3.175 \(\int (d \cos (e+f x))^m (b \tan ^2(e+f x))^p \, dx\)

Optimal. Leaf size=99 \[ \frac{\tan (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \cos (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (-m+2 p+1)} \text{Hypergeometric2F1}\left (\frac{1}{2} (2 p+1),\frac{1}{2} (-m+2 p+1),\frac{1}{2} (2 p+3),\sin ^2(e+f x)\right )}{f (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.13651, antiderivative size = 99, 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 = {3658, 2603, 2617} \[ \frac{\tan (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \cos (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (-m+2 p+1)} \, _2F_1\left (\frac{1}{2} (2 p+1),\frac{1}{2} (-m+2 p+1);\frac{1}{2} (2 p+3);\sin ^2(e+f x)\right )}{f (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]

[Out]

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

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, 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 \tan ^2(e+f x)\right )^p \, dx &=\left (\tan ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p\right ) \int (d \cos (e+f x))^m \tan ^{2 p}(e+f x) \, dx\\ &=\left ((d \cos (e+f x))^m \left (\frac{\sec (e+f x)}{d}\right )^m \tan ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p\right ) \int \left (\frac{\sec (e+f x)}{d}\right )^{-m} \tan ^{2 p}(e+f x) \, dx\\ &=\frac{(d \cos (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (1-m+2 p)} \, _2F_1\left (\frac{1}{2} (1+2 p),\frac{1}{2} (1-m+2 p);\frac{1}{2} (3+2 p);\sin ^2(e+f x)\right ) \tan (e+f x) \left (b \tan ^2(e+f x)\right )^p}{f (1+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.500143, size = 81, normalized size = 0.82 \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} \left (b \tan ^2(e+f x)\right )^p (d \cos (e+f x))^m \text{Hypergeometric2F1}\left (\frac{m}{2}+1,p+\frac{1}{2},p+\frac{3}{2},-\tan ^2(e+f x)\right )}{f (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]

[Out]

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

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Maple [F]  time = 0.706, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \left ( fx+e \right ) \right ) ^{m} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)

[Out]

int((d*cos(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )^{2}\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*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2)^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 (b \tan \left (f x + e\right )^{2}\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*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2)^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*tan(f*x+e)**2)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )^{2}\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*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2)^p*(d*cos(f*x + e))^m, x)