\(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx\) [602]

Optimal result

Integrand size = 25, antiderivative size = 215 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {2 b \left (3 a^2-b^2\right ) d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)}}{5 f}+\frac {2 b^3 d^2 \sec ^4(e+f x) \sqrt {d \sec (e+f x)}}{9 f}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {6 a b^2 d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \tan (e+f x)}{7 f} \] Output:

2/5*b*(3*a^2-b^2)*d^2*sec(f*x+e)^2*(d*sec(f*x+e))^(1/2)/f+2/9*b^3*d^2*sec( 
f*x+e)^4*(d*sec(f*x+e))^(1/2)/f+2/21*a*(7*a^2-6*b^2)*d^2*InverseJacobiAM(1 
/2*arctan(tan(f*x+e)),2^(1/2))*(d*sec(f*x+e))^(1/2)/f/(sec(f*x+e)^2)^(1/4) 
+2/21*a*(7*a^2-6*b^2)*d^2*(d*sec(f*x+e))^(1/2)*tan(f*x+e)/f+6/7*a*b^2*d^2* 
sec(f*x+e)^2*(d*sec(f*x+e))^(1/2)*tan(f*x+e)/f
 

Mathematica [A] (verified)

Time = 2.36 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.73 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=-\frac {2 d (d \sec (e+f x))^{3/2} \left (63 b \left (-3 a^2+b^2\right ) \cos ^2(e+f x)-15 a \left (7 a^2-6 b^2\right ) \cos ^{\frac {9}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-15 a \left (7 a^2-6 b^2\right ) \cos ^3(e+f x) \sin (e+f x)-\frac {5}{2} b^2 (14 b+27 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{315 f (a \cos (e+f x)+b \sin (e+f x))^3} \] Input:

Integrate[(d*Sec[e + f*x])^(5/2)*(a + b*Tan[e + f*x])^3,x]
 

Output:

(-2*d*(d*Sec[e + f*x])^(3/2)*(63*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 - 15*a*(7 
*a^2 - 6*b^2)*Cos[e + f*x]^(9/2)*EllipticF[(e + f*x)/2, 2] - 15*a*(7*a^2 - 
 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - (5*b^2*(14*b + 27*a*Sin[2*(e + f*x)] 
))/2)*(a + b*Tan[e + f*x])^3)/(315*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3994, 497, 27, 25, 676, 211, 229}

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.

\(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3dx\)

\(\Big \downarrow \) 3994

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \int (a+b \tan (e+f x))^3 \sqrt [4]{\tan ^2(e+f x)+1}d(b \tan (e+f x))}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 497

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {2}{9} b^2 \int -\frac {(a+b \tan (e+f x)) \left (\left (4-\frac {9 a^2}{b^2}\right ) b^2-13 a b \tan (e+f x)\right ) \sqrt [4]{\tan ^2(e+f x)+1}}{2 b^2}d(b \tan (e+f x))+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2-\frac {1}{9} \int -\left ((a+b \tan (e+f x)) \left (9 a^2+13 b \tan (e+f x) a-4 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}\right )d(b \tan (e+f x))\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \int (a+b \tan (e+f x)) \left (9 a^2+13 b \tan (e+f x) a-4 b^2\right ) \sqrt [4]{\tan ^2(e+f x)+1}d(b \tan (e+f x))+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \left (\frac {9}{7} a \left (7 a^2-6 b^2\right ) \int \sqrt [4]{\tan ^2(e+f x)+1}d(b \tan (e+f x))+\frac {4}{5} b^2 \left (11 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{5/4}+\frac {26}{7} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{5/4}\right )+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \left (\frac {9}{7} a \left (7 a^2-6 b^2\right ) \left (\frac {1}{3} \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{3/4}}d(b \tan (e+f x))+\frac {2}{3} b \sqrt [4]{\tan ^2(e+f x)+1} \tan (e+f x)\right )+\frac {4}{5} b^2 \left (11 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{5/4}+\frac {26}{7} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{5/4}\right )+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {d^2 \sqrt {d \sec (e+f x)} \left (\frac {1}{9} \left (\frac {9}{7} a \left (7 a^2-6 b^2\right ) \left (\frac {2}{3} b \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )+\frac {2}{3} b \tan (e+f x) \sqrt [4]{\tan ^2(e+f x)+1}\right )+\frac {4}{5} b^2 \left (11 a^2-2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{5/4}+\frac {26}{7} a b^3 \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{5/4}\right )+\frac {2}{9} b^2 \left (\tan ^2(e+f x)+1\right )^{5/4} (a+b \tan (e+f x))^2\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\)

Input:

Int[(d*Sec[e + f*x])^(5/2)*(a + b*Tan[e + f*x])^3,x]
 

Output:

