\(\int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx\) [272]

Optimal result

Integrand size = 21, antiderivative size = 231 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b) d (1+n)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b) d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)} \]

[Out]

1/4*hypergeom([1, 1+n],[2+n],(a+b*sec(d*x+c))/(a-b))*(a+b*sec(d*x+c))^(1+n)/(a-b)/d/(1+n)-1/4*hypergeom([1, 1+
n],[2+n],(a+b*sec(d*x+c))/(a+b))*(a+b*sec(d*x+c))^(1+n)/(a+b)/d/(1+n)+1/4*b*hypergeom([2, 1+n],[2+n],(a+b*sec(
d*x+c))/(a-b))*(a+b*sec(d*x+c))^(1+n)/(a-b)^2/d/(1+n)+1/4*b*hypergeom([2, 1+n],[2+n],(a+b*sec(d*x+c))/(a+b))*(
a+b*sec(d*x+c))^(1+n)/(a+b)^2/d/(1+n)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3959, 186, 70, 726} \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)}-\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)}+\frac {b (a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)^2}+\frac {b (a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)^2} \]

[In]

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

[Out]

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

Rule 70

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

Rule 186

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 3959

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[-f^(-1), Subs
t[Int[(-1 + x)^((p - 1)/2)*(1 + x)^((p - 1)/2)*((a + b*x)^m/x^(p + 1)), x], x, Csc[e + f*x]], x] /; FreeQ[{a,
b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^2 (a-b x)^n}{(-1+x)^2 (1+x)^2} \, dx,x,-\sec (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {(a-b x)^n}{4 (-1+x)^2}+\frac {(a-b x)^n}{4 (1+x)^2}+\frac {(a-b x)^n}{2 \left (-1+x^2\right )}\right ) \, dx,x,-\sec (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {(a-b x)^n}{(-1+x)^2} \, dx,x,-\sec (c+d x)\right )}{4 d}-\frac {\text {Subst}\left (\int \frac {(a-b x)^n}{(1+x)^2} \, dx,x,-\sec (c+d x)\right )}{4 d}-\frac {\text {Subst}\left (\int \frac {(a-b x)^n}{-1+x^2} \, dx,x,-\sec (c+d x)\right )}{2 d} \\ & = \frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)}-\frac {\text {Subst}\left (\int \left (-\frac {(a-b x)^n}{2 (1-x)}-\frac {(a-b x)^n}{2 (1+x)}\right ) \, dx,x,-\sec (c+d x)\right )}{2 d} \\ & = \frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)}+\frac {\text {Subst}\left (\int \frac {(a-b x)^n}{1-x} \, dx,x,-\sec (c+d x)\right )}{4 d}+\frac {\text {Subst}\left (\int \frac {(a-b x)^n}{1+x} \, dx,x,-\sec (c+d x)\right )}{4 d} \\ & = \frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b) d (1+n)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b) d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1520\) vs. \(2(231)=462\).

Time = 16.83 (sec) , antiderivative size = 1520, normalized size of antiderivative = 6.58 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\frac {(a+b) (b+a \cos (c+d x))^{-n} \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^n \left (\frac {(-a+b) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^n \left (\cos (c+d x) \sec ^4\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} (a+b \sec (c+d x))^n \left (\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^n \left (a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (1-\tan ^4\left (\frac {1}{2} (c+d x)\right )\right )^n \left (b+\frac {a-a \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^n \left (-\frac {\operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{2 b}\right ) \left (2-2 \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{b}\right )^n \left (a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{1+n}}{(a-b) (1+n)}+\frac {\operatorname {Hypergeometric2F1}\left (1,-n,1-n,\frac {(a+b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right ) \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^n}{n}-\frac {\cot ^2\left (\frac {1}{2} (c+d x)\right ) \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{1-n} \left (a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{1+n}}{a+b}-\frac {2^n \operatorname {Hypergeometric2F1}\left (-n,-n,1-n,\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{2 b}\right ) \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^n \left (\frac {a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^{-n}}{n}-\frac {\frac {\operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{2 b}\right ) \left (2-2 \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{b}\right )^n \left (a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{1+n}}{1+n}+\frac {(a-b) \operatorname {Hypergeometric2F1}\left (1,-n,1-n,\frac {(a+b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right ) \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^n}{n}-\frac {2^n (a-b) \operatorname {Hypergeometric2F1}\left (-n,-n,1-n,\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{2 b}\right ) \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^n \left (\frac {a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^{-n}}{n}-\frac {2 (a+b n) \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{-n} \left (a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^n \left (\frac {a+b+(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^{-n} \left (-2^n \operatorname {Hypergeometric2F1}\left (-n,-n,1-n,\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{2 b}\right )+\operatorname {Hypergeometric2F1}\left (1,-n,1-n,\frac {(a+b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right ) \left (\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^n\right )}{n}}{a+b}\right )}{2 d \left (a \left (\frac {(-a+b) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^n+4 b \left (\frac {(-a+b) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^n+3 a \left (\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{b}\right )^n+4 a \cos (c+d x) \left (\left (\frac {(-a+b) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^n-\left (\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{b}\right )^n\right )-a \cos (2 (c+d x)) \left (\left (\frac {(-a+b) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^n-\left (\frac {(a-b) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{b}\right )^n\right )\right )} \]

[In]

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

[Out]

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

Maple [F]

\[\int \csc \left (d x +c \right )^{3} \left (a +b \sec \left (d x +c \right )\right )^{n}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\sin \left (c+d\,x\right )}^3} \,d x \]

[In]

int((a + b/cos(c + d*x))^n/sin(c + d*x)^3,x)

[Out]

int((a + b/cos(c + d*x))^n/sin(c + d*x)^3, x)