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

Optimal. Leaf size=72 \[ \frac {\tan (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \tan (e+f x))^m \, _2F_1\left (1,\frac {1}{2} (m+2 p+1);\frac {1}{2} (m+2 p+3);-\tan ^2(e+f x)\right )}{f (m+2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3578, 20, 3476, 364} \[ \frac {\tan (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \tan (e+f x))^m \, _2F_1\left (1,\frac {1}{2} (m+2 p+1);\frac {1}{2} (m+2 p+3);-\tan ^2(e+f x)\right )}{f (m+2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3578

Int[((c_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol]
:> Dist[(c^IntPart[n]*(c*(d*Tan[e + f*x])^p)^FracPart[n])/(d*Tan[e + f*x])^(p*FracPart[n]), Int[(a + b*Tan[e
+ f*x])^m*(d*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[n] &&  !Intege
rQ[m]

Rubi steps

\begin {align*} \int (d \tan (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 \tan ^{2 p}(e+f x) (d \tan (e+f x))^m \, dx\\ &=\left (\tan ^{-m-2 p}(e+f x) (d \tan (e+f x))^m \left (b \tan ^2(e+f x)\right )^p\right ) \int \tan ^{m+2 p}(e+f x) \, dx\\ &=\frac {\left (\tan ^{-m-2 p}(e+f x) (d \tan (e+f x))^m \left (b \tan ^2(e+f x)\right )^p\right ) \operatorname {Subst}\left (\int \frac {x^{m+2 p}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\, _2F_1\left (1,\frac {1}{2} (1+m+2 p);\frac {1}{2} (3+m+2 p);-\tan ^2(e+f x)\right ) \tan (e+f x) (d \tan (e+f x))^m \left (b \tan ^2(e+f x)\right )^p}{f (1+m+2 p)}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 74, normalized size = 1.03 \[ \frac {\tan (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \tan (e+f x))^m \, _2F_1\left (1,\frac {1}{2} (m+2 p+1);\frac {1}{2} (m+2 p+1)+1;-\tan ^2(e+f x)\right )}{f (m+2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \tan \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

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

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maple [F]  time = 3.06, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (f x +e \right )\right )^{m} \left (b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*tan(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 \[ \int \left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*tan(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 \[ \int \left (b \tan ^{2}{\left (e + f x \right )}\right )^{p} \left (d \tan {\left (e + f x \right )}\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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