3.150 \(\int \csc ^3(c+d x) (a+a \sec (c+d x))^n \, dx\)

Optimal. Leaf size=112 \[ -\frac{(n+2) (a \sec (c+d x)+a)^n \text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (\sec (c+d x)+1)\right )}{8 d n}-\frac{a (2-n) (a \sec (c+d x)+a)^{n-1}}{4 d (1-n)}+\frac{a (a \sec (c+d x)+a)^{n-1}}{2 d (1-\sec (c+d x))} \]

[Out]

-(a*(2 - n)*(a + a*Sec[c + d*x])^(-1 + n))/(4*d*(1 - n)) + (a*(a + a*Sec[c + d*x])^(-1 + n))/(2*d*(1 - Sec[c +
d*x])) - ((2 + n)*Hypergeometric2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^n)/(8*d*n)

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Rubi [A]  time = 0.0968238, antiderivative size = 112, 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 = {3873, 89, 79, 68} \[ -\frac{(n+2) (a \sec (c+d x)+a)^n \, _2F_1\left (1,n;n+1;\frac{1}{2} (\sec (c+d x)+1)\right )}{8 d n}-\frac{a (2-n) (a \sec (c+d x)+a)^{n-1}}{4 d (1-n)}+\frac{a (a \sec (c+d x)+a)^{n-1}}{2 d (1-\sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*(a + a*Sec[c + d*x])^n,x]

[Out]

-(a*(2 - n)*(a + a*Sec[c + d*x])^(-1 + n))/(4*d*(1 - n)) + (a*(a + a*Sec[c + d*x])^(-1 + n))/(2*d*(1 - Sec[c +
d*x])) - ((2 + n)*Hypergeometric2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^n)/(8*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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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 ^3(c+d x) (a+a \sec (c+d x))^n \, dx &=-\frac{a^4 \operatorname{Subst}\left (\int \frac{x^2 (a-a x)^{-2+n}}{(-a-a x)^2} \, dx,x,-\sec (c+d x)\right )}{d}\\ &=\frac{a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))}+\frac{\operatorname{Subst}\left (\int \frac{(a-a x)^{-2+n} \left (-a^3 n+2 a^3 x\right )}{-a-a x} \, dx,x,-\sec (c+d x)\right )}{2 d}\\ &=-\frac{a (2-n) (a+a \sec (c+d x))^{-1+n}}{4 d (1-n)}+\frac{a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))}-\frac{\left (a^2 (2+n)\right ) \operatorname{Subst}\left (\int \frac{(a-a x)^{-1+n}}{-a-a x} \, dx,x,-\sec (c+d x)\right )}{4 d}\\ &=-\frac{a (2-n) (a+a \sec (c+d x))^{-1+n}}{4 d (1-n)}+\frac{a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))}-\frac{(2+n) \, _2F_1\left (1,n;1+n;\frac{1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^n}{8 d n}\\ \end{align*}

Mathematica [A]  time = 1.68421, size = 123, normalized size = 1.1 \[ \frac{2^{n-4} (\sec (c+d x)+1)^{-n} (a (\sec (c+d x)+1))^n \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{n-1} \left (2 (n+2) \text{Hypergeometric2F1}\left (1,1-n,2-n,\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )-\csc ^2\left (\frac{1}{2} (c+d x)\right ) (n \cos (c+d x)+n-2)\right )}{d (n-1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^3*(a + a*Sec[c + d*x])^n,x]

[Out]

(2^(-4 + n)*(-((-2 + n + n*Cos[c + d*x])*Csc[(c + d*x)/2]^2) + 2*(2 + n)*Hypergeometric2F1[1, 1 - n, 2 - n, Co
s[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 = 0.275, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( dx+c \right ) \right ) ^{3} \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)^3*(a+a*sec(d*x+c))^n,x)

[Out]

int(csc(d*x+c)^3*(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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+a*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^3, 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 )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+a*sec(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*csc(d*x + c)^3, 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)**3*(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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+a*sec(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^3, x)