3.145 \(\int \csc (c+d x) (a+a \sin (c+d x))^n \, dx\)
Optimal. Leaf size=85 \[ -\frac{2^{n+\frac{1}{2}} \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac{1}{2}} (a \sin (c+d x)+a)^n F_1\left (\frac{1}{2};1,\frac{1}{2}-n;\frac{3}{2};1-\sin (c+d x),\frac{1}{2} (1-\sin (c+d x))\right )}{d} \]
[Out]
-((2^(1/2 + n)*AppellF1[1/2, 1, 1/2 - n, 3/2, 1 - Sin[c + d*x], (1 - Sin[c + d*x])/2]*Cos[c + d*x]*(1 + Sin[c
+ d*x])^(-1/2 - n)*(a + a*Sin[c + d*x])^n)/d)
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Rubi [A] time = 0.110809, antiderivative size = 85, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used =
{2787, 2785, 130, 429} \[ -\frac{2^{n+\frac{1}{2}} \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac{1}{2}} (a \sin (c+d x)+a)^n F_1\left (\frac{1}{2};1,\frac{1}{2}-n;\frac{3}{2};1-\sin (c+d x),\frac{1}{2} (1-\sin (c+d x))\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Csc[c + d*x]*(a + a*Sin[c + d*x])^n,x]
[Out]
-((2^(1/2 + n)*AppellF1[1/2, 1, 1/2 - n, 3/2, 1 - Sin[c + d*x], (1 - Sin[c + d*x])/2]*Cos[c + d*x]*(1 + Sin[c
+ d*x])^(-1/2 - n)*(a + a*Sin[c + d*x])^n)/d)
Rule 2787
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Sin[e + f*x])^FracPart[m])/(1 + (b*Sin[e + f*x])/a)^FracPart[m], Int[(1 + (b*Sin[e + f*x])/a)^
m*(d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && !GtQ
[a, 0]
Rule 2785
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Dist[(b*(d
/b)^n*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a - x)^n*(2*a - x)^(m -
1/2))/Sqrt[x], x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !In
tegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]
Rule 130
Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]
Rule 429
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rubi steps
\begin{align*} \int \csc (c+d x) (a+a \sin (c+d x))^n \, dx &=\left ((1+\sin (c+d x))^{-n} (a+a \sin (c+d x))^n\right ) \int \csc (c+d x) (1+\sin (c+d x))^n \, dx\\ &=-\frac{\left (\cos (c+d x) (1+\sin (c+d x))^{-\frac{1}{2}-n} (a+a \sin (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(2-x)^{-\frac{1}{2}+n}}{(1-x) \sqrt{x}} \, dx,x,1-\sin (c+d x)\right )}{d \sqrt{1-\sin (c+d x)}}\\ &=-\frac{\left (2 \cos (c+d x) (1+\sin (c+d x))^{-\frac{1}{2}-n} (a+a \sin (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{\left (2-x^2\right )^{-\frac{1}{2}+n}}{1-x^2} \, dx,x,\sqrt{1-\sin (c+d x)}\right )}{d \sqrt{1-\sin (c+d x)}}\\ &=-\frac{2^{\frac{1}{2}+n} F_1\left (\frac{1}{2};1,\frac{1}{2}-n;\frac{3}{2};1-\sin (c+d x),\frac{1}{2} (1-\sin (c+d x))\right ) \cos (c+d x) (1+\sin (c+d x))^{-\frac{1}{2}-n} (a+a \sin (c+d x))^n}{d}\\ \end{align*}
