Integrand size = 23, antiderivative size = 245 \[ \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx=\frac {3 a b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {a^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {3 a^2 b \cos ^2(e+f x)^{\frac {2+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {2+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) \sec (e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {b^3 \cos ^2(e+f x)^{\frac {4+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {4+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) \sec ^3(e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)} \]
3*a*b^2*(d*tan(f*x+e))^(1+n)/d/f/(1+n)+a^3*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)+3*a^2*b*(cos(f*x+e)^2) ^(1+1/2*n)*hypergeom([1+1/2*n, 1/2+1/2*n],[3/2+1/2*n],sin(f*x+e)^2)*sec(f* x+e)*(d*tan(f*x+e))^(1+n)/d/f/(1+n)+b^3*(cos(f*x+e)^2)^(2+1/2*n)*hypergeom ([2+1/2*n, 1/2+1/2*n],[3/2+1/2*n],sin(f*x+e)^2)*sec(f*x+e)^3*(d*tan(f*x+e) )^(1+n)/d/f/(1+n)
Time = 3.16 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.97 \[ \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx=\frac {d (d \tan (e+f x))^{-1+n} \left (-\tan ^2(e+f x)\right )^{-n/2} \left (9 a^2 b (1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3}{2},\sec ^2(e+f x)\right ) \sec (e+f x) \sqrt {-\tan ^2(e+f x)}+b^3 (1+n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1-n}{2},\frac {5}{2},\sec ^2(e+f x)\right ) \sec ^3(e+f x) \sqrt {-\tan ^2(e+f x)}-3 a^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{\frac {2+n}{2}}+9 a b^2 \left (\sqrt {-\tan ^2(e+f x)}-\left (-\tan ^2(e+f x)\right )^{\frac {2+n}{2}}\right )\right )}{3 f (1+n)} \]
(d*(d*Tan[e + f*x])^(-1 + n)*(9*a^2*b*(1 + n)*Hypergeometric2F1[1/2, (1 - n)/2, 3/2, Sec[e + f*x]^2]*Sec[e + f*x]*Sqrt[-Tan[e + f*x]^2] + b^3*(1 + n )*Hypergeometric2F1[3/2, (1 - n)/2, 5/2, Sec[e + f*x]^2]*Sec[e + f*x]^3*Sq rt[-Tan[e + f*x]^2] - 3*a^3*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Ta n[e + f*x]^2]*(-Tan[e + f*x]^2)^((2 + n)/2) + 9*a*b^2*(Sqrt[-Tan[e + f*x]^ 2] - (-Tan[e + f*x]^2)^((2 + n)/2))))/(3*f*(1 + n)*(-Tan[e + f*x]^2)^(n/2) )
Time = 0.52 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4374, 2009}
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.
| \(\displaystyle \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3 \left (-d \cot \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \int \left (a^3 (d \tan (e+f x))^n+3 a^2 b \sec (e+f x) (d \tan (e+f x))^n+3 a b^2 \sec ^2(e+f x) (d \tan (e+f x))^n+b^3 \sec ^3(e+f x) (d \tan (e+f x))^n\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {3 a^2 b \sec (e+f x) \cos ^2(e+f x)^{\frac {n+2}{2}} (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {n+2}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{d f (n+1)}+\frac {3 a b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac {b^3 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac {n+4}{2}} (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {n+4}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{d f (n+1)}\) |
(3*a*b^2*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + (a^3*Hypergeometric2F1[ 1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(d*f*( 1 + n)) + (3*a^2*b*(Cos[e + f*x]^2)^((2 + n)/2)*Hypergeometric2F1[(1 + n)/ 2, (2 + n)/2, (3 + n)/2, Sin[e + f*x]^2]*Sec[e + f*x]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + (b^3*(Cos[e + f*x]^2)^((4 + n)/2)*Hypergeometric2F1[ (1 + n)/2, (4 + n)/2, (3 + n)/2, Sin[e + f*x]^2]*Sec[e + f*x]^3*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n))
3.4.44.3.1 Defintions of rubi rules used
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
\[\int \left (a +b \sec \left (f x +e \right )\right )^{3} \left (d \tan \left (f x +e \right )\right )^{n}d x\]
\[ \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
integral((b^3*sec(f*x + e)^3 + 3*a*b^2*sec(f*x + e)^2 + 3*a^2*b*sec(f*x + e) + a^3)*(d*tan(f*x + e))^n, x)
\[ \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \sec {\left (e + f x \right )}\right )^{3}\, dx \]
\[ \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^3 \,d x \]