Integrand size = 43, antiderivative size = 193 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {(3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{8 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(3 A+4 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sin ^3(c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \] Output:
1/8*(3*A+4*C)*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+ 1/4*A*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)+1/8*(3*A+4*C)*sin (d*x+c)/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+B*sin(d*x+c)/d/cos(d*x+c)^ (1/2)/(b*cos(d*x+c))^(1/2)+1/3*B*sin(d*x+c)^3/d/cos(d*x+c)^(5/2)/(b*cos(d* x+c))^(1/2)
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.57 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {3 (3 A+4 C) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\sin (c+d x) \left (6 A+3 (3 A+4 C) \cos ^2(c+d x)+24 B \cos ^3(c+d x)+8 B \cos (c+d x) \sin ^2(c+d x)\right )}{24 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \] Input:
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt [b*Cos[c + d*x]]),x]
Output:
(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]^(7/2)*Sqrt[b*Cos[c + d*x]])
Time = 0.66 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.62, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 2032 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^5(c+d x)dx}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \int (4 B+(3 A+4 C) \cos (c+d x)) \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \int \frac {4 B+(3 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \sec ^3(c+d x)dx+4 B \int \sec ^4(c+d x)dx\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 B \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
Input:
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt[b*Cos [c + d*x]]),x]
Output:
(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))/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 = 0.75 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04
| method | result | size |
| default | \(\frac {-9 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}-12 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}+9 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}+12 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}+\left (9 \cos \left (d x +c \right )^{2}+6\right ) \sin \left (d x +c \right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (16 \cos \left (d x +c \right )^{2}+8\right ) B +12 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{24 d \cos \left (d x +c \right )^{\frac {7}{2}} \sqrt {b \cos \left (d x +c \right )}}\) | \(200\) |
| parts | \(-\frac {A \left (3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}-3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}-3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{\frac {7}{2}} \sqrt {b \cos \left (d x +c \right )}}+\frac {B \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}+1\right )}{3 d \cos \left (d x +c \right )^{\frac {5}{2}} \sqrt {b \cos \left (d x +c \right )}}+\frac {C \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{\frac {3}{2}} \sqrt {b \cos \left (d x +c \right )}}\) | \(231\) |
| risch | \(-\frac {i \left (9 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 C \,{\mathrm e}^{6 i \left (d x +c \right )}+33 A \,{\mathrm e}^{4 i \left (d x +c \right )}+12 C \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,{\mathrm e}^{3 i \left (d x +c \right )}-33 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A -12 C -80 B \cos \left (d x +c \right )-48 i B \sin \left (d x +c \right )\right )}{24 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}-\frac {\sqrt {\cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {b \cos \left (d x +c \right )}\, d}+\frac {\sqrt {\cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {b \cos \left (d x +c \right )}\, d}\) | \(235\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2), x,method=_RETURNVERBOSE)
Output:
1/24/d*(-9*A*ln(-cot(d*x+c)+csc(d*x+c)-1)*cos(d*x+c)^4-12*C*ln(-cot(d*x+c) +csc(d*x+c)-1)*cos(d*x+c)^4+9*A*ln(-cot(d*x+c)+csc(d*x+c)+1)*cos(d*x+c)^4+ 12*C*ln(-cot(d*x+c)+csc(d*x+c)+1)*cos(d*x+c)^4+(9*cos(d*x+c)^2+6)*sin(d*x+ c)*A+sin(d*x+c)*cos(d*x+c)*(16*cos(d*x+c)^2+8)*B+12*C*cos(d*x+c)^2*sin(d*x +c))/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)
Time = 0.13 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.58 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\left [\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, b d \cos \left (d x + c\right )^{5}}, -\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, b d \cos \left (d x + c\right )^{5}}\right ] \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^ (1/2),x, algorithm="fricas")
Output:
[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*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*d*cos(d*x + c)^5)]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2)/(b*cos(d*x+c) )**(1/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 2611 vs. \(2 (165) = 330\).
Time = 0.35 (sec) , antiderivative size = 2611, normalized size of antiderivative = 13.53 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^ (1/2),x, algorithm="maxima")
Output:
-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*...
Exception generated. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^ (1/2),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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(9/2)*(b*cos(c + d*x))^(1/2)),x)
Output:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(9/2)*(b*cos(c + d*x))^(1/2)), x)
Time = 0.17 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.81 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {b}\, \left (-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c +18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c -18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -9 \sin \left (d x +c \right )^{3} a -12 \sin \left (d x +c \right )^{3} c +15 \sin \left (d x +c \right ) a +12 \sin \left (d x +c \right ) c \right )}{24 b d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2), x)
Output:
(sqrt(b)*( - 16*cos(c + d*x)*sin(c + d*x)**3*b + 24*cos(c + d*x)*sin(c + d *x)*b - 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 12*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**4*c + 18*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 2*a + 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 9*log(tan((c + d*x) /2) - 1)*a - 12*log(tan((c + d*x)/2) - 1)*c + 9*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**4*a + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 18*lo g(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 24*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**2*c + 9*log(tan((c + d*x)/2) + 1)*a + 12*log(tan((c + d*x)/2) + 1)*c - 9*sin(c + d*x)**3*a - 12*sin(c + d*x)**3*c + 15*sin(c + d*x)*a + 12*sin(c + d*x)*c))/(24*b*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))