3.155 \(\int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx\)

Optimal. Leaf size=349 \[ \frac{n \left (-n^2-3 n+7\right ) \sin (c+d x) \cos (c+d x) \left (\frac{\cos (c+d x)+1}{1-\cos (c+d x)}\right )^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n \text{Hypergeometric2F1}\left (-n-\frac{1}{2},1-n,2-n,-\frac{2 \cos (c+d x)}{1-\cos (c+d x)}\right )}{d (1-2 n) (3-2 n) (1-n) (2 n+1) (1-\cos (c+d x))^2}-\frac{a^3 (4-n) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d \left (4 n^2-8 n+3\right ) (a-a \cos (c+d x))^2 (a \cos (c+d x)+a)}-\frac{a^4 \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d (3-2 n) (a-a \cos (c+d x))^2 (a \cos (c+d x)+a)^2}+\frac{\left (n^2-n+2\right ) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d (3-2 n) \left (1-4 n^2\right ) (1-\cos (c+d x))^2} \]

[Out]

((2 - n + n^2)*Cos[c + d*x]*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 2*n)*(1 - 4*n^2)*(1 - Cos[c + d*x])^2
) - (a^4*Cos[c + d*x]*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 2*n)*(a - a*Cos[c + d*x])^2*(a + a*Cos[c +
d*x])^2) - (a^3*(4 - n)*Cos[c + d*x]*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 8*n + 4*n^2)*(a - a*Cos[c +
d*x])^2*(a + a*Cos[c + d*x])) + (n*(7 - 3*n - n^2)*Cos[c + d*x]*((1 + Cos[c + d*x])/(1 - Cos[c + d*x]))^(-1/2
- n)*Hypergeometric2F1[-1/2 - n, 1 - n, 2 - n, (-2*Cos[c + d*x])/(1 - Cos[c + d*x])]*(a + a*Sec[c + d*x])^n*Si
n[c + d*x])/(d*(1 - 2*n)*(3 - 2*n)*(1 - n)*(1 + 2*n)*(1 - Cos[c + d*x])^2)

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Rubi [A]  time = 0.541432, antiderivative size = 349, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3876, 2883, 129, 155, 12, 132} \[ -\frac{a^3 (4-n) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d \left (4 n^2-8 n+3\right ) (a-a \cos (c+d x))^2 (a \cos (c+d x)+a)}-\frac{a^4 \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d (3-2 n) (a-a \cos (c+d x))^2 (a \cos (c+d x)+a)^2}+\frac{n \left (-n^2-3 n+7\right ) \sin (c+d x) \cos (c+d x) \left (\frac{\cos (c+d x)+1}{1-\cos (c+d x)}\right )^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n \, _2F_1\left (-n-\frac{1}{2},1-n;2-n;-\frac{2 \cos (c+d x)}{1-\cos (c+d x)}\right )}{d (1-2 n) (3-2 n) (1-n) (2 n+1) (1-\cos (c+d x))^2}+\frac{\left (n^2-n+2\right ) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d (3-2 n) \left (1-4 n^2\right ) (1-\cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*(a + a*Sec[c + d*x])^n,x]

[Out]

((2 - n + n^2)*Cos[c + d*x]*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 2*n)*(1 - 4*n^2)*(1 - Cos[c + d*x])^2
) - (a^4*Cos[c + d*x]*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 2*n)*(a - a*Cos[c + d*x])^2*(a + a*Cos[c +
d*x])^2) - (a^3*(4 - n)*Cos[c + d*x]*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 8*n + 4*n^2)*(a - a*Cos[c +
d*x])^2*(a + a*Cos[c + d*x])) + (n*(7 - 3*n - n^2)*Cos[c + d*x]*((1 + Cos[c + d*x])/(1 - Cos[c + d*x]))^(-1/2
- n)*Hypergeometric2F1[-1/2 - n, 1 - n, 2 - n, (-2*Cos[c + d*x])/(1 - Cos[c + d*x])]*(a + a*Sec[c + d*x])^n*Si
n[c + d*x])/(d*(1 - 2*n)*(3 - 2*n)*(1 - n)*(1 + 2*n)*(1 - Cos[c + d*x])^2)

Rule 3876

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(Sin[
e + f*x]^FracPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(b + a*Sin[e + f*x])^FracPart[m], Int[((g*Cos[e + f*x])
^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])

