Integrand size = 23, antiderivative size = 409 \[ \int \frac {(d \csc (e+f x))^n}{(3+b \sin (e+f x))^3} \, dx=-\frac {9 b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1+n),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^4(e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-1+n)}}{\left (9-b^2\right )^3 d^3 f}+\frac {b^3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-2+n),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}}{\left (9-b^2\right )^3 d^3 f}+\frac {27 b \operatorname {AppellF1}\left (\frac {1}{2},\frac {n}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^3(e+f x) \sin ^2(e+f x)^{n/2}}{\left (9-b^2\right )^3 d^3 f}-\frac {27 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1+n}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) (d \csc (e+f x))^{3+n} \sin ^2(e+f x)^{\frac {3+n}{2}}}{\left (9-b^2\right )^3 d^3 f} \]
-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)*sin(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
Leaf count is larger than twice the leaf count of optimal. \(2297\) vs. \(2(409)=818\).
Time = 20.15 (sec) , antiderivative size = 2297, normalized size of antiderivative = 5.62 \[ \int \frac {(d \csc (e+f x))^n}{(3+b \sin (e+f x))^3} \, dx=\text {Result too large to show} \]
-1/81*((d*Csc[e + f*x])^n*(Cot[e + f*x]*Sqrt[Sec[e + f*x]^2])^n*Tan[e + f* x]*(-9*(3 + b^2)*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 2, (3 - n)/2, -Tan [e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] + b*(12*b*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x] ^2)/9] + (-1 + n)*((27 + b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2, 2 - n/2, -T an[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] - 4*b^2*AppellF1[1 - n/2, (- 1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9])*Tan [e + f*x])))/((-9 + b^2)*f*(-2 + n)*(-1 + n)*(Sec[e + f*x]^2)^(n/2)*(3 + b *Sin[e + f*x])^3*(-1/81*((Sec[e + f*x]^2)^(1 - n/2)*(Cot[e + f*x]*Sqrt[Sec [e + f*x]^2])^n*(-9*(3 + b^2)*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 2, (3 - n)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] + b*(12*b*(-2 + n )*AppellF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e + f*x]^2, ((-9 + b^2) *Tan[e + f*x]^2)/9] + (-1 + n)*((27 + b^2)*AppellF1[1 - n/2, (-1 - n)/2, 2 , 2 - n/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] - 4*b^2*AppellF 1[1 - n/2, (-1 - n)/2, 3, 2 - n/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f* x]^2)/9])*Tan[e + f*x])))/((-9 + 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*Sq rt[Sec[e + f*x]^2])*Tan[e + f*x]*(-9*(3 + b^2)*(-2 + n)*AppellF1[(1 - n)/2 , -1 - n/2, 2, (3 - n)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] + b*(12*b*(-2 + n)*AppellF1[(1 - n)/2, -1 - n/2, 3, (3 - n)/2, -Tan[e +...
Time = 1.01 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3717, 3042, 4356, 3042, 3303, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(d \csc (e+f x))^n}{(a+b \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \csc (e+f x))^n}{(a+b \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3717 |
| \(\displaystyle \frac {\int \frac {(d \csc (e+f x))^{n+3}}{(b+a \csc (e+f x))^3}dx}{d^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(d \csc (e+f x))^{n+3}}{(b+a \csc (e+f x))^3}dx}{d^3}\) |
| \(\Big \downarrow \) 4356 |
| \(\displaystyle \frac {\sin ^{n+3}(e+f x) (d \csc (e+f x))^{n+3} \int \frac {\sin ^{-n}(e+f x)}{(a+b \sin (e+f x))^3}dx}{d^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^{n+3}(e+f x) (d \csc (e+f x))^{n+3} \int \frac {\sin (e+f x)^{-n}}{(a+b \sin (e+f x))^3}dx}{d^3}\) |
| \(\Big \downarrow \) 3303 |
| \(\displaystyle \frac {\sin ^{n+3}(e+f x) (d \csc (e+f x))^{n+3} \int \left (-\frac {3 a^2 b \sin ^{1-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac {3 a b^2 \sin ^{2-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac {b^3 \sin ^{3-n}(e+f x)}{\left (b^2 \sin ^2(e+f x)-a^2\right )^3}+\frac {a^3 \sin ^{-n}(e+f x)}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}\right )dx}{d^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sin ^{n+3}(e+f x) (d \csc (e+f x))^{n+3} \left (-\frac {3 a b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {n-1}{2}} \sin ^{1-n}(e+f x) \operatorname {AppellF1}\left (\frac {1}{2},\frac {n-1}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}+\frac {3 a^2 b \cos (e+f x) \sin ^2(e+f x)^{n/2} \sin ^{-n}(e+f x) \operatorname {AppellF1}\left (\frac {1}{2},\frac {n}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}+\frac {b^3 \cos (e+f x) \sin ^2(e+f x)^{n/2} \sin ^{-n}(e+f x) \operatorname {AppellF1}\left (\frac {1}{2},\frac {n-2}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}-\frac {a^3 \cos (e+f x) \sin ^2(e+f x)^{\frac {n+1}{2}} \sin ^{-n-1}(e+f x) \operatorname {AppellF1}\left (\frac {1}{2},\frac {n+1}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}\right )}{d^3}\) |
((d*Csc[e + f*x])^(3 + n)*Sin[e + f*x]^(3 + n)*((-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]*Sin[e + f*x]^(1 - n)*(Sin[e + f*x]^2)^((-1 + n)/2))/((a^2 - b^2) ^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]*(Sin[e + f*x]^2)^(n/2))/((a^2 - b^2 )^3*f*Sin[e + f*x]^n) + (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]*(Sin[e + f*x]^2)^(n/2) )/((a^2 - b^2)^3*f*Sin[e + f*x]^n) - (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]*Sin[e + f*x]^(-1 - n)*(Sin[e + f*x]^2)^((1 + n)/2))/((a^2 - b^2)^3*f)))/d^3
3.9.30.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[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]
\[\int \frac {\left (d \csc \left (f x +e \right )\right )^{n}}{\left (a +b \sin \left (f x +e \right )\right )^{3}}d x\]
\[ \int \frac {(d \csc (e+f x))^n}{(3+b \sin (e+f x))^3} \, dx=\int { \frac {\left (d \csc \left (f x + e\right )\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
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)*sin(f*x + e)), x)
\[ \int \frac {(d \csc (e+f x))^n}{(3+b \sin (e+f x))^3} \, dx=\int \frac {\left (d \csc {\left (e + f x \right )}\right )^{n}}{\left (a + b \sin {\left (e + f x \right )}\right )^{3}}\, dx \]
\[ \int \frac {(d \csc (e+f x))^n}{(3+b \sin (e+f x))^3} \, dx=\int { \frac {\left (d \csc \left (f x + e\right )\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(d \csc (e+f x))^n}{(3+b \sin (e+f x))^3} \, dx=\int { \frac {\left (d \csc \left (f x + e\right )\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(d \csc (e+f x))^n}{(3+b \sin (e+f x))^3} \, dx=\int \frac {{\left (\frac {d}{\sin \left (e+f\,x\right )}\right )}^n}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]