Integrand size = 35, antiderivative size = 122 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {(3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{8 d \sqrt {\cos (c+d x)}}+\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {(3 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:
1/8*(3*A+4*C)*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+ 1/4*A*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(9/2)+1/8*(3*A+4*C)*(b* cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {b \cos (c+d x)} \left ((3 A+4 C) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\left (2 A+(3 A+4 C) \cos ^2(c+d x)\right ) \sin (c+d x)\right )}{8 d \cos ^{\frac {9}{2}}(c+d x)} \] Input:
Integrate[(Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2 ),x]
Output:
(Sqrt[b*Cos[c + d*x]]*((3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (2*A + (3*A + 4*C)*Cos[c + d*x]^2)*Sin[c + d*x]))/(8*d*Cos[c + d*x]^(9/2))
Time = 0.42 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2031, 3042, 3491, 3042, 4255, 3042, 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.
| \(\displaystyle \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+A\right ) \sec ^5(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3491 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \int \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[(Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]
Output:
(Sqrt[b*Cos[c + d*x]]*((A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((3*A + 4*C )*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/S qrt[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_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x _)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.70 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.43
| method | result | size |
| default | \(-\frac {\left (3 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}+4 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}-3 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}-4 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}+\left (-3 \cos \left (d x +c \right )^{2}-2\right ) \sin \left (d x +c \right ) A -4 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(174\) |
| parts | \(\frac {C \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}-\frac {A \left (3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}-3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}-3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(189\) |
| risch | \(-\frac {i \sqrt {b \cos \left (d x +c \right )}\, \left (3 A \,{\mathrm e}^{7 i \left (d x +c \right )}+4 C \,{\mathrm e}^{7 i \left (d x +c \right )}+11 A \,{\mathrm e}^{5 i \left (d x +c \right )}+4 C \,{\mathrm e}^{5 i \left (d x +c \right )}-11 A \,{\mathrm e}^{3 i \left (d x +c \right )}-4 C \,{\mathrm e}^{3 i \left (d x +c \right )}-3 A \,{\mathrm e}^{i \left (d x +c \right )}-4 C \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\sqrt {b \cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}-\frac {\sqrt {b \cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(222\) |
Input:
int((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x,method=_RE TURNVERBOSE)
Output:
-1/8/d*(3*A*ln(-cot(d*x+c)+csc(d*x+c)-1)*cos(d*x+c)^4+4*C*ln(-cot(d*x+c)+c sc(d*x+c)-1)*cos(d*x+c)^4-3*A*ln(-cot(d*x+c)+csc(d*x+c)+1)*cos(d*x+c)^4-4* C*ln(-cot(d*x+c)+csc(d*x+c)+1)*cos(d*x+c)^4+(-3*cos(d*x+c)^2-2)*sin(d*x+c) *A-4*C*cos(d*x+c)^2*sin(d*x+c))*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2)
Time = 0.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.09 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\left [\frac {{\left (3 \, A + 4 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{5}}, -\frac {{\left (3 \, A + 4 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{5}}\right ] \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, alg orithm="fricas")
Output:
[1/16*((3*A + 4*C)*sqrt(b)*cos(d*x + c)^5*log(-(b*cos(d*x + c)^3 - 2*sqrt( b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c) )/cos(d*x + c)^3) + 2*((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5), -1/8*((3*A + 4*C) *sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d* x + c))))*cos(d*x + c)^5 - ((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*sqrt(b*cos(d *x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)]
Timed out. \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(1/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(11/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 2318 vs. \(2 (104) = 208\).
Time = 0.39 (sec) , antiderivative size = 2318, normalized size of antiderivative = 19.00 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, alg orithm="maxima")
Output:
-1/16*((12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4 *sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d *x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(sin( 8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c ))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(1 /2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d *x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c ) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 1 6*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d *x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*log(cos(1/2*ar ctan2(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) + 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2* d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x ...
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (104) = 208\).
Time = 0.27 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.76 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {{\left ({\left (3 \, A + 4 \, C\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - {\left (3 \, A + 4 \, C\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )} \sqrt {b}}{8 \, d} \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, alg orithm="giac")
Output:
1/8*((3*A + 4*C)*log(tan(1/2*d*x + 1/2*c) + 1) - (3*A + 4*C)*log(tan(1/2*d *x + 1/2*c) - 1) + 2*(5*A*tan(1/2*d*x + 1/2*c)^7 + 4*C*tan(1/2*d*x + 1/2*c )^7 + 3*A*tan(1/2*d*x + 1/2*c)^5 - 4*C*tan(1/2*d*x + 1/2*c)^5 + 3*A*tan(1/ 2*d*x + 1/2*c)^3 - 4*C*tan(1/2*d*x + 1/2*c)^3 + 5*A*tan(1/2*d*x + 1/2*c) + 4*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^8 - 4*tan(1/2*d*x + 1/2*c )^6 + 6*tan(1/2*d*x + 1/2*c)^4 - 4*tan(1/2*d*x + 1/2*c)^2 + 1))*sqrt(b)/d
Timed out. \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{11/2}} \,d x \] Input:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2))/cos(c + d*x)^(11/2),x)
Output:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2))/cos(c + d*x)^(11/2), x )
Time = 0.16 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.57 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, \left (-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -3 \sin \left (d x +c \right )^{3} a -4 \sin \left (d x +c \right )^{3} c +5 \sin \left (d x +c \right ) a +4 \sin \left (d x +c \right ) c \right )}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x)
Output:
(sqrt(b)*( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x )**2*a + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 3*log(tan((c + d* x)/2) - 1)*a - 4*log(tan((c + d*x)/2) - 1)*c + 3*log(tan((c + d*x)/2) + 1) *sin(c + d*x)**4*a + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 6*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 8*log(tan((c + d*x)/2) + 1)*sin (c + d*x)**2*c + 3*log(tan((c + d*x)/2) + 1)*a + 4*log(tan((c + d*x)/2) + 1)*c - 3*sin(c + d*x)**3*a - 4*sin(c + d*x)**3*c + 5*sin(c + d*x)*a + 4*si n(c + d*x)*c))/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))