Integrand size = 23, antiderivative size = 140 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {\left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a b \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)} \]
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)
Time = 0.51 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (\left (a^2-b^2\right ) (2+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )+b \left (b (2+n)+2 a (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{f (1+n) (2+n)} \]
(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)*Hyper geometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(f* (1 + n)*(2 + n))
Time = 0.50 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4026, 3042, 4021, 3042, 3957, 278}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (a+b \tan (e+f x))^2 (d \tan (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (e+f x))^2 (d \tan (e+f x))^ndx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle \int (d \tan (e+f x))^n \left (a^2+2 b \tan (e+f x) a-b^2\right )dx+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \tan (e+f x))^n \left (a^2+2 b \tan (e+f x) a-b^2\right )dx+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}\) |
\(\Big \downarrow \) 4021 |
\(\displaystyle \left (a^2-b^2\right ) \int (d \tan (e+f x))^ndx+\frac {2 a b \int (d \tan (e+f x))^{n+1}dx}{d}+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (a^2-b^2\right ) \int (d \tan (e+f x))^ndx+\frac {2 a b \int (d \tan (e+f x))^{n+1}dx}{d}+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {d \left (a^2-b^2\right ) \int \frac {(d \tan (e+f x))^n}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}+\frac {2 a b \int \frac {(d \tan (e+f x))^{n+1}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\left (a^2-b^2\right ) (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {2 a b (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}\) |
(b^2*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + ((a^2 - b^2)*Hypergeometric 2F1[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))
3.7.98.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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]
Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Tan[e + f*x])^m, x], x] + Simp[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]
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])
\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{2}d x\]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]