\(\int \sin ^m(e+f x) (1+\sin (e+f x))^{3/2} \, dx\) [116]

Optimal result

Integrand size = 21, antiderivative size = 96 \[ \int \sin ^m(e+f x) (1+\sin (e+f x))^{3/2} \, dx=-\frac {2 (5+4 m) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1-\sin (e+f x)\right )}{f (3+2 m) \sqrt {1+\sin (e+f x)}}-\frac {2 \cos (e+f x) \sin ^{1+m}(e+f x)}{f (3+2 m) \sqrt {1+\sin (e+f x)}} \] Output:

-2*(5+4*m)*cos(f*x+e)*hypergeom([1/2, -m],[3/2],1-sin(f*x+e))/f/(3+2*m)/(1 
+sin(f*x+e))^(1/2)-2*cos(f*x+e)*sin(f*x+e)^(1+m)/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 = 22.83 (sec) , antiderivative size = 5109, normalized size of antiderivative = 53.22 \[ \int \sin ^m(e+f x) (1+\sin (e+f x))^{3/2} \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3242, 27, 2011, 3042, 3255, 75}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3255

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

(-2*(5 + 4*m)*Cos[e + f*x]*Hypergeometric2F1[1/2, -m, 3/2, 1 - Sin[e + f*x 
]])/(f*(3 + 2*m)*Sqrt[1 + Sin[e + f*x]]) - (2*Cos[e + f*x]*Sin[e + f*x]^(1 
 + m))/(f*(3 + 2*m)*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 75

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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)