3.5.84 \(\int \cos (e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [484]

Optimal. Leaf size=79 \[ \frac {\cos ^2(e+f x)^{\frac {n p}{2}} \, _2F_1\left (\frac {n p}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sin ^2(e+f x)\right ) \sin (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+n p)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3740, 2697} \begin {gather*} \frac {\sin (e+f x) \cos ^2(e+f x)^{\frac {n p}{2}} \, _2F_1\left (\frac {n p}{2},\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

((Cos[e + f*x]^2)^((n*p)/2)*Hypergeometric2F1[(n*p)/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]*
(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + n*p))

Rule 2697

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*Sec[e + f
*x])^m*(b*Tan[e + f*x])^(n + 1)*((Cos[e + f*x]^2)^((m + n + 1)/2)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2,
(m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rule 3740

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 (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 (e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\frac {\cos ^2(e+f x)^{\frac {n p}{2}} \, _2F_1\left (\frac {n p}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sin ^2(e+f x)\right ) \sin (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] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 2.29, size = 482, normalized size = 6.10 \begin {gather*} \frac {(3+n p) \left (F_1\left (\frac {1}{2} (1+n p);n p,1;\frac {1}{2} (3+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 F_1\left (\frac {1}{2} (1+n p);n p,2;\frac {1}{2} (3+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (2 (e+f x)) \left (b (c \tan (e+f x))^n\right )^p}{2 f (1+n p) \left ((3+n p) F_1\left (\frac {1}{2} (1+n p);n p,1;\frac {1}{2} (3+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((3+n p) F_1\left (\frac {1}{2} (1+n p);n p,2;\frac {1}{2} (3+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\left (F_1\left (\frac {1}{2} (3+n p);n p,2;\frac {1}{2} (5+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 F_1\left (\frac {1}{2} (3+n p);n p,3;\frac {1}{2} (5+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n p F_1\left (\frac {1}{2} (3+n p);1+n p,1;\frac {1}{2} (5+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+2 n p F_1\left (\frac {1}{2} (3+n p);1+n p,2;\frac {1}{2} (5+n p);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

((3 + 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] - 2*AppellF1[(
1 + n*p)/2, n*p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sin[2*(e + f*x)]*(b*(c*Tan[e + f*x]
)^n)^p)/(2*f*(1 + n*p)*((3 + 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] - 2*((3 + n*p)*AppellF1[(1 + n*p)/2, n*p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (
AppellF1[(3 + n*p)/2, n*p, 2, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*AppellF1[(3 + n*p)/2,
n*p, 3, (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] + 2*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])*Tan[(e + f*x)/2]^2)))

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \cos \left (f x +e \right ) \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(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), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(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), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(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), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \cos \left (e+f\,x\right )\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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