3.830 \(\int \frac{(d \csc (e+f x))^n}{(a+b \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=432 \[ -\frac{3 a b^2 \sin ^4(e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac{n-1}{2}} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n-1}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3}+\frac{b^3 \sin ^3(e+f x) \cos (e+f x) \sin ^2(e+f x)^{n/2} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n-2}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3}+\frac{3 a^2 b \sin ^3(e+f x) \cos (e+f x) \sin ^2(e+f x)^{n/2} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3}-\frac{a^3 \cos (e+f x) \sin ^2(e+f x)^{\frac{n+3}{2}} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n+1}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3} \]
[Out]
(-3*a*b^2*AppellF1[1/2, (-1 + n)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*
(d*Csc[e + f*x])^(3 + n)*Sin[e + f*x]^4*(Sin[e + f*x]^2)^((-1 + n)/2))/((a^2 - b^2)^3*d^3*f) + (b^3*AppellF1[1
/2, (-2 + n)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Csc[e + f*x])^(3
+ n)*Sin[e + f*x]^3*(Sin[e + f*x]^2)^(n/2))/((a^2 - b^2)^3*d^3*f) + (3*a^2*b*AppellF1[1/2, n/2, 3, 3/2, Cos[e
+ f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Csc[e + f*x])^(3 + n)*Sin[e + f*x]^3*(Sin[e + f
*x]^2)^(n/2))/((a^2 - b^2)^3*d^3*f) - (a^3*AppellF1[1/2, (1 + n)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x
]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Csc[e + f*x])^(3 + n)*(Sin[e + f*x]^2)^((3 + n)/2))/((a^2 - b^2)^3*d^3*f)
________________________________________________________________________________________
Rubi [A] time = 0.702643, antiderivative size = 432, normalized size of antiderivative = 1.,
number of steps used = 12, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used
= {3238, 3869, 2824, 3189, 429} \[ -\frac{3 a b^2 \sin ^4(e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac{n-1}{2}} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n-1}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3}+\frac{b^3 \sin ^3(e+f x) \cos (e+f x) \sin ^2(e+f x)^{n/2} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n-2}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3}+\frac{3 a^2 b \sin ^3(e+f x) \cos (e+f x) \sin ^2(e+f x)^{n/2} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3}-\frac{a^3 \cos (e+f x) \sin ^2(e+f x)^{\frac{n+3}{2}} (d \csc (e+f x))^{n+3} F_1\left (\frac{1}{2};\frac{n+1}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d^3 f \left (a^2-b^2\right )^3} \]
Antiderivative was successfully verified.
[In]
Int[(d*Csc[e + f*x])^n/(a + b*Sin[e + f*x])^3,x]
[Out]
(-3*a*b^2*AppellF1[1/2, (-1 + n)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*
(d*Csc[e + f*x])^(3 + n)*Sin[e + f*x]^4*(Sin[e + f*x]^2)^((-1 + n)/2))/((a^2 - b^2)^3*d^3*f) + (b^3*AppellF1[1
/2, (-2 + n)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Csc[e + f*x])^(3
+ n)*Sin[e + f*x]^3*(Sin[e + f*x]^2)^(n/2))/((a^2 - b^2)^3*d^3*f) + (3*a^2*b*AppellF1[1/2, n/2, 3, 3/2, Cos[e
+ f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Csc[e + f*x])^(3 + n)*Sin[e + f*x]^3*(Sin[e + f
*x]^2)^(n/2))/((a^2 - b^2)^3*d^3*f) - (a^3*AppellF1[1/2, (1 + n)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x
]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Csc[e + f*x])^(3 + n)*(Sin[e + f*x]^2)^((3 + n)/2))/((a^2 - b^2)^3*d^3*f)
Rule 3238
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] && !IntegerQ[m] && IntegersQ[n, p]
Rule 3869
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Dist[Sin[
e + f*x]^n*(d*Csc[e + f*x])^n, Int[(b + a*Sin[e + f*x])^m/Sin[e + f*x]^(m + n), x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m]
Rule 2824
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(d*sin[e + f*x])^n/((a - b*sin[e + f*x])^m/(a^2 - b^2*sin[e + f*x]^2)^m), x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m, -1]
Rule 3189
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Cos[e + f*x], x]}, -Dist[(ff*d^(2*IntPart[(m - 1)/2] + 1)*(d*Sin[e + f*x])^(2*FracPart[(m - 1)/
2]))/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x]
