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

Optimal. Leaf size=240 \[ \frac {a^2 \left (n^2+9 n+12\right ) (a \sec (c+d x)+a)^{n-2} \, _2F_1\left (1,n-2;n-1;\frac {1}{2} (\sec (c+d x)+1)\right )}{16 d (2-n)}-\frac {a^2 \left (-2 (1-n) (n+6) \sec (c+d x)-n^3-7 n^2+4 n+12\right ) (a \sec (c+d x)+a)^{n-2}}{8 d \left (n^2-3 n+2\right ) (1-\sec (c+d x))}-\frac {a^2 \sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{d (1-n) (1-\sec (c+d x))^2}+\frac {a^2 (n+3) \sec ^2(c+d x) (a \sec (c+d x)+a)^{n-2}}{4 d (1-n) (1-\sec (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 240, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [B]  time = 6.53, size = 492, normalized size = 2.05 \[ -\frac {\cos (c+d x) (\sec (c+d x)+1)^{-n} (a (\sec (c+d x)+1))^n \left (-2^n \left (n^2+7 n-18\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{n-1} \, _2F_1\left (2,1-n;2-n;\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )-3\ 2^{n+2} (n-2) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{n-1} \, _2F_1\left (1,1-n;2-n;\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )+2^n n^2 \cot ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{n-1}+2 n \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (\sec (c+d x)+1)^n-2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (\sec (c+d x)+1)^n+2 n \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (\sec (c+d x)+1)^n-12 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (\sec (c+d x)+1)^n-12 n \sec (c+d x) (\sec (c+d x)+1)^n+32 \sec (c+d x) (\sec (c+d x)+1)^n+2^{n+1} \cot ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{n-1}-3\ 2^n n \cot ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{n-1}\right )}{64 d (n-2) (n-1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/64*(Cos[c + d*x]*(a*(1 + Sec[c + d*x]))^n*(2^(1 + n)*Cot[(c + d*x)/2]^4*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Se
c[c + d*x])^(-1 + n) - 3*2^n*n*Cot[(c + d*x)/2]^4*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) + 2^
n*n^2*Cot[(c + d*x)/2]^4*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) - 3*2^(2 + n)*(-2 + n)*Hyperg
eometric2F1[1, 1 - n, 2 - n, Cos[c + d*x]*Sec[(c + d*x)/2]^2]*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(
-1 + n) - 2^n*(-18 + 7*n + n^2)*Hypergeometric2F1[2, 1 - n, 2 - n, Cos[c + d*x]*Sec[(c + d*x)/2]^2]*Sec[c + d*
x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) + 32*Sec[c + d*x]*(1 + Sec[c + d*x])^n - 12*n*Sec[c + d*x]*(1 +
Sec[c + d*x])^n - 12*Sec[(c + d*x)/2]^2*Sec[c + d*x]*(1 + Sec[c + d*x])^n + 2*n*Sec[(c + d*x)/2]^2*Sec[c + d*x
]*(1 + Sec[c + d*x])^n - 2*Sec[(c + d*x)/2]^4*Sec[c + d*x]*(1 + Sec[c + d*x])^n + 2*n*Sec[(c + d*x)/2]^4*Sec[c
+ d*x]*(1 + Sec[c + d*x])^n))/(d*(-2 + n)*(-1 + n)*(1 + Sec[c + d*x])^n)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{5}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{5}\,{d x} \]

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)

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maple [F]  time = 1.30, size = 0, normalized size = 0.00 \[ \int \left (\csc ^{5}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{n}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{5}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{{\sin \left (c+d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^n/sin(c + d*x)^5,x)

[Out]

int((a + a/cos(c + d*x))^n/sin(c + d*x)^5, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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