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

Optimal. Leaf size=92 \[ -\frac{\csc ^2(e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p+1)} \text{Hypergeometric2F1}\left (\frac{1}{2} (n p-2),\frac{1}{2} (n p+1),\frac{n p}{2},\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.145556, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3659, 2601, 2577} \[ -\frac{\csc ^2(e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p+1)} \, _2F_1\left (\frac{1}{2} (n p-2),\frac{1}{2} (n p+1);\frac{n p}{2};\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((Cos[e + f*x]^2)^((1 + n*p)/2)*Csc[e + f*x]^2*Hypergeometric2F1[(-2 + n*p)/2, (1 + n*p)/2, (n*p)/2, Sin[e +
f*x]^2]*Sec[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(2 - 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 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

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

Mathematica [B]  time = 7.30732, size = 217, normalized size = 2.36 \[ \frac{\tan ^2\left (\frac{1}{2} (e+f x)\right ) \left (\cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )\right )^{n p} \left (b (c \tan (e+f x))^n\right )^p \left (2 \left (n^2 p^2-4\right ) \cot ^2\left (\frac{1}{2} (e+f x)\right ) \text{Hypergeometric2F1}\left (\frac{n p}{2},n p,\frac{n p}{2}+1,\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+n p \left ((n p-2) \text{Hypergeometric2F1}\left (n p,\frac{n p}{2}+1,\frac{n p}{2}+2,\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+(n p+2) \cot ^4\left (\frac{1}{2} (e+f x)\right ) \text{Hypergeometric2F1}\left (n p,\frac{n p}{2}-1,\frac{n p}{2},\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{4 f n p \left (n^2 p^2-4\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*(-4 + n^2*p^2)*Cot[(e + f*x)/2]^2*Hypergeometric2F1[(n*p)/2, n*p, 1 + (n*p)/2, Tan[(e + f*x)/2]^2] + n*p*(
(2 + n*p)*Cot[(e + f*x)/2]^4*Hypergeometric2F1[n*p, -1 + (n*p)/2, (n*p)/2, Tan[(e + f*x)/2]^2] + (-2 + n*p)*Hy
pergeometric2F1[n*p, 1 + (n*p)/2, 2 + (n*p)/2, Tan[(e + f*x)/2]^2]))*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n*p)*T
an[(e + f*x)/2]^2*(b*(c*Tan[e + f*x])^n)^p)/(4*f*n*p*(-4 + n^2*p^2))

________________________________________________________________________________________

Maple [F]  time = 3.131, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \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(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

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

________________________________________________________________________________________

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} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(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*csc(f*x + e)^3, x)

________________________________________________________________________________________

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} \csc \left (f x + e\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(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*csc(f*x + e)^3, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(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*csc(f*x + e)^3, x)