3.177 \(\int (d \cos (e+f x))^m (b (c \tan (e+f x))^n)^p \, dx\)

Optimal. Leaf size=101 \[ \frac{\tan (e+f x) (d \cos (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (-m+n p+1)} \left (b (c \tan (e+f x))^n\right )^p \text{Hypergeometric2F1}\left (\frac{1}{2} (n p+1),\frac{1}{2} (-m+n p+1),\frac{1}{2} (n p+3),\sin ^2(e+f x)\right )}{f (n p+1)} \]

[Out]

((d*Cos[e + f*x])^m*(Cos[e + f*x]^2)^((1 - m + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (1 - m + n*p)/2, (3 + n*
p)/2, Sin[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 = 0.147662, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3659, 2603, 2617} \[ \frac{\tan (e+f x) (d \cos (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (-m+n p+1)} \left (b (c \tan (e+f x))^n\right )^p \, _2F_1\left (\frac{1}{2} (n p+1),\frac{1}{2} (-m+n p+1);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right )}{f (n p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Cos[e + f*x])^m*(Cos[e + f*x]^2)^((1 - m + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (1 - m + n*p)/2, (3 + n*
p)/2, Sin[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 2603

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f*
x])^FracPart[m]*(Sec[e + f*x]/a)^FracPart[m], Int[(b*Tan[e + f*x])^n/(Sec[e + f*x]/a)^m, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

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 (d \cos (e+f x))^m \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 (d \cos (e+f x))^m (c \tan (e+f x))^{n p} \, dx\\ &=\left ((d \cos (e+f x))^m \left (\frac{\sec (e+f x)}{d}\right )^m (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \left (\frac{\sec (e+f x)}{d}\right )^{-m} (c \tan (e+f x))^{n p} \, dx\\ &=\frac{(d \cos (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (1-m+n p)} \, _2F_1\left (\frac{1}{2} (1+n p),\frac{1}{2} (1-m+n p);\frac{1}{2} (3+n p);\sin ^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 [A]  time = 0.463351, size = 91, normalized size = 0.9 \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \text{Hypergeometric2F1}\left (\frac{m+2}{2},\frac{1}{2} (n p+1),\frac{1}{2} (n p+3),-\tan ^2(e+f x)\right )}{f (n p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.224, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \left ( fx+e \right ) \right ) ^{m} \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((d*cos(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int((d*cos(f*x+e))^m*(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} \left (d \cos \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*(d*cos(f*x + e))^m, 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} \left (d \cos \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")

[Out]

integral(((c*tan(f*x + e))^n*b)^p*(d*cos(f*x + e))^m, 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((d*cos(f*x+e))**m*(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} \left (d \cos \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")

[Out]

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