\(\int \frac {(a+a \sec (c+d x))^n}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [232]

Optimal result

Integrand size = 23, antiderivative size = 112 \[ \int \frac {(a+a \sec (c+d x))^n}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2^{\frac {1}{2}+n} \operatorname {AppellF1}\left (-\frac {1}{4},-\frac {3}{2}+n,1,\frac {3}{4},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{-\frac {1}{2}+n} (a+a \sec (c+d x))^n}{d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2164\) vs. \(2(112)=224\).

Time = 16.58 (sec) , antiderivative size = 2164, normalized size of antiderivative = 19.32 \[ \int \frac {(a+a \sec (c+d x))^n}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

-1/21*(2^(1/2 + n)*Cot[c + d*x]^2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*(Cos 
[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*(21*Hypergeometri 
c2F1[-1/4, -1/2 + n, 3/4, Tan[(c + d*x)/2]^2] + 7*AppellF1[3/4, -1/2 + n, 
1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 + 7*Hy 
pergeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 
 - 3*AppellF1[7/4, -1/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2 
]^2]*Tan[(c + d*x)/2]^4))/(d*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*((2^(-1 
/2 + n)*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*Sec[c + d*x]^2*(Cos[(c + d*x)/ 
2]^2*Sec[c + d*x])^n*(21*Hypergeometric2F1[-1/4, -1/2 + n, 3/4, Tan[(c + d 
*x)/2]^2] + 7*AppellF1[3/4, -1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c 
+ d*x)/2]^2]*Tan[(c + d*x)/2]^2 + 7*Hypergeometric2F1[3/4, -1/2 + n, 7/4, 
Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 - 3*AppellF1[7/4, -1/2 + n, 1, 11/4 
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^4))/(21*Sqrt[C 
os[c + d*x]/(1 + Cos[c + d*x])]*Tan[c + d*x]^(3/2)) + (2^(-1/2 + n)*(Cos[c 
 + d*x]*Sec[(c + d*x)/2]^2)^n*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*((Cos[c 
+ d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x] 
))*(21*Hypergeometric2F1[-1/4, -1/2 + n, 3/4, Tan[(c + d*x)/2]^2] + 7*Appe 
llF1[3/4, -1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[( 
c + d*x)/2]^2 + 7*Hypergeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c + d*x)/2]^2 
]*Tan[(c + d*x)/2]^2 - 3*AppellF1[7/4, -1/2 + n, 1, 11/4, Tan[(c + d*x)...
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 112, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4377

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

Input:

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

Output:

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

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]

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

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]

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

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

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

Output:

Integral((a*(sec(c + d*x) + 1))**n/tan(c + d*x)**(3/2), x)
 

Maxima [F]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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