3.5.96 \(\int \csc ^3(e+f x) (b \sec (e+f x))^n \, dx\) [496]

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

[Out]

hypergeom([2, 3/2+1/2*n],[5/2+1/2*n],sec(f*x+e)^2)*(b*sec(f*x+e))^(3+n)/b^3/f/(3+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 = {2702, 371} \begin {gather*} \frac {(b \sec (e+f x))^{n+3} \, _2F_1\left (2,\frac {n+3}{2};\frac {n+5}{2};\sec ^2(e+f x)\right )}{b^3 f (n+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*(b*Sec[e + f*x])^n,x]

[Out]

(Hypergeometric2F1[2, (3 + n)/2, (5 + n)/2, Sec[e + f*x]^2]*(b*Sec[e + f*x])^(3 + n))/(b^3*f*(3 + 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 2702

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

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(48)=96\).
time = 3.09, size = 201, normalized size = 4.19 \begin {gather*} \frac {b (b \sec (e+f x))^{-1+n} \left (2 \, _2F_1(1,1-n;2-n;\cos (e+f x))+2 \, _2F_1(2,1-n;2-n;\cos (e+f x))+2^n \, _2F_1\left (1-n,-n;2-n;\frac {1}{2} \cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )^{1-n}+2^n \, _2F_1\left (1-n,1-n;2-n;\frac {1}{2} \cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sec ^{1-n}(e+f x) \left (\cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)\right )^{-1+n}\right )}{8 f (-1+n)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^3*(b*Sec[e + f*x])^n,x]

[Out]

(b*(b*Sec[e + f*x])^(-1 + n)*(2*Hypergeometric2F1[1, 1 - n, 2 - n, Cos[e + f*x]] + 2*Hypergeometric2F1[2, 1 -
n, 2 - n, Cos[e + f*x]] + 2^n*Hypergeometric2F1[1 - n, -n, 2 - n, (Cos[e + f*x]*Sec[(e + f*x)/2]^2)/2]*(Sec[(e
+ f*x)/2]^2)^(1 - n) + 2^n*Hypergeometric2F1[1 - n, 1 - n, 2 - n, (Cos[e + f*x]*Sec[(e + f*x)/2]^2)/2]*Sec[e
+ f*x]^(1 - n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1 + n)))/(8*f*(-1 + n))

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (\csc ^{3}\left (f x +e \right )\right ) \left (b \sec \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(b*sec(f*x+e))^n,x)

[Out]

int(csc(f*x+e)^3*(b*sec(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(csc(f*x+e)^3*(b*sec(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^n*csc(f*x + e)^3, 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(csc(f*x+e)^3*(b*sec(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e))^n*csc(f*x + e)^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \sec {\left (e + f x \right )}\right )^{n} \csc ^{3}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3*(b*sec(f*x+e))**n,x)

[Out]

Integral((b*sec(e + f*x))**n*csc(e + f*x)**3, 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(csc(f*x+e)^3*(b*sec(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n*csc(f*x + e)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n}{{\sin \left (e+f\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(e + f*x))^n/sin(e + f*x)^3,x)

[Out]

int((b/cos(e + f*x))^n/sin(e + f*x)^3, x)

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