3.463 \(\int (a+b (c \sec (e+f x))^n)^p \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=226 \[ \frac {\sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \, _2F_1\left (\frac {4}{n},-p;\frac {n+4}{n};-\frac {b (c \sec (e+f x))^n}{a}\right )}{4 f}-\frac {\sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \, _2F_1\left (\frac {2}{n},-p;\frac {n+2}{n};-\frac {b (c \sec (e+f x))^n}{a}\right )}{f}-\frac {\left (a+b (c \sec (e+f x))^n\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b (c \sec (e+f x))^n}{a}+1\right )}{a f n (p+1)} \]

[Out]

-hypergeom([1, 1+p],[2+p],1+b*(c*sec(f*x+e))^n/a)*(a+b*(c*sec(f*x+e))^n)^(1+p)/a/f/n/(1+p)-hypergeom([-p, 2/n]
,[(2+n)/n],-b*(c*sec(f*x+e))^n/a)*sec(f*x+e)^2*(a+b*(c*sec(f*x+e))^n)^p/f/((1+b*(c*sec(f*x+e))^n/a)^p)+1/4*hyp
ergeom([-p, 4/n],[(4+n)/n],-b*(c*sec(f*x+e))^n/a)*sec(f*x+e)^4*(a+b*(c*sec(f*x+e))^n)^p/f/((1+b*(c*sec(f*x+e))
^n/a)^p)

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Rubi [A]  time = 0.52, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4139, 6742, 367, 12, 266, 65, 365, 364} \[ \frac {\sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \, _2F_1\left (\frac {4}{n},-p;\frac {n+4}{n};-\frac {b (c \sec (e+f x))^n}{a}\right )}{4 f}-\frac {\sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \, _2F_1\left (\frac {2}{n},-p;\frac {n+2}{n};-\frac {b (c \sec (e+f x))^n}{a}\right )}{f}-\frac {\left (a+b (c \sec (e+f x))^n\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b (c \sec (e+f x))^n}{a}+1\right )}{a f n (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c*Sec[e + f*x])^n)^p*Tan[e + f*x]^5,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))) - (Hypergeometric2F1[2/n, -p, (2 + n)/n, -((b*(c*Sec[e + f*x])^n)/a)]*Sec[e + f*x]^2*(a + b*(c*Sec[
e + f*x])^n)^p)/(f*(1 + (b*(c*Sec[e + f*x])^n)/a)^p) + (Hypergeometric2F1[4/n, -p, (4 + n)/n, -((b*(c*Sec[e +
f*x])^n)/a)]*Sec[e + f*x]^4*(a + b*(c*Sec[e + f*x])^n)^p)/(4*f*(1 + (b*(c*Sec[e + f*x])^n)/a)^p)

Rule 12

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

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])

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 364

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

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 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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \left (a+b (c x)^n\right )^p}{x} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {\left (a+b (c x)^n\right )^p}{x}-2 x \left (a+b (c x)^n\right )^p+x^3 \left (a+b (c x)^n\right )^p\right ) \, dx,x,\sec (e+f x)\right )}{f}\\ &=\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 x^3 \left (a+b (c x)^n\right )^p \, dx,x,\sec (e+f x)\right )}{f}-\frac {2 \operatorname {Subst}\left (\int x \left (a+b (c x)^n\right )^p \, 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 {x^3 \left (a+b x^n\right )^p}{c^3} \, dx,x,c \sec (e+f x)\right )}{c f}-\frac {2 \operatorname {Subst}\left (\int \frac {x \left (a+b x^n\right )^p}{c} \, 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 x^3 \left (a+b x^n\right )^p \, dx,x,c \sec (e+f x)\right )}{c^4 f}-\frac {2 \operatorname {Subst}\left (\int x \left (a+b x^n\right )^p \, dx,x,c \sec (e+f x)\right )}{c^2 f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,(c \sec (e+f x))^n\right )}{f n}+\frac {\left (\left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b (c \sec (e+f x))^n}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int x^3 \left (1+\frac {b x^n}{a}\right )^p \, dx,x,c \sec (e+f x)\right )}{c^4 f}-\frac {\left (2 \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b (c \sec (e+f x))^n}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int x \left (1+\frac {b x^n}{a}\right )^p \, dx,x,c \sec (e+f x)\right )}{c^2 f}\\ &=-\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)}-\frac {\, _2F_1\left (\frac {2}{n},-p;\frac {2+n}{n};-\frac {b (c \sec (e+f x))^n}{a}\right ) \sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b (c \sec (e+f x))^n}{a}\right )^{-p}}{f}+\frac {\, _2F_1\left (\frac {4}{n},-p;\frac {4+n}{n};-\frac {b (c \sec (e+f x))^n}{a}\right ) \sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b (c \sec (e+f x))^n}{a}\right )^{-p}}{4 f}\\ \end {align*}

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Mathematica [A]  time = 10.12, size = 245, normalized size = 1.08 \[ \frac {\left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}+1\right )^{-p} \left (-4 \left (a+b \left (c \sqrt {\sec ^2(e+f x)}\right )^n\right ) \left (\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}+1\right )^p \, _2F_1\left (1,p+1;p+2;\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}+1\right )-4 a n (p+1) \sec ^2(e+f x) \, _2F_1\left (\frac {2}{n},-p;\frac {n+2}{n};-\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right )+a n (p+1) \sec ^4(e+f x) \, _2F_1\left (\frac {4}{n},-p;\frac {n+4}{n};-\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right )\right )}{4 a f n (p+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*(c*Sec[e + f*x])^n)^p*(-4*a*n*(1 + p)*Hypergeometric2F1[2/n, -p, (2 + n)/n, -((b*(c*Sqrt[Sec[e + f*x]^
2])^n)/a)]*Sec[e + f*x]^2 + a*n*(1 + p)*Hypergeometric2F1[4/n, -p, (4 + n)/n, -((b*(c*Sqrt[Sec[e + f*x]^2])^n)
/a)]*Sec[e + f*x]^4 - 4*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*(c*Sqrt[Sec[e + f*x]^2])^n)/a]*(a + b*(c*Sqr
t[Sec[e + f*x]^2])^n)*(1 + (b*(c*Sqrt[Sec[e + f*x]^2])^n)/a)^p))/(4*a*f*n*(1 + p)*(1 + (b*(c*Sqrt[Sec[e + f*x]
^2])^n)/a)^p)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5}, x\right ) \]

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)^5,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{d x} \]

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)^5,x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.92, size = 0, normalized size = 0.00 \[ \int \left (a +b \left (c \sec \left (f x +e \right )\right )^{n}\right )^{p} \left (\tan ^{5}\left (f x +e \right )\right )\, dx \]

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)^5,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{d x} \]

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)^5,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (a+b\,{\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^n\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**5,x)

[Out]

Timed out

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