\(\int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx\) [640]

Optimal result

Integrand size = 23, antiderivative size = 173 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx=-\frac {a \left (3 b^2-a^2 (1+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-\frac {m}{2},\frac {3}{2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x)}{f (1+m)}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))^2}{f (2+m)}-\frac {b (d \sec (e+f x))^m \left (2 (1+m) \left (b^2-a^2 (3+m)\right )-a b m (4+m) \tan (e+f x)\right )}{f m \left (2+3 m+m^2\right )} \]

[Out]

-a*(3*b^2-a^2*(1+m))*hypergeom([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))+b*(d*sec(f*x+e))^m*(a+b*tan(f*x+e))^2/f/(2+m)-b*(d*sec(f*x+e))^m*(2*(1+m)*(b^2-a^2*(3+m))-a*
b*m*(4+m)*tan(f*x+e))/f/m/(m^2+3*m+2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3593, 757, 794, 251} \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx=\frac {a \left (a^2-\frac {3 b^2}{m+1}\right ) \tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-\frac {m}{2},\frac {3}{2},-\tan ^2(e+f x)\right )}{f}-\frac {b (d \sec (e+f x))^m \left (2 (m+1) \left (b^2-a^2 (m+3)\right )-a b m (m+4) \tan (e+f x)\right )}{f m \left (m^2+3 m+2\right )}+\frac {b (a+b \tan (e+f x))^2 (d \sec (e+f x))^m}{f (m+2)} \]

[In]

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

[Out]

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

Rule 251

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

Rule 757

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

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(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 3593

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[d^(2*
IntPart[m/2])*((d*Sec[e + f*x])^(2*FracPart[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/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \text {Subst}\left (\int (a+x)^3 \left (1+\frac {x^2}{b^2}\right )^{-1+\frac {m}{2}} \, dx,x,b \tan (e+f x)\right )}{b f} \\ & = \frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))^2}{f (2+m)}+\frac {\left (b (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \text {Subst}\left (\int (a+x) \left (-2+\frac {a^2 (2+m)}{b^2}+\frac {a (4+m) x}{b^2}\right ) \left (1+\frac {x^2}{b^2}\right )^{-1+\frac {m}{2}} \, dx,x,b \tan (e+f x)\right )}{f (2+m)} \\ & = \frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))^2}{f (2+m)}-\frac {b (d \sec (e+f x))^m \left (2 (1+m) \left (b^2-a^2 (3+m)\right )-a b m (4+m) \tan (e+f x)\right )}{f m \left (2+3 m+m^2\right )}-\frac {\left (a \left (3 b^2-a^2 (1+m)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \text {Subst}\left (\int \left (1+\frac {x^2}{b^2}\right )^{-1+\frac {m}{2}} \, dx,x,b \tan (e+f x)\right )}{b 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)}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))^2}{f (2+m)}-\frac {b (d \sec (e+f x))^m \left (2 (1+m) \left (b^2-a^2 (3+m)\right )-a b m (4+m) \tan (e+f x)\right )}{f m \left (2+3 m+m^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.15 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.92 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^3 \, dx=\frac {(d \sec (e+f x))^m \left (3 a b^2 (2+m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)-a^3 (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)+b \left (\left (3 a^2-b^2\right ) (2+m)+b^2 m \sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f m (2+m) \sqrt {-\tan ^2(e+f x)}} \]

[In]

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

[Out]

((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])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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