Integrand size = 23, antiderivative size = 74 \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {b \text {arctanh}(\sin (c+d x)) \sqrt {b \sec (c+d x)}}{2 d \sqrt {\sec (c+d x)}}+\frac {b \sec ^{\frac {3}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin (c+d x)}{2 d} \] Output:
1/2*b*arctanh(sin(d*x+c))*(b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)+1/2*b*se c(d*x+c)^(3/2)*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/d
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {(b \sec (c+d x))^{3/2} (\text {arctanh}(\sin (c+d x))+\sec (c+d x) \tan (c+d x))}{2 d \sec ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(3/2),x]
Output:
((b*Sec[c + d*x])^(3/2)*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x] ))/(2*d*Sec[c + d*x]^(3/2))
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.
\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \int \sec ^3(c+d x)dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\sec (c+d x)}}\) |
Input:
Int[Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(3/2),x]
Output:
(b*Sqrt[b*Sec[c + d*x]]*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/Sqrt[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 = 0.85 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {b \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {b \sec \left (d x +c \right )}\, \left (\cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-\cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{2 d}\) | \(92\) |
risch | \(-\frac {i b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \cos \left (d x +c \right )}{d}-\frac {b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \cos \left (d x +c \right )}{d}\) | \(235\) |
Input:
int(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2*b/d*sec(d*x+c)^(5/2)*(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 = 204, normalized size of antiderivative = 2.76 \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\left [\frac {b^{\frac {3}{2}} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, b \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, d \cos \left (d x + c\right )}, -\frac {\sqrt {-b} b \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{b \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - \frac {b \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, d \cos \left (d x + c\right )}\right ] \] Input:
integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
[1/4*(b^(3/2)*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*b*sqrt (b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)), -1/2*( sqrt(-b)*b*arctan(sqrt(-b)*sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))/(b*sin( d*x + c)))*cos(d*x + c) - b*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(3/2)*(b*sec(d*x+c))**(3/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 691 vs. \(2 (62) = 124\).
Time = 0.21 (sec) , antiderivative size = 691, normalized size of antiderivative = 9.34 \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-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 )
Time = 0.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26 \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {b^{\frac {3}{2}} {\left (\frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{2 \, d} \] Input:
integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
1/2*b^(3/2)*(2*(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))/(tan(1/2*d* x + 1/2*c)^4 - 2*tan(1/2*d*x + 1/2*c)^2 + 1) + log(tan(1/2*d*x + 1/2*c) + 1) - log(tan(1/2*d*x + 1/2*c) - 1))*sgn(cos(d*x + c))/d
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int((b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(3/2),x)
Output:
int((b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.32 \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {\sqrt {b}\, b \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )\right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^(3/2),x)
Output:
(sqrt(b)*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*d*(sin(c + d*x)**2 - 1))