3.2.90 \(\int (a \tan (e+f x))^m (b \tan (e+f x))^n \, dx\) [190]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, 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, 3557, 371} \begin {gather*} \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)} \end {gather*}

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 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 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]

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 ) \text {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*}

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Mathematica [A]
time = 0.08, size = 66, normalized size = 1.05 \begin {gather*} \frac {\, _2F_1\left (1,\frac {1}{2} (1+m+n);1+\frac {1}{2} (1+m+n);-\tan ^2(e+f x)\right ) \tan (e+f x) (a \tan (e+f x))^m (b \tan (e+f x))^n}{f (1+m+n)} \end {gather*}

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 = 0.22, size = 0, normalized size = 0.00 \[\int \left (a \tan \left (f x +e \right )\right )^{m} \left (b \tan \left (f x +e \right )\right )^{n}\, dx\]

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 = 0.00, 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((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 = 0.00, 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((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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \tan {\left (e + f x \right )}\right )^{m} \left (b \tan {\left (e + f x \right )}\right )^{n}\, dx \end {gather*}

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 = 0.00, 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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (a\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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