3.2.59 \(\int \frac {(a+a \sec (c+d x))^n}{\sin ^{\frac {3}{2}}(c+d x)} \, dx\) [159]

3.2.59.1 Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \frac {(a+a \sec (c+d x))^n}{\sin ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\operatorname {AppellF1}\left (1-n,\frac {5}{4},\frac {5}{4}-n,2-n,\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{5/4} \cos (c+d x) (1+\cos (c+d x))^{\frac {5}{4}-n} (a+a \sec (c+d x))^n}{d (1-n) \sin ^{\frac {5}{2}}(c+d x)} \]

output

-AppellF1(1-n,5/4-n,5/4,2-n,-cos(d*x+c),cos(d*x+c))*(1-cos(d*x+c))^(5/4)*c 
os(d*x+c)*(1+cos(d*x+c))^(5/4-n)*(a+a*sec(d*x+c))^n/d/(1-n)/sin(d*x+c)^(5/ 
2)
 

3.2.59.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(212\) vs. \(2(105)=210\).

Time = 4.38 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.02 \[ \int \frac {(a+a \sec (c+d x))^n}{\sin ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {6 \operatorname {AppellF1}\left (-\frac {1}{4},n,-\frac {1}{2},\frac {3}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n}{d \left (-2 \left (\operatorname {AppellF1}\left (\frac {3}{4},n,\frac {1}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+2 n \operatorname {AppellF1}\left (\frac {3}{4},1+n,-\frac {1}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+3 \operatorname {AppellF1}\left (-\frac {1}{4},n,-\frac {1}{2},\frac {3}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right ) \sqrt {\sin (c+d x)}} \]

input

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

output

(-6*AppellF1[-1/4, n, -1/2, 3/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]* 
(1 + Cos[c + d*x])*(a*(1 + Sec[c + d*x]))^n)/(d*(-2*(AppellF1[3/4, n, 1/2, 
 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*n*AppellF1[3/4, 1 + n, 
-1/2, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 
 3*AppellF1[-1/4, n, -1/2, 3/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*( 
1 + Cos[c + d*x]))*Sqrt[Sin[c + d*x]])
 

3.2.59.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.27, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4364, 3042, 3365, 152, 152, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4364

\(\displaystyle (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^n}{\sin ^{\frac {3}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3365

\(\displaystyle -\frac {(a \cos (c+d x)-a)^{5/4} (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{\frac {5}{4}-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^{n-\frac {5}{4}}}{(a \cos (c+d x)-a)^{5/4}}d\cos (c+d x)}{d \sin ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 152

\(\displaystyle \frac {(a (-\cos (c+d x))-a) (a \cos (c+d x)-a)^{5/4} (-\cos (c+d x))^n (\cos (c+d x)+1)^{\frac {1}{4}-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{n-\frac {5}{4}}}{(a \cos (c+d x)-a)^{5/4}}d\cos (c+d x)}{a d \sin ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 152

\(\displaystyle -\frac {\sqrt [4]{1-\cos (c+d x)} (a (-\cos (c+d x))-a) (a \cos (c+d x)-a) (-\cos (c+d x))^n (\cos (c+d x)+1)^{\frac {1}{4}-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{n-\frac {5}{4}}}{(1-\cos (c+d x))^{5/4}}d\cos (c+d x)}{a^2 d \sin ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {\sqrt [4]{1-\cos (c+d x)} \cos (c+d x) (a (-\cos (c+d x))-a) (a \cos (c+d x)-a) (\cos (c+d x)+1)^{\frac {1}{4}-n} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (1-n,\frac {5}{4},\frac {5}{4}-n,2-n,\cos (c+d x),-\cos (c+d x)\right )}{a^2 d (1-n) \sin ^{\frac {5}{2}}(c+d x)}\)

input

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

output

-((AppellF1[1 - n, 5/4, 5/4 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - 
Cos[c + d*x])^(1/4)*Cos[c + d*x]*(1 + Cos[c + d*x])^(1/4 - n)*(-a - a*Cos[ 
c + d*x])*(-a + a*Cos[c + d*x])*(a + a*Sec[c + d*x])^n)/(a^2*d*(1 - n)*Sin 
[c + d*x]^(5/2)))
 

3.2.59.3.1 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 152

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

rule 3042

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

rule 3365

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*((g*Cos 
[e + f*x])^(p - 1)/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e + f*x]) 
^((p - 1)/2)))   Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p 
- 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

rule 4364

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[Sin[e + f*x]^FracPart[m]*((a + b*Csc[e + f*x] 
)^FracPart[m]/(b + a*Sin[e + f*x])^FracPart[m])   Int[(g*Cos[e + f*x])^p*(( 
b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x], x] /; FreeQ[{a, b, e, f, g, m, p 
}, x] && (EqQ[a^2 - b^2, 0] || IntegersQ[2*m, p])
 

3.2.59.4 Maple [F]

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

input

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

output

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

3.2.59.5 Fricas [F]

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

input

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

output

integral(-(a*sec(d*x + c) + a)^n*sqrt(sin(d*x + c))/(cos(d*x + c)^2 - 1), 
x)
 

3.2.59.6 Sympy [F]

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

input

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

output

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

3.2.59.7 Maxima [F]

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

input

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

output

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

3.2.59.8 Giac [F]

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

input

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

output

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

3.2.59.9 Mupad [F(-1)]

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

input

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

output

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