3.3.84 \(\int (b \csc (e+f x))^n \sec ^5(e+f x) \, dx\) [284]

Optimal. Leaf size=48 \[ \frac {(b \csc (e+f x))^{5+n} \, _2F_1\left (3,\frac {5+n}{2};\frac {7+n}{2};\csc ^2(e+f x)\right )}{b^5 f (5+n)} \]

[Out]

(b*csc(f*x+e))^(5+n)*hypergeom([3, 5/2+1/2*n],[7/2+1/2*n],csc(f*x+e)^2)/b^5/f/(5+n)

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Rubi [A]
time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2701, 371} \begin {gather*} \frac {(b \csc (e+f x))^{n+5} \, _2F_1\left (3,\frac {n+5}{2};\frac {n+7}{2};\csc ^2(e+f x)\right )}{b^5 f (n+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Csc[e + f*x])^n*Sec[e + f*x]^5,x]

[Out]

((b*Csc[e + f*x])^(5 + n)*Hypergeometric2F1[3, (5 + n)/2, (7 + n)/2, Csc[e + f*x]^2])/(b^5*f*(5 + 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 2701

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

Rubi steps

\begin {align*} \int (b \csc (e+f x))^n \sec ^5(e+f x) \, dx &=-\frac {\text {Subst}\left (\int \frac {x^{4+n}}{\left (-1+\frac {x^2}{b^2}\right )^3} \, dx,x,b \csc (e+f x)\right )}{b^5 f}\\ &=\frac {(b \csc (e+f x))^{5+n} \, _2F_1\left (3,\frac {5+n}{2};\frac {7+n}{2};\csc ^2(e+f x)\right )}{b^5 f (5+n)}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 51, normalized size = 1.06 \begin {gather*} -\frac {b (b \csc (e+f x))^{-1+n} \, _2F_1\left (3,\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{f (-1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Csc[e + f*x])^n*Sec[e + f*x]^5,x]

[Out]

-((b*(b*Csc[e + f*x])^(-1 + n)*Hypergeometric2F1[3, (1 - n)/2, (3 - n)/2, Sin[e + f*x]^2])/(f*(-1 + n)))

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Maple [F]
time = 0.30, size = 0, normalized size = 0.00 \[\int \left (b \csc \left (f x +e \right )\right )^{n} \left (\sec ^{5}\left (f x +e \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*csc(f*x+e))^n*sec(f*x+e)^5,x)

[Out]

int((b*csc(f*x+e))^n*sec(f*x+e)^5,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))^n*sec(f*x+e)^5,x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^n*sec(f*x + e)^5, 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))^n*sec(f*x+e)^5,x, algorithm="fricas")

[Out]

integral((b*csc(f*x + e))^n*sec(f*x + e)^5, 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 )^{n} \sec ^{5}{\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))**n*sec(f*x+e)**5,x)

[Out]

Integral((b*csc(e + f*x))**n*sec(e + f*x)**5, 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))^n*sec(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^n*sec(f*x + e)^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n}{{\cos \left (e+f\,x\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/sin(e + f*x))^n/cos(e + f*x)^5,x)

[Out]

int((b/sin(e + f*x))^n/cos(e + f*x)^5, x)

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