b*lo g(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1))*C*sqrt(b) + 8*B*b ^(3/2)*sin(2*d*x + 2*c)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2 *d*x + 2*c) + 1) - (4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*cos(3/2* arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(b*sin(4*d*x + 4*c) + 2*b *sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*sin(4*d*x + 4*c)^2 + 4* b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2*b*cos( 2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*log(cos(1/2 *arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d* x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c))) + 1) + (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*si n(4*d*x + 4*c)^2 + 4*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin (1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/2*arctan2(si n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(b*cos(4*d*x + 4*c) + 2*b*cos( 2*d*x + 2*c) + b)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4 *(b*cos(4*d*x + 4*c) + 2*b*cos(2*d*x + 2*c) + b)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A*sqrt(b)/(2*(2*cos(2*d*x + 2*c) + 1)*cos(...
Time = 0.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.11 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {{\left ({\left (A + 2 \, C\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - {\left (A + 2 \, C\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )} b^{\frac {3}{2}}}{2 \, d} \] Input:
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (9/2),x, algorithm="giac")
Output:
1/2*((A + 2*C)*log(tan(1/2*d*x + 1/2*c) + 1) - (A + 2*C)*log(tan(1/2*d*x + 1/2*c) - 1) + 2*(A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + A*tan(1/2*d*x + 1/2*c) + 2*B*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 2*tan(1/2*d*x + 1/2*c)^2 + 1))*b^(3/2)/d
Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \] Input:
int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(9/2),x)
Output:
int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(9/2), x)
Time = 0.18 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.71 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b \left (-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\sin \left (d x +c \right ) a \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2), x)
Output:
(sqrt(b)*b*( - 2*cos(c + d*x)*sin(c + d*x)*b - log(tan((c + d*x)/2) - 1)*s in(c + d*x)**2*a - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + log(tan ((c + d*x)/2) - 1)*a + 2*log(tan((c + d*x)/2) - 1)*c + log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**2*a + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - log(tan((c + d*x)/2) + 1)*a - 2*log(tan((c + d*x)/2) + 1)*c - sin(c + d*x )*a))/(2*d*(sin(c + d*x)**2 - 1))