3.2.41 \(\int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx\) [141]

3.2.41.1 Optimal result

Integrand size = 25, antiderivative size = 107 \[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=-\frac {2 e \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-\frac {m}{2},\frac {3}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos (c+d x) (1+\cos (c+d x))^{-m/2} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}}{d} \]

output

-2*e*AppellF1(1/2,-1/2*m,-1/2*m+1/2,3/2,-cos(d*x+c),cos(d*x+c))*(1-cos(d*x 
+c))^(-1/2*m+1/2)*cos(d*x+c)*(e*sin(d*x+c))^(-1+m)*(a+a*sec(d*x+c))^(1/2)/ 
d/((1+cos(d*x+c))^(1/2*m))
 

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

Leaf count is larger than twice the leaf count of optimal. \(433\) vs. \(2(107)=214\).

Time = 2.37 (sec) , antiderivative size = 433, normalized size of antiderivative = 4.05 \[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=\frac {4 (3+m) \left (\operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left (\left (2 (1+m) \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},2+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+(1+2 m) \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-\operatorname {AppellF1}\left (\frac {3+m}{2},\frac {3}{2},m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))+(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \]

input

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

output

(4*(3 + m)*(AppellF1[(1 + m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2 
, -Tan[(c + d*x)/2]^2] + AppellF1[(1 + m)/2, 1/2, m, (3 + m)/2, Tan[(c + d 
*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[(c + d*x)/2]^3*Sqrt[a*(1 + Sec[c + d*x 
])]*Sin[(c + d*x)/2]*(e*Sin[c + d*x])^m)/(d*(1 + m)*((2*(1 + m)*AppellF1[( 
3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] 
 + (1 + 2*m)*AppellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2 
, -Tan[(c + d*x)/2]^2] - AppellF1[(3 + m)/2, 3/2, m, (5 + m)/2, Tan[(c + d 
*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + (3 + m)*AppellF1[(1 
+ m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*( 
1 + Cos[c + d*x]) + (3 + m)*AppellF1[(1 + m)/2, 1/2, m, (3 + m)/2, Tan[(c 
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))
 

3.2.41.3 Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.58, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4364

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3365

\(\displaystyle -\frac {e \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}} (e \sin (c+d x))^{m-1} \int \frac {(-\cos (c+d x) a-a)^{m/2} (a \cos (c+d x)-a)^{\frac {m-1}{2}}}{\sqrt {-\cos (c+d x)}}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 152

\(\displaystyle -\frac {e \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}} (e \sin (c+d x))^{m-1} \int \frac {(\cos (c+d x)+1)^{m/2} (a \cos (c+d x)-a)^{\frac {m-1}{2}}}{\sqrt {-\cos (c+d x)}}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 152

\(\displaystyle -\frac {e \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} (e \sin (c+d x))^{m-1} \int \frac {(1-\cos (c+d x))^{\frac {m-1}{2}} (\cos (c+d x)+1)^{m/2}}{\sqrt {-\cos (c+d x)}}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {2 e \cos (c+d x) \sqrt {a \sec (c+d x)+a} (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-\frac {m}{2},\frac {3}{2},\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{d}\)

input

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

output

(-2*e*AppellF1[1/2, (1 - m)/2, -1/2*m, 3/2, Cos[c + d*x], -Cos[c + d*x]]*( 
1 - Cos[c + d*x])^((1 - m)/2)*Cos[c + d*x]*(-a - a*Cos[c + d*x])^(-1/2 + ( 
1 - m)/2 + m/2)*(-a + a*Cos[c + d*x])^((1 - m)/2 + (-1 + m)/2)*Sqrt[a + a* 
Sec[c + d*x]]*(e*Sin[c + d*x])^(-1 + m))/(d*(1 + Cos[c + d*x])^(m/2))
 

3.2.41.3.1 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])
 

3.2.41.4 Maple [F]

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

input

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

output

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

3.2.41.5 Fricas [F]

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

input

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

output

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

3.2.41.6 Sympy [F]

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

input

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

output

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

3.2.41.7 Maxima [F]

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

input

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

output

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

3.2.41.8 Giac [F]

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

input

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

output

sage0*x
 

3.2.41.9 Mupad [F(-1)]

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

input

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

output

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