3.154 \(\int \csc ^2(c+d x) (a+a \sec (c+d x))^n \, dx\)

Optimal. Leaf size=98 \[ \frac{2^{n-\frac{1}{2}} n \tan (c+d x) (\sec (c+d x)+1)^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{2}-n,\frac{3}{2},\frac{1}{2} (1-\sec (c+d x))\right )}{d}-\frac{\cot (c+d x) (a \sec (c+d x)+a)^n}{d} \]

[Out]

-((Cot[c + d*x]*(a + a*Sec[c + d*x])^n)/d) + (2^(-1/2 + n)*n*Hypergeometric2F1[1/2, 3/2 - n, 3/2, (1 - Sec[c +
d*x])/2]*(1 + Sec[c + d*x])^(-1/2 - n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x])/d

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Rubi [A]  time = 0.132263, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3875, 3828, 3827, 69} \[ \frac{2^{n-\frac{1}{2}} n \tan (c+d x) (\sec (c+d x)+1)^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n \, _2F_1\left (\frac{1}{2},\frac{3}{2}-n;\frac{3}{2};\frac{1}{2} (1-\sec (c+d x))\right )}{d}-\frac{\cot (c+d x) (a \sec (c+d x)+a)^n}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2*(a + a*Sec[c + d*x])^n,x]

[Out]

-((Cot[c + d*x]*(a + a*Sec[c + d*x])^n)/d) + (2^(-1/2 + n)*n*Hypergeometric2F1[1/2, 3/2 - n, 3/2, (1 - Sec[c +
d*x])/2]*(1 + Sec[c + d*x])^(-1/2 - n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x])/d

Rule 3875

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)/cos[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[(Tan[e + f*x]*(a
+ b*Csc[e + f*x])^m)/f, x] + Dist[b*m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e
, f, m}, x]

Rule 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 3827

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \csc ^2(c+d x) (a+a \sec (c+d x))^n \, dx &=-\frac{\cot (c+d x) (a+a \sec (c+d x))^n}{d}+(a n) \int \sec (c+d x) (a+a \sec (c+d x))^{-1+n} \, dx\\ &=-\frac{\cot (c+d x) (a+a \sec (c+d x))^n}{d}+\left (n (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec (c+d x) (1+\sec (c+d x))^{-1+n} \, dx\\ &=-\frac{\cot (c+d x) (a+a \sec (c+d x))^n}{d}-\frac{\left (n (1+\sec (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(1+x)^{-\frac{3}{2}+n}}{\sqrt{1-x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)}}\\ &=-\frac{\cot (c+d x) (a+a \sec (c+d x))^n}{d}+\frac{2^{-\frac{1}{2}+n} n \, _2F_1\left (\frac{1}{2},\frac{3}{2}-n;\frac{3}{2};\frac{1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 1.05219, size = 87, normalized size = 0.89 \[ -\frac{\tan \left (\frac{1}{2} (c+d x)\right ) (a (\sec (c+d x)+1))^n \left (-2 n \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^n \text{Hypergeometric2F1}\left (\frac{1}{2},n,\frac{3}{2},\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+\cot ^2\left (\frac{1}{2} (c+d x)\right )-1\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2*(a + a*Sec[c + d*x])^n,x]

[Out]

-((-1 + Cot[(c + d*x)/2]^2 - 2*n*Hypergeometric2F1[1/2, n, 3/2, Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x
)/2]^2)^n)*(a*(1 + Sec[c + d*x]))^n*Tan[(c + d*x)/2])/(2*d)

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Maple [F]  time = 0.276, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( dx+c \right ) \right ) ^{2} \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)^2*(a+a*sec(d*x+c))^n,x)

[Out]

int(csc(d*x+c)^2*(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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^2, 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 )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*csc(d*x + c)^2, 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)**2*(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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^2, x)