\(\int (a+b \sec ^2(e+f x))^p \tan ^5(e+f x) \, dx\) [442]

Optimal result

Integrand size = 23, antiderivative size = 122 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=-\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)} \]

[Out]

-1/2*(a+2*b)*(a+b*sec(f*x+e)^2)^(p+1)/b^2/f/(p+1)-1/2*hypergeom([1, p+1],[2+p],1+b*sec(f*x+e)^2/a)*(a+b*sec(f*
x+e)^2)^(p+1)/a/f/(p+1)+1/2*(a+b*sec(f*x+e)^2)^(2+p)/b^2/f/(2+p)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4224, 457, 90, 67} \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=-\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{2 b^2 f (p+1)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{p+2}}{2 b^2 f (p+2)}-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)} \]

[In]

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

[Out]

-1/2*((a + 2*b)*(a + b*Sec[e + f*x]^2)^(1 + p))/(b^2*f*(1 + p)) - (Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*S
ec[e + f*x]^2)/a]*(a + b*Sec[e + f*x]^2)^(1 + p))/(2*a*f*(1 + p)) + (a + b*Sec[e + f*x]^2)^(2 + p)/(2*b^2*f*(2
+ p))

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4224

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x)
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \left (a+b x^2\right )^p}{x} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(-1+x)^2 (a+b x)^p}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {(-a-2 b) (a+b x)^p}{b}+\frac {(a+b x)^p}{x}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)}+\frac {\text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.77 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=-\frac {\left (a+b \sec ^2(e+f x)\right )^{1+p} \left (b^2 (2+p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right )+a \left (a+2 b (2+p)-b (1+p) \sec ^2(e+f x)\right )\right )}{2 a b^2 f (1+p) (2+p)} \]

[In]

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

[Out]

-1/2*((a + b*Sec[e + f*x]^2)^(1 + p)*(b^2*(2 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sec[e + f*x]^2)/a]
+ a*(a + 2*b*(2 + p) - b*(1 + p)*Sec[e + f*x]^2)))/(a*b^2*f*(1 + p)*(2 + p))

Maple [F]

\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \tan \left (f x +e \right )^{5}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b*sec(f*x + e)^2 + a)^p*tan(f*x + e)^5, x)

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*sec(f*x+e)**2)**p*tan(f*x+e)**5,x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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