(d^2*Sqrt[d*Sec[e + f*x]]*((2*b^2*(a + b*Tan[e + f*x])^2*(1 + Tan[e + f*x] 
^2)^(5/4))/9 + ((4*b^2*(11*a^2 - 2*b^2)*(1 + Tan[e + f*x]^2)^(5/4))/5 + (2 
6*a*b^3*Tan[e + f*x]*(1 + Tan[e + f*x]^2)^(5/4))/7 + (9*a*(7*a^2 - 6*b^2)* 
((2*b*EllipticF[ArcTan[Tan[e + f*x]]/2, 2])/3 + (2*b*Tan[e + f*x]*(1 + Tan 
[e + f*x]^2)^(1/4))/3))/7)/9))/(b*f*(Sec[e + f*x]^2)^(1/4))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 497

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b 
*(n + 2*p + 1))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 
 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n 
, p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p 
+ 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 3042

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

rule 3994

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP 
art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]))   Subst[Int[(a + x)^n*(1 + 
x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, 
n}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[m] && IntegerQ[n]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 922.67 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {d^{2} \left (\left (\frac {2 \sec \left (f x +e \right )^{4}}{9}-\frac {2 \sec \left (f x +e \right )^{2}}{5}\right ) b^{3}+\frac {2 \tan \left (f x +e \right ) a^{3}}{3}+\frac {6 \sec \left (f x +e \right )^{2} a^{2} b}{5}+\left (\frac {6 \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{7}-\frac {4 \tan \left (f x +e \right )}{7}\right ) b^{2} a -\frac {2 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{3}}{3}+\frac {4 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) b^{2} a}{7}\right ) \sqrt {d \sec \left (f x +e \right )}}{f}\)	\(232\)
parts	\(\frac {a^{3} \left (-\frac {2 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )}{3}+\frac {2 \tan \left (f x +e \right )}{3}\right ) d^{2} \sqrt {d \sec \left (f x +e \right )}}{f}+\frac {2 b^{3} \left (\frac {\left (d \sec \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {d^{2} \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f \,d^{2}}+\frac {6 a^{2} b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}+\frac {3 a \,b^{2} \left (-\frac {4 \tan \left (f x +e \right )}{21}+\frac {2 \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{7}+\frac {4 i \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )}{21}\right ) d^{2} \sqrt {d \sec \left (f x +e \right )}}{f}\)	\(262\)

Input:

int((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

d^2/f*((2/9*sec(f*x+e)^4-2/5*sec(f*x+e)^2)*b^3+2/3*tan(f*x+e)*a^3+6/5*sec( 
f*x+e)^2*a^2*b+(6/7*tan(f*x+e)*sec(f*x+e)^2-4/7*tan(f*x+e))*b^2*a-2/3*I*(1 
+cos(f*x+e))*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*El 
lipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a^3+4/7*I*(1+cos(f*x+e))*(1/(1+cos(f* 
x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot 
(f*x+e)),I)*b^2*a)*(d*sec(f*x+e))^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.94 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {-15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (35 \, b^{3} d^{2} + 63 \, {\left (3 \, a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right )^{2} + 15 \, {\left (9 \, a b^{2} d^{2} \cos \left (f x + e\right ) + {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{2} \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{315 \, f \cos \left (f x + e\right )^{4}} \] Input:

integrate((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

1/315*(-15*I*sqrt(2)*(7*a^3 - 6*a*b^2)*d^(5/2)*cos(f*x + e)^4*weierstrassP 
Inverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 15*I*sqrt(2)*(7*a^3 - 6*a* 
b^2)*d^(5/2)*cos(f*x + e)^4*weierstrassPInverse(-4, 0, cos(f*x + e) - I*si 
n(f*x + e)) + 2*(35*b^3*d^2 + 63*(3*a^2*b - b^3)*d^2*cos(f*x + e)^2 + 15*( 
9*a*b^2*d^2*cos(f*x + e) + (7*a^3 - 6*a*b^2)*d^2*cos(f*x + e)^3)*sin(f*x + 
 e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^4)
 

Sympy [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))**(5/2)*(a+b*tan(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \] Input:

int((d/cos(e + f*x))^(5/2)*(a + b*tan(e + f*x))^3,x)
 

Output:

int((d/cos(e + f*x))^(5/2)*(a + b*tan(e + f*x))^3, x)
 

Reduce [F]

\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {\sqrt {d}\, d^{2} \left (10 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2} b^{3}+54 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} a^{2} b -8 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} b^{3}+135 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2}d x \right ) a \,b^{2} f +45 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2}d x \right ) a^{3} f \right )}{45 f} \] Input:

int((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x)
 

Output:

(sqrt(d)*d**2*(10*sqrt(sec(e + f*x))*sec(e + f*x)**2*tan(e + f*x)**2*b**3 
+ 54*sqrt(sec(e + f*x))*sec(e + f*x)**2*a**2*b - 8*sqrt(sec(e + f*x))*sec( 
e + f*x)**2*b**3 + 135*int(sqrt(sec(e + f*x))*sec(e + f*x)**2*tan(e + f*x) 
**2,x)*a*b**2*f + 45*int(sqrt(sec(e + f*x))*sec(e + f*x)**2,x)*a**3*f))/(4 
5*f)