3.167 \(\int \csc ^2(e+f x) (b (c \tan (e+f x))^n)^p \, dx\)

Optimal. Leaf size=33 \[ -\frac {\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1-n p)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3659, 2591, 30} \[ -\frac {\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1-n p)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2*(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)))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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 \csc ^2(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 ^2(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^{-2+n p} \, 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)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 31, normalized size = 0.94 \[ \frac {\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (n p-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2*(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))

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fricas [A]  time = 0.45, size = 51, normalized size = 1.55 \[ \frac {\cos \left (f x + e\right ) e^{\left (n p \log \left (\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + p \log \relax (b)\right )}}{{\left (f n p - f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(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)^2, x)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (\csc ^{2}\left (f x +e \right )\right ) \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int(csc(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x)

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maxima [A]  time = 0.77, size = 37, normalized size = 1.12 \[ \frac {b^{p} c^{n p} {\left (\tan \left (f x + e\right )^{n}\right )}^{p}}{{\left (n p - 1\right )} f \tan \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(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)*f*tan(f*x + e))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\sin \left (e+f\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^2,x)

[Out]

int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \csc ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**2*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*csc(e + f*x)**2, x)

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