\(\int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx\) [125]

Optimal result

Integrand size = 23, antiderivative size = 63 \[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=-\frac {2 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1-\sin (e+f x)\right ) \sin ^{-m}(e+f x) (d \sin (e+f x))^m}{f \sqrt {1+\sin (e+f x)}} \] Output:

-2*cos(f*x+e)*hypergeom([1/2, -m],[3/2],1-sin(f*x+e))*(d*sin(f*x+e))^m/f/( 
sin(f*x+e)^m)/(1+sin(f*x+e))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 6.35 (sec) , antiderivative size = 2282, normalized size of antiderivative = 36.22 \[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=\text {Result too large to show} \] Input:

Integrate[(d*Sin[e + f*x])^m*Sqrt[1 + Sin[e + f*x]],x]
 

Output:

(2^(2 + m)*(Sec[(e + f*x)/4]^2)^(2*m)*(Cos[(e + f*x)/4]*(-Sin[(e + f*x)/4] 
 + Sin[(3*(e + f*x))/4]))^m*(d*Sin[e + f*x])^m*Sqrt[1 + Sin[e + f*x]]*(Cos 
[(e + f*x)/2]*Sin[e + f*x]^m + Sin[(e + f*x)/2]*Sin[e + f*x]^m)*Tan[(e + f 
*x)/4]*(2*(2 + m)*AppellF1[(1 + m)/2, -m, 2*(1 + m), (3 + m)/2, Tan[(e + f 
*x)/4]^2, -Tan[(e + f*x)/4]^2] - (2 + m)*AppellF1[(1 + m)/2, -m, 1 + 2*m, 
(3 + m)/2, Tan[(e + f*x)/4]^2, -Tan[(e + f*x)/4]^2] + 2*(1 + m)*AppellF1[1 
 + m/2, -m, 2 + 2*m, 2 + m/2, Tan[(e + f*x)/4]^2, -Tan[(e + f*x)/4]^2]*Tan 
[(e + f*x)/4]))/(f*(1 + m)*(2 + m)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*S 
in[e + f*x]^m*(1 - Tan[(e + f*x)/4]^2)^m*((2^(1 + m)*m*(Sec[(e + f*x)/4]^2 
)^(1 + 2*m)*(Cos[(e + f*x)/4]*(-Sin[(e + f*x)/4] + Sin[(3*(e + f*x))/4]))^ 
m*Tan[(e + f*x)/4]^2*(2*(2 + m)*AppellF1[(1 + m)/2, -m, 2*(1 + m), (3 + m) 
/2, Tan[(e + f*x)/4]^2, -Tan[(e + f*x)/4]^2] - (2 + m)*AppellF1[(1 + m)/2, 
 -m, 1 + 2*m, (3 + m)/2, Tan[(e + f*x)/4]^2, -Tan[(e + f*x)/4]^2] + 2*(1 + 
 m)*AppellF1[1 + m/2, -m, 2 + 2*m, 2 + m/2, Tan[(e + f*x)/4]^2, -Tan[(e + 
f*x)/4]^2]*Tan[(e + f*x)/4])*(1 - Tan[(e + f*x)/4]^2)^(-1 - m))/((1 + m)*( 
2 + m)) + (2^m*(Sec[(e + f*x)/4]^2)^(1 + 2*m)*(Cos[(e + f*x)/4]*(-Sin[(e + 
 f*x)/4] + Sin[(3*(e + f*x))/4]))^m*(2*(2 + m)*AppellF1[(1 + m)/2, -m, 2*( 
1 + m), (3 + m)/2, Tan[(e + f*x)/4]^2, -Tan[(e + f*x)/4]^2] - (2 + m)*Appe 
llF1[(1 + m)/2, -m, 1 + 2*m, (3 + m)/2, Tan[(e + f*x)/4]^2, -Tan[(e + f*x) 
/4]^2] + 2*(1 + m)*AppellF1[1 + m/2, -m, 2 + 2*m, 2 + m/2, Tan[(e + f*x...
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3255, 74}

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.

\(\displaystyle \int \sqrt {\sin (e+f x)+1} (d \sin (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\sin (e+f x)+1} (d \sin (e+f x))^mdx\)

\(\Big \downarrow \) 3255

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

\(\Big \downarrow \) 74

\(\displaystyle \frac {\cos (e+f x) (d \sin (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},m+1,m+2,\sin (e+f x)\right )}{d f (m+1) \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

Input:

Int[(d*Sin[e + f*x])^m*Sqrt[1 + Sin[e + f*x]],x]
 

Output:

(Cos[e + f*x]*Hypergeometric2F1[1/2, 1 + m, 2 + m, Sin[e + f*x]]*(d*Sin[e 
+ f*x])^(1 + m))/(d*f*(1 + m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[1 + Sin[e + f*x] 
])
 

Defintions of rubi rules used

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 3042

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

rule 3255

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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[(c + d*x)^n/Sqrt[a - b*x], x] 
, x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 
 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !IntegerQ[2*n]
 

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{m} \sqrt {1+\sin \left (f x +e \right )}d x\]

Input:

int((d*sin(f*x+e))^m*(1+sin(f*x+e))^(1/2),x)
 

Output:

int((d*sin(f*x+e))^m*(1+sin(f*x+e))^(1/2),x)
 

Fricas [F]

\[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{m} \sqrt {\sin \left (f x + e\right ) + 1} \,d x } \] Input:

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

Output:

integral((d*sin(f*x + e))^m*sqrt(sin(f*x + e) + 1), x)
 

Sympy [F]

\[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=\int \left (d \sin {\left (e + f x \right )}\right )^{m} \sqrt {\sin {\left (e + f x \right )} + 1}\, dx \] Input:

integrate((d*sin(f*x+e))**m*(1+sin(f*x+e))**(1/2),x)
 

Output:

Integral((d*sin(e + f*x))**m*sqrt(sin(e + f*x) + 1), x)
 

Maxima [F]

\[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{m} \sqrt {\sin \left (f x + e\right ) + 1} \,d x } \] Input:

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

Output:

integrate((d*sin(f*x + e))^m*sqrt(sin(f*x + e) + 1), x)
 

Giac [F]

\[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{m} \sqrt {\sin \left (f x + e\right ) + 1} \,d x } \] Input:

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

Output:

integrate((d*sin(f*x + e))^m*sqrt(sin(f*x + e) + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,\sqrt {\sin \left (e+f\,x\right )+1} \,d x \] Input:

int((d*sin(e + f*x))^m*(sin(e + f*x) + 1)^(1/2),x)
 

Output:

int((d*sin(e + f*x))^m*(sin(e + f*x) + 1)^(1/2), x)
 

Reduce [F]

\[ \int (d \sin (e+f x))^m \sqrt {1+\sin (e+f x)} \, dx=d^{m} \left (\int \sin \left (f x +e \right )^{m} \sqrt {\sin \left (f x +e \right )+1}d x \right ) \] Input:

int((d*sin(f*x+e))^m*(1+sin(f*x+e))^(1/2),x)
 

Output:

d**m*int(sin(e + f*x)**m*sqrt(sin(e + f*x) + 1),x)