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

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

Optimal result

Integrand size = 23, antiderivative size = 130 \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=-\frac {2 \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {1+\sin (e+f x)}}+\frac {(5+4 n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+n,2+n,\sin (e+f x)\right ) (d \sin (e+f x))^{1+n}}{d f (1+n) (3+2 n) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \]

[Out]

-2*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(3+2*n)/(1+sin(f*x+e))^(1/2)+(5+4*n)*cos(f*x+e)*hypergeom([1/2, 1+n],[2
+n],sin(f*x+e))*(d*sin(f*x+e))^(1+n)/d/f/(1+n)/(3+2*n)/(1-sin(f*x+e))^(1/2)/(1+sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2842, 21, 2855, 66} \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\frac {(4 n+5) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n+1,n+2,\sin (e+f x)\right )}{d f (n+1) (2 n+3) \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {\sin (e+f x)+1}} \]

[In]

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

[Out]

(-2*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(3 + 2*n)*Sqrt[1 + Sin[e + f*x]]) + ((5 + 4*n)*Cos[e + f*x]*Hy
pergeometric2F1[1/2, 1 + n, 2 + n, Sin[e + f*x]]*(d*Sin[e + f*x])^(1 + n))/(d*f*(1 + n)*(3 + 2*n)*Sqrt[1 - Sin
[e + f*x]]*Sqrt[1 + Sin[e + f*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-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 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] + Dist[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 2855

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {1+\sin (e+f x)}}+\frac {2 \int \frac {(d \sin (e+f x))^n \left (\frac {1}{2} d (5+4 n)+\frac {1}{2} d (5+4 n) \sin (e+f x)\right )}{\sqrt {1+\sin (e+f x)}} \, dx}{d (3+2 n)} \\ & = -\frac {2 \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {1+\sin (e+f x)}}+\frac {(5+4 n) \int (d \sin (e+f x))^n \sqrt {1+\sin (e+f x)} \, dx}{3+2 n} \\ & = -\frac {2 \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {1+\sin (e+f x)}}+\frac {((5+4 n) \cos (e+f x)) \text {Subst}\left (\int \frac {(d x)^n}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {2 \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {1+\sin (e+f x)}}+\frac {(5+4 n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+n,2+n,\sin (e+f x)\right ) (d \sin (e+f x))^{1+n}}{d f (1+n) (3+2 n) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ \end{align*}

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 = 5129, normalized size of antiderivative = 39.45 \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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