3.697 \(\int (d \tan (e+f x))^n (a+b \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=198 \[ \frac{b \left (3 a^2-b^2\right ) (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac{a \left (a^2-3 b^2\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac{a b^2 (2 n+5) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2)}+\frac{b^2 (a+b \tan (e+f x)) (d \tan (e+f x))^{n+1}}{d f (n+2)} \]

[Out]

(a*b^2*(5 + 2*n)*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(2 + n)) + (a*(a^2 - 3*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)) + (b*(3*a^2 - b^2)*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)) + (b^2*(d*Tan[e + f*x])^(
1 + n)*(a + b*Tan[e + f*x]))/(d*f*(2 + n))

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Rubi [A]  time = 0.281034, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3566, 3630, 3538, 3476, 364} \[ \frac{b \left (3 a^2-b^2\right ) (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac{a \left (a^2-3 b^2\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac{a b^2 (2 n+5) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2)}+\frac{b^2 (a+b \tan (e+f x)) (d \tan (e+f x))^{n+1}}{d f (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2*(5 + 2*n)*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(2 + n)) + (a*(a^2 - 3*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)) + (b*(3*a^2 - b^2)*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)) + (b^2*(d*Tan[e + f*x])^(
1 + n)*(a + b*Tan[e + f*x]))/(d*f*(2 + n))

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

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 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 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])

Rubi steps

\begin{align*} \int (d \tan (e+f x))^n (a+b \tan (e+f x))^3 \, dx &=\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\frac{\int (d \tan (e+f x))^n \left (-a d \left (b^2 (1+n)-a^2 (2+n)\right )+b \left (3 a^2-b^2\right ) d (2+n) \tan (e+f x)+a b^2 d (5+2 n) \tan ^2(e+f x)\right ) \, dx}{d (2+n)}\\ &=\frac{a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\frac{\int (d \tan (e+f x))^n \left (a \left (a^2-3 b^2\right ) d (2+n)+b \left (3 a^2-b^2\right ) d (2+n) \tan (e+f x)\right ) \, dx}{d (2+n)}\\ &=\frac{a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\left (a \left (a^2-3 b^2\right )\right ) \int (d \tan (e+f x))^n \, dx+\frac{\left (b \left (3 a^2-b^2\right )\right ) \int (d \tan (e+f x))^{1+n} \, dx}{d}\\ &=\frac{a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\frac{\left (b \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac{\left (a \left (a^2-3 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{a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac{a \left (a^2-3 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{b \left (3 a^2-b^2\right ) \, _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)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}\\ \end{align*}

Mathematica [A]  time = 0.714548, size = 141, normalized size = 0.71 \[ \frac{\tan (e+f x) (d \tan (e+f x))^n \left (a (n+2) \left (a^2-3 b^2\right ) \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )+b \left ((n+1) \left (3 a^2-b^2\right ) \tan (e+f x) \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )+b (3 a (n+2)+b (n+1) \tan (e+f x))\right )\right )}{f (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.242, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,x)

[Out]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,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 ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^3*(d*tan(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b \tan \left (f x + e\right ) + a^{3}\right )} \left (d \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan{\left (e + f x \right )}\right )^{n} \left (a + b \tan{\left (e + f x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**n*(a+b*tan(f*x+e))**3,x)

[Out]

Integral((d*tan(e + f*x))**n*(a + b*tan(e + f*x))**3, 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 ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^3*(d*tan(f*x + e))^n, x)