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

Optimal. Leaf size=231 \[ \frac{(a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)}-\frac{(a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)}+\frac{b (a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (2,n+1,n+2,\frac{a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)^2}+\frac{b (a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (2,n+1,n+2,\frac{a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)^2} \]

[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))

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Rubi [A]  time = 0.1947, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3874, 180, 68, 712} \[ \frac{(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)}-\frac{(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)}+\frac{b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)^2}+\frac{b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)^2} \]

Antiderivative was successfully verified.

[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 3874

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]

Rule 180

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 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 712

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]

Rubi steps

\begin{align*} \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx &=-\frac{\operatorname{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{\operatorname{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{\operatorname{Subst}\left (\int \frac{(a-b x)^n}{(-1+x)^2} \, dx,x,-\sec (c+d x)\right )}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{(a-b x)^n}{(1+x)^2} \, dx,x,-\sec (c+d x)\right )}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{(a-b x)^n}{-1+x^2} \, dx,x,-\sec (c+d x)\right )}{2 d}\\ &=\frac{b \, _2F_1\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 \, _2F_1\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{\operatorname{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 \, _2F_1\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 \, _2F_1\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{\operatorname{Subst}\left (\int \frac{(a-b x)^n}{1-x} \, dx,x,-\sec (c+d x)\right )}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{(a-b x)^n}{1+x} \, dx,x,-\sec (c+d x)\right )}{4 d}\\ &=\frac{\, _2F_1\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{\, _2F_1\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 \, _2F_1\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 \, _2F_1\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]  time = 17.198, size = 710, normalized size = 3.07 \[ \frac{\left (\frac{1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}\right )^n \left (1-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^{-2 n} \left (1-\tan ^4\left (\frac{1}{2} (c+d x)\right )\right )^n \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\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 \cos (c+d x)+b)^{-n} \left (\frac{a-a \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}+b\right )^n (a+b \sec (c+d x))^n \left (2 (a+b n+b) \left (1-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^n \text{Hypergeometric2F1}\left (1,-n,1-n,\frac{(a+b) \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )}{a \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )-b \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )}\right )-\frac{2^{-n} \cot ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{b}\right )^{-n} \left (n \left (1-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^n \left (a \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )-b \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )\right ) \left (2^n (n+1) (a-b) \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) \left (\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{b}\right )^n-2 a \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) \left (a^2 \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )-2 a b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b^2 \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{b^2}\right )^n \text{Hypergeometric2F1}\left (n,n+1,n+2,\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{2 b}\right )\right )+2^{n+1} (n+1) (a-b) (a+b n+b) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (2-2 \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^n \text{Hypergeometric2F1}\left (-n,-n,1-n,\frac{(a-b) \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )}{2 b}\right )\right )}{(n+1) (a-b)}\right )}{8 d n (a+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((Cos[c + d*x]*Sec[(c + d*x)/2]^2)^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*(2*(a + b + b*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))]*(1 -
Tan[(c + d*x)/2]^2)^n - (Cot[(c + d*x)/2]^2*(2^(1 + n)*(a - b)*(1 + n)*(a + b + b*n)*Hypergeometric2F1[-n, -n
, 1 - n, ((a - b)*(-1 + Tan[(c + d*x)/2]^2))/(2*b)]*Tan[(c + d*x)/2]^2*(2 - 2*Tan[(c + d*x)/2]^2)^n + n*(1 - T
an[(c + d*x)/2]^2)^n*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2))*(2^n*(a - b)*(1 + n)*(-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 - 2*a*Hypergeometric2F1[n, 1 + n,
2 + n, (a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(2*b)]*Tan[(c + d*x)/2]^2*(-(((-1 + Tan[(c + d*x
)/2]^2)*(-2*a*b*Tan[(c + d*x)/2]^2 + a^2*(-1 + Tan[(c + d*x)/2]^2) + b^2*(1 + Tan[(c + d*x)/2]^2)))/b^2))^n)))
/(2^n*(a - b)*(1 + n)*((a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/b)^n)))/(8*(a + b)*d*n*(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*(1 - Tan[(c + d*x)/2]^2)^
(2*n))

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Maple [F]  time = 0.261, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( dx+c \right ) \right ) ^{3} \left ( a+b\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+b*sec(d*x+c))^n,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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+b*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

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

[Out]

integral((b*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+b*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 (b \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+b*sec(d*x+c))^n,x, algorithm="giac")

[Out]

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