3.160 \(\int (a+b \tan ^2(e+f x))^p \, dx\)

Optimal. Leaf size=78 \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};1,-p;\frac{3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f} \]

[Out]

(AppellF1[1/2, 1, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]*(a + b*Tan[e + f*x]^2)^p)/(f
*(1 + (b*Tan[e + f*x]^2)/a)^p)

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Rubi [A]  time = 0.064155, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3661, 430, 429} \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};1,-p;\frac{3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x]^2)^p,x]

[Out]

(AppellF1[1/2, 1, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]*(a + b*Tan[e + f*x]^2)^p)/(f
*(1 + (b*Tan[e + f*x]^2)/a)^p)

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (\left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a}\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{F_1\left (\frac{1}{2};1,-p;\frac{3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a}\right )^{-p}}{f}\\ \end{align*}

Mathematica [B]  time = 0.552796, size = 192, normalized size = 2.46 \[ \frac{3 a \sin (2 (e+f x)) \left (a+b \tan ^2(e+f x)\right )^p F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b \tan ^2(e+f x)}{a},-\tan ^2(e+f x)\right )}{4 f \tan ^2(e+f x) \left (b p F_1\left (\frac{3}{2};1-p,1;\frac{5}{2};-\frac{b \tan ^2(e+f x)}{a},-\tan ^2(e+f x)\right )-a F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b \tan ^2(e+f x)}{a},-\tan ^2(e+f x)\right )\right )+6 a f F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b \tan ^2(e+f x)}{a},-\tan ^2(e+f x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Tan[e + f*x]^2)^p,x]

[Out]

(3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sin[2*(e + f*x)]*(a + b*Tan[e + f*x]^
2)^p)/(6*a*f*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 4*f*(b*p*AppellF1[3/2, 1 -
p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] - a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -T
an[e + f*x]^2])*Tan[e + f*x]^2)

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Maple [F]  time = 0.072, size = 0, normalized size = 0. \begin{align*} \int \left ( a+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((a+b*tan(f*x+e)^2)^p,x)

[Out]

int((a+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} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e)^2 + a)^p, 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} + a\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*tan(f*x + e)^2 + a)^p, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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