Integrand size = 43, antiderivative size = 208 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {(3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{8 b d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{4 b d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(3 A+4 C) \sin (c+d x)}{8 b d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sin ^3(c+d x)}{3 b d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]
1/4*A*sin(d*x+c)/b/d/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)+1/8*(3*A+4*C)*s in(d*x+c)/b/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+1/3*B*sin(d*x+c)^3/b/d /cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2)+B*sin(d*x+c)/b/d/cos(d*x+c)^(1/2)/( b*cos(d*x+c))^(1/2)+1/8*(3*A+4*C)*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/b/d /(b*cos(d*x+c))^(1/2)
Time = 0.17 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.53 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {3 (3 A+4 C) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\sin (c+d x) \left (6 A+3 (3 A+4 C) \cos ^2(c+d x)+24 B \cos ^3(c+d x)+8 B \cos (c+d x) \sin ^2(c+d x)\right )}{24 d \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(b*C os[c + d*x])^(3/2)),x]
(3*(3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + Sin[c + d*x]*(6*A + 3*(3*A + 4*C)*Cos[c + d*x]^2 + 24*B*Cos[c + d*x]^3 + 8*B*Cos[c + d*x]*Sin[ c + d*x]^2))/(24*d*Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(3/2))
Time = 0.65 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.59, 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, 2009, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2032 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^5(c+d x)dx}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \int (4 B+(3 A+4 C) \cos (c+d x)) \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \int \frac {4 B+(3 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \sec ^3(c+d x)dx+4 B \int \sec ^4(c+d x)dx\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 B \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{b \sqrt {b \cos (c+d x)}}\) |
(Sqrt[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*B*( -Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/4))/(b*Sqrt[b*Cos[c + d*x]])
3.4.30.3.1 Defintions of rubi rules used
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 = 9.45 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.03
| method | result | size |
| default | \(-\frac {9 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-9 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+12 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-12 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-16 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-9 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-12 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 B \sin \left (d x +c \right ) \cos \left (d x +c \right )-6 A \sin \left (d x +c \right )}{24 b d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {7}{2}}}\) | \(215\) |
| parts | \(\frac {A \left (-3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 \sin \left (d x +c \right )\right )}{8 d b \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {B \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \sin \left (d x +c \right )}{3 d b \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \left (-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )\right )}{2 d b \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) | \(240\) |
| risch | \(-\frac {i \left (9 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 C \,{\mathrm e}^{6 i \left (d x +c \right )}+33 A \,{\mathrm e}^{4 i \left (d x +c \right )}+12 C \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,{\mathrm e}^{3 i \left (d x +c \right )}-33 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A -12 C -80 B \cos \left (d x +c \right )-48 i B \sin \left (d x +c \right )\right )}{24 b \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 b \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 b \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(244\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(cos(d*x+c)*b)^(3/2), x,method=_RETURNVERBOSE)
-1/24/b/d*(9*A*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)-1)-9*A*cos(d*x+c)^4* ln(-cot(d*x+c)+csc(d*x+c)+1)+12*C*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)-1 )-12*C*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)+1)-16*B*sin(d*x+c)*cos(d*x+c )^3-9*A*sin(d*x+c)*cos(d*x+c)^2-12*C*cos(d*x+c)^2*sin(d*x+c)-8*B*sin(d*x+c )*cos(d*x+c)-6*A*sin(d*x+c))/(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(7/2)
Time = 0.31 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.47 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\left [\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, b^{2} d \cos \left (d x + c\right )^{5}}, -\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, b^{2} d \cos \left (d x + c\right )^{5}}\right ] \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^ (3/2),x, algorithm="fricas")
[1/48*(3*(3*A + 4*C)*sqrt(b)*cos(d*x + c)^5*log(-(b*cos(d*x + c)^3 - 2*sqr t(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*(16*B*cos(d*x + c)^3 + 3*(3*A + 4*C)*cos(d*x + c)^ 2 + 8*B*cos(d*x + c) + 6*A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d* x + c))/(b^2*d*cos(d*x + c)^5), -1/24*(3*(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 - (16*B*cos(d*x + c)^3 + 3*(3*A + 4*C)*cos(d*x + c)^2 + 8*B*cos(d*x + c ) + 6*A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2*d*cos( d*x + c)^5)]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2660 vs. \(2 (180) = 360\).
Time = 0.57 (sec) , antiderivative size = 2660, normalized size of antiderivative = 12.79 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^ (3/2),x, algorithm="maxima")
-1/48*(3*(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*(si n(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) + 16*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* 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) + 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*...
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^ (3/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c))^(3/2)* cos(d*x + c)^(7/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]