3.482 \(\int \sec ^3(e+f x) (b (c \tan (e+f x))^n)^p \, dx\)

Optimal. Leaf size=93 \[ \frac{\tan (e+f x) \sec ^3(e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p+4)} \text{Hypergeometric2F1}\left (\frac{1}{2} (n p+1),\frac{1}{2} (n p+4),\frac{1}{2} (n p+3),\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]

[Out]

((Cos[e + f*x]^2)^((4 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (4 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sec[e
+ f*x]^3*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + n*p))

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Rubi [A]  time = 0.0968147, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3659, 2617} \[ \frac{\tan (e+f x) \sec ^3(e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p+4)} \, _2F_1\left (\frac{1}{2} (n p+1),\frac{1}{2} (n p+4);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

((Cos[e + f*x]^2)^((4 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (4 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sec[e
+ f*x]^3*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + n*p))

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]])

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]

Rubi steps

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

Mathematica [A]  time = 0.111356, size = 81, normalized size = 0.87 \[ \frac{\csc (e+f x) \sec ^2(e+f x) \left (-\tan ^2(e+f x)\right )^{\frac{1}{2} (1-n p)} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2} (1-n p),\frac{5}{2},\sec ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(Csc[e + f*x]*Hypergeometric2F1[3/2, (1 - n*p)/2, 5/2, Sec[e + f*x]^2]*Sec[e + f*x]^2*(-Tan[e + f*x]^2)^((1 -
n*p)/2)*(b*(c*Tan[e + f*x])^n)^p)/(3*f)

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Maple [F]  time = 0.537, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{3} \left ( b \left ( c\tan \left ( fx+e \right ) \right ) ^{n} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int(sec(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \sec \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \sec \left (f x + e\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \left (c \tan{\left (e + f x \right )}\right )^{n}\right )^{p} \sec ^{3}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**3*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*sec(e + f*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \sec \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)