\(\int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx\) [229]

Optimal result

Integrand size = 23, antiderivative size = 114 \[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\frac {2^{\frac {7}{2}+n} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2}+n,1,\frac {9}{4},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {5}{2}+n} (a+a \sec (c+d x))^n \tan ^{\frac {5}{2}}(c+d x)}{5 d} \] Output:

1/5*2^(7/2+n)*AppellF1(5/4,1,3/2+n,9/4,(a-a*sec(d*x+c))/(a+a*sec(d*x+c)),- 
(a-a*sec(d*x+c))/(a+a*sec(d*x+c)))*(1/(1+sec(d*x+c)))^(5/2+n)*(a+a*sec(d*x 
+c))^n*tan(d*x+c)^(5/2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2072\) vs. \(2(114)=228\).

Time = 26.37 (sec) , antiderivative size = 2072, normalized size of antiderivative = 18.18 \[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\text {Result too large to show} \] Input:

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^(3/2),x]
 

Output:

(2^(1 + n)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*(- 
1 + Tan[(c + d*x)/2])^(-1/2 - n)*(-2*AppellF1[1/4, 1/2 + n, 1, 5/4, Tan[(c 
 + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(1/2 
+ n)*(-1 + Tan[(c + d*x)/2])^(1/2 + n) + (AppellF1[1/2, 1/2 + n, 3/2 + n, 
3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]] + AppellF1[1/2, 3/2 + n, 1/2 + n 
, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]])*(1 - Tan[(c + d*x)/2])^(1/2 + 
 n)*(-1 + Tan[(c + d*x)/2]^2)^(1/2 + n))*Tan[c + d*x]^2)/(d*((2^n*Sec[c + 
d*x]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(-1 + Tan[(c + d*x)/2])^(-1/2 - 
 n)*(-2*AppellF1[1/4, 1/2 + n, 1, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/ 
2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(1/2 + n)*(-1 + Tan[(c + d*x)/2])^ 
(1/2 + n) + (AppellF1[1/2, 1/2 + n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[( 
c + d*x)/2]] + AppellF1[1/2, 3/2 + n, 1/2 + n, 3/2, Tan[(c + d*x)/2], -Tan 
[(c + d*x)/2]])*(1 - Tan[(c + d*x)/2])^(1/2 + n)*(-1 + Tan[(c + d*x)/2]^2) 
^(1/2 + n)))/Sqrt[Tan[c + d*x]] + 2^n*(-1/2 - n)*Sec[(c + d*x)/2]^2*(Cos[( 
c + d*x)/2]^2*Sec[c + d*x])^n*(-1 + Tan[(c + d*x)/2])^(-3/2 - n)*(-2*Appel 
lF1[1/4, 1/2 + n, 1, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Cos[c 
+ d*x]*Sec[(c + d*x)/2]^2)^(1/2 + n)*(-1 + Tan[(c + d*x)/2])^(1/2 + n) + ( 
AppellF1[1/2, 1/2 + n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]] 
+ AppellF1[1/2, 3/2 + n, 1/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2] 
])*(1 - Tan[(c + d*x)/2])^(1/2 + n)*(-1 + Tan[(c + d*x)/2]^2)^(1/2 + n)...
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3042, 4377}

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.

\(\displaystyle \int \tan ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (-\cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^ndx\)

\(\Big \downarrow \) 4377

\(\displaystyle \frac {2^{n+\frac {7}{2}} \tan ^{\frac {5}{2}}(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {5}{2}} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (\frac {5}{4},n+\frac {3}{2},1,\frac {9}{4},-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{5 d}\)

Input:

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^(3/2),x]
 

Output:

(2^(7/2 + n)*AppellF1[5/4, 3/2 + n, 1, 9/4, -((a - a*Sec[c + d*x])/(a + a* 
Sec[c + d*x])), (a - a*Sec[c + d*x])/(a + a*Sec[c + d*x])]*((1 + Sec[c + d 
*x])^(-1))^(5/2 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^(5/2))/(5*d)
 

