Integrand size = 25, antiderivative size = 75 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}} \] Output:
2/3*(A+3*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(b*sec (d*x+c))^(1/2)/b^2/d+2/3*A*tan(d*x+c)/d/(b*sec(d*x+c))^(3/2)
Time = 0.37 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\sec ^2(c+d x) \left (2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+A \sin (2 (c+d x))\right )}{3 d (b \sec (c+d x))^{3/2}} \] Input:
Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(3/2),x]
Output:
(Sec[c + d*x]^2*(2*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + A*Sin[2*(c + d*x)]))/(3*d*(b*Sec[c + d*x])^(3/2))
Time = 0.38 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4533, 3042, 4258, 3042, 3120}
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.
| \(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {(A+3 C) \int \sqrt {b \sec (c+d x)}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+3 C) \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {(A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 A \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}}\) |
Input:
Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(3/2),x]
Output:
(2*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d *x]])/(3*b^2*d) + (2*A*Tan[c + d*x])/(3*d*(b*Sec[c + d*x])^(3/2))
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Result contains complex when optimal does not.
Time = 3.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.11
| method | result | size |
| default | \(\frac {-\frac {2 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (1+\sec \left (d x +c \right )\right )}{3}+\frac {2 A \sin \left (d x +c \right )}{3}-\frac {2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (3+3 \sec \left (d x +c \right )\right )}{3}}{d \sqrt {b \sec \left (d x +c \right )}\, b}\) | \(158\) |
| parts | \(\frac {A \left (-\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1+\sec \left (d x +c \right )\right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {b \sec \left (d x +c \right )}\, b}-\frac {2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{d b \sqrt {b \sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\) | \(164\) |
Input:
int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/d*(-2/3*I*A*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*E llipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1+sec(d*x+c))+2/3*A*sin(d*x+c)-2/3* I*C*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I *(csc(d*x+c)-cot(d*x+c)),I)*(3+3*sec(d*x+c)))/(b*sec(d*x+c))^(1/2)/b
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 \, A \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, b^{2} d} \] Input:
integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/3*(2*A*sqrt(b/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + sqrt(2)*(-I*A - 3*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*A + 3*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I *sin(d*x + c)))/(b^2*d)
\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(3/2),x)
Output:
Integral((A + C*sec(c + d*x)**2)/(b*sec(c + d*x))**(3/2), x)
\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(3/2), x)
\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(3/2), x)
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(3/2),x)
Output:
int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(3/2), x)
\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) c \right )}{b^{2}} \] Input:
int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(3/2),x)
Output:
(sqrt(b)*(int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*a + int(sqrt(sec(c + d *x)),x)*c))/b**2