Integrand size = 23, antiderivative size = 92 \[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=-\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \csc ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2+n p),\frac {1}{2} (1+n p),\frac {n p}{2},\sin ^2(e+f x)\right ) \sec (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)} \] Output:
-(cos(f*x+e)^2)^(1/2*n*p+1/2)*csc(f*x+e)^2*hypergeom([1/2*n*p-1, 1/2*n*p+1 /2],[1/2*n*p],sin(f*x+e)^2)*sec(f*x+e)*(b*(c*tan(f*x+e))^n)^p/f/(-n*p+2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 13.65 (sec) , antiderivative size = 1399, normalized size of antiderivative = 15.21 \[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx =\text {Too large to display} \] Input:
Integrate[Csc[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
Output:
(Cot[(e + f*x)/2]^2*Hypergeometric2F1[n*p, -1 + (n*p)/2, (n*p)/2, Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n*p)*(b*(c*Tan[e + f*x])^n) ^p)/(f*(-8 + 4*n*p)) + ((4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n*p)/ 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[e + f*x]^2*(b*(c*Tan[e + f *x])^n)^p)/(8*f*(2 + n*p)*((4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n* p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + Appell F1[2 + (n*p)/2, n*p, 2, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2] ^2]*(-1 + Cos[e + f*x]) + 2*n*p*AppellF1[2 + (n*p)/2, 1 + n*p, 1, 3 + (n*p )/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[(e + f*x)/2]^2)) + (Hype rgeometric2F1[n*p, 1 + (n*p)/2, 2 + (n*p)/2, Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n*p)*Tan[(e + f*x)/2]^2*(b*(c*Tan[e + f*x])^n)^p )/(f*(8 + 4*n*p)) + (Cot[(e + f*x)/2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n *p)*((2 + n*p)*Hypergeometric2F1[(n*p)/2, n*p, 1 + (n*p)/2, Tan[(e + f*x)/ 2]^2] - n*p*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*Tan[e + f*x]^(n*p)*(b*(c*Tan[e + f*x])^n)^p)/(8*f*n*p*(2 + n*p)*(((Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1 + n*p)*(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e + f*x]*Sec[(e + f*x)/2]^2 *Tan[(e + f*x)/2])*((2 + n*p)*Hypergeometric2F1[(n*p)/2, n*p, 1 + (n*p)/2, Tan[(e + f*x)/2]^2] - n*p*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n*p)/2, Tan[ (e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*Tan[e + f*x]^(...
Time = 0.75 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4142, 3042, 3081, 3042, 3057}
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.
\(\displaystyle \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b (c \tan (e+f x))^n\right )^p}{\sin (e+f x)^3}dx\) |
\(\Big \downarrow \) 4142 |
\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \csc ^3(e+f x) (c \tan (e+f x))^{n p}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \frac {(c \tan (e+f x))^{n p}}{\sin (e+f x)^3}dx\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos ^{-n p}(e+f x) \sin ^{n p-3}(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos (e+f x)^{-n p} \sin (e+f x)^{n p-3}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle -\frac {\csc ^2(e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (n p-2),\frac {1}{2} (n p+1),\frac {n p}{2},\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)}\) |
Input:
Int[Csc[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
Output:
-(((Cos[e + f*x]^2)^((1 + n*p)/2)*Csc[e + f*x]^2*Hypergeometric2F1[(-2 + n *p)/2, (1 + n*p)/2, (n*p)/2, Sin[e + f*x]^2]*Sec[e + f*x]*(b*(c*Tan[e + f* x])^n)^p)/(f*(2 - n*p)))
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> S imp[b^IntPart[p]*((b*(c*Tan[e + f*x])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*Fr acPart[p])) Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{ b, c, e, f, n, p}, x] && !IntegerQ[p] && !IntegerQ[n] && (EqQ[u, 1] || Ma tchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
\[\int \csc \left (f x +e \right )^{3} \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]
Input:
int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
Output:
int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:
integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")
Output:
integral(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \csc ^{3}{\left (e + f x \right )}\, dx \] Input:
integrate(csc(f*x+e)**3*(b*(c*tan(f*x+e))**n)**p,x)
Output:
Integral((b*(c*tan(e + f*x))**n)**p*csc(e + f*x)**3, x)
\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:
integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")
Output:
integrate(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:
integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")
Output:
integrate(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
Timed out. \[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \frac {{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\sin \left (e+f\,x\right )}^3} \,d x \] Input:
int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^3,x)
Output:
int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^3, x)
\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=c^{n p} b^{p} \left (\int \tan \left (f x +e \right )^{n p} \csc \left (f x +e \right )^{3}d x \right ) \] Input:
int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
Output:
c**(n*p)*b**p*int(tan(e + f*x)**(n*p)*csc(e + f*x)**3,x)