Integrand size = 35, antiderivative size = 125 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\frac {b (3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{8 d \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {b (3 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)} \]
1/4*A*b*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(9/2)+1/8*b*(3*A+4*C) *sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+1/8*b*(3*A+4*C)*arctan h(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.65 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\frac {b \sqrt {b \cos (c+d x)} \left ((3 A+4 C) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\left (2 A+(3 A+4 C) \cos ^2(c+d x)\right ) \sin (c+d x)\right )}{8 d \cos ^{\frac {9}{2}}(c+d x)} \]
(b*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 = 0.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.74, 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.
| \(\displaystyle \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+A\right ) \sec ^5(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3491 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \int \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b*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)) /Sqrt[Cos[c + d*x]]
3.2.6.3.1 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[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 = 8.69 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.45
| method | result | size |
| default | \(\frac {b \left (3 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-3 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+4 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-4 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+3 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+4 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 A \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(181\) |
| parts | \(-\frac {A \left (3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}\, b}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}}-\frac {C b \left (\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-\sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(193\) |
| risch | \(-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (3 A \,{\mathrm e}^{7 i \left (d x +c \right )}+4 C \,{\mathrm e}^{7 i \left (d x +c \right )}+11 A \,{\mathrm e}^{5 i \left (d x +c \right )}+4 C \,{\mathrm e}^{5 i \left (d x +c \right )}-11 A \,{\mathrm e}^{3 i \left (d x +c \right )}-4 C \,{\mathrm e}^{3 i \left (d x +c \right )}-3 A \,{\mathrm e}^{i \left (d x +c \right )}-4 C \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}-\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(225\) |
1/8*b/d*(3*A*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)+1)-3*A*cos(d*x+c)^4*ln (-cot(d*x+c)+csc(d*x+c)-1)+4*C*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)+1)-4 *C*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)-1)+3*A*sin(d*x+c)*cos(d*x+c)^2+4 *C*cos(d*x+c)^2*sin(d*x+c)+2*A*sin(d*x+c))*(cos(d*x+c)*b)^(1/2)/cos(d*x+c) ^(9/2)
Time = 0.31 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.08 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\left [\frac {{\left (3 \, A + 4 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left ({\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 2 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{5}}, -\frac {{\left (3 \, A + 4 \, C\right )} \sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left ({\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 2 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{5}}\right ] \]
[1/16*((3*A + 4*C)*b^(3/2)*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)*b*cos(d*x + c)^2 + 2*A*b)*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)*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)*b*cos(d*x + c)^2 + 2*A*b)*sq rt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)]
Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2434 vs. \(2 (107) = 214\).
Time = 0.52 (sec) , antiderivative size = 2434, normalized size of antiderivative = 19.47 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\text {Too large to display} \]
-1/16*((12*(b*sin(8*d*x + 8*c) + 4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x + 4* c) + 4*b*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 *c))) + 44*(b*sin(8*d*x + 8*c) + 4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x + 4* c) + 4*b*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 *c))) - 44*(b*sin(8*d*x + 8*c) + 4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x + 4* c) + 4*b*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 *c))) - 12*(b*sin(8*d*x + 8*c) + 4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x + 4* c) + 4*b*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 *c))) - 3*(b*cos(8*d*x + 8*c)^2 + 16*b*cos(6*d*x + 6*c)^2 + 36*b*cos(4*d*x + 4*c)^2 + 16*b*cos(2*d*x + 2*c)^2 + b*sin(8*d*x + 8*c)^2 + 16*b*sin(6*d* x + 6*c)^2 + 36*b*sin(4*d*x + 4*c)^2 + 48*b*sin(4*d*x + 4*c)*sin(2*d*x + 2 *c) + 16*b*sin(2*d*x + 2*c)^2 + 2*(4*b*cos(6*d*x + 6*c) + 6*b*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*cos(8*d*x + 8*c) + 8*(6*b*cos(4*d*x + 4*c ) + 4*b*cos(2*d*x + 2*c) + b)*cos(6*d*x + 6*c) + 12*(4*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 8*b*cos(2*d*x + 2*c) + 4*(2*b*sin(6*d*x + 6*c) + 3 *b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*b*sin (4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + b)*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*(b*cos(8*d*x + 8*c)^2 + 16*b*cos(6*d*x + 6*c)^2 + 3...
\[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {13}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^{13/2}} \,d x \]