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

Optimal. Leaf size=99 \[ \frac {\tan (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \cos (e+f x))^m \cos ^2(e+f x)^{\frac {1}{2} (-m+2 p+1)} \, _2F_1\left (\frac {1}{2} (2 p+1),\frac {1}{2} (-m+2 p+1);\frac {1}{2} (2 p+3);\sin ^2(e+f x)\right )}{f (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 99, 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 = {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 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]

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
]])

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*}

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Mathematica [A]  time = 0.52, size = 81, normalized size = 0.82 \[ \frac {\tan (e+f x) \sec ^2(e+f x)^{m/2} \left (b \tan ^2(e+f x)\right )^p (d \cos (e+f x))^m \, _2F_1\left (\frac {m}{2}+1,p+\frac {1}{2};p+\frac {3}{2};-\tan ^2(e+f x)\right )}{f (2 p+1)} \]

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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \cos \left (f x + e\right )\right )^{m}, x\right ) \]

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

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)

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maple [F]  time = 4.90, size = 0, normalized size = 0.00 \[ \int \left (d \cos \left (f x +e \right )\right )^{m} \left (b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,\cos \left (e+f\,x\right )\right )}^m\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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