3.5.23 \(\int (d \cot (e+f x))^m (b (c \tan (e+f x))^n)^p \, dx\) [423]

Optimal. Leaf size=80 \[ \frac {(d \cot (e+f x))^m \, _2F_1\left (1,\frac {1}{2} (1-m+n p);\frac {1}{2} (3-m+n p);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1-m+n p)} \]

[Out]

(d*cot(f*x+e))^m*hypergeom([1, 1/2*n*p-1/2*m+1/2],[1/2*n*p-1/2*m+3/2],-tan(f*x+e)^2)*tan(f*x+e)*(b*(c*tan(f*x+
e))^n)^p/f/(n*p-m+1)

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Rubi [A]
time = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3740, 2684, 3557, 371} \begin {gather*} \frac {\tan (e+f x) (d \cot (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, _2F_1\left (1,\frac {1}{2} (-m+n p+1);\frac {1}{2} (-m+n p+3);-\tan ^2(e+f x)\right )}{f (-m+n p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Cot[e + f*x])^m*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

((d*Cot[e + f*x])^m*Hypergeometric2F1[1, (1 - m + n*p)/2, (3 - m + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(b*(c
*Tan[e + f*x])^n)^p)/(f*(1 - m + n*p))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2684

Int[(cot[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cot[e + f*
x])^m*(b*Tan[e + f*x])^m, Int[(b*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[m
] &&  !IntegerQ[n]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

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 (d \cot (e+f x))^m \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 (d \cot (e+f x))^m (c \tan (e+f x))^{n p} \, dx\\ &=\left ((d \cot (e+f x))^m (c \tan (e+f x))^{m-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int (c \tan (e+f x))^{-m+n p} \, dx\\ &=\frac {\left (c (d \cot (e+f x))^m (c \tan (e+f x))^{m-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \text {Subst}\left (\int \frac {x^{-m+n p}}{c^2+x^2} \, dx,x,c \tan (e+f x)\right )}{f}\\ &=\frac {(d \cot (e+f x))^m \, _2F_1\left (1,\frac {1}{2} (1-m+n p);\frac {1}{2} (3-m+n p);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1-m+n p)}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 77, normalized size = 0.96 \begin {gather*} \frac {d (d \cot (e+f x))^{-1+m} \, _2F_1\left (1,\frac {1}{2} (1-m+n p);\frac {1}{2} (3-m+n p);-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (1-m+n p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Cot[e + f*x])^m*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(d*(d*Cot[e + f*x])^(-1 + m)*Hypergeometric2F1[1, (1 - m + n*p)/2, (3 - m + n*p)/2, -Tan[e + f*x]^2]*(b*(c*Tan
[e + f*x])^n)^p)/(f*(1 - m + n*p))

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Maple [F]
time = 0.36, size = 0, normalized size = 0.00 \[\int \left (d \cot \left (f x +e \right )\right )^{m} \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cot(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int((d*cot(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*(d*cot(f*x + e))^m, x)

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Fricas [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((d*cot(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")

[Out]

integral(((c*tan(f*x + e))^n*b)^p*(d*cot(f*x + e))^m, x)

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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} \left (d \cot {\left (e + f x \right )}\right )^{m}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))**m*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*(d*cot(e + f*x))**m, 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((d*cot(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*(d*cot(f*x + e))^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,\mathrm {cot}\left (e+f\,x\right )\right )}^m\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cot(e + f*x))^m*(b*(c*tan(e + f*x))^n)^p,x)

[Out]

int((d*cot(e + f*x))^m*(b*(c*tan(e + f*x))^n)^p, x)

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