Integrand size = 23, antiderivative size = 78 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\text {arctanh}(\sin (c+d x)) \sqrt {\sec (c+d x)}}{2 b d \sqrt {b \sec (c+d x)}}+\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b d \sqrt {b \sec (c+d x)}} \] Output:
1/2*arctanh(sin(d*x+c))*sec(d*x+c)^(1/2)/b/d/(b*sec(d*x+c))^(1/2)+1/2*sec( d*x+c)^(5/2)*sin(d*x+c)/b/d/(b*sec(d*x+c))^(1/2)
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\sec ^{\frac {3}{2}}(c+d x) (\text {arctanh}(\sin (c+d x))+\sec (c+d x) \tan (c+d x))}{2 d (b \sec (c+d x))^{3/2}} \] Input:
Integrate[Sec[c + d*x]^(9/2)/(b*Sec[c + d*x])^(3/2),x]
Output:
(Sec[c + d*x]^(3/2)*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x]))/( 2*d*(b*Sec[c + d*x])^(3/2))
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77, 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.
\(\displaystyle \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \sec ^3(c+d x)dx}{b \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{b \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{b \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{b \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{b \sqrt {b \sec (c+d x)}}\) |
Input:
Int[Sec[c + d*x]^(9/2)/(b*Sec[c + d*x])^(3/2),x]
Output:
(Sqrt[Sec[c + d*x]]*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d *x])/(2*d)))/(b*Sqrt[b*Sec[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[(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 = 0.83 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21
method | result | size |
default | \(\frac {\sec \left (d x +c \right )^{\frac {7}{2}} \left (\cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-\cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{2 b d \sqrt {b \sec \left (d x +c \right )}}\) | \(94\) |
risch | \(-\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{4 i \left (d x +c \right )} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )-{\mathrm e}^{4 i \left (d x +c \right )} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )+2 i {\mathrm e}^{3 i \left (d x +c \right )}-2 i {\mathrm e}^{i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )-2 \,{\mathrm e}^{2 i \left (d x +c \right )} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )+\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )-\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )\right )}{2 b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}\) | \(221\) |
Input:
int(sec(d*x+c)^(9/2)/(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/b/d*sec(d*x+c)^(7/2)/(b*sec(d*x+c))^(1/2)*(cos(d*x+c)^3*ln(-cot(d*x+c) +csc(d*x+c)+1)-cos(d*x+c)^3*ln(-cot(d*x+c)+csc(d*x+c)-1)+sin(d*x+c)*cos(d* x+c))
Time = 0.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.65 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\left [\frac {\sqrt {b} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, b^{2} d \cos \left (d x + c\right )}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{b \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, b^{2} d \cos \left (d x + c\right )}\right ] \] Input:
integrate(sec(d*x+c)^(9/2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
[1/4*(sqrt(b)*cos(d*x + c)*log(-(b*cos(d*x + c)^2 - 2*sqrt(b)*sqrt(b/cos(d *x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b)/cos(d*x + c)^2) + 2*sqrt(b /cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(b^2*d*cos(d*x + c)), -1/2 *(sqrt(-b)*arctan(sqrt(-b)*sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))/(b*sin( d*x + c)))*cos(d*x + c) - sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(b^2*d*cos(d*x + c))]
Timed out. \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(9/2)/(b*sec(d*x+c))**(3/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 670 vs. \(2 (66) = 132\).
Time = 0.19 (sec) , antiderivative size = 670, normalized size of antiderivative = 8.59 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(9/2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-1/4*(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*s in(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), co s(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) - 4*(cos(4 *d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), co s(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*a rctan2(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)*sqrt(b)*d)
Time = 0.23 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\frac {4 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} - 4} + \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) - \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{4 \, b^{\frac {3}{2}} d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate(sec(d*x+c)^(9/2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
1/4*(4*(1/tan(1/2*d*x + 1/2*c) + tan(1/2*d*x + 1/2*c))/((1/tan(1/2*d*x + 1 /2*c) + tan(1/2*d*x + 1/2*c))^2 - 4) + log(abs(1/tan(1/2*d*x + 1/2*c) + ta n(1/2*d*x + 1/2*c) + 2)) - log(abs(1/tan(1/2*d*x + 1/2*c) + tan(1/2*d*x + 1/2*c) - 2)))/(b^(3/2)*d*sgn(cos(d*x + c)))
Timed out. \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((1/cos(c + d*x))^(9/2)/(b/cos(c + d*x))^(3/2),x)
Output:
int((1/cos(c + d*x))^(9/2)/(b/cos(c + d*x))^(3/2), x)
Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.28 \[ \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )\right )}{2 b^{2} d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^(9/2)/(b*sec(d*x+c))^(3/2),x)
Output:
(sqrt(b)*( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + log(tan((c + d*x) /2) - 1) + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - log(tan((c + d*x)/2 ) + 1) - sin(c + d*x)))/(2*b**2*d*(sin(c + d*x)**2 - 1))