\(\int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx\) [124]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=-\frac {2 (5+4 m) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1-\sin (e+f x)\right ) \sin ^{-m}(e+f x) (d \sin (e+f x))^m}{f (3+2 m) \sqrt {1+\sin (e+f x)}}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{1+m}}{d f (3+2 m) \sqrt {1+\sin (e+f x)}} \] Output:

-2*(5+4*m)*cos(f*x+e)*hypergeom([1/2, -m],[3/2],1-sin(f*x+e))*(d*sin(f*x+e 
))^m/f/(3+2*m)/(sin(f*x+e)^m)/(1+sin(f*x+e))^(1/2)-2*cos(f*x+e)*(d*sin(f*x 
+e))^(1+m)/d/f/(3+2*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 = 6.60 (sec) , antiderivative size = 5129, normalized size of antiderivative = 42.39 \[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=\text {Result too large to show} \] Input:

Integrate[(d*Sin[e + f*x])^m*(1 + Sin[e + f*x])^(3/2),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3242, 27, 2011, 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.

\(\displaystyle \int (\sin (e+f x)+1)^{3/2} (d \sin (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (\sin (e+f x)+1)^{3/2} (d \sin (e+f x))^mdx\)

\(\Big \downarrow \) 3242

\(\displaystyle \frac {2 \int \frac {(d \sin (e+f x))^m (d (4 m+5)+d \sin (e+f x) (4 m+5))}{2 \sqrt {\sin (e+f x)+1}}dx}{d (2 m+3)}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (2 m+3) \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d \sin (e+f x))^m (d (4 m+5)+d \sin (e+f x) (4 m+5))}{\sqrt {\sin (e+f x)+1}}dx}{d (2 m+3)}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (2 m+3) \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {(4 m+5) \int (d \sin (e+f x))^m \sqrt {\sin (e+f x)+1}dx}{2 m+3}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (2 m+3) \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(4 m+5) \int (d \sin (e+f x))^m \sqrt {\sin (e+f x)+1}dx}{2 m+3}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (2 m+3) \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 3255

\(\displaystyle \frac {(4 m+5) \cos (e+f x) \int \frac {(d \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (2 m+3) \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (2 m+3) \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {(4 m+5) \cos (e+f x) (d \sin (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},m+1,m+2,\sin (e+f x)\right )}{d f (m+1) (2 m+3) \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (2 m+3) \sqrt {\sin (e+f x)+1}}\)

Input:

Int[(d*Sin[e + f*x])^m*(1 + Sin[e + f*x])^(3/2),x]
 

Output:

(-2*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + m))/(d*f*(3 + 2*m)*Sqrt[1 + Sin[e + 
 f*x]]) + ((5 + 4*m)*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)*(3 + 2*m)*Sqrt[1 - Sin[e 
 + f*x]]*Sqrt[1 + Sin[e + f*x]])
 

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 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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], 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] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

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]

\[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (1+\sin \left (f x +e \right )\right )^{\frac {3}{2}}d x\]

Input:

int((d*sin(f*x+e))^m*(1+sin(f*x+e))^(3/2),x)
 

Output:

int((d*sin(f*x+e))^m*(1+sin(f*x+e))^(3/2),x)
 

Fricas [F]

\[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{m} {\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((d*sin(f*x+e))^m*(1+sin(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

integral((d*sin(f*x + e))^m*(sin(f*x + e) + 1)^(3/2), x)
 

Sympy [F]

\[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=\int \left (d \sin {\left (e + f x \right )}\right )^{m} \left (\sin {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}\, dx \] Input:

integrate((d*sin(f*x+e))**m*(1+sin(f*x+e))**(3/2),x)
 

Output:

Integral((d*sin(e + f*x))**m*(sin(e + f*x) + 1)**(3/2), x)
 

Maxima [F]

\[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{m} {\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((d*sin(f*x+e))^m*(1+sin(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*sin(f*x + e))^m*(sin(f*x + e) + 1)^(3/2), x)
 

Giac [F]

\[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{m} {\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((d*sin(f*x+e))^m*(1+sin(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((d*sin(f*x + e))^m*(sin(f*x + e) + 1)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (\sin \left (e+f\,x\right )+1\right )}^{3/2} \,d x \] Input:

int((d*sin(e + f*x))^m*(sin(e + f*x) + 1)^(3/2),x)
 

Output:

int((d*sin(e + f*x))^m*(sin(e + f*x) + 1)^(3/2), x)
 

Reduce [F]

\[ \int (d \sin (e+f x))^m (1+\sin (e+f x))^{3/2} \, dx=d^{m} \left (\int \sin \left (f x +e \right )^{m} \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x +\int \sin \left (f x +e \right )^{m} \sqrt {\sin \left (f x +e \right )+1}d x \right ) \] Input:

int((d*sin(f*x+e))^m*(1+sin(f*x+e))^(3/2),x)
                                                                                    
                                                                                    
 

Output:

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