3.173 \(\int \csc ^3(e+f x) (b (c \tan (e+f x))^n)^p \, dx\)
Optimal. Leaf size=92 \[ -\frac {\csc ^2(e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \, _2F_1\left (\frac {1}{2} (n p-2),\frac {1}{2} (n p+1);\frac {n p}{2};\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)} \]
[Out]
-(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)
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Rubi [A] time = 0.15, antiderivative size = 92, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used =
{3659, 2601, 2577} \[ -\frac {\csc ^2(e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \, _2F_1\left (\frac {1}{2} (n p-2),\frac {1}{2} (n p+1);\frac {n p}{2};\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
[Out]
-(((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)))
Rule 2577
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*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]
Rule 2601
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(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])
Rule 3659
Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Tan[e + f*x
])^n)^FracPart[p])/(c*Tan[e + f*x])^(n*FracPart[p]), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] && !IntegerQ[p] && !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Rubi steps
\begin {align*} \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx &=\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \csc ^3(e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\left (\cos ^{n p}(e+f x) \sin ^{-n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p\right ) \int \cos ^{-n p}(e+f x) \sin ^{-3+n p}(e+f x) \, dx\\ &=-\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \csc ^2(e+f x) \, _2F_1\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)}\\ \end {align*}
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Mathematica [C] time = 16.28, size = 1399, normalized size = 15.21 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Csc[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
[Out]
(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 + Ap
pellF1[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)) +
(Hypergeometric2F1[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]^(n*p))/(2*(2 + n*p)) + ((Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n*p)*
(-(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]*Sec[(e + f*x)/2]^2*
Tan[(e + f*x)/2]) - n*p*Tan[(e + f*x)/2]^2*(-(((1 + (n*p)/2)*AppellF1[2 + (n*p)/2, n*p, 2, 3 + (n*p)/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(2 + (n*p)/2)) + (n*p*(1 + (n*p)/2)*Ap
pellF1[2 + (n*p)/2, 1 + n*p, 1, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(
e + f*x)/2])/(2 + (n*p)/2)) + (n*p*(2 + n*p)*Csc[(e + f*x)/2]*Sec[(e + f*x)/2]*(-Hypergeometric2F1[(n*p)/2, n*
p, 1 + (n*p)/2, Tan[(e + f*x)/2]^2] + (1 - Tan[(e + f*x)/2]^2)^(-(n*p))))/2)*Tan[e + f*x]^(n*p))/(2*n*p*(2 + n
*p)) + ((Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n*p)*Sec[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]^(-1 + n*p))/(2*(2 + n*p))))
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")
[Out]
integral(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")
[Out]
integrate(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
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maple [F] time = 21.94, size = 0, normalized size = 0.00 \[ \int \left (\csc ^{3}\left (f x +e \right )\right ) \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
[Out]
int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")
[Out]
integrate(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\sin \left (e+f\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^3,x)
[Out]
int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^3, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \csc ^{3}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)**3*(b*(c*tan(f*x+e))**n)**p,x)
[Out]
Integral((b*(c*tan(e + f*x))**n)**p*csc(e + f*x)**3, x)
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