3.149 \(\int \csc (c+d x) (a+a \sec (c+d x))^n \, dx\)

Optimal. Leaf size=40 \[ -\frac{(a \sec (c+d x)+a)^n \text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (\sec (c+d x)+1)\right )}{2 d n} \]

[Out]

-(Hypergeometric2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^n)/(2*d*n)

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Rubi [A]  time = 0.0459735, antiderivative size = 40, normalized size of antiderivative = 1., 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 = {3873, 68} \[ -\frac{(a \sec (c+d x)+a)^n \, _2F_1\left (1,n;n+1;\frac{1}{2} (\sec (c+d x)+1)\right )}{2 d n} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]*(a + a*Sec[c + d*x])^n,x]

[Out]

-(Hypergeometric2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^n)/(2*d*n)

Rule 3873

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[(f*b^(p - 1)
)^(-1), Subst[Int[((-a + b*x)^((p - 1)/2)*(a + b*x)^(m + (p - 1)/2))/x^(p + 1), x], x, Csc[e + f*x]], x] /; Fr
eeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \csc (c+d x) (a+a \sec (c+d x))^n \, dx &=-\frac{a^2 \operatorname{Subst}\left (\int \frac{(a-a x)^{-1+n}}{-a-a x} \, dx,x,-\sec (c+d x)\right )}{d}\\ &=-\frac{\, _2F_1\left (1,n;1+n;\frac{1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^n}{2 d n}\\ \end{align*}

Mathematica [B]  time = 0.768554, size = 92, normalized size = 2.3 \[ \frac{2^{n-1} (\sec (c+d x)+1)^{-n} (a (\sec (c+d x)+1))^n \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{n-1} \text{Hypergeometric2F1}\left (1,1-n,2-n,\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )}{d (n-1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]*(a + a*Sec[c + d*x])^n,x]

[Out]

(2^(-1 + n)*Hypergeometric2F1[1, 1 - n, 2 - n, Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*
x])^(-1 + n)*(a*(1 + Sec[c + d*x]))^n)/(d*(-1 + n)*(1 + Sec[c + d*x])^n)

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Maple [F]  time = 0.254, size = 0, normalized size = 0. \begin{align*} \int \csc \left ( dx+c \right ) \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*(a+a*sec(d*x+c))^n,x)

[Out]

int(csc(d*x+c)*(a+a*sec(d*x+c))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+a*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+a*sec(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*csc(d*x + c), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+a*sec(d*x+c))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+a*sec(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c), x)