3.45 \(\int \tan ^2(c+d x) (b \tan (c+d x))^n (A+B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=132 \[ \frac{(A-C) (b \tan (c+d x))^{n+3} \text{Hypergeometric2F1}\left (1,\frac{n+3}{2},\frac{n+5}{2},-\tan ^2(c+d x)\right )}{b^3 d (n+3)}+\frac{B (b \tan (c+d x))^{n+4} \text{Hypergeometric2F1}\left (1,\frac{n+4}{2},\frac{n+6}{2},-\tan ^2(c+d x)\right )}{b^4 d (n+4)}+\frac{C (b \tan (c+d x))^{n+3}}{b^3 d (n+3)} \]

[Out]

(C*(b*Tan[c + d*x])^(3 + n))/(b^3*d*(3 + n)) + ((A - C)*Hypergeometric2F1[1, (3 + n)/2, (5 + n)/2, -Tan[c + d*
x]^2]*(b*Tan[c + d*x])^(3 + n))/(b^3*d*(3 + n)) + (B*Hypergeometric2F1[1, (4 + n)/2, (6 + n)/2, -Tan[c + d*x]^
2]*(b*Tan[c + d*x])^(4 + n))/(b^4*d*(4 + n))

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Rubi [A]  time = 0.155361, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {16, 3630, 3538, 3476, 364} \[ \frac{(A-C) (b \tan (c+d x))^{n+3} \, _2F_1\left (1,\frac{n+3}{2};\frac{n+5}{2};-\tan ^2(c+d x)\right )}{b^3 d (n+3)}+\frac{B (b \tan (c+d x))^{n+4} \, _2F_1\left (1,\frac{n+4}{2};\frac{n+6}{2};-\tan ^2(c+d x)\right )}{b^4 d (n+4)}+\frac{C (b \tan (c+d x))^{n+3}}{b^3 d (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(b*Tan[c + d*x])^n*(A + B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(C*(b*Tan[c + d*x])^(3 + n))/(b^3*d*(3 + n)) + ((A - C)*Hypergeometric2F1[1, (3 + n)/2, (5 + n)/2, -Tan[c + d*
x]^2]*(b*Tan[c + d*x])^(3 + n))/(b^3*d*(3 + n)) + (B*Hypergeometric2F1[1, (4 + n)/2, (6 + n)/2, -Tan[c + d*x]^
2]*(b*Tan[c + d*x])^(4 + n))/(b^4*d*(4 + n))

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 \tan ^2(c+d x) (b \tan (c+d x))^n \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx &=\frac{\int (b \tan (c+d x))^{2+n} \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac{C (b \tan (c+d x))^{3+n}}{b^3 d (3+n)}+\frac{\int (b \tan (c+d x))^{2+n} (A-C+B \tan (c+d x)) \, dx}{b^2}\\ &=\frac{C (b \tan (c+d x))^{3+n}}{b^3 d (3+n)}+\frac{B \int (b \tan (c+d x))^{3+n} \, dx}{b^3}+\frac{(A-C) \int (b \tan (c+d x))^{2+n} \, dx}{b^2}\\ &=\frac{C (b \tan (c+d x))^{3+n}}{b^3 d (3+n)}+\frac{B \operatorname{Subst}\left (\int \frac{x^{3+n}}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{b^2 d}+\frac{(A-C) \operatorname{Subst}\left (\int \frac{x^{2+n}}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{C (b \tan (c+d x))^{3+n}}{b^3 d (3+n)}+\frac{(A-C) \, _2F_1\left (1,\frac{3+n}{2};\frac{5+n}{2};-\tan ^2(c+d x)\right ) (b \tan (c+d x))^{3+n}}{b^3 d (3+n)}+\frac{B \, _2F_1\left (1,\frac{4+n}{2};\frac{6+n}{2};-\tan ^2(c+d x)\right ) (b \tan (c+d x))^{4+n}}{b^4 d (4+n)}\\ \end{align*}

Mathematica [A]  time = 0.401928, size = 110, normalized size = 0.83 \[ \frac{\tan ^3(c+d x) (b \tan (c+d x))^n \left ((n+4) (A-C) \text{Hypergeometric2F1}\left (1,\frac{n+3}{2},\frac{n+5}{2},-\tan ^2(c+d x)\right )+B (n+3) \tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{n+4}{2},\frac{n+6}{2},-\tan ^2(c+d x)\right )+C (n+4)\right )}{d (n+3) (n+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2*(b*Tan[c + d*x])^n*(A + B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(Tan[c + d*x]^3*(b*Tan[c + d*x])^n*(C*(4 + n) + (A - C)*(4 + n)*Hypergeometric2F1[1, (3 + n)/2, (5 + n)/2, -Ta
n[c + d*x]^2] + B*(3 + n)*Hypergeometric2F1[1, (4 + n)/2, (6 + n)/2, -Tan[c + d*x]^2]*Tan[c + d*x]))/(d*(3 + n
)*(4 + n))

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Maple [F]  time = 0.326, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{2} \left ( b\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) +C \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

int(tan(d*x+c)^2*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (d x + c\right )^{2} + B \tan \left (d x + c\right ) + A\right )} \left (b \tan \left (d x + c\right )\right )^{n} \tan \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*tan(d*x + c)^2 + B*tan(d*x + c) + A)*(b*tan(d*x + c))^n*tan(d*x + c)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \tan \left (d x + c\right )^{4} + B \tan \left (d x + c\right )^{3} + A \tan \left (d x + c\right )^{2}\right )} \left (b \tan \left (d x + c\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*tan(d*x + c)^4 + B*tan(d*x + c)^3 + A*tan(d*x + c)^2)*(b*tan(d*x + c))^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan{\left (c + d x \right )}\right )^{n} \left (A + B \tan{\left (c + d x \right )} + C \tan ^{2}{\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(b*tan(d*x+c))**n*(A+B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Integral((b*tan(c + d*x))**n*(A + B*tan(c + d*x) + C*tan(c + d*x)**2)*tan(c + d*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (d x + c\right )^{2} + B \tan \left (d x + c\right ) + A\right )} \left (b \tan \left (d x + c\right )\right )^{n} \tan \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*tan(d*x + c)^2 + B*tan(d*x + c) + A)*(b*tan(d*x + c))^n*tan(d*x + c)^2, x)