3.168 \(\int \csc ^4(e+f x) (b (c \tan (e+f x))^n)^p \, dx\)

Optimal. Leaf size=69 \[ -\frac{\cot ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (3-n p)}-\frac{\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1-n p)} \]

[Out]

-((Cot[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 - n*p))) - (Cot[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p)/(f*(3 - n
*p))

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Rubi [A]  time = 0.116107, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3659, 2591, 14} \[ -\frac{\cot ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (3-n p)}-\frac{\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1-n p)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

-((Cot[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 - n*p))) - (Cot[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p)/(f*(3 - 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 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.15638, size = 59, normalized size = 0.86 \[ \frac{\cot (e+f x) \csc ^2(e+f x) (\cos (2 (e+f x))+n p-2) \left (b (c \tan (e+f x))^n\right )^p}{f (n p-3) (n p-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

((-2 + n*p + Cos[2*(e + f*x)])*Cot[e + f*x]*Csc[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p)/(f*(-3 + n*p)*(-1 + n*p))

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( fx+e \right ) \right ) ^{4} \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)^4*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int(csc(f*x+e)^4*(b*(c*tan(f*x+e))^n)^p,x)

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Maxima [A]  time = 1.24483, size = 99, normalized size = 1.43 \begin{align*} \frac{\frac{b^{p}{\left (c^{n}\right )}^{p}{\left (\tan \left (f x + e\right )^{n}\right )}^{p}}{{\left (n p - 1\right )} \tan \left (f x + e\right )} + \frac{b^{p}{\left (c^{n}\right )}^{p}{\left (\tan \left (f x + e\right )^{n}\right )}^{p}}{{\left (n p - 3\right )} \tan \left (f x + e\right )^{3}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")

[Out]

(b^p*(c^n)^p*(tan(f*x + e)^n)^p/((n*p - 1)*tan(f*x + e)) + b^p*(c^n)^p*(tan(f*x + e)^n)^p/((n*p - 3)*tan(f*x +
e)^3))/f

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Fricas [A]  time = 1.08657, size = 250, normalized size = 3.62 \begin{align*} \frac{{\left (2 \, \cos \left (f x + e\right )^{3} +{\left (n p - 3\right )} \cos \left (f x + e\right )\right )} e^{\left (n p \log \left (\frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + p \log \left (b\right )\right )}}{{\left (f n^{2} p^{2} - 4 \, f n p -{\left (f n^{2} p^{2} - 4 \, f n p + 3 \, f\right )} \cos \left (f x + e\right )^{2} + 3 \, f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")

[Out]

(2*cos(f*x + e)^3 + (n*p - 3)*cos(f*x + e))*e^(n*p*log(c*sin(f*x + e)/cos(f*x + e)) + p*log(b))/((f*n^2*p^2 -
4*f*n*p - (f*n^2*p^2 - 4*f*n*p + 3*f)*cos(f*x + e)^2 + 3*f)*sin(f*x + e))

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

[Out]

Timed out

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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} \csc \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(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)^4, x)