\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx\) [342]

Optimal result
Mathematica [F]
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 37, antiderivative size = 272 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=-\frac {2^{\frac {1}{2}+m} a (A-B) (c+d) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2}-m,-\frac {3}{2},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-1+m} \sqrt {c+d \sin (e+f x)}}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2^{\frac {3}{2}+m} B (c+d) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2}-m,-\frac {3}{2},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)}}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \] Output:

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

Mathematica [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.77 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {3042, 3466, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3466

\(\displaystyle (A-B) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{3/2}dx+\frac {B \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{3/2}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{3/2}dx+\frac {B \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{3/2}dx}{a}\)

\(\Big \downarrow \) 3267

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

\(\Big \downarrow \) 157

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

Input:

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

Output:

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

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]
 

rule 3466
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A*b - a*B)/b   Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x 
] + Simp[B/b   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^(3/2),x, al 
gorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^(3/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^(3/2),x, al 
gorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{2}d x \right ) b d +\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )d x \right ) a d +\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )d x \right ) b c +\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m}d x \right ) a c \] Input:

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

Output:

int(sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a)**m*sin(e + f*x)**2,x)*b* 
d + int(sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a)**m*sin(e + f*x),x)*a 
*d + int(sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a)**m*sin(e + f*x),x)* 
b*c + int(sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a)**m,x)*a*c