3.485 \(\int \cos ^3(e+f x) (b (c \tan (e+f x))^n)^p \, dx\)
Optimal. Leaf size=82 \[ \frac{\sin (e+f x) \cos ^2(e+f x)^{\frac{n p}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2} (n p-2),\frac{1}{2} (n p+1),\frac{1}{2} (n p+3),\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]
[Out]
((Cos[e + f*x]^2)^((n*p)/2)*Hypergeometric2F1[(-2 + 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))
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Rubi [A] time = 0.0918643, antiderivative size = 82, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used =
{3659, 2617} \[ \frac{\sin (e+f x) \cos ^2(e+f x)^{\frac{n p}{2}} \, _2F_1\left (\frac{1}{2} (n p-2),\frac{1}{2} (n p+1);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1)} \]
Antiderivative was successfully verified.
[In]
Int[Cos[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
[Out]
((Cos[e + f*x]^2)^((n*p)/2)*Hypergeometric2F1[(-2 + 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 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 2617
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)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[(n - 1)/2] && !In
tegerQ[m/2]
Rubi steps
\begin{align*} \int \cos ^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 \cos ^3(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{1}{2} (-2+n p),\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*}
Mathematica [C] time = 6.56998, size = 1552, normalized size = 18.93 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
[Out]
((6 + 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] - 6*AppellF1
[(1 + n*p)/2, n*p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 12*AppellF1[(1 + n*p)/2, n*p, 3,
(3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 8*AppellF1[(1 + n*p)/2, n*p, 4, (3 + n*p)/2, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cos[(e + f*x)/2]^3*Cos[e + f*x]^3*Sin[(e + f*x)/2]*(b*(c*Tan[e + f*x])^n)^p)
/(f*(1 + n*p)*(-AppellF1[(3 + n*p)/2, n*p, 2, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 12*Appel
lF1[(3 + n*p)/2, n*p, 3, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 36*AppellF1[(3 + n*p)/2, n*p,
4, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 32*AppellF1[(3 + n*p)/2, n*p, 5, (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] - 6*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] + 12*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] - 8*n*
p*AppellF1[(3 + n*p)/2, 1 + n*p, 4, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (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]*Cos[(e + f*x)/2]^2 - 18*AppellF1[(
1 + n*p)/2, n*p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 - 6*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]*Cos[(e + f*x)/2]^2 + 36*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]^2 + 12*n*p*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]^2 - 8*(3 + n*p)*Ap
pellF1[(1 + n*p)/2, n*p, 4, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + AppellF
1[(3 + n*p)/2, n*p, 2, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x] - 12*AppellF1[(3 + n
*p)/2, n*p, 3, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x] + 36*AppellF1[(3 + n*p)/2, n
*p, 4, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x] - 32*AppellF1[(3 + n*p)/2, n*p, 5, (
5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x] - 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]*Cos[e + f*x] + 6*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]*Cos[e + f*x] - 12*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]*Cos[e + f*x] + 8*n*p*AppellF1[(3 + n*p)/2, 1 + n*p, 4, (5 + n
*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]))
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Maple [F] time = 6.417, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{3} \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)^3*(b*(c*tan(f*x+e))^n)^p,x)
[Out]
int(cos(f*x+e)^3*(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 )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(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*cos(f*x + e)^3, 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 )^{3}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(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*cos(f*x + e)^3, 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)**3*(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 )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(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*cos(f*x + e)^3, x)