Integrand size = 33, antiderivative size = 145 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {A \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)} \]
1/2*B*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+A*sin(d*x+c)*(b*c os(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)+1/3*A*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2 )/d/cos(d*x+c)^(7/2)+1/2*B*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos( d*x+c)^(1/2)
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {b \cos (c+d x)} \left (3 B \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+3 B \sin (c+d x)+2 A (2+\cos (2 (c+d x))) \tan (c+d x)\right )}{6 d \cos ^{\frac {5}{2}}(c+d x)} \]
(Sqrt[b*Cos[c + d*x]]*(3*B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + 3*B*Sin[ c + d*x] + 2*A*(2 + Cos[2*(c + d*x)])*Tan[c + d*x]))/(6*d*Cos[c + d*x]^(5/ 2))
Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.60, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2031, 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.
| \(\displaystyle \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int (A+B \cos (c+d x)) \sec ^4(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (A \int \sec ^4(c+d x)dx+B \int \sec ^3(c+d x)dx\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {A \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (B \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (B \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (B \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
(Sqrt[b*Cos[c + d*x]]*(B*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[ c + d*x])/(2*d)) - (A*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d))/Sqrt[Cos[c + d*x]]
3.9.49.3.1 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_)*((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[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 = 4.93 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.84
| method | result | size |
| default | \(\frac {\left (-3 B \left (\cos ^{3}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+3 B \left (\cos ^{3}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+4 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 A \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{6 d \cos \left (d x +c \right )^{\frac {7}{2}}}\) | \(122\) |
| parts | \(\frac {A \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{3 d \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {B \left (-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(128\) |
| risch | \(-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-4 A \right )}{3 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}-\frac {\sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(153\) |
1/6/d*(-3*B*cos(d*x+c)^3*ln(-cot(d*x+c)+csc(d*x+c)-1)+3*B*cos(d*x+c)^3*ln( -cot(d*x+c)+csc(d*x+c)+1)+4*A*sin(d*x+c)*cos(d*x+c)^2+3*B*sin(d*x+c)*cos(d *x+c)+2*A*sin(d*x+c))*(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(7/2)
Time = 0.33 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.74 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\left [\frac {3 \, B \sqrt {b} \cos \left (d x + c\right )^{4} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (4 \, A \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{4}}, -\frac {3 \, B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{4} - {\left (4 \, A \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{4}}\right ] \]
[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*(4*A*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))/(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*sqrt(cos(d*x + c ))))*cos(d*x + c)^4 - (4*A*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))/(d*cos(d*x + c)^4)]
Timed out. \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 957 vs. \(2 (123) = 246\).
Time = 0.72 (sec) , antiderivative size = 957, normalized size of antiderivative = 6.60 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/12*(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*sqrt(b)/(2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(s in(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d* x + 2*c)^2 + 6*cos(2*d*x + 2*c) + 1) - 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 *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) + (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*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...
\[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \]