3.190 \(\int (a \tan (e+f x))^m (b \tan (e+f x))^n \, dx\)

Optimal. Leaf size=63 \[ \frac{(a \tan (e+f x))^{m+1} (b \tan (e+f x))^n \, _2F_1\left (1,\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);-\tan ^2(e+f x)\right )}{a f (m+n+1)} \]

[Out]

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

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Rubi [A]  time = 0.0362079, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {20, 3476, 364} \[ \frac{(a \tan (e+f x))^{m+1} (b \tan (e+f x))^n \, _2F_1\left (1,\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);-\tan ^2(e+f x)\right )}{a f (m+n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Tan[e + f*x])^m*(b*Tan[e + f*x])^n,x]

[Out]

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 3476

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.0715552, size = 66, normalized size = 1.05 \[ \frac{\tan (e+f x) (a \tan (e+f x))^m (b \tan (e+f x))^n \, _2F_1\left (1,\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+1)+1;-\tan ^2(e+f x)\right )}{f (m+n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Tan[e + f*x])^m*(b*Tan[e + f*x])^n,x]

[Out]

(Hypergeometric2F1[1, (1 + m + n)/2, 1 + (1 + m + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(a*Tan[e + f*x])^m*(b*Ta
n[e + f*x])^n)/(f*(1 + m + n))

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Maple [F]  time = 0.393, size = 0, normalized size = 0. \begin{align*} \int \left ( a\tan \left ( fx+e \right ) \right ) ^{m} \left ( b\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*tan(f*x+e))^m*(b*tan(f*x+e))^n,x)

[Out]

int((a*tan(f*x+e))^m*(b*tan(f*x+e))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \tan \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tan(f*x+e))^m*(b*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((a*tan(f*x + e))^m*(b*tan(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (a \tan \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tan(f*x+e))^m*(b*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((a*tan(f*x + e))^m*(b*tan(f*x + e))^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \tan{\left (e + f x \right )}\right )^{m} \left (b \tan{\left (e + f x \right )}\right )^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tan(f*x+e))**m*(b*tan(f*x+e))**n,x)

[Out]

Integral((a*tan(e + f*x))**m*(b*tan(e + f*x))**n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tan(f*x+e))^m*(b*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((a*tan(f*x + e))^m*(b*tan(f*x + e))^n, x)