Integrand size = 23, antiderivative size = 261 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=-\frac {b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac {\left (a^4-6 a^2 b^2+b^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {4 a b \left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)} \]
[Out]
Time = 0.85 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3647, 3718, 3711, 3619, 3557, 371} \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\frac {4 a b \left (a^2-b^2\right ) (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{d^2 f (n+2)}-\frac {b^2 \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+3)}+\frac {\left (a^4-6 a^2 b^2+b^4\right ) (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{d f (n+2) (n+3)}+\frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)} \]
[In]
[Out]
Rule 371
Rule 3557
Rule 3619
Rule 3647
Rule 3711
Rule 3718
Rubi steps \begin{align*} \text {integral}& = \frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}+\frac {\int (d \tan (e+f x))^n (a+b \tan (e+f x)) \left (-a d \left (b^2 (1+n)-a^2 (3+n)\right )+b \left (3 a^2-b^2\right ) d (3+n) \tan (e+f x)+2 a b^2 d (4+n) \tan ^2(e+f x)\right ) \, dx}{d (3+n)} \\ & = \frac {2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}-\frac {\int (d \tan (e+f x))^n \left (a^2 d^2 (2+n) \left (b^2 (1+n)-a^2 (3+n)\right )-4 a b \left (a^2-b^2\right ) d^2 (2+n) (3+n) \tan (e+f x)+b^2 d^2 (2+n) \left (b^2 (3+n)-a^2 (17+5 n)\right ) \tan ^2(e+f x)\right ) \, dx}{d^2 (2+n) (3+n)} \\ & = -\frac {b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac {2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}-\frac {\int (d \tan (e+f x))^n \left (-\left (\left (a^4-6 a^2 b^2+b^4\right ) d^2 (2+n) (3+n)\right )-4 a b \left (a^2-b^2\right ) d^2 (2+n) (3+n) \tan (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)} \\ & = -\frac {b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac {2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}-\left (-a^4+6 a^2 b^2-b^4\right ) \int (d \tan (e+f x))^n \, dx+\frac {\left (4 a b \left (a^2-b^2\right )\right ) \int (d \tan (e+f x))^{1+n} \, dx}{d} \\ & = -\frac {b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac {2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}+\frac {\left (4 a b \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac {\left (\left (a^4-6 a^2 b^2+b^4\right ) d\right ) \text {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f} \\ & = -\frac {b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac {\left (a^4-6 a^2 b^2+b^4\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^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {4 a b \left (a^2-b^2\right ) \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)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)} \\ \end{align*}
Time = 1.87 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.73 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (\frac {-b^4 (3+n)+a^2 b^2 (17+5 n)}{1+n}+\frac {\left (a^4-6 a^2 b^2+b^4\right ) (3+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )}{1+n}+\frac {2 a b^3 (4+n) \tan (e+f x)}{2+n}+\frac {4 a (a-b) b (a+b) (3+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{2+n}+b^2 (a+b \tan (e+f x))^2\right )}{f (3+n)} \]
[In]
[Out]
\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{4}d x\]
[In]
[Out]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
[In]
[Out]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{4}\, dx \]
[In]
[Out]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
[In]
[Out]
\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
[In]
[Out]
Timed out. \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^4 \,d x \]
[In]
[Out]