3.698 \(\int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=140 \[ \frac {\left (a^2-b^2\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {2 a b (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3543, 3538, 3476, 364} \[ \frac {\left (a^2-b^2\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {2 a b (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + ((a^2 - b^2)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e
+ f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + (2*a*b*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e +
f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(d^2*f*(2 + 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 3538

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx &=\frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\int (d \tan (e+f x))^n \left (a^2-b^2+2 a b \tan (e+f x)\right ) \, dx\\ &=\frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\left (a^2-b^2\right ) \int (d \tan (e+f x))^n \, dx+\frac {(2 a b) \int (d \tan (e+f x))^{1+n} \, dx}{d}\\ &=\frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {(2 a b) \operatorname {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac {\left (\left (a^2-b^2\right ) d\right ) \operatorname {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {\left (a^2-b^2\right ) \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a b \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 116, normalized size = 0.83 \[ \frac {\tan (e+f x) (d \tan (e+f x))^n \left ((n+2) \left (a^2-b^2\right ) \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )+b \left (2 a (n+1) \tan (e+f x) \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};-\tan ^2(e+f x)\right )+b (n+2)\right )\right )}{f (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[e + f*x]*(d*Tan[e + f*x])^n*((a^2 - b^2)*(2 + n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]
^2] + b*(b*(2 + n) + 2*a*(1 + n)*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(
f*(1 + n)*(2 + n))

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \left (d \tan \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)*(d*tan(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.05, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^2,x)

[Out]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x))^2,x)

[Out]

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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