\(\int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx\) [660]

Optimal result

Integrand size = 21, antiderivative size = 159 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) \sec (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (1+n) \sqrt {1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}}} \] Output:

AppellF1(1+n,-1/2,-1/2,2+n,(a+b*tan(d*x+c))/(a-(-b^2)^(1/2)),(a+b*tan(d*x+ 
c))/(a+(-b^2)^(1/2)))*sec(d*x+c)*(a+b*tan(d*x+c))^(1+n)/b/d/(1+n)/(1-(a+b* 
tan(d*x+c))/(a-(-b^2)^(1/2)))^(1/2)/(1-(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)))^ 
(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 12.55 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.92 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {2 (a-i b) (a+i b) (2+n) \operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right ) \sec (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (1+n) \left (2 \left (a^2+b^2\right ) (2+n) \operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )-\left ((a-i b) \operatorname {AppellF1}\left (2+n,-\frac {1}{2},\frac {1}{2},3+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )+(a+i b) \operatorname {AppellF1}\left (2+n,\frac {1}{2},-\frac {1}{2},3+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))\right )} \] Input:

Integrate[Sec[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]
 

Output:

(2*(a - I*b)*(a + I*b)*(2 + n)*AppellF1[1 + n, -1/2, -1/2, 2 + n, (a + b*T 
an[c + d*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)]*Sec[c + d*x]*(a + 
b*Tan[c + d*x])^(1 + n))/(b*d*(1 + n)*(2*(a^2 + b^2)*(2 + n)*AppellF1[1 + 
n, -1/2, -1/2, 2 + n, (a + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c + d*x]) 
/(a + I*b)] - ((a - I*b)*AppellF1[2 + n, -1/2, 1/2, 3 + n, (a + b*Tan[c + 
d*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)] + (a + I*b)*AppellF1[2 + 
n, 1/2, -1/2, 3 + n, (a + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c + d*x])/ 
(a + I*b)])*(a + b*Tan[c + d*x])))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3992, 514, 150}

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.

\(\displaystyle \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^3 (a+b \tan (c+d x))^ndx\)

\(\Big \downarrow \) 3992

\(\displaystyle \frac {\sec (c+d x) \int (a+b \tan (c+d x))^n \sqrt {\tan ^2(c+d x)+1}d(b \tan (c+d x))}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {\sqrt {\tan ^2(c+d x)+1} \sec (c+d x) \int (a+b \tan (c+d x))^n \sqrt {1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}}d(a+b \tan (c+d x))}{b d \sqrt {\sec ^2(c+d x)} \sqrt {1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\sqrt {\tan ^2(c+d x)+1} \sec (c+d x) (a+b \tan (c+d x))^{n+1} \operatorname {AppellF1}\left (n+1,-\frac {1}{2},-\frac {1}{2},n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{b d (n+1) \sqrt {\sec ^2(c+d x)} \sqrt {1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}}}\)

Input:

Int[Sec[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]
 

Output:

(AppellF1[1 + n, -1/2, -1/2, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]), 
 (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*Sec[c + d*x]*(a + b*Tan[c + d*x])^ 
(1 + n)*Sqrt[1 + Tan[c + d*x]^2])/(b*d*(1 + n)*Sqrt[Sec[c + d*x]^2]*Sqrt[1 
 - (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])]*Sqrt[1 - (a + b*Tan[c + d*x])/(a 
 + Sqrt[-b^2])])
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 514

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[-a/b, 2]}, Simp[(a + b*x^2)^p/(d*(1 - (c + d*x)/(c - d*q))^p*(1 - ( 
c + d*x)/(c + d*q))^p)   Subst[Int[x^n*Simp[1 - x/(c + d*q), x]^p*Simp[1 - 
x/(c - d*q), x]^p, x], x, c + d*x], x]] /; FreeQ[{a, b, c, d, n, p}, x] && 
NeQ[b*c^2 + a*d^2, 0]
 

rule 3042

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

rule 3992

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[Sec[e + f*x]/(b*f*Sqrt[Sec[e + f*x]^2])   Subst[Int[( 
a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b 
, e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2]
 

Maple [F]

\[\int \sec \left (d x +c \right )^{3} \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

Input:

int(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x)
 

Output:

int(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x)
 

Fricas [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \] Input:

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="fricas")
 

Output:

integral((b*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)
 

Sympy [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)**3*(a+b*tan(d*x+c))**n,x)
 

Output:

Integral((a + b*tan(c + d*x))**n*sec(c + d*x)**3, x)
 

Maxima [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \] Input:

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)
 

Giac [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \] Input:

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((b*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:

int((a + b*tan(c + d*x))^n/cos(c + d*x)^3,x)
 

Output:

int((a + b*tan(c + d*x))^n/cos(c + d*x)^3, x)
 

Reduce [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (\tan \left (d x +c \right ) b +a \right )^{n} \sec \left (d x +c \right )^{3}d x \] Input:

int(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x)
 

Output:

int((tan(c + d*x)*b + a)**n*sec(c + d*x)**3,x)