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

Optimal. Leaf size=53 \[ -\frac{b^3 (b \tan (e+f x))^{n-3}}{f (3-n)}-\frac{b (b \tan (e+f x))^{n-1}}{f (1-n)} \]

[Out]

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

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Rubi [A]  time = 0.0550273, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2591, 14} \[ -\frac{b^3 (b \tan (e+f x))^{n-3}}{f (3-n)}-\frac{b (b \tan (e+f x))^{n-1}}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.147351, size = 46, normalized size = 0.87 \[ \frac{b \csc ^2(e+f x) (\cos (2 (e+f x))+n-2) (b \tan (e+f x))^{n-1}}{f (n-3) (n-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.641, size = 13019, normalized size = 245.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [A]  time = 0.97252, size = 74, normalized size = 1.4 \begin{align*} \frac{\frac{b^{n} \tan \left (f x + e\right )^{n}}{{\left (n - 1\right )} \tan \left (f x + e\right )} + \frac{b^{n} \tan \left (f x + e\right )^{n}}{{\left (n - 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*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.62222, size = 204, normalized size = 3.85 \begin{align*} \frac{{\left (2 \, \cos \left (f x + e\right )^{3} +{\left (n - 3\right )} \cos \left (f x + e\right )\right )} \left (\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right )^{n}}{{\left (f n^{2} -{\left (f n^{2} - 4 \, f n + 3 \, f\right )} \cos \left (f x + e\right )^{2} - 4 \, f n + 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*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

(2*cos(f*x + e)^3 + (n - 3)*cos(f*x + e))*(b*sin(f*x + e)/cos(f*x + e))^n/((f*n^2 - (f*n^2 - 4*f*n + 3*f)*cos(
f*x + e)^2 - 4*f*n + 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*tan(f*x+e))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )\right )^{n} \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*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^n*csc(f*x + e)^4, x)