3.7.14 \(\int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx\) [614]

3.7.14.1 Optimal result

Integrand size = 27, antiderivative size = 138 \[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx=\frac {\sqrt {2} (c-d)^2 \operatorname {AppellF1}\left (\frac {1}{2}+m,\frac {1}{2},-\frac {5}{2},\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (3+3 \sin (e+f x))^m \sqrt {c+d \sin (e+f x)}}{f (1+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}} \]

output

(c-d)^2*AppellF1(1/2+m,-5/2,1/2,3/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin( 
f*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^m*2^(1/2)*(c+d*sin(f*x+e))^(1/2)/f/(1+ 
2*m)/(1-sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c-d))^(1/2)
 

3.7.14.2 Mathematica [F]

\[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx=\int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx \]

input

Integrate[(3 + 3*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(5/2),x]
 

output

Integrate[(3 + 3*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(5/2), x]
 

3.7.14.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3267, 157, 27, 156, 155}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{5/2}dx\)

\(\Big \downarrow \) 3267

\(\displaystyle \frac {a^2 \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{5/2}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 157

\(\displaystyle \frac {a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{5/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{5/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {a^2 (c-d)^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \sqrt {c+d \sin (e+f x)} \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^{5/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} a (c-d)^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},-\frac {5}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (a-a \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\)

input

Int[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(5/2),x]
 

output

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

3.7.14.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 157

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
&  !GtQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !Si 
mplerQ[e + f*x, a + b*x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3267

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e 
+ f*x]]*Sqrt[a - b*Sin[e + f*x]]))   Subst[Int[(a + b*x)^(m - 1/2)*((c + d* 
x)^n/Sqrt[a - b*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m 
, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && 
 !IntegerQ[m]
 

3.7.14.4 Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}d x\]

input

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(5/2),x)
 

output

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(5/2),x)
 

3.7.14.5 Fricas [F]

\[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

input

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")
 

output

integral(-(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)*sqrt(d*sin 
(f*x + e) + c)*(a*sin(f*x + e) + a)^m, x)
 

3.7.14.6 Sympy [F(-1)]

Timed out. \[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]

input

integrate((a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**(5/2),x)
 

output

Timed out
 

3.7.14.7 Maxima [F]

\[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

input

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")
 

output

integrate((d*sin(f*x + e) + c)^(5/2)*(a*sin(f*x + e) + a)^m, x)
 

3.7.14.8 Giac [F]

\[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

input

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")
 

output

integrate((d*sin(f*x + e) + c)^(5/2)*(a*sin(f*x + e) + a)^m, x)
 

3.7.14.9 Mupad [F(-1)]

Timed out. \[ \int (3+3 \sin (e+f x))^m (c+d \sin (e+f x))^{5/2} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

input

int((a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^(5/2),x)
 

output

int((a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^(5/2), x)