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

3.2.57.1 Optimal result

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

output

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

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

Leaf count is larger than twice the leaf count of optimal. \(214\) vs. \(2(105)=210\).

Time = 3.04 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.04 \[ \int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx=\frac {14 \operatorname {AppellF1}\left (\frac {3}{4},n,\frac {3}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n \sin ^{\frac {3}{2}}(c+d x)}{d \left (6 \left (3 \operatorname {AppellF1}\left (\frac {7}{4},n,\frac {5}{2},\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n \operatorname {AppellF1}\left (\frac {7}{4},1+n,\frac {3}{2},\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+21 \operatorname {AppellF1}\left (\frac {3}{4},n,\frac {3}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \]

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])))
 

3.2.57.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, 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]]))
 

3.2.57.3.1 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])
 

3.2.57.4 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)
 

3.2.57.5 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)
 

3.2.57.6 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)
 

3.2.57.7 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)
 

3.2.57.8 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)
 

3.2.57.9 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)