Integrand size = 23, antiderivative size = 74 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {b \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {b \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)} \]
1/2*b*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+1/2*b*arctanh(sin (d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.70 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{3/2} \left (\text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+\sin (c+d x)\right )}{2 d \cos ^{\frac {7}{2}}(c+d x)} \]
((b*Cos[c + d*x])^(3/2)*(ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + Sin[c + d* x]))/(2*d*Cos[c + d*x]^(7/2))
Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2031, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \sec ^3(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b*Sqrt[b*Cos[c + d*x]]*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/Sqrt[Cos[c + d*x]]
3.2.57.3.1 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[(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 = 2.87 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15
method | result | size |
default | \(\frac {b \left (-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(85\) |
risch | \(-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(135\) |
1/2/d*b*(-cos(d*x+c)^2*ln(-cot(d*x+c)+csc(d*x+c)-1)+cos(d*x+c)^2*ln(-cot(d *x+c)+csc(d*x+c)+1)+sin(d*x+c))*(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(5/2)
Time = 0.34 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.76 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\left [\frac {b^{\frac {3}{2}} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{3}}, -\frac {\sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt {b \cos \left (d x + c\right )} b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{3}}\right ] \]
[1/4*(b^(3/2)*cos(d*x + c)^3*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*sqrt(b*cos(d*x + c))*b*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d* x + c)^3), -1/2*(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)^3 - sqrt(b*cos(d*x + c))*b*sqrt(c os(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)]
Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 691 vs. \(2 (62) = 124\).
Time = 0.44 (sec) , antiderivative size = 691, normalized size of antiderivative = 9.34 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]
-1/4*(4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2* d*x + 2*c), cos(2*d*x + 2*c))) - 4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2 *c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*sin(4*d*x + 4*c)^2 + 4*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*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) + (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*sin(4*d*x + 4*c) ^2 + 4*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2 *b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*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*c) , cos(2*d*x + 2*c))) + 1) - 4*(b*cos(4*d*x + 4*c) + 2*b*cos(2*d*x + 2*c) + b)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(b*cos(4*d*x + 4*c) + 2*b*cos(2*d*x + 2*c) + b)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c))))*sqrt(b)/((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)*d )
\[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \]