3.8.78 \(\int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx\) [778]

3.8.78.1 Optimal result

Integrand size = 23, antiderivative size = 181 \[ \int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx=-\frac {d \left (b^2 n+a^2 (1+n)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{-1+n} \sin (e+f x)}{f \left (1-n^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {2 a b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right ) (d \sec (e+f x))^n \sin (e+f x)}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+n)} \]

output

-d*(b^2*n+a^2*(1+n))*hypergeom([1/2, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)* 
(d*sec(f*x+e))^(-1+n)*sin(f*x+e)/f/(-n^2+1)/(sin(f*x+e)^2)^(1/2)+2*a*b*hyp 
ergeom([1/2, -1/2*n],[1-1/2*n],cos(f*x+e)^2)*(d*sec(f*x+e))^n*sin(f*x+e)/f 
/n/(sin(f*x+e)^2)^(1/2)+b^2*(d*sec(f*x+e))^n*tan(f*x+e)/f/(1+n)
 

3.8.78.2 Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.94 \[ \int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx=\frac {\csc (e+f x) \left (a^2 \left (2+3 n+n^2\right ) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(e+f x)\right )+b n \left (2 a (2+n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(e+f x)\right )+b (1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sec ^2(e+f x)\right )\right )\right ) \sec (e+f x) (d \sec (e+f x))^n \sqrt {-\tan ^2(e+f x)}}{f n (1+n) (2+n)} \]

input

Integrate[(d*Sec[e + f*x])^n*(a + b*Sec[e + f*x])^2,x]
 

output

(Csc[e + f*x]*(a^2*(2 + 3*n + n^2)*Cos[e + f*x]^2*Hypergeometric2F1[1/2, n 
/2, (2 + n)/2, Sec[e + f*x]^2] + b*n*(2*a*(2 + n)*Cos[e + f*x]*Hypergeomet 
ric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[e + f*x]^2] + b*(1 + n)*Hypergeometr 
ic2F1[1/2, (2 + n)/2, (4 + n)/2, Sec[e + f*x]^2]))*Sec[e + f*x]*(d*Sec[e + 
 f*x])^n*Sqrt[-Tan[e + f*x]^2])/(f*n*(1 + n)*(2 + n))
 

3.8.78.3 Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4275, 3042, 4259, 3042, 3122, 4534, 3042, 4259, 3042, 3122}

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.

\(\displaystyle \int (a+b \sec (e+f x))^2 (d \sec (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2 \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\)

\(\Big \downarrow \) 4275

\(\displaystyle \int (d \sec (e+f x))^n \left (a^2+b^2 \sec ^2(e+f x)\right )dx+\frac {2 a b \int (d \sec (e+f x))^{n+1}dx}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n \left (a^2+b^2 \csc \left (e+f x+\frac {\pi }{2}\right )^2\right )dx+\frac {2 a b \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{n+1}dx}{d}\)

\(\Big \downarrow \) 4259

\(\displaystyle \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n \left (a^2+b^2 \csc \left (e+f x+\frac {\pi }{2}\right )^2\right )dx+\frac {2 a b \left (\frac {\cos (e+f x)}{d}\right )^n (d \sec (e+f x))^n \int \left (\frac {\cos (e+f x)}{d}\right )^{-n-1}dx}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n \left (a^2+b^2 \csc \left (e+f x+\frac {\pi }{2}\right )^2\right )dx+\frac {2 a b \left (\frac {\cos (e+f x)}{d}\right )^n (d \sec (e+f x))^n \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}\right )^{-n-1}dx}{d}\)

\(\Big \downarrow \) 3122

\(\displaystyle \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n \left (a^2+b^2 \csc \left (e+f x+\frac {\pi }{2}\right )^2\right )dx+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}\)

\(\Big \downarrow \) 4534

\(\displaystyle \left (a^2+\frac {b^2 n}{n+1}\right ) \int (d \sec (e+f x))^ndx+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 \tan (e+f x) (d \sec (e+f x))^n}{f (n+1)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (a^2+\frac {b^2 n}{n+1}\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 \tan (e+f x) (d \sec (e+f x))^n}{f (n+1)}\)

\(\Big \downarrow \) 4259

\(\displaystyle \left (a^2+\frac {b^2 n}{n+1}\right ) \left (\frac {\cos (e+f x)}{d}\right )^n (d \sec (e+f x))^n \int \left (\frac {\cos (e+f x)}{d}\right )^{-n}dx+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 \tan (e+f x) (d \sec (e+f x))^n}{f (n+1)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (a^2+\frac {b^2 n}{n+1}\right ) \left (\frac {\cos (e+f x)}{d}\right )^n (d \sec (e+f x))^n \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}\right )^{-n}dx+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 \tan (e+f x) (d \sec (e+f x))^n}{f (n+1)}\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {d \left (a^2+\frac {b^2 n}{n+1}\right ) \sin (e+f x) (d \sec (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}}+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 \tan (e+f x) (d \sec (e+f x))^n}{f (n+1)}\)

input

Int[(d*Sec[e + f*x])^n*(a + b*Sec[e + f*x])^2,x]
 

output

-((d*(a^2 + (b^2*n)/(1 + n))*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, 
Cos[e + f*x]^2]*(d*Sec[e + f*x])^(-1 + n)*Sin[e + f*x])/(f*(1 - n)*Sqrt[Si 
n[e + f*x]^2])) + (2*a*b*Hypergeometric2F1[1/2, -1/2*n, (2 - n)/2, Cos[e + 
 f*x]^2]*(d*Sec[e + f*x])^n*Sin[e + f*x])/(f*n*Sqrt[Sin[e + f*x]^2]) + (b^ 
2*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + n))
 

3.8.78.3.1 Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

rule 4275

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^2, x_Symbol] :> Simp[2*a*(b/d)   Int[(d*Csc[e + f*x])^(n + 1), x], x] 
 + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, 
 e, f, n}, x]
 

rule 4534

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

3.8.78.4 Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{n} \left (a +b \sec \left (f x +e \right )\right )^{2}d x\]

input

int((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x)
 

output

int((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x)
 

3.8.78.5 Fricas [F]

\[ \int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{n} \,d x } \]

input

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x, algorithm="fricas")
 

output

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

3.8.78.6 Sympy [F]

\[ \int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{n} \left (a + b \sec {\left (e + f x \right )}\right )^{2}\, dx \]

input

integrate((d*sec(f*x+e))**n*(a+b*sec(f*x+e))**2,x)
 

output

Integral((d*sec(e + f*x))**n*(a + b*sec(e + f*x))**2, x)
 

3.8.78.7 Maxima [F]

\[ \int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{n} \,d x } \]

input

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x, algorithm="maxima")
 

output

integrate((b*sec(f*x + e) + a)^2*(d*sec(f*x + e))^n, x)
 

3.8.78.8 Giac [F]

\[ \int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{n} \,d x } \]

input

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x, algorithm="giac")
 

output

integrate((b*sec(f*x + e) + a)^2*(d*sec(f*x + e))^n, x)
 

3.8.78.9 Mupad [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx=\int {\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^2\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

input

int((a + b/cos(e + f*x))^2*(d/cos(e + f*x))^n,x)
 

output

int((a + b/cos(e + f*x))^2*(d/cos(e + f*x))^n, x)