\(\int (a+a \sec (c+d x))^n \sin ^{\frac {3}{2}}(c+d x) \, dx\) [156]

Optimal result

Integrand size = 23, antiderivative size = 109 \[ \int (a+a \sec (c+d x))^n \sin ^{\frac {3}{2}}(c+d x) \, dx=\frac {2^{\frac {9}{4}+n} \operatorname {AppellF1}\left (\frac {5}{4},n,-\frac {1}{4}-n,\frac {9}{4},1-\cos (c+d x),\frac {1}{2} (1-\cos (c+d x))\right ) (1-\cos (c+d x)) \cos ^n(c+d x) (1+\cos (c+d x))^{-\frac {1}{4}-n} (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)}}{5 d} \] Output:

1/5*2^(9/4+n)*AppellF1(5/4,n,-1/4-n,9/4,1-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*sin( 
d*x+c)^(1/2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(382\) vs. \(2(109)=218\).

Time = 14.79 (sec) , antiderivative size = 382, normalized size of antiderivative = 3.50 \[ \int (a+a \sec (c+d x))^n \sin ^{\frac {3}{2}}(c+d x) \, dx=\frac {10 \left (\operatorname {AppellF1}\left (\frac {1}{4},n,\frac {3}{2},\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-\operatorname {AppellF1}\left (\frac {1}{4},n,\frac {5}{2},\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n \sin ^{\frac {5}{2}}(c+d x)}{d \left (2 \left (3 \operatorname {AppellF1}\left (\frac {5}{4},n,\frac {5}{2},\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-5 \operatorname {AppellF1}\left (\frac {5}{4},n,\frac {7}{2},\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n \operatorname {AppellF1}\left (\frac {5}{4},1+n,\frac {3}{2},\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+2 n \operatorname {AppellF1}\left (\frac {5}{4},1+n,\frac {5}{2},\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+5 \operatorname {AppellF1}\left (\frac {1}{4},n,\frac {3}{2},\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))-5 \operatorname {AppellF1}\left (\frac {1}{4},n,\frac {5}{2},\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \] Input:

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

Output:

(10*(AppellF1[1/4, n, 3/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 
 AppellF1[1/4, n, 5/2, 5/4, 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]^(5/2))/(d*(2*(3*Appel 
lF1[5/4, n, 5/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 5*AppellF 
1[5/4, n, 7/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*n*AppellF 
1[5/4, 1 + n, 3/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*n*App 
ellF1[5/4, 1 + n, 5/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 
+ Cos[c + d*x]) + 5*AppellF1[1/4, n, 3/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c 
 + d*x)/2]^2]*(1 + Cos[c + d*x]) - 5*AppellF1[1/4, n, 5/2, 5/4, Tan[(c + d 
*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))
 

Rubi [A] (verified)

Time = 0.53 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4364

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3365

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

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

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]

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

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)
 

Fricas [F]

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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