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

Optimal. Leaf size=91 \[ \frac {\tan (e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+2)} \, _2F_1\left (\frac {1}{2} (n p+1),\frac {1}{2} (n p+2);\frac {1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]

[Out]

(cos(f*x+e)^2)^(1/2*n*p+1)*hypergeom([1/2*n*p+1, 1/2*n*p+1/2],[1/2*n*p+3/2],sin(f*x+e)^2)*sec(f*x+e)*tan(f*x+e
)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+1)

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Rubi [A]  time = 0.06, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3659, 2617} \[ \frac {\tan (e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+2)} \, _2F_1\left (\frac {1}{2} (n p+1),\frac {1}{2} (n p+2);\frac {1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^((2 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (2 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sec[e
+ f*x]*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + n*p))

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rule 3659

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Tan[e + f*x
])^n)^FracPart[p])/(c*Tan[e + f*x])^(n*FracPart[p]), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 70, normalized size = 0.77 \[ \frac {\csc (e+f x) \left (-\tan ^2(e+f x)\right )^{\frac {1}{2} (1-n p)} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {3}{2};\sec ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e + f*x]*Hypergeometric2F1[1/2, (1 - n*p)/2, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^((1 - n*p)/2)*(b*(c*T
an[e + f*x])^n)^p)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 12.00, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{\cos \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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