3.151 \(\int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx\)

Optimal. Leaf size=240 \[ \frac{a^2 \left (n^2+9 n+12\right ) (a \sec (c+d x)+a)^{n-2} \text{Hypergeometric2F1}\left (1,n-2,n-1,\frac{1}{2} (\sec (c+d x)+1)\right )}{16 d (2-n)}-\frac{a^2 \left (-2 (1-n) (n+6) \sec (c+d x)-n^3-7 n^2+4 n+12\right ) (a \sec (c+d x)+a)^{n-2}}{8 d \left (n^2-3 n+2\right ) (1-\sec (c+d x))}-\frac{a^2 \sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{d (1-n) (1-\sec (c+d x))^2}+\frac{a^2 (n+3) \sec ^2(c+d x) (a \sec (c+d x)+a)^{n-2}}{4 d (1-n) (1-\sec (c+d x))^2} \]

[Out]

(a^2*(12 + 9*n + n^2)*Hypergeometric2F1[1, -2 + n, -1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(-2 + n)
)/(16*d*(2 - n)) + (a^2*(3 + n)*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(-2 + n))/(4*d*(1 - n)*(1 - Sec[c + d*x])^
2) - (a^2*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(-2 + n))/(d*(1 - n)*(1 - Sec[c + d*x])^2) - (a^2*(a + a*Sec[c +
d*x])^(-2 + n)*(12 + 4*n - 7*n^2 - n^3 - 2*(1 - n)*(6 + n)*Sec[c + d*x]))/(8*d*(2 - 3*n + n^2)*(1 - Sec[c + d
*x]))

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Rubi [A]  time = 0.223967, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3873, 100, 149, 146, 68} \[ \frac{a^2 \left (n^2+9 n+12\right ) (a \sec (c+d x)+a)^{n-2} \, _2F_1\left (1,n-2;n-1;\frac{1}{2} (\sec (c+d x)+1)\right )}{16 d (2-n)}-\frac{a^2 \left (-2 (1-n) (n+6) \sec (c+d x)-n^3-7 n^2+4 n+12\right ) (a \sec (c+d x)+a)^{n-2}}{8 d \left (n^2-3 n+2\right ) (1-\sec (c+d x))}-\frac{a^2 \sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{d (1-n) (1-\sec (c+d x))^2}+\frac{a^2 (n+3) \sec ^2(c+d x) (a \sec (c+d x)+a)^{n-2}}{4 d (1-n) (1-\sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5*(a + a*Sec[c + d*x])^n,x]

[Out]

(a^2*(12 + 9*n + n^2)*Hypergeometric2F1[1, -2 + n, -1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(-2 + n)
)/(16*d*(2 - n)) + (a^2*(3 + n)*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(-2 + n))/(4*d*(1 - n)*(1 - Sec[c + d*x])^
2) - (a^2*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(-2 + n))/(d*(1 - n)*(1 - Sec[c + d*x])^2) - (a^2*(a + a*Sec[c +
d*x])^(-2 + n)*(12 + 4*n - 7*n^2 - n^3 - 2*(1 - n)*(6 + n)*Sec[c + d*x]))/(8*d*(2 - 3*n + n^2)*(1 - Sec[c + d
*x]))

