Integrand size = 43, antiderivative size = 208 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\frac {b^2 (3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{8 d \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {b^2 (3 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)} \] Output:
1/8*b^2*(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^2*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(9/2)+1/8*b^2*( 3*A+4*C)*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+b^2*B*(b*cos(d *x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+1/3*b^2*B*(b*cos(d*x+c))^(1/2)* sin(d*x+c)^3/d/cos(d*x+c)^(7/2)
Time = 0.41 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.53 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} \left (3 (3 A+4 C) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\sin (c+d x) \left (6 A+3 (3 A+4 C) \cos ^2(c+d x)+24 B \cos ^3(c+d x)+8 B \cos (c+d x) \sin ^2(c+d x)\right )\right )}{24 d \cos ^{\frac {13}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(15/2),x]
Output:
((b*Cos[c + d*x])^(5/2)*(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]^(13/2))
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 = {2031, 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.
| \(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^5(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \int (4 B+(3 A+4 C) \cos (c+d x)) \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \int \frac {4 B+(3 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \sec ^3(c+d x)dx+4 B \int \sec ^4(c+d x)dx\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 B \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(15/2),x]
Output:
(b^2*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*B*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/4))/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_)]*(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 = 0.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(-\frac {b^{2} \left (9 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}+12 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}-9 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}-12 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}+\left (-9 \cos \left (d x +c \right )^{2}-6\right ) \sin \left (d x +c \right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-16 \cos \left (d x +c \right )^{2}-8\right ) B -12 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{24 d \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(203\) |
| parts | \(\frac {A \left (-3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}+3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+2 \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}\, b^{2}}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}}+\frac {B \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}+1\right ) \sqrt {b \cos \left (d x +c \right )}\, b^{2}}{3 d \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {C \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}\, b^{2}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(240\) |
| risch | \(-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (9 A \,{\mathrm e}^{7 i \left (d x +c \right )}+12 C \,{\mathrm e}^{7 i \left (d x +c \right )}+33 A \,{\mathrm e}^{5 i \left (d x +c \right )}+12 C \,{\mathrm e}^{5 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-33 A \,{\mathrm e}^{3 i \left (d x +c \right )}-12 C \,{\mathrm e}^{3 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A \,{\mathrm e}^{i \left (d x +c \right )}-12 C \,{\mathrm e}^{i \left (d x +c \right )}-16 B \right )}{12 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(258\) |
Input:
int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2) ,x,method=_RETURNVERBOSE)
Output:
-1/24*b^2/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*s in(d*x+c))*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2)
Time = 0.12 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.57 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\left [\frac {3 \, {\left (3 \, A + 4 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (16 \, B b^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 8 \, B b^{2} \cos \left (d x + c\right ) + 6 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{5}}, -\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left (16 \, B b^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 8 \, B b^{2} \cos \left (d x + c\right ) + 6 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )^{5}}\right ] \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (15/2),x, algorithm="fricas")
Output:
[1/48*(3*(3*A + 4*C)*b^(5/2)*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*b^2*cos(d*x + c)^3 + 3*(3*A + 4*C)*b^2*cos(d *x + c)^2 + 8*B*b^2*cos(d*x + c) + 6*A*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos( d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5), -1/24*(3*(3*A + 4*C)*sqrt(-b)* b^2*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*b^2*cos(d*x + c)^3 + 3*(3*A + 4*C)*b^2*cos(d*x + c)^2 + 8*B*b^2*cos(d*x + c) + 6*A*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c )**(15/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 2972 vs. \(2 (180) = 360\).
Time = 0.38 (sec) , antiderivative size = 2972, normalized size of antiderivative = 14.29 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (15/2),x, algorithm="maxima")
Output:
-1/48*(3*(12*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*sin(4* d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c))) + 44*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^ 2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2 *c), cos(2*d*x + 2*c))) - 44*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c ) + 6*b^2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2 *d*x + 2*c), cos(2*d*x + 2*c))) - 12*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d *x + 6*c) + 6*b^2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(1/2*arcta n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(b^2*cos(8*d*x + 8*c)^2 + 16*b ^2*cos(6*d*x + 6*c)^2 + 36*b^2*cos(4*d*x + 4*c)^2 + 16*b^2*cos(2*d*x + 2*c )^2 + b^2*sin(8*d*x + 8*c)^2 + 16*b^2*sin(6*d*x + 6*c)^2 + 36*b^2*sin(4*d* x + 4*c)^2 + 48*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*b^2*sin(2*d*x + 2*c)^2 + 8*b^2*cos(2*d*x + 2*c) + b^2 + 2*(4*b^2*cos(6*d*x + 6*c) + 6*b^2 *cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*cos(8*d*x + 8*c) + 8*(6* b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*cos(6*d*x + 6*c) + 12 *(4*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c) + 4*(2*b^2*sin(6*d*x + 6* c) + 3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 1 6*(3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*log( cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2...
Time = 0.53 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.53 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\frac {{\left (3 \, {\left (3 \, A b^{2} + 4 \, C b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 3 \, {\left (3 \, A b^{2} + 4 \, C b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (15 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )} \sqrt {b}}{24 \, d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (15/2),x, algorithm="giac")
Output:
1/24*(3*(3*A*b^2 + 4*C*b^2)*log(tan(1/2*d*x + 1/2*c) + 1) - 3*(3*A*b^2 + 4 *C*b^2)*log(tan(1/2*d*x + 1/2*c) - 1) + 2*(15*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*b^2*tan(1/2*d*x + 1/2*c)^7 + 12*C*b^2*tan(1/2*d*x + 1/2*c)^7 + 9*A *b^2*tan(1/2*d*x + 1/2*c)^5 + 40*B*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*C*b^2*t an(1/2*d*x + 1/2*c)^5 + 9*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 40*B*b^2*tan(1/2* d*x + 1/2*c)^3 - 12*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*A*b^2*tan(1/2*d*x + 1/2*c) + 24*B*b^2*tan(1/2*d*x + 1/2*c) + 12*C*b^2*tan(1/2*d*x + 1/2*c))/(t an(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. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{15/2}} \,d x \] Input:
int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(15/2),x)
Output:
int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(15/2), x)
Time = 0.16 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.68 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b^{2} \left (-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c +18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c -18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -9 \sin \left (d x +c \right )^{3} a -12 \sin \left (d x +c \right )^{3} c +15 \sin \left (d x +c \right ) a +12 \sin \left (d x +c \right ) c \right )}{24 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2) ,x)
Output:
(sqrt(b)*b**2*( - 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*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*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*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))