3.9.30 \(\int \frac {(d \csc (e+f x))^n}{(a+b \sin (e+f x))^3} \, dx\) [830]

Optimal. Leaf size=432 \[ -\frac {3 a b^2 F_1\left (\frac {1}{2};\frac {1}{2} (-1+n),3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^4(e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-1+n)}}{\left (a^2-b^2\right )^3 d^3 f}+\frac {b^3 F_1\left (\frac {1}{2};\frac {1}{2} (-2+n),3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}}{\left (a^2-b^2\right )^3 d^3 f}+\frac {3 a^2 b F_1\left (\frac {1}{2};\frac {n}{2},3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}}{\left (a^2-b^2\right )^3 d^3 f}-\frac {a^3 F_1\left (\frac {1}{2};\frac {1+n}{2},3;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^2(e+f x)^{\frac {3+n}{2}}}{\left (a^2-b^2\right )^3 d^3 f} \]

[Out]

-3*a*b^2*AppellF1(1/2,-1/2+1/2*n,3,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*cos(f*x+e)*(d*csc(f*x+e))^(3+
n)*sin(f*x+e)^4*(sin(f*x+e)^2)^(-1/2+1/2*n)/(a^2-b^2)^3/d^3/f+b^3*AppellF1(1/2,-1+1/2*n,3,3/2,cos(f*x+e)^2,-b^
2*cos(f*x+e)^2/(a^2-b^2))*cos(f*x+e)*(d*csc(f*x+e))^(3+n)*sin(f*x+e)^3*(sin(f*x+e)^2)^(1/2*n)/(a^2-b^2)^3/d^3/
f+3*a^2*b*AppellF1(1/2,1/2*n,3,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*cos(f*x+e)*(d*csc(f*x+e))^(3+n)*s
in(f*x+e)^3*(sin(f*x+e)^2)^(1/2*n)/(a^2-b^2)^3/d^3/f-a^3*AppellF1(1/2,1/2+1/2*n,3,3/2,cos(f*x+e)^2,-b^2*cos(f*
x+e)^2/(a^2-b^2))*cos(f*x+e)*(d*csc(f*x+e))^(3+n)*(sin(f*x+e)^2)^(3/2+1/2*n)/(a^2-b^2)^3/d^3/f

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Rubi [A]
time = 0.48, antiderivative size = 432, normalized size of antiderivative = 1.00, 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 = {3317, 3954, 2903, 3268, 440} \begin {gather*} -\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 {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 {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 {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} \end {gather*}

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 440

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])

Rule 2903

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*(1/((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 3268

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 3317

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 3954

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]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(2406\) vs. \(2(432)=864\).
time = 16.67, size = 2406, normalized size = 5.57 \begin {gather*} \text {Result too large to show} \end {gather*}

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, (-1 + b^2/a^2)*Tan[e + f*x]^2]) + b*(4*a*b*(-2 + n)*AppellF1
[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (-1 + n)*((3*a^2 + b^2)*
AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 4*b^2*AppellF1[1 -
n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^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, (-1 + b^2/a^2)*Tan[e + f*x]^2]) + b*(4*a*b*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f
*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (-1 + n)*((3*a^2 + b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -Tan
[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 4*b^2*AppellF1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2,
(-1 + b^2/a^2)*Tan[e + f*x]^2])*Tan[e + f*x])))/(a^4*(a^2 - b^2)*(-2 + n)*(-1 + n)) + (n*(Cot[e + f*x]*Sqrt[S
ec[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, (-1 + b^2/a^2)*Tan[e + f*x]^2])
+ b*(4*a*b*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2
] + (-1 + n)*((3*a^2 + b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e +
f*x]^2] - 4*b^2*AppellF1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^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, (
-1 + b^2/a^2)*Tan[e + f*x]^2]) + b*(4*a*b*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f*x]^2
, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (-1 + n)*((3*a^2 + b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -Tan[e +
f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 4*b^2*AppellF1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, (-1
+ b^2/a^2)*Tan[e + f*x]^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*(-1 + b^2/a^2)*(1 - n)*AppellF1
[1 + (1 - n)/2, -1 - n/2, 3, 1 + (3 - n)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan
[e + f*x])/(3 - n) - (2*(1 - n)*(-1 - n/2)*AppellF1[1 + (1 - n)/2, -1/2*n, 2, 1 + (3 - n)/2, -Tan[e + f*x]^2,
(-1 + b^2/a^2)*Tan[e + f*x]^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, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 4*b^2*AppellF1[1 - n/2, (-1 -
n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2])*Sec[e + f*x]^2 + 4*a*b*(-2 + n)*((6*(-1 +
b^2/a^2)*(1 - n)*AppellF1[1 + (1 - n)/2, -1 - n/2, 4, 1 + (3 - n)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f
*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 - n) - (2*(1 - n)*(-1 - n/2)*AppellF1[1 + (1 - n)/2, -1/2*n, 3, 1 + (3
- n)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^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, (
-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 - n/2)) + (4*(-1 + b^2/a^2)*(1 - n/2)*AppellF1[2
- n/2, (-1 - n)/2, 3, 3 - n/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(
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, (-1 +
b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 - n/2)) + (6*(-1 + b^2/a^2)*(1 - n/2)*AppellF1[2 - n/
2, (-1 - n)/2, 4, 3 - n/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 - n
/2))))))/(a^4*(a^2 - b^2)*(-2 + n)*(-1 + n)*(Sec[e + f*x]^2)^(n/2))))

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Maple [F]
time = 0.66, size = 0, normalized size = 0.00 \[\int \frac {\left (d \csc \left (f x +e \right )\right )^{n}}{\left (a +b \sin \left (f x +e \right )\right )^{3}}\, dx\]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

integrate((d*csc(f*x + e))^n/(b*sin(f*x + e) + a)^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (\frac {d}{\sin \left (e+f\,x\right )}\right )}^n}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/sin(e + f*x))^n/(a + b*sin(e + f*x))^3,x)

[Out]

int((d/sin(e + f*x))^n/(a + b*sin(e + f*x))^3, x)

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