3.2.67 \(\int \csc ^2(e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [167]

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 = 0.06, 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 = {3740, 2671, 30} \begin {gather*} -\frac {\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1-n p)} \end {gather*}

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 2671

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]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 3740

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 ) \text {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 = 0.04, size = 31, normalized size = 0.94 \begin {gather*} \frac {\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (-1+n p)} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 6.62, size = 10285, normalized size = 311.67

method	result	size
risch	\(\text {Expression too large to display}\)	\(10285\)

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,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [A]
time = 0.31, size = 39, normalized size = 1.18 \begin {gather*} \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 )} \end {gather*}

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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Fricas [A]
time = 1.44, size = 55, normalized size = 1.67 \begin {gather*} \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 \left (b\right )\right )}}{{\left (f n p - f\right )} \sin \left (f x + e\right )} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \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 \end {gather*}

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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