\(\int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x)) \, dx\) [588]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x)) \, dx=-\frac {2 a d^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 b (d \sec (e+f x))^{3/2}}{3 f}+\frac {2 a d \sqrt {d \sec (e+f x)} \sin (e+f x)}{f} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.66 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x)) \, dx=\frac {(d \sec (e+f x))^{3/2} \left (2 b-6 a \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+3 a \sin (2 (e+f x))\right )}{3 f} \] Input:

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

Output:

((d*Sec[e + f*x])^(3/2)*(2*b - 6*a*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/ 
2, 2] + 3*a*Sin[2*(e + f*x)]))/(3*f)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 89, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3967

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3119

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]
 

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 4255

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]
 

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 = 12.40 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.26

method	result	size
default	\(\frac {2 d \sqrt {d \sec \left (f x +e \right )}\, \left (i \left (3 \cos \left (f x +e \right )^{2}+6 \cos \left (f x +e \right )+3\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, a +i \left (-3 \cos \left (f x +e \right )^{2}-6 \cos \left (f x +e \right )-3\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, a +3 a \sin \left (f x +e \right )+b \left (1+\sec \left (f x +e \right )\right )\right )}{3 f \left (1+\cos \left (f x +e \right )\right )}\)	\(199\)
parts	\(\frac {2 a \left (i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+\sin \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d}{f \left (1+\cos \left (f x +e \right )\right )}+\frac {2 b \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}\)	\(200\)

Input:

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

Output:

2/3/f*d*(d*sec(f*x+e))^(1/2)/(1+cos(f*x+e))*(I*(3*cos(f*x+e)^2+6*cos(f*x+e 
)+3)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)) 
,I)*(1/(1+cos(f*x+e)))^(1/2)*a+I*(-3*cos(f*x+e)^2-6*cos(f*x+e)-3)*(cos(f*x 
+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(1+cos 
(f*x+e)))^(1/2)*a+3*a*sin(f*x+e)+b*(1+sec(f*x+e)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x)) \, dx=\frac {-3 i \, \sqrt {2} a d^{\frac {3}{2}} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {2} a d^{\frac {3}{2}} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (3 \, a d \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b d\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, f \cos \left (f x + e\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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