\(\int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx\) [648]

Optimal result

Integrand size = 23, antiderivative size = 169 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx=\frac {b \left (3 a^2-b^2\right ) (d \sec (e+f x))^m}{f m}+\frac {b^3 \sec ^2(e+f x) (d \sec (e+f x))^m}{f (2+m)}+\frac {3 a b^2 (d \sec (e+f x))^m \tan (e+f x)}{f (1+m)}-\frac {a \left (3 b^2-a^2 (1+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-\frac {m}{2},\frac {3}{2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x)}{f (1+m)} \] Output:

b*(3*a^2-b^2)*(d*sec(f*x+e))^m/f/m+b^3*sec(f*x+e)^2*(d*sec(f*x+e))^m/f/(2+ 
m)+3*a*b^2*(d*sec(f*x+e))^m*tan(f*x+e)/f/(1+m)-a*(3*b^2-a^2*(1+m))*hyperge 
om([1/2, 1-1/2*m],[3/2],-tan(f*x+e)^2)*(d*sec(f*x+e))^m*tan(f*x+e)/f/(1+m) 
/((sec(f*x+e)^2)^(1/2*m))
 

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.94 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx=\frac {(d \sec (e+f x))^m \left (3 a b^2 (2+m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)-a^3 (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)+b \left (\left (3 a^2-b^2\right ) (2+m)+b^2 m \sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f m (2+m) \sqrt {-\tan ^2(e+f x)}} \] Input:

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

Output:

((d*Sec[e + f*x])^m*(3*a*b^2*(2 + m)*Hypergeometric2F1[-1/2, m/2, (2 + m)/ 
2, Sec[e + f*x]^2]*Tan[e + f*x] - a^3*(2 + m)*Hypergeometric2F1[1/2, m/2, 
(2 + m)/2, Sec[e + f*x]^2]*Tan[e + f*x] + b*((3*a^2 - b^2)*(2 + m) + b^2*m 
*Sec[e + f*x]^2)*Sqrt[-Tan[e + f*x]^2]))/(f*m*(2 + m)*Sqrt[-Tan[e + f*x]^2 
])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3994, 497, 25, 27, 676, 237}

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.

\(\displaystyle \int (a+b \tan (e+f x))^3 (d \sec (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x))^3 (d \sec (e+f x))^mdx\)

\(\Big \downarrow \) 3994

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

\(\Big \downarrow \) 497

\(\displaystyle \frac {\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (\frac {b^2 \int -\frac {(a+b \tan (e+f x)) \left (b^2 \left (2-\frac {a^2 (m+2)}{b^2}\right )-a b (m+4) \tan (e+f x)\right ) \left (\tan ^2(e+f x)+1\right )^{\frac {m-2}{2}}}{b^2}d(b \tan (e+f x))}{m+2}+\frac {b^2 \left (\tan ^2(e+f x)+1\right )^{m/2} (a+b \tan (e+f x))^2}{m+2}\right )}{b f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (\frac {b^2 \left (\tan ^2(e+f x)+1\right )^{m/2} (a+b \tan (e+f x))^2}{m+2}-\frac {b^2 \int \frac {(a+b \tan (e+f x)) \left (-\left ((m+2) a^2\right )-b (m+4) \tan (e+f x) a+2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{\frac {m-2}{2}}}{b^2}d(b \tan (e+f x))}{m+2}\right )}{b f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (\frac {b^2 \left (\tan ^2(e+f x)+1\right )^{m/2} (a+b \tan (e+f x))^2}{m+2}-\frac {\int (a+b \tan (e+f x)) \left (-\left ((m+2) a^2\right )-b (m+4) \tan (e+f x) a+2 b^2\right ) \left (\tan ^2(e+f x)+1\right )^{\frac {m-2}{2}}d(b \tan (e+f x))}{m+2}\right )}{b f}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (\frac {b^2 \left (\tan ^2(e+f x)+1\right )^{m/2} (a+b \tan (e+f x))^2}{m+2}-\frac {\frac {a (m+2) \left (3 b^2-a^2 (m+1)\right ) \int \left (\tan ^2(e+f x)+1\right )^{\frac {m-2}{2}}d(b \tan (e+f x))}{m+1}+\frac {2 b^2 \left (b^2-a^2 (m+3)\right ) \left (\tan ^2(e+f x)+1\right )^{m/2}}{m}-\frac {a b^3 (m+4) \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{m/2}}{m+1}}{m+2}\right )}{b f}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (\frac {b^2 \left (\tan ^2(e+f x)+1\right )^{m/2} (a+b \tan (e+f x))^2}{m+2}-\frac {\frac {a b (m+2) \left (3 b^2-a^2 (m+1)\right ) \tan (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {3}{2},-\tan ^2(e+f x)\right )}{m+1}+\frac {2 b^2 \left (b^2-a^2 (m+3)\right ) \left (\tan ^2(e+f x)+1\right )^{m/2}}{m}-\frac {a b^3 (m+4) \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{m/2}}{m+1}}{m+2}\right )}{b f}\)

Input:

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

Output:

((d*Sec[e + f*x])^m*((b^2*(a + b*Tan[e + f*x])^2*(1 + Tan[e + f*x]^2)^(m/2 
))/(2 + m) - ((a*b*(2 + m)*(3*b^2 - a^2*(1 + m))*Hypergeometric2F1[1/2, (2 
 - m)/2, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(1 + m) + (2*b^2*(b^2 - a^2*( 
3 + m))*(1 + Tan[e + f*x]^2)^(m/2))/m - (a*b^3*(4 + m)*Tan[e + f*x]*(1 + T 
an[e + f*x]^2)^(m/2))/(1 + m))/(2 + m)))/(b*f*(Sec[e + f*x]^2)^(m/2))
 

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 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

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 [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx=\frac {d^{m} \left (\sec \left (f x +e \right )^{m} \tan \left (f x +e \right )^{2} b^{3} m +3 \sec \left (f x +e \right )^{m} a^{2} b m +6 \sec \left (f x +e \right )^{m} a^{2} b -2 \sec \left (f x +e \right )^{m} b^{3}+\left (\int \sec \left (f x +e \right )^{m}d x \right ) a^{3} f \,m^{2}+2 \left (\int \sec \left (f x +e \right )^{m}d x \right ) a^{3} f m +3 \left (\int \sec \left (f x +e \right )^{m} \tan \left (f x +e \right )^{2}d x \right ) a \,b^{2} f \,m^{2}+6 \left (\int \sec \left (f x +e \right )^{m} \tan \left (f x +e \right )^{2}d x \right ) a \,b^{2} f m \right )}{f m \left (m +2\right )} \] Input:

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

Output:

(d**m*(sec(e + f*x)**m*tan(e + f*x)**2*b**3*m + 3*sec(e + f*x)**m*a**2*b*m 
 + 6*sec(e + f*x)**m*a**2*b - 2*sec(e + f*x)**m*b**3 + int(sec(e + f*x)**m 
,x)*a**3*f*m**2 + 2*int(sec(e + f*x)**m,x)*a**3*f*m + 3*int(sec(e + f*x)** 
m*tan(e + f*x)**2,x)*a*b**2*f*m**2 + 6*int(sec(e + f*x)**m*tan(e + f*x)**2 
,x)*a*b**2*f*m))/(f*m*(m + 2))