3.2.40 \(\int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx\) [140]

3.2.40.1 Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\frac {2 a e \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-m}{2},\frac {1}{2} (-2-m),\frac {1}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} (1+\cos (c+d x))^{-m/2} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}}{d} \]

output

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

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

Leaf count is larger than twice the leaf count of optimal. \(1243\) vs. \(2(106)=212\).

Time = 9.56 (sec) , antiderivative size = 1243, normalized size of antiderivative = 11.73 \[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx =\text {Too large to display} \]

input

Integrate[(a + a*Sec[c + d*x])^(3/2)*(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] + 2*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, 
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[(c + d*x)/2]^3*(a*(1 + Sec[c 
 + d*x]))^(3/2)*Sin[(c + d*x)/2]*(e*Sin[c + d*x])^m)/(d*(1 + m)*(6*AppellF 
1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 
 2*m*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + 
d*x)/2]^2] - 2*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2 
]^2, -Tan[(c + d*x)/2]^2] - 2*m*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2 
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[(3 + m)/2, 1/2, 1 + 
m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 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] + A 
ppellF1[(3 + m)/2, 3/2, m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2 
]^2] - 4*m*AppellF1[(3 + m)/2, 3/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, 
-Tan[(c + d*x)/2]^2] + 6*AppellF1[(3 + m)/2, 5/2, m, (5 + m)/2, Tan[(c + d 
*x)/2]^2, -Tan[(c + d*x)/2]^2] + 6*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, 
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] + 2*m*AppellF1[(1 + 
m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + 
d*x] + 2*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, - 
Tan[(c + d*x)/2]^2]*Cos[c + d*x] + 2*m*AppellF1[(3 + m)/2, -1/2, 2 + m,...
 

3.2.40.3 Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 168, 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.

\(\displaystyle \int (a \sec (c+d x)+a)^{3/2} (e \sin (c+d x))^m \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4364

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3365

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

input

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

output

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

3.2.40.3.1 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])
 

3.2.40.4 Maple [F]

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

input

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

output

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

3.2.40.5 Fricas [F]

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

input

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

output

integral((a*sec(d*x + c) + a)^(3/2)*(e*sin(d*x + c))^m, x)
 

3.2.40.6 Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.2.40.7 Maxima [F]

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

input

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

output

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

3.2.40.8 Giac [F]

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

input

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

output

sage0*x
 

3.2.40.9 Mupad [F(-1)]

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

input

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

output

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