Mathematica [C] time = 15.841, size = 2560, normalized size = 30.12 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Csc[c + d*x]*(a + a*Sin[c + d*x])^n,x]
[Out]
-(Csc[c + d*x]*(a + a*Sin[c + d*x])^n*(AppellF1[2*n, n, n, 1 + 2*n, (-1 - I)/(-1 + Tan[(-c + Pi/2 - d*x)/2]),
(-1 + I)/(-1 + Tan[(-c + Pi/2 - d*x)/2])]*((-I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^n*
((I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^n - AppellF1[2*n, n, n, 1 + 2*n, (1 - I)/(1 +
Tan[(-c + Pi/2 - d*x)/2]), (1 + I)/(1 + Tan[(-c + Pi/2 - d*x)/2])]*((-I + Tan[(-c + Pi/2 - d*x)/2])/(1 + Tan[
(-c + Pi/2 - d*x)/2]))^n*((I + Tan[(-c + Pi/2 - d*x)/2])/(1 + Tan[(-c + Pi/2 - d*x)/2]))^n))/(2*d*n*(Sec[(-c +
Pi/2 - d*x)/2]^2)^n*(-(Tan[(-c + Pi/2 - d*x)/2]*(AppellF1[2*n, n, n, 1 + 2*n, (-1 - I)/(-1 + Tan[(-c + Pi/2 -
d*x)/2]), (-1 + I)/(-1 + Tan[(-c + Pi/2 - d*x)/2])]*((-I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d
*x)/2]))^n*((I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^n - AppellF1[2*n, n, n, 1 + 2*n, (
1 - I)/(1 + Tan[(-c + Pi/2 - d*x)/2]), (1 + I)/(1 + Tan[(-c + Pi/2 - d*x)/2])]*((-I + Tan[(-c + Pi/2 - d*x)/2]
)/(1 + Tan[(-c + Pi/2 - d*x)/2]))^n*((I + Tan[(-c + Pi/2 - d*x)/2])/(1 + Tan[(-c + Pi/2 - d*x)/2]))^n))/(2*(Se
c[(-c + Pi/2 - d*x)/2]^2)^n) + ((((1 - I)*n^2*AppellF1[1 + 2*n, n, 1 + n, 2 + 2*n, (-1 - I)/(-1 + Tan[(-c + Pi
/2 - d*x)/2]), (-1 + I)/(-1 + Tan[(-c + Pi/2 - d*x)/2])]*Sec[(-c + Pi/2 - d*x)/2]^2)/((1 + 2*n)*(-1 + Tan[(-c
+ Pi/2 - d*x)/2])^2) + ((1 + I)*n^2*AppellF1[1 + 2*n, 1 + n, n, 2 + 2*n, (-1 - I)/(-1 + Tan[(-c + Pi/2 - d*x)/
2]), (-1 + I)/(-1 + Tan[(-c + Pi/2 - d*x)/2])]*Sec[(-c + Pi/2 - d*x)/2]^2)/((1 + 2*n)*(-1 + Tan[(-c + Pi/2 - d
*x)/2])^2))*((-I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^n*((I + Tan[(-c + Pi/2 - d*x)/2]
)/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^n + n*AppellF1[2*n, n, n, 1 + 2*n, (-1 - I)/(-1 + Tan[(-c + Pi/2 - d*x)/2])
, (-1 + I)/(-1 + Tan[(-c + Pi/2 - d*x)/2])]*((-I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^
(-1 + n)*((I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^n*(Sec[(-c + Pi/2 - d*x)/2]^2/(2*(-1
+ Tan[(-c + Pi/2 - d*x)/2])) - (Sec[(-c + Pi/2 - d*x)/2]^2*(-I + Tan[(-c + Pi/2 - d*x)/2]))/(2*(-1 + Tan[(-c
+ Pi/2 - d*x)/2])^2)) + n*AppellF1[2*n, n, n, 1 + 2*n, (-1 - I)/(-1 + Tan[(-c + Pi/2 - d*x)/2]), (-1 + I)/(-1
+ Tan[(-c + Pi/2 - d*x)/2])]*((-I + Tan[(-c + Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^n*((I + Tan[(-c