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 100

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 149

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*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 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 ^5(c+d x) (a+a \sec (c+d x))^n \, dx &=-\frac{a^6 \operatorname{Subst}\left (\int \frac{x^4 (a-a x)^{-3+n}}{(-a-a x)^3} \, dx,x,-\sec (c+d x)\right )}{d}\\ &=-\frac{a^2 \sec ^3(c+d x) (a+a \sec (c+d x))^{-2+n}}{d (1-n) (1-\sec (c+d x))^2}+\frac{a^4 \operatorname{Subst}\left (\int \frac{x^2 (a-a x)^{-3+n} \left (3 a^2-a^2 n x\right )}{(-a-a x)^3} \, dx,x,-\sec (c+d x)\right )}{d (1-n)}\\ &=\frac{a^2 (3+n) \sec ^2(c+d x) (a+a \sec (c+d x))^{-2+n}}{4 d (1-n) (1-\sec (c+d x))^2}-\frac{a^2 \sec ^3(c+d x) (a+a \sec (c+d x))^{-2+n}}{d (1-n) (1-\sec (c+d x))^2}+\frac{a \operatorname{Subst}\left (\int \frac{x (a-a x)^{-3+n} \left (-2 a^4 (3+n)-a^4 (1-n) (6+n) x\right )}{(-a-a x)^2} \, dx,x,-\sec (c+d x)\right )}{4 d (1-n)}\\ &=\frac{a^2 (3+n) \sec ^2(c+d x) (a+a \sec (c+d x))^{-2+n}}{4 d (1-n) (1-\sec (c+d x))^2}-\frac{a^2 \sec ^3(c+d x) (a+a \sec (c+d x))^{-2+n}}{d (1-n) (1-\sec (c+d x))^2}-\frac{a^2 (a+a \sec (c+d x))^{-2+n} \left (12+4 n-7 n^2-n^3-2 (1-n) (6+n) \sec (c+d x)\right )}{8 d \left (2-3 n+n^2\right ) (1-\sec (c+d x))}-\frac{\left (a^4 \left (12+9 n+n^2\right )\right ) \operatorname{Subst}\left (\int \frac{(a-a x)^{-3+n}}{-a-a x} \, dx,x,-\sec (c+d x)\right )}{8 d}\\ &=\frac{a^2 \left (12+9 n+n^2\right ) \, _2F_1\left (1,-2+n;-1+n;\frac{1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^{-2+n}}{16 d (2-n)}+\frac{a^2 (3+n) \sec ^2(c+d x) (a+a \sec (c+d x))^{-2+n}}{4 d (1-n) (1-\sec (c+d x))^2}-\frac{a^2 \sec ^3(c+d x) (a+a \sec (c+d x))^{-2+n}}{d (1-n) (1-\sec (c+d x))^2}-\frac{a^2 (a+a \sec (c+d x))^{-2+n} \left (12+4 n-7 n^2-n^3-2 (1-n) (6+n) \sec (c+d x)\right )}{8 d \left (2-3 n+n^2\right ) (1-\sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.72126, size = 316, normalized size = 1.32 \[ -\frac{2^{n-6} \tan ^4\left (\frac{1}{2} (c+d x)\right ) \left (\cot ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) (\sec (c+d x)+1)^{-n} (a (\sec (c+d x)+1))^n \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^n \left (-2 \left (n^3+4 n^2-15 n+6\right ) \cot ^2\left (\frac{1}{2} (c+d x)\right ) \text{Hypergeometric2F1}\left (1,1-n,2-n,\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )+\left (n^3+2 n^2-21 n+18\right ) \cos (c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right ) \text{Hypergeometric2F1}\left (1,2-n,3-n,\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )-\frac{1}{8} \csc ^6\left (\frac{1}{2} (c+d x)\right ) \left (n^2 \cos (3 (c+d x))+\left (-5 n^2+2 n+9\right ) \cos (c+d x)+\left (n^3+2 n^2-21 n+30\right ) \cos (2 (c+d x))+2 n \cos (3 (c+d x))-9 \cos (3 (c+d x))-n^3-6 n^2+41 n-46\right )\right )}{d (n-2) (n-1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^5*(a + a*Sec[c + d*x])^n,x]

[Out]

-((2^(-6 + n)*(-1 + Cot[(c + d*x)/2]^2)*(-((-46 + 41*n - 6*n^2 - n^3 + (9 + 2*n - 5*n^2)*Cos[c + d*x] + (30 -
21*n + 2*n^2 + n^3)*Cos[2*(c + d*x)] - 9*Cos[3*(c + d*x)] + 2*n*Cos[3*(c + d*x)] + n^2*Cos[3*(c + d*x)])*Csc[(
c + d*x)/2]^6)/8 - 2*(6 - 15*n + 4*n^2 + n^3)*Cot[(c + d*x)/2]^2*Hypergeometric2F1[1, 1 - n, 2 - n, Cos[c + d*
x]*Sec[(c + d*x)/2]^2] + (18 - 21*n + 2*n^2 + n^3)*Cos[c + d*x]*Csc[(c + d*x)/2]^2*Hypergeometric2F1[1, 2 - n,
3 - n, Cos[c + d*x]*Sec[(c + d*x)/2]^2])*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*Tan[(c
+ d*x)/2]^4)/(d*(-2 + n)*(-1 + n)*(1 + Sec[c + d*x])^n))

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Maple [F]  time = 0.296, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( dx+c \right ) \right ) ^{5} \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)^5*(a+a*sec(d*x+c))^n,x)

[Out]

int(csc(d*x+c)^5*(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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+a*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^5, 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 )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+a*sec(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*csc(d*x + c)^5, 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)**5*(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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+a*sec(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*csc(d*x + c)^5, x)