Rule 2883

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[Cos[e + f*x]/(a^(p - 2)*f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int
[(d*x)^n*(a + b*x)^(m + p/2 - 1/2)*(a - b*x)^(p/2 - 1/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m,
n}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[p/2] &&  !IntegerQ[m]

Rule 129

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

Rule 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx &=\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n \csc ^4(c+d x) \, dx\\ &=-\frac{\left (a^6 (-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-n} (-a-a x)^{-\frac{5}{2}+n}}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d \sqrt{-a+a \cos (c+d x)}}\\ &=-\frac{a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac{\left (a^3 (-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-n} (-a-a x)^{-\frac{3}{2}+n} \left (-a^2 (2-n)+2 a^2 x\right )}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d (3-2 n) \sqrt{-a+a \cos (c+d x)}}\\ &=-\frac{a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac{a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}-\frac{\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-n} (-a-a x)^{-\frac{1}{2}+n} \left (-a^4 \left (2-n^2\right )-a^4 (4-n) x\right )}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d (1-2 n) (3-2 n) \sqrt{-a+a \cos (c+d x)}}\\ &=\frac{\left (2-n+n^2\right ) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1+2 n) (1-\cos (c+d x))^2}-\frac{a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac{a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}-\frac{\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{a^6 n \left (7-3 n-n^2\right ) (-x)^{-n} (-a-a x)^{\frac{1}{2}+n}}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{a^3 d (1-2 n) (3-2 n) (1+2 n) \sqrt{-a+a \cos (c+d x)}}\\ &=\frac{\left (2-n+n^2\right ) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1+2 n) (1-\cos (c+d x))^2}-\frac{a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac{a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}-\frac{\left (a^3 n \left (7-3 n-n^2\right ) (-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-n} (-a-a x)^{\frac{1}{2}+n}}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d (1-2 n) (3-2 n) (1+2 n) \sqrt{-a+a \cos (c+d x)}}\\ &=\frac{\left (2-n+n^2\right ) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1+2 n) (1-\cos (c+d x))^2}-\frac{a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac{a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}+\frac{n \left (7-3 n-n^2\right ) \cos (c+d x) \left (\frac{1+\cos (c+d x)}{1-\cos (c+d x)}\right )^{-\frac{1}{2}-n} \, _2F_1\left (-\frac{1}{2}-n,1-n;2-n;-\frac{2 \cos (c+d x)}{1-\cos (c+d x)}\right ) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1-n) (1+2 n) (1-\cos (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 6.76317, size = 214, normalized size = 0.61 \[ -\frac{2^{n-3} \tan ^3\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^{-n} (a (\sec (c+d x)+1))^n \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^n \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^n \left (-\text{Hypergeometric2F1}\left (\frac{3}{2},n,\frac{5}{2},\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+\cot ^6\left (\frac{1}{2} (c+d x)\right ) \text{Hypergeometric2F1}\left (-\frac{3}{2},n,-\frac{1}{2},\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+9 \cot ^4\left (\frac{1}{2} (c+d x)\right ) \text{Hypergeometric2F1}\left (-\frac{1}{2},n,\frac{1}{2},\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-9 \cot ^2\left (\frac{1}{2} (c+d x)\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},n,\frac{3}{2},\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{3 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^4*(a + a*Sec[c + d*x])^n,x]

[Out]

-(2^(-3 + n)*(Cot[(c + d*x)/2]^6*Hypergeometric2F1[-3/2, n, -1/2, Tan[(c + d*x)/2]^2] + 9*Cot[(c + d*x)/2]^4*H
ypergeometric2F1[-1/2, n, 1/2, Tan[(c + d*x)/2]^2] - 9*Cot[(c + d*x)/2]^2*Hypergeometric2F1[1/2, n, 3/2, Tan[(
c + d*x)/2]^2] - Hypergeometric2F1[3/2, n, 5/2, Tan[(c + d*x)/2]^2])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*(Cos[
(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*Tan[(c + d*x)/2]^3)/(3*d*(1 + Sec[c + d*x])^n)

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Maple [F]  time = 0.321, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( dx+c \right ) \right ) ^{4} \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)^4*(a+a*sec(d*x+c))^n,x)

[Out]

int(csc(d*x+c)^4*(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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^4, 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 )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*csc(d*x + c)^4, 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)**4*(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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^4, x)