+ Pi/2 - d*x)/2])/(-1 + Tan[(-c + Pi/2 - d*x)/2]))^(-1 + n)*(Sec[(-c + Pi/2 - d*x)/2]^2/(2*(-1 + Tan[(-c + Pi
/2 - d*x)/2])) - (Sec[(-c + Pi/2 - d*x)/2]^2*(I + Tan[(-c + Pi/2 - d*x)/2]))/(2*(-1 + Tan[(-c + Pi/2 - d*x)/2]
)^2)) - ((-I + Tan[(-c + Pi/2 - d*x)/2])/(1 + Tan[(-c + Pi/2 - d*x)/2]))^n*((I + Tan[(-c + Pi/2 - d*x)/2])/(1
+ Tan[(-c + Pi/2 - d*x)/2]))^n*(((-1 - I)*n^2*AppellF1[1 + 2*n, n, 1 + n, 2 + 2*n, (1 - I)/(1 + Tan[(-c + Pi/2
- d*x)/2]), (1 + I)/(1 + Tan[(-c + Pi/2 - d*x)/2])]*Sec[(-c + Pi/2 - d*x)/2]^2)/((1 + 2*n)*(1 + Tan[(-c + Pi/
2 - d*x)/2])^2) - ((1 - I)*n^2*AppellF1[1 + 2*n, 1 + n, n, 2 + 2*n, (1 - I)/(1 + Tan[(-c + Pi/2 - d*x)/2]), (1
+ I)/(1 + Tan[(-c + Pi/2 - d*x)/2])]*Sec[(-c + Pi/2 - d*x)/2]^2)/((1 + 2*n)*(1 + Tan[(-c + Pi/2 - d*x)/2])^2)
) - n*AppellF1[2*n, n, n, 1 + 2*n, (1 - I)/(1 + Tan[(-c + Pi/2 - d*x)/2]), (1 + I)/(1 + Tan[(-c + Pi/2 - d*x)/
2])]*((-I + Tan[(-c + Pi/2 - d*x)/2])/(1 + Tan[(-c + Pi/2 - d*x)/2]))^(-1 + n)*((I + Tan[(-c + Pi/2 - d*x)/2])
/(1 + Tan[(-c + Pi/2 - d*x)/2]))^n*(-(Sec[(-c + Pi/2 - d*x)/2]^2*(-I + Tan[(-c + Pi/2 - d*x)/2]))/(2*(1 + Tan[
(-c + Pi/2 - d*x)/2])^2) + Sec[(-c + Pi/2 - d*x)/2]^2/(2*(1 + Tan[(-c + Pi/2 - d*x)/2]))) - n*AppellF1[2*n, n,
n, 1 + 2*n, (1 - I)/(1 + Tan[(-c + Pi/2 - d*x)/2]), (1 + I)/(1 + Tan[(-c + Pi/2 - d*x)/2])]*((-I + Tan[(-c +
Pi/2 - d*x)/2])/(1 + Tan[(-c + Pi/2 - d*x)/2]))^n*((I + Tan[(-c + Pi/2 - d*x)/2])/(1 + Tan[(-c + Pi/2 - d*x)/2
]))^(-1 + n)*(-(Sec[(-c + Pi/2 - d*x)/2]^2*(I + Tan[(-c + Pi/2 - d*x)/2]))/(2*(1 + Tan[(-c + Pi/2 - d*x)/2])^2
) + Sec[(-c + Pi/2 - d*x)/2]^2/(2*(1 + Tan[(-c + Pi/2 - d*x)/2]))))/(2*n*(Sec[(-c + Pi/2 - d*x)/2]^2)^n)))
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Maple [F] time = 0.736, size = 0, normalized size = 0. \begin{align*} \int \csc \left ( dx+c \right ) \left ( a+a\sin \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)*(a+a*sin(d*x+c))^n,x)
[Out]
int(csc(d*x+c)*(a+a*sin(d*x+c))^n,x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))^n,x, algorithm="maxima")
[Out]
integrate((a*sin(d*x + c) + a)^n*csc(d*x + c), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))^n,x, algorithm="fricas")
[Out]
integral((a*sin(d*x + c) + a)^n*csc(d*x + c), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{n} \csc{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))**n,x)
[Out]
Integral((a*(sin(c + d*x) + 1))**n*csc(c + d*x), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)*(a+a*sin(d*x+c))^n,x, algorithm="giac")
[Out]
integrate((a*sin(d*x + c) + a)^n*csc(d*x + c), x)