, x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && !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 \frac{(d \csc (e+f x))^n}{(a+b \sin (e+f x))^3} \, dx &=\frac{\int \frac{(d \csc (e+f x))^{3+n}}{(b+a \csc (e+f x))^3} \, dx}{d^3}\\ &=\frac{\left ((d \csc (e+f x))^{3+n} \sin ^{3+n}(e+f x)\right ) \int \frac{\sin ^{-n}(e+f x)}{(a+b \sin (e+f x))^3} \, dx}{d^3}\\ &=\frac{\left ((d \csc (e+f x))^{3+n} \sin ^{3+n}(e+f x)\right ) \int \left (-\frac{3 a^2 b \sin ^{1-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac{3 a b^2 \sin ^{2-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac{a^3 \sin ^{-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac{b^3 \sin ^{3-n}(e+f x)}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^3}\right ) \, dx}{d^3}\\ &=\frac{\left (a^3 (d \csc (e+f x))^{3+n} \sin ^{3+n}(e+f x)\right ) \int \frac{\sin ^{-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx}{d^3}-\frac{\left (3 a^2 b (d \csc (e+f x))^{3+n} \sin ^{3+n}(e+f x)\right ) \int \frac{\sin ^{1-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx}{d^3}+\frac{\left (3 a b^2 (d \csc (e+f x))^{3+n} \sin ^{3+n}(e+f x)\right ) \int \frac{\sin ^{2-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3} \, dx}{d^3}+\frac{\left (b^3 (d \csc (e+f x))^{3+n} \sin ^{3+n}(e+f x)\right ) \int \frac{\sin ^{3-n}(e+f x)}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^3} \, dx}{d^3}\\ &=-\frac{\left (3 a b^2 (d \csc (e+f x))^{3+n} \sin ^{3+2 \left (\frac{1}{2}-\frac{n}{2}\right )+n}(e+f x) \sin ^2(e+f x)^{-\frac{1}{2}+\frac{n}{2}}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{1-n}{2}}}{\left (a^2-b^2+b^2 x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{d^3 f}-\frac{\left (a^3 (d \csc (e+f x))^{3+n} \sin ^{3+2 \left (-\frac{1}{2}-\frac{n}{2}\right )+n}(e+f x) \sin ^2(e+f x)^{\frac{1}{2}+\frac{n}{2}}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{1}{2} (-1-n)}}{\left (a^2-b^2+b^2 x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{d^3 f}+\frac{\left (3 a^2 b (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{-n/2}}{\left (a^2-b^2+b^2 x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{d^3 f}-\frac{\left (b^3 (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{2-n}{2}}}{\left (-a^2+b^2-b^2 x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{d^3 f}\\ &=-\frac{3 a b^2 F_1\left (\frac{1}{2};\frac{1}{2} (-1+n),3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^4(e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-1+n)}}{\left (a^2-b^2\right )^3 d^3 f}+\frac{b^3 F_1\left (\frac{1}{2};\frac{1}{2} (-2+n),3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}}{\left (a^2-b^2\right )^3 d^3 f}+\frac{3 a^2 b F_1\left (\frac{1}{2};\frac{n}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}}{\left (a^2-b^2\right )^3 d^3 f}-\frac{a^3 F_1\left (\frac{1}{2};\frac{1+n}{2},3;\frac{3}{2};\cos ^2(e+f x),-\frac{b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^2(e+f x)^{1+\frac{1+n}{2}}}{\left (a^2-b^2\right )^3 d^3 f}\\ \end{align*}
Mathematica [B] time = 20.3277, size = 2496, normalized size = 5.78 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*Csc[e + f*x])^n/(a + b*Sin[e + f*x])^3,x]
[Out]
((d*Csc[e + f*x])^n*(Cot[e + f*x]*Sqrt[Sec[e + f*x]^2])^n*Tan[e + f*x]*(-(a*(a^2 + 3*b^2)*(-2 + n)*AppellF1[(1
- n)/2, -1 - n/2, 2, (3 - n)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]) + b*(4*a*b*(-2 + n)*Appe
llF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] + (-1 + n)*((3*a^2
+ b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - 4*b^2*
AppellF1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2])*Tan[e + f*x])))
/(a^4*(a^2 - b^2)*f*(-2 + n)*(-1 + n)*(Sec[e + f*x]^2)^(n/2)*(a + b*Sin[e + f*x])^3*(((Sec[e + f*x]^2)^(1 - n/
2)*(Cot[e + f*x]*Sqrt[Sec[e + f*x]^2])^n*(-(a*(a^2 + 3*b^2)*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 2, (3 - n)/
2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]) + b*(4*a*b*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 3, (
3 - n)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] + (-1 + n)*((3*a^2 + b^2)*AppellF1[1 - n/2, (-1
- n)/2, 2, 2 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - 4*b^2*AppellF1[1 - n/2, (-1 - n)/2,
3, 2 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2])*Tan[e + f*x])))/(a^4*(a^2 - b^2)*(-2 + n)*(-1
+ n)) + (n*(Cot[e + f*x]*Sqrt[Sec[e + f*x]^2])^(-1 + n)*(Sqrt[Sec[e + f*x]^2] - Csc[e + f*x]^2*Sqrt[Sec[e + f
