\(\int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx\) [591]

Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 a \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}} \] Output:

-2/3*b/f/(d*sec(f*x+e))^(3/2)+2/3*a*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f 
*x+1/2*e,2^(1/2))*(d*sec(f*x+e))^(1/2)/d^2/f+2/3*a*sin(f*x+e)/d/f/(d*sec(f 
*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {\sqrt {d \sec (e+f x)} \left (b+b \cos (2 (e+f x))-2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-a \sin (2 (e+f x))\right )}{3 d^2 f} \] Input:

Integrate[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(3/2),x]
 

Output:

-1/3*(Sqrt[d*Sec[e + f*x]]*(b + b*Cos[2*(e + f*x)] - 2*a*Sqrt[Cos[e + f*x] 
]*EllipticF[(e + f*x)/2, 2] - a*Sin[2*(e + f*x)]))/(d^2*f)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3967, 3042, 4256, 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.

\(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3967

\(\displaystyle a \int \frac {1}{(d \sec (e+f x))^{3/2}}dx-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \frac {1}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle a \left (\frac {\int \sqrt {d \sec (e+f x)}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {\int \sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle a \left (\frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\)

Input:

Int[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(3/2),x]
 

Output:

(-2*b)/(3*f*(d*Sec[e + f*x])^(3/2)) + a*((2*Sqrt[Cos[e + f*x]]*EllipticF[( 
e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*d^2*f) + (2*Sin[e + f*x])/(3*d*f*S 
qrt[d*Sec[e + f*x]]))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3120

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]
 

rule 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 10.36 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {-\frac {2 i a \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right ) \left (-1-\sec \left (f x +e \right )\right )}{3}+\frac {2 a \sin \left (f x +e \right )}{3}-\frac {2 b \cos \left (f x +e \right )}{3}}{f \sqrt {d \sec \left (f x +e \right )}\, d}\)	\(103\)
parts	\(\frac {a \left (-\frac {2 i \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (1+\sec \left (f x +e \right )\right )}{3}+\frac {2 \sin \left (f x +e \right )}{3}\right )}{f \sqrt {d \sec \left (f x +e \right )}\, d}-\frac {2 b}{3 f \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\)	\(108\)

Input:

int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(-2/3*I*a*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*E 
llipticF(I*(-csc(f*x+e)+cot(f*x+e)),I)*(-1-sec(f*x+e))+2/3*a*sin(f*x+e)-2/ 
3*b*cos(f*x+e))/(d*sec(f*x+e))^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\frac {-i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (b \cos \left (f x + e\right )^{2} - a \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d^{2} f} \] Input:

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

1/3*(-I*sqrt(2)*a*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin( 
f*x + e)) + I*sqrt(2)*a*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) - 
I*sin(f*x + e)) - 2*(b*cos(f*x + e)^2 - a*cos(f*x + e)*sin(f*x + e))*sqrt( 
d/cos(f*x + e)))/(d^2*f)
 

Sympy [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(3/2),x)
 

Output:

Integral((a + b*tan(e + f*x))/(d*sec(e + f*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(3/2), x)
 

Giac [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(3/2),x)
 

Output:

int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{2}}d x \right ) b \right )}{d^{2}} \] Input:

int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x)
 

Output:

(sqrt(d)*(int(sqrt(sec(e + f*x))/sec(e + f*x)**2,x)*a + int((sqrt(sec(e + 
f*x))*tan(e + f*x))/sec(e + f*x)**2,x)*b))/d**2