3.4.75 \(\int (b \csc (e+f x))^m \tan (e+f x) \, dx\) [375]

Optimal. Leaf size=39 \[ \frac {(b \csc (e+f x))^m \, _2F_1\left (1,\frac {m}{2};\frac {2+m}{2};\csc ^2(e+f x)\right )}{f m} \]

[Out]

(b*csc(f*x+e))^m*hypergeom([1, 1/2*m],[1+1/2*m],csc(f*x+e)^2)/f/m

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Rubi [A]
time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2686, 371} \begin {gather*} \frac {(b \csc (e+f x))^m \, _2F_1\left (1,\frac {m}{2};\frac {m+2}{2};\csc ^2(e+f x)\right )}{f m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Csc[e + f*x])^m*Hypergeometric2F1[1, m/2, (2 + m)/2, Csc[e + f*x]^2])/(f*m)

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 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 52, normalized size = 1.33 \begin {gather*} -\frac {(b \csc (e+f x))^m \, _2F_1\left (1,1-\frac {m}{2};2-\frac {m}{2};\sin ^2(e+f x)\right ) \sin ^2(e+f x)}{f (-2+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*Csc[e + f*x])^m*Hypergeometric2F1[1, 1 - m/2, 2 - m/2, Sin[e + f*x]^2]*Sin[e + f*x]^2)/(f*(-2 + m)))

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (b \csc \left (f x +e \right )\right )^{m} \tan \left (f x +e \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \csc {\left (e + f x \right )}\right )^{m} \tan {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \mathrm {tan}\left (e+f\,x\right )\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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