Integrand size = 43, antiderivative size = 164 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 b d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(2 A+3 C) \sin (c+d x)}{3 b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \] Output:
1/2*B*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/b/d/(b*cos(d*x+c))^(1/2)+1/3*A* sin(d*x+c)/b/d/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2)+1/2*B*sin(d*x+c)/b/d/ cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+1/3*(2*A+3*C)*sin(d*x+c)/b/d/cos(d*x +c)^(1/2)/(b*cos(d*x+c))^(1/2)
Time = 0.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {3 B \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+(4 A+3 C+3 B \cos (c+d x)+(2 A+3 C) \cos (2 (c+d x))) \tan (c+d x)}{6 d \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \] Input:
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(b*C os[c + d*x])^(3/2)),x]
Output:
(3*B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + (4*A + 3*C + 3*B*Cos[c + d*x] + (2*A + 3*C)*Cos[2*(c + d*x)])*Tan[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(b*C os[c + d*x])^(3/2))
Time = 0.63 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.66, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {2032, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2032 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^4(c+d x)dx}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \int (3 B+(2 A+3 C) \cos (c+d x)) \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \int \frac {3 B+(2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left ((2 A+3 C) \int \sec ^2(c+d x)dx+3 B \int \sec ^3(c+d x)dx\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left ((2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {(2 A+3 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {(2 A+3 C) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(2 A+3 C) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(2 A+3 C) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (\frac {(2 A+3 C) \tan (c+d x)}{d}+3 B \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
Input:
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(3/2)),x]
Output:
(Sqrt[Cos[c + d*x]]*((A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (((2*A + 3*C) *Tan[c + d*x])/d + 3*B*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/3))/(b*Sqrt[b*Cos[c + d*x]])
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m - 1/ 2)*b^(n + 1/2)*(Sqrt[a*v]/Sqrt[b*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && ILtQ[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_)]*(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 = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {-3 B \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{3}+3 B \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{3}+\left (4 \cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right ) A +6 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+3 B \sin \left (d x +c \right ) \cos \left (d x +c \right )}{6 b d \cos \left (d x +c \right )^{\frac {5}{2}} \sqrt {b \cos \left (d x +c \right )}}\) | \(136\) |
parts | \(\frac {A \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}+1\right )}{3 d \cos \left (d x +c \right )^{\frac {5}{2}} b \sqrt {b \cos \left (d x +c \right )}}+\frac {B \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{\frac {3}{2}} b \sqrt {b \cos \left (d x +c \right )}}+\frac {C \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}\, b \sqrt {b \cos \left (d x +c \right )}}\) | \(166\) |
risch | \(-\frac {i \left (3 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{3 i \left (d x +c \right )}-3 B +\left (-16 A -18 C \right ) \cos \left (d x +c \right )+i \left (-8 A -6 C \right ) \sin \left (d x +c \right )\right )}{6 b \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}+\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b \sqrt {b \cos \left (d x +c \right )}\, d}-\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b \sqrt {b \cos \left (d x +c \right )}\, d}\) | \(180\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2), x,method=_RETURNVERBOSE)
Output:
1/6/b/d*(-3*B*ln(-cot(d*x+c)+csc(d*x+c)-1)*cos(d*x+c)^3+3*B*ln(-cot(d*x+c) +csc(d*x+c)+1)*cos(d*x+c)^3+(4*cos(d*x+c)^2+2)*sin(d*x+c)*A+6*C*cos(d*x+c) ^2*sin(d*x+c)+3*B*sin(d*x+c)*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^( 1/2)
Time = 0.12 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\left [\frac {3 \, B \sqrt {b} \cos \left (d x + c\right )^{4} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b^{2} d \cos \left (d x + c\right )^{4}}, -\frac {3 \, B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{4} - {\left (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b^{2} d \cos \left (d x + c\right )^{4}}\right ] \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^ (3/2),x, algorithm="fricas")
Output:
[1/12*(3*B*sqrt(b)*cos(d*x + c)^4*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*(2*(2*A + 3*C)*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*A)*sqrt (b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c)^4), -1/6*(3*B*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sq rt(cos(d*x + c))))*cos(d*x + c)^4 - (2*(2*A + 3*C)*cos(d*x + c)^2 + 3*B*co s(d*x + c) + 2*A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b ^2*d*cos(d*x + c)^4)]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(b*cos(d*x+c) )**(3/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1048 vs. \(2 (140) = 280\).
Time = 0.39 (sec) , antiderivative size = 1048, normalized size of antiderivative = 6.39 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^ (3/2),x, algorithm="maxima")
Output:
1/12*(24*C*sqrt(b)*sin(2*d*x + 2*c)/(b^2*cos(2*d*x + 2*c)^2 + b^2*sin(2*d* x + 2*c)^2 + 2*b^2*cos(2*d*x + 2*c) + b^2) + 16*((3*cos(2*d*x + 2*c) + 1)* sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) - 3*cos(6*d *x + 6*c)*sin(2*d*x + 2*c) - 9*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*A/((b*co s(6*d*x + 6*c)^2 + 9*b*cos(4*d*x + 4*c)^2 + 9*b*cos(2*d*x + 2*c)^2 + b*sin (6*d*x + 6*c)^2 + 9*b*sin(4*d*x + 4*c)^2 + 18*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*b*sin(2*d*x + 2*c)^2 + 2*(3*b*cos(4*d*x + 4*c) + 3*b*cos(2*d*x + 2*c) + b)*cos(6*d*x + 6*c) + 6*(3*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4 *c) + 6*b*cos(2*d*x + 2*c) + 6*(b*sin(4*d*x + 4*c) + b*sin(2*d*x + 2*c))*s in(6*d*x + 6*c) + b)*sqrt(b)) - 3*(4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c ))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(4*d*x + 4 *c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2* c))) - (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan 2(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) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4 *c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin (2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(...
Exception generated. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^ (3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(3/2)),x)
Output:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(3/2)), x)
Time = 0.19 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +4 \sin \left (d x +c \right )^{3} a +6 \sin \left (d x +c \right )^{3} c -6 \sin \left (d x +c \right ) a -6 \sin \left (d x +c \right ) c \right )}{6 \cos \left (d x +c \right ) b^{2} d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2), x)
Output:
(sqrt(b)*( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 3*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*sin(c + d*x)**2*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 3*cos(c + d*x)*sin(c + d*x)*b + 4*sin(c + d*x)**3*a + 6*sin(c + d*x)**3 *c - 6*sin(c + d*x)*a - 6*sin(c + d*x)*c))/(6*cos(c + d*x)*b**2*d*(sin(c + d*x)**2 - 1))