*x]^2])*Tan[e + f*x]*(-(a*(a^2 + 3*b^2)*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 2, (3 - n)/2, -Tan[e + f*x]^2,
((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]) + b*(4*a*b*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f
*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] + (-1 + n)*((3*a^2 + b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2,
-Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - 4*b^2*AppellF1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e +
f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2])*Tan[e + f*x])))/(a^4*(a^2 - b^2)*(-2 + n)*(-1 + n)*(Sec[e + f*x]^
2)^(n/2)) - (n*(Cot[e + f*x]*Sqrt[Sec[e + f*x]^2])^n*Tan[e + f*x]^2*(-(a*(a^2 + 3*b^2)*(-2 + n)*AppellF1[(1 -
n)/2, -1 - n/2, 2, (3 - n)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]) + b*(4*a*b*(-2 + n)*AppellF
1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] + (-1 + n)*((3*a^2 +
b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - 4*b^2*App
ellF1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2])*Tan[e + f*x])))/(a
^4*(a^2 - b^2)*(-2 + n)*(-1 + n)*(Sec[e + f*x]^2)^(n/2)) + ((Cot[e + f*x]*Sqrt[Sec[e + f*x]^2])^n*Tan[e + f*x]
*(-(a*(a^2 + 3*b^2)*(-2 + n)*((4*(-a^2 + b^2)*(1 - n)*AppellF1[1 + (1 - n)/2, -1 - n/2, 3, 1 + (3 - n)/2, -Tan
[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(a^2*(3 - n)) - (2*(1 - n)*(-1 -
n/2)*AppellF1[1 + (1 - n)/2, -n/2, 2, 1 + (3 - n)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e
+ f*x]^2*Tan[e + f*x])/(3 - n))) + b*((-1 + n)*((3*a^2 + b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -Tan[
e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] - 4*b^2*AppellF1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]
^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2])*Sec[e + f*x]^2 + 4*a*b*(-2 + n)*((6*(-a^2 + b^2)*(1 - n)*AppellF1[1 +
(1 - n)/2, -1 - n/2, 4, 1 + (3 - n)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[
e + f*x])/(a^2*(3 - n)) - (2*(1 - n)*(-1 - n/2)*AppellF1[1 + (1 - n)/2, -n/2, 3, 1 + (3 - n)/2, -Tan[e + f*x]^
2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 - n)) + (-1 + n)*Tan[e + f*x]*((3*a^2 +
b^2)*(-(((-1 - n)*(1 - n/2)*AppellF1[2 - n/2, 1 + (-1 - n)/2, 2, 3 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e
+ f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 - n/2)) + (4*(-a^2 + b^2)*(1 - n/2)*AppellF1[2 - n/2, (-1 - n)
/2, 3, 3 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(a^2*(2 - n/2
))) - 4*b^2*(-(((-1 - n)*(1 - n/2)*AppellF1[2 - n/2, 1 + (-1 - n)/2, 3, 3 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2
)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 - n/2)) + (6*(-a^2 + b^2)*(1 - n/2)*AppellF1[2 - n/2, (
-1 - n)/2, 4, 3 - n/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(a^2*(
2 - n/2)))))))/(a^4*(a^2 - b^2)*(-2 + n)*(-1 + n)*(Sec[e + f*x]^2)^(n/2))))
________________________________________________________________________________________
Maple [F] time = 1.06, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\csc \left ( fx+e \right ) \right ) ^{n}}{ \left ( a+b\sin \left ( fx+e \right ) \right ) ^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*csc(f*x+e))^n/(a+b*sin(f*x+e))^3,x)
[Out]
int((d*csc(f*x+e))^n/(a+b*sin(f*x+e))^3,x)
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Maxima [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((d*csc(f*x+e))^n/(a+b*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\left (d \csc \left (f x + e\right )\right )^{n}}{3 \, a b^{2} \cos \left (f x + e\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*csc(f*x+e))^n/(a+b*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
integral(-(d*csc(f*x + e))^n/(3*a*b^2*cos(f*x + e)^2 - a^3 - 3*a*b^2 + (b^3*cos(f*x + e)^2 - 3*a^2*b - b^3)*si
n(f*x + e)), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc{\left (e + f x \right )}\right )^{n}}{\left (a + b \sin{\left (e + f x \right )}\right )^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*csc(f*x+e))**n/(a+b*sin(f*x+e))**3,x)
[Out]
Integral((d*csc(e + f*x))**n/(a + b*sin(e + f*x))**3, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (f x + e\right )\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*csc(f*x+e))^n/(a+b*sin(f*x+e))^3,x, algorithm="giac")
[Out]
integrate((d*csc(f*x + e))^n/(b*sin(f*x + e) + a)^3, x)