3.481 \(\int \cos ^2(e+f x) (b (c \tan (e+f x))^n)^p \, dx\)
Optimal. Leaf size=61 \[ \frac{\tan (e+f x) \text{Hypergeometric2F1}\left (2,\frac{1}{2} (n p+1),\frac{1}{2} (n p+3),-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]
[Out]
(Hypergeometric2F1[2, (1 + n*p)/2, (3 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1
+ n*p))
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Rubi [A] time = 0.103805, antiderivative size = 61, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used =
{3659, 2607, 364} \[ \frac{\tan (e+f x) \, _2F_1\left (2,\frac{1}{2} (n p+1);\frac{1}{2} (n p+3);-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]
Antiderivative was successfully verified.
[In]
Int[Cos[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p,x]
[Out]
(Hypergeometric2F1[2, (1 + n*p)/2, (3 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1
+ n*p))
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]])
Rule 2607
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])
Rule 364
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])
Rubi steps
\begin{align*} \int \cos ^2(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 \cos ^2(e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\frac{\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \operatorname{Subst}\left (\int \frac{(c x)^{n p}}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\, _2F_1\left (2,\frac{1}{2} (1+n p);\frac{1}{2} (3+n p);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+n p)}\\ \end{align*}
Mathematica [C] time = 5.5656, size = 1060, normalized size = 17.38 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p,x]
[Out]
(2*(AppellF1[(1 + n*p)/2, n*p, 1, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*AppellF1[(1 + n*p)
/2, n*p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*AppellF1[(1 + n*p)/2, n*p, 3, (3 + n*p)/
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cos[e + f*x]^2*Tan[(e + f*x)/2]*(b*(c*Tan[e + f*x])^n)^p)/(f*((Ap
pellF1[(1 + n*p)/2, n*p, 1, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*AppellF1[(1 + n*p)/2, n*
p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*AppellF1[(1 + n*p)/2, n*p, 3, (3 + n*p)/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2 + n*p*(AppellF1[(1 + n*p)/2, n*p, 1, (3 + n*p)/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*AppellF1[(1 + n*p)/2, n*p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan
[(e + f*x)/2]^2] + 4*AppellF1[(1 + n*p)/2, n*p, 3, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[
(e + f*x)/2]^2*Sec[e + f*x] + (2*(1 + n*p)*(-AppellF1[(3 + n*p)/2, n*p, 2, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2] + 8*AppellF1[(3 + n*p)/2, n*p, 3, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 1
2*AppellF1[(3 + n*p)/2, n*p, 4, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*p*AppellF1[(3 + n*p)
/2, 1 + n*p, 1, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*n*p*AppellF1[(3 + n*p)/2, 1 + n*p, 2
, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*n*p*AppellF1[(3 + n*p)/2, 1 + n*p, 3, (5 + 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)/(3 + n*p) - 2*n*p*(AppellF1[
(1 + n*p)/2, n*p, 1, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*AppellF1[(1 + n*p)/2, n*p, 2, (
3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*AppellF1[(1 + n*p)/2, n*p, 3, (3 + n*p)/2, Tan[(e + f
*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[e + f*x]*Tan[(e + f*x)/2]^2))
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Maple [F] time = 5.872, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( b \left ( c\tan \left ( fx+e \right ) \right ) ^{n} \right ) ^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x)
[Out]
int(cos(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")
[Out]
integrate(((c*tan(f*x + e))^n*b)^p*cos(f*x + e)^2, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \cos \left (f x + e\right )^{2}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")
[Out]
integral(((c*tan(f*x + e))^n*b)^p*cos(f*x + e)^2, x)
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Sympy [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(cos(f*x+e)**2*(b*(c*tan(f*x+e))**n)**p,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")
[Out]
integrate(((c*tan(f*x + e))^n*b)^p*cos(f*x + e)^2, x)