Integrand size = 12, antiderivative size = 98 \[ \int (b \sec (c+d x))^{7/2} \, dx=-\frac {6 b^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 b^3 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \] Output:
-6/5*b^4*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b*sec(d *x+c))^(1/2)+6/5*b^3*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/d+2/5*b*(b*sec(d*x+c) )^(5/2)*sin(d*x+c)/d
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.63 \[ \int (b \sec (c+d x))^{7/2} \, dx=\frac {b (b \sec (c+d x))^{5/2} \left (-12 \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+7 \sin (c+d x)+3 \sin (3 (c+d x))\right )}{10 d} \] Input:
Integrate[(b*Sec[c + d*x])^(7/2),x]
Output:
(b*(b*Sec[c + d*x])^(5/2)*(-12*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2 ] + 7*Sin[c + d*x] + 3*Sin[3*(c + d*x)]))/(10*d)
Time = 0.44 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119}
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.
\(\displaystyle \int (b \sec (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {3}{5} b^2 \int (b \sec (c+d x))^{3/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} b^2 \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\) |
Input:
Int[(b*Sec[c + d*x])^(7/2),x]
Output:
(2*b*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + (3*b^2*((-2*b^2*Elliptic E[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*Sqrt [b*Sec[c + d*x]]*Sin[c + d*x])/d))/5
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
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_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Result contains complex when optimal does not.
Time = 3.10 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.14
method | result | size |
default | \(-\frac {2 \sqrt {b \sec \left (d x +c \right )}\, b^{3} \left (-3 \sin \left (d x +c \right )-\tan \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )+i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \operatorname {EllipticE}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}\) | \(210\) |
Input:
int((b*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/5/d*(b*sec(d*x+c))^(1/2)*b^3/(cos(d*x+c)+1)*(-3*sin(d*x+c)-tan(d*x+c)-s ec(d*x+c)*tan(d*x+c)+I*(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*EllipticE(I*(-cot(d *x+c)+csc(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^ (1/2)+I*(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2) *EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2))
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.28 \[ \int (b \sec (c+d x))^{7/2} \, dx=\frac {-3 i \, \sqrt {2} b^{\frac {7}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} b^{\frac {7}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((b*sec(d*x+c))^(7/2),x, algorithm="fricas")
Output:
1/5*(-3*I*sqrt(2)*b^(7/2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstras sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*b^(7/2)*cos (d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*b^3*cos(d*x + c)^2 + b^3)*sqrt(b/cos(d*x + c))* sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (b \sec (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate((b*sec(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int (b \sec (c+d x))^{7/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((b*sec(d*x+c))^(7/2),x, algorithm="maxima")
Output:
integrate((b*sec(d*x + c))^(7/2), x)
\[ \int (b \sec (c+d x))^{7/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((b*sec(d*x+c))^(7/2),x, algorithm="giac")
Output:
integrate((b*sec(d*x + c))^(7/2), x)
Timed out. \[ \int (b \sec (c+d x))^{7/2} \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2} \,d x \] Input:
int((b/cos(c + d*x))^(7/2),x)
Output:
int((b/cos(c + d*x))^(7/2), x)
\[ \int (b \sec (c+d x))^{7/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b^{3} \] Input:
int((b*sec(d*x+c))^(7/2),x)
Output:
sqrt(b)*int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*b**3