3.465 \(\int (a+b (c \sec (e+f x))^n)^p \tan (e+f x) \, dx\)

Optimal. Leaf size=59 \[ -\frac{\left (a+b (c \sec (e+f x))^n\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b (c \sec (e+f x))^n}{a}+1\right )}{a f n (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0769761, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4139, 367, 12, 266, 65} \[ -\frac{\left (a+b (c \sec (e+f x))^n\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b (c \sec (e+f x))^n}{a}+1\right )}{a f n (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4139

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])

Rule 367

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 266

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

Rule 65

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

Rubi steps

\begin{align*} \int \left (a+b (c \sec (e+f x))^n\right )^p \tan (e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b (c x)^n\right )^p}{x} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{c \left (a+b x^n\right )^p}{x} \, dx,x,c \sec (e+f x)\right )}{c f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^n\right )^p}{x} \, dx,x,c \sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,(c \sec (e+f x))^n\right )}{f n}\\ &=-\frac{\, _2F_1\left (1,1+p;2+p;1+\frac{b (c \sec (e+f x))^n}{a}\right ) \left (a+b (c \sec (e+f x))^n\right )^{1+p}}{a f n (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0895108, size = 59, normalized size = 1. \[ -\frac{\left (a+b (c \sec (e+f x))^n\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b (c \sec (e+f x))^n}{a}+1\right )}{a f n (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.553, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( c\sec \left ( fx+e \right ) \right ) ^{n} \right ) ^{p}\tan \left ( fx+e \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e),x)

[Out]

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \left (c \sec{\left (e + f x \right )}\right )^{n}\right )^{p} \tan{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*sec(f*x+e))**n)**p*tan(f*x+e),x)

[Out]

Integral((a + b*(c*sec(e + f*x))**n)**p*tan(e + f*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e), x)