3.6.84 \(\int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx\) [584]

3.6.84.1 Optimal result
3.6.84.2 Mathematica [A] (verified)
3.6.84.3 Rubi [A] (verified)
3.6.84.4 Maple [C] (verified)
3.6.84.5 Fricas [C] (verification not implemented)
3.6.84.6 Sympy [F]
3.6.84.7 Maxima [F]
3.6.84.8 Giac [F]
3.6.84.9 Mupad [F(-1)]

3.6.84.1 Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac {6 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}} \]

output
-2/5*b/f/(d*sec(f*x+e))^(5/2)+2/5*a*sin(f*x+e)/d/f/(d*sec(f*x+e))^(3/2)+6/ 
5*a*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+ 
1/2*e),2^(1/2))/d^2/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)
 
3.6.84.2 Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 \sqrt {d \sec (e+f x)} \left (3 a \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\cos ^2(e+f x) (-b \cos (e+f x)+a \sin (e+f x))\right )}{5 d^3 f} \]

input
Integrate[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(5/2),x]
 
output
(2*Sqrt[d*Sec[e + f*x]]*(3*a*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] 
+ Cos[e + f*x]^2*(-(b*Cos[e + f*x]) + a*Sin[e + f*x])))/(5*d^3*f)
 
3.6.84.3 Rubi [A] (verified)

Time = 0.44 (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, 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.

\(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3967

\(\displaystyle a \int \frac {1}{(d \sec (e+f x))^{5/2}}dx-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \frac {1}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle a \left (\frac {3 \int \frac {1}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {3 \int \frac {1}{\sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{5 d^2}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle a \left (\frac {3 \int \sqrt {\cos (e+f x)}dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {3 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle a \left (\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\)

input
Int[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(5/2),x]
 
output
(-2*b)/(5*f*(d*Sec[e + f*x])^(5/2)) + a*((6*EllipticE[(e + f*x)/2, 2])/(5* 
d^2*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*Sin[e + f*x])/(5*d*f*( 
d*Sec[e + f*x])^(3/2)))
 

3.6.84.3.1 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 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]
 
3.6.84.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 6.15 (sec) , antiderivative size = 438, normalized size of antiderivative = 4.66

method result size
default \(\frac {2 a \left (3 i \cos \left (f x +e \right ) E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-3 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-6 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+3 i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-3 i \sec \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+3 \sin \left (f x +e \right )\right )}{5 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}-\frac {2 b}{5 f \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\) \(438\)
parts \(\frac {2 a \left (3 i \cos \left (f x +e \right ) E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-3 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-6 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+3 i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-3 i \sec \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+3 \sin \left (f x +e \right )\right )}{5 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}-\frac {2 b}{5 f \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\) \(438\)

input
int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 
output
2/5*a/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/d^2*(3*I*cos(f*x+e)*EllipticE( 
I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x 
+e)+1))^(1/2)-3*I*cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(co 
s(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+6*I*(1/(cos(f*x+e)+1) 
)^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+ 
e)),I)-6*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f 
*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)+3*I*sec(f*x+e)*EllipticE(I*(csc(f*x+e)- 
cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)- 
3*I*sec(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^( 
1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+sin(f*x+e)*cos(f*x+e)^2+sin(f*x+e)* 
cos(f*x+e)+3*sin(f*x+e))-2/5*b/f/(d*sec(f*x+e))^(5/2)
 
3.6.84.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\frac {3 i \, \sqrt {2} a \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 3 i \, \sqrt {2} a \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (b \cos \left (f x + e\right )^{3} - a \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \]

input
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="fricas")
 
output
1/5*(3*I*sqrt(2)*a*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(f*x + e) + I*sin(f*x + e))) - 3*I*sqrt(2)*a*sqrt(d)*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(b* 
cos(f*x + e)^3 - a*cos(f*x + e)^2*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^3 
*f)
 
3.6.84.6 Sympy [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

input
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(5/2),x)
 
output
Integral((a + b*tan(e + f*x))/(d*sec(e + f*x))**(5/2), x)
 
3.6.84.7 Maxima [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="maxima")
 
output
integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(5/2), x)
 
3.6.84.8 Giac [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="giac")
 
output
integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(5/2), x)
 
3.6.84.9 Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

input
int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(5/2),x)
 
output
int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(5/2), x)