3.362 \(\int \tan ^3(e+f x) (a+b \tan ^2(e+f x))^p \, dx\)

Optimal. Leaf size=95 \[ \frac {\left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a}{a-b}\right )}{2 f (p+1) (a-b)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{p+1}}{2 b f (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 95, 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 = {3670, 446, 80, 68} \[ \frac {\left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a}{a-b}\right )}{2 f (p+1) (a-b)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{p+1}}{2 b f (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 446

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((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[(((d*ff*x)/c)^m*(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, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 73, normalized size = 0.77 \[ -\frac {\left (a+b \tan ^2(e+f x)\right )^{p+1} \left (b \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a}{a-b}\right )+a-b\right )}{2 b f (p+1) (b-a)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

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

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2 + a)^p*tan(f*x + e)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*tan(f*x + e)^3, x)

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maple [F]  time = 0.97, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{3}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(tan(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*tan(f*x + e)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^3*(a + b*tan(e + f*x)^2)^p,x)

[Out]

int(tan(e + f*x)^3*(a + b*tan(e + f*x)^2)^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{p} \tan ^{3}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**3*(a+b*tan(f*x+e)**2)**p,x)

[Out]

Integral((a + b*tan(e + f*x)**2)**p*tan(e + f*x)**3, x)

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