\(\int (a \sin ^4(x))^{5/2} \, dx\) [16]

Optimal result

Integrand size = 10, antiderivative size = 132 \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=-\frac {63}{256} a^2 \cot (x) \sqrt {a \sin ^4(x)}+\frac {63}{256} a^2 x \csc ^2(x) \sqrt {a \sin ^4(x)}-\frac {21}{128} a^2 \cos (x) \sin (x) \sqrt {a \sin ^4(x)}-\frac {21}{160} a^2 \cos (x) \sin ^3(x) \sqrt {a \sin ^4(x)}-\frac {9}{80} a^2 \cos (x) \sin ^5(x) \sqrt {a \sin ^4(x)}-\frac {1}{10} a^2 \cos (x) \sin ^7(x) \sqrt {a \sin ^4(x)} \] Output:

-63/256*a^2*cot(x)*(a*sin(x)^4)^(1/2)+63/256*a^2*x*csc(x)^2*(a*sin(x)^4)^( 
1/2)-21/128*a^2*cos(x)*sin(x)*(a*sin(x)^4)^(1/2)-21/160*a^2*cos(x)*sin(x)^ 
3*(a*sin(x)^4)^(1/2)-9/80*a^2*cos(x)*sin(x)^5*(a*sin(x)^4)^(1/2)-1/10*a^2* 
cos(x)*sin(x)^7*(a*sin(x)^4)^(1/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.40 \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=\frac {a \csc ^6(x) \left (a \sin ^4(x)\right )^{3/2} (2520 x-2100 \sin (2 x)+600 \sin (4 x)-150 \sin (6 x)+25 \sin (8 x)-2 \sin (10 x))}{10240} \] Input:

Integrate[(a*Sin[x]^4)^(5/2),x]
 

Output:

(a*Csc[x]^6*(a*Sin[x]^4)^(3/2)*(2520*x - 2100*Sin[2*x] + 600*Sin[4*x] - 15 
0*Sin[6*x] + 25*Sin[8*x] - 2*Sin[10*x]))/10240
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.70, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.300, Rules used = {3042, 3686, 3042, 3115, 3042, 3115, 3042, 3115, 3042, 3115, 3042, 3115, 24}

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.

\(\displaystyle \int \left (a \sin ^4(x)\right )^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \sin (x)^4\right )^{5/2}dx\)

\(\Big \downarrow \) 3686

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \int \sin ^{10}(x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \int \sin (x)^{10}dx\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \int \sin ^8(x)dx-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \int \sin (x)^8dx-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \int \sin ^6(x)dx-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \int \sin (x)^6dx-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \left (\frac {5}{6} \int \sin ^4(x)dx-\frac {1}{6} \sin ^5(x) \cos (x)\right )-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \left (\frac {5}{6} \int \sin (x)^4dx-\frac {1}{6} \sin ^5(x) \cos (x)\right )-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sin ^2(x)dx-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sin (x)^2dx-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

\(\Big \downarrow \) 24

\(\displaystyle a^2 \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {9}{10} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {x}{2}-\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )-\frac {1}{8} \sin ^7(x) \cos (x)\right )-\frac {1}{10} \sin ^9(x) \cos (x)\right )\)

Input:

Int[(a*Sin[x]^4)^(5/2),x]
 

Output:

a^2*Csc[x]^2*Sqrt[a*Sin[x]^4]*(-1/10*(Cos[x]*Sin[x]^9) + (9*(-1/8*(Cos[x]* 
Sin[x]^7) + (7*(-1/6*(Cos[x]*Sin[x]^5) + (5*(-1/4*(Cos[x]*Sin[x]^3) + (3*( 
x/2 - (Cos[x]*Sin[x])/2))/4))/6))/8))/10)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3686

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ 
n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Si 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 

Maple [A] (verified)

Time = 11.00 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.41

method	result	size
default	\(-\frac {\sqrt {a \sin \left (x \right )^{4}}\, a^{2} \left (\cot \left (x \right ) \left (128 \cos \left (x \right )^{8}-656 \cos \left (x \right )^{6}+1368 \cos \left (x \right )^{4}-1490 \cos \left (x \right )^{2}+965\right )-315 \csc \left (x \right )^{2} x \right ) \sqrt {16}}{5120}\)	\(54\)
risch	\(-\frac {63 a^{2} {\mathrm e}^{2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}\, x}{256 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {i a^{2} {\mathrm e}^{12 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{10240 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {5 i a^{2} {\mathrm e}^{10 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{4096 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {105 i a^{2} {\mathrm e}^{4 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{1024 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {105 i a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{1024 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {15 i a^{2} {\mathrm e}^{-2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{512 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {15 i a^{2} {\mathrm e}^{-4 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{2048 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {37 i a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}\, \cos \left (8 x \right )}{5120 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {19 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}\, \sin \left (8 x \right )}{2560 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {115 i a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}\, \cos \left (6 x \right )}{4096 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {125 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}\, \sin \left (6 x \right )}{4096 \left ({\mathrm e}^{2 i x}-1\right )^{2}}\)	\(409\)

Input:

int((a*sin(x)^4)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/5120*(a*sin(x)^4)^(1/2)*a^2*(cot(x)*(128*cos(x)^8-656*cos(x)^6+1368*cos 
(x)^4-1490*cos(x)^2+965)-315*csc(x)^2*x)*16^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.62 \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=-\frac {\sqrt {a \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a} {\left (315 \, a^{2} x - {\left (128 \, a^{2} \cos \left (x\right )^{9} - 656 \, a^{2} \cos \left (x\right )^{7} + 1368 \, a^{2} \cos \left (x\right