\(\int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx\) [157]

Optimal result

Integrand size = 23, antiderivative size = 109 \[ \int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx=\frac {2^{\frac {7}{4}+n} \operatorname {AppellF1}\left (\frac {3}{4},n,\frac {1}{4}-n,\frac {7}{4},1-\cos (c+d x),\frac {1}{2} (1-\cos (c+d x))\right ) (1-\cos (c+d x)) \cos ^n(c+d x) (1+\cos (c+d x))^{\frac {1}{4}-n} (a+a \sec (c+d x))^n}{3 d \sqrt {\sin (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 2.73 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.96 \[ \int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx=\frac {14 \operatorname {AppellF1}\left (\frac {3}{4},n,\frac {3}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n \sin ^{\frac {3}{2}}(c+d x)}{d \left (6 \left (3 \operatorname {AppellF1}\left (\frac {7}{4},n,\frac {5}{2},\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n \operatorname {AppellF1}\left (\frac {7}{4},1+n,\frac {3}{2},\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+21 \operatorname {AppellF1}\left (\frac {3}{4},n,\frac {3}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \] Input:

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

Output:

(14*AppellF1[3/4, n, 3/2, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 
 + Cos[c + d*x])*(a*(1 + Sec[c + d*x]))^n*Sin[c + d*x]^(3/2))/(d*(6*(3*App 
ellF1[7/4, n, 5/2, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*n*Ap 
pellF1[7/4, 1 + n, 3/2, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(- 
1 + Cos[c + d*x]) + 21*AppellF1[3/4, n, 3/2, 7/4, Tan[(c + d*x)/2]^2, -Tan 
[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.96, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4364

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3365

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

-((AppellF1[1 - n, 1/4, 1/4 - 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*Sec[c 
 + d*x])^n)/(d*(1 - n)*Sqrt[Sin[c + d*x]]))
 

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])
 

Maple [F]

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

Input:

int((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x)
 

Output:

int((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x)
 

Fricas [F]

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

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

integral((a*sec(d*x + c) + a)^n*sqrt(sin(d*x + c)), x)
 

Sympy [F]

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

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

Output:

Integral((a*(sec(c + d*x) + 1))**n*sqrt(sin(c + d*x)), x)
 

Maxima [F]

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

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^n*sqrt(sin(d*x + c)), x)
 

Giac [F]

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

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^n*sqrt(sin(d*x + c)), x)
 

Mupad [F(-1)]

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

int(sin(c + d*x)^(1/2)*(a + a/cos(c + d*x))^n,x)
 

Output:

int(sin(c + d*x)^(1/2)*(a + a/cos(c + d*x))^n, x)
 

Reduce [F]

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

int((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x)
 

Output:

int(sqrt(sin(c + d*x))*(sec(c + d*x)*a + a)**n,x)