3.2.89 \(\int (a \cos (e+f x))^m (b \tan (e+f x))^n \, dx\) [189]

Optimal. Leaf size=86 \[ \frac {(a \cos (e+f x))^m \cos ^2(e+f x)^{\frac {1}{2} (1-m+n)} \, _2F_1\left (\frac {1+n}{2},\frac {1}{2} (1-m+n);\frac {3+n}{2};\sin ^2(e+f x)\right ) (b \tan (e+f x))^{1+n}}{b f (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2683, 2697} \begin {gather*} \frac {(a \cos (e+f x))^m (b \tan (e+f x))^{n+1} \cos ^2(e+f x)^{\frac {1}{2} (-m+n+1)} \, _2F_1\left (\frac {n+1}{2},\frac {1}{2} (-m+n+1);\frac {n+3}{2};\sin ^2(e+f x)\right )}{b f (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2683

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

Rule 2697

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*Sec[e + f
*x])^m*(b*Tan[e + f*x])^(n + 1)*((Cos[e + f*x]^2)^((m + n + 1)/2)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2,
(m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

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

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Mathematica [A]
time = 0.54, size = 81, normalized size = 0.94 \begin {gather*} \frac {(a \cos (e+f x))^m \, _2F_1\left (\frac {2+m}{2},\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{m/2} \tan (e+f x) (b \tan (e+f x))^n}{f (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \left (a \cos \left (f x +e \right )\right )^{m} \left (b \tan \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((a*cos(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 \cos {\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*cos(f*x+e))**m*(b*tan(f*x+e))**n,x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a\,\cos \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*cos(e + f*x))^m*(b*tan(e + f*x))^n,x)

[Out]

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

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