Defintions of rubi rules used

rule 3042

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

rule 4377

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Simp[(-2^(m + n + 1))*(e*Cot[c + d*x])^(m + 1)*((a 
+ b*Csc[c + d*x])^n/(d*e*(m + 1)))*(a/(a + b*Csc[c + d*x]))^(m + n + 1)*App 
ellF1[(m + 1)/2, m + n, 1, (m + 3)/2, -(a - b*Csc[c + d*x])/(a + b*Csc[c + 
d*x]), (a - b*Csc[c + d*x])/(a + b*Csc[c + d*x])], x] /; FreeQ[{a, b, c, d, 
 e, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]
 

Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{\frac {3}{2}}d x\]

Input:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2),x)
 

Output:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2),x)
 

Fricas [F]

\[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^(3/2), x)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^(3/2), x)
 

Giac [F]

\[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \] Input:

int(tan(c + d*x)^(3/2)*(a + a/cos(c + d*x))^n,x)
 

Output:

int(tan(c + d*x)^(3/2)*(a + a/cos(c + d*x))^n, x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 \sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n}-2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n} \sec \left (d x +c \right )}{2 \sec \left (d x +c \right ) \tan \left (d x +c \right ) n +\sec \left (d x +c \right ) \tan \left (d x +c \right )+2 \tan \left (d x +c \right ) n +\tan \left (d x +c \right )}d x \right ) d n -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n} \sec \left (d x +c \right )}{2 \sec \left (d x +c \right ) \tan \left (d x +c \right ) n +\sec \left (d x +c \right ) \tan \left (d x +c \right )+2 \tan \left (d x +c \right ) n +\tan \left (d x +c \right )}d x \right ) d +4 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )}{2 \sec \left (d x +c \right ) n +\sec \left (d x +c \right )+2 n +1}d x \right ) d \,n^{2}+2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )}{2 \sec \left (d x +c \right ) n +\sec \left (d x +c \right )+2 n +1}d x \right ) d n -2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n}}{2 \sec \left (d x +c \right ) \tan \left (d x +c \right ) n +\sec \left (d x +c \right ) \tan \left (d x +c \right )+2 \tan \left (d x +c \right ) n +\tan \left (d x +c \right )}d x \right ) d n -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n}}{2 \sec \left (d x +c \right ) \tan \left (d x +c \right ) n +\sec \left (d x +c \right ) \tan \left (d x +c \right )+2 \tan \left (d x +c \right ) n +\tan \left (d x +c \right )}d x \right ) d}{d \left (2 n +1\right )} \] Input:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2),x)
 

Output:

(2*sqrt(tan(c + d*x))*(sec(c + d*x)*a + a)**n - 2*int((sqrt(tan(c + d*x))* 
(sec(c + d*x)*a + a)**n*sec(c + d*x))/(2*sec(c + d*x)*tan(c + d*x)*n + sec 
(c + d*x)*tan(c + d*x) + 2*tan(c + d*x)*n + tan(c + d*x)),x)*d*n - int((sq 
rt(tan(c + d*x))*(sec(c + d*x)*a + a)**n*sec(c + d*x))/(2*sec(c + d*x)*tan 
(c + d*x)*n + sec(c + d*x)*tan(c + d*x) + 2*tan(c + d*x)*n + tan(c + d*x)) 
,x)*d + 4*int((sqrt(tan(c + d*x))*(sec(c + d*x)*a + a)**n*tan(c + d*x))/(2 
*sec(c + d*x)*n + sec(c + d*x) + 2*n + 1),x)*d*n**2 + 2*int((sqrt(tan(c + 
d*x))*(sec(c + d*x)*a + a)**n*tan(c + d*x))/(2*sec(c + d*x)*n + sec(c + d* 
x) + 2*n + 1),x)*d*n - 2*int((sqrt(tan(c + d*x))*(sec(c + d*x)*a + a)**n)/ 
(2*sec(c + d*x)*tan(c + d*x)*n + sec(c + d*x)*tan(c + d*x) + 2*tan(c + d*x 
)*n + tan(c + d*x)),x)*d*n - int((sqrt(tan(c + d*x))*(sec(c + d*x)*a + a)* 
*n)/(2*sec(c + d*x)*tan(c + d*x)*n + sec(c + d*x)*tan(c + d*x) + 2*tan(c + 
 d*x)*n + tan(c + d*x)),x)*d)/(d*(2*n + 1))