3.4.87 \(\int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx\) [387]

Optimal. Leaf size=89 \[ \frac {\cos ^2(e+f x)^{\frac {1+n}{2}} (a \csc (e+f x))^m \, _2F_1\left (\frac {1+n}{2},\frac {1}{2} (1-m+n);\frac {1}{2} (3-m+n);\sin ^2(e+f x)\right ) (b \tan (e+f x))^{1+n}}{b f (1-m+n)} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 89, 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 = {2698, 2682, 2657} \begin {gather*} \frac {\cos ^2(e+f x)^{\frac {n+1}{2}} (a \csc (e+f x))^m (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac {n+1}{2},\frac {1}{2} (-m+n+1);\frac {1}{2} (-m+n+3);\sin ^2(e+f x)\right )}{b f (-m+n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2682

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[a*Cos[e + f*
x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b*(a*Sin[e + f*x])^(n + 1))), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rule 2698

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 2.12, size = 287, normalized size = 3.22 \begin {gather*} -\frac {a (-3+m-n) F_1\left (\frac {1}{2} (1-m+n);n,1-m;\frac {1}{2} (3-m+n);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (a \csc (e+f x))^{-1+m} (b \tan (e+f x))^n}{f (-1+m-n) \left ((-3+m-n) F_1\left (\frac {1}{2} (1-m+n);n,1-m;\frac {1}{2} (3-m+n);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((-1+m) F_1\left (\frac {1}{2} (3-m+n);n,2-m;\frac {1}{2} (5-m+n);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n F_1\left (\frac {1}{2} (3-m+n);1+n,1-m;\frac {1}{2} (5-m+n);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a*(-3 + m - n)*AppellF1[(1 - m + n)/2, n, 1 - m, (3 - m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(a
*Csc[e + f*x])^(-1 + m)*(b*Tan[e + f*x])^n)/(f*(-1 + m - n)*((-3 + m - n)*AppellF1[(1 - m + n)/2, n, 1 - m, (3
- m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((-1 + m)*AppellF1[(3 - m + n)/2, n, 2 - m, (5 - m +
n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[(3 - m + n)/2, 1 + n, 1 - m, (5 - m + n)/2, Tan[(
e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)))

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Maple [F]
time = 0.26, size = 0, normalized size = 0.00 \[\int \left (a \csc \left (f x +e \right )\right )^{m} \left (b \tan \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

integrate((a*csc(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 (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (\frac {a}{\sin \left (e+f\,x\right )}\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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