3.481 \(\int \cos ^2(e+f x) (b (c \tan (e+f x))^n)^p \, dx\)

Optimal. Leaf size=61 \[ \frac {\tan (e+f x) \, _2F_1\left (2,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]

[Out]

hypergeom([2, 1/2*n*p+1/2],[1/2*n*p+3/2],-tan(f*x+e)^2)*tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+1)

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Rubi [A]  time = 0.10, antiderivative size = 61, 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, 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 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])

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 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 \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*}

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Mathematica [C]  time = 5.51, 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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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \cos \left (f x + e\right )^{2}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \cos \left (f x + e\right )^{2}\,{d x} \]

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)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \cos \left (f x + e\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\cos \left (e+f\,x\right )}^2\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^2*(b*(c*tan(e + f*x))^n)^p,x)

[Out]

int(cos(e + f*x)^2*(b*(c*tan(e + f*x))^n)^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \cos ^{2}{\left (e + f x \right )}\, dx \]

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]

Integral((b*(c*tan(e + f*x))**n)**p*cos(e + f*x)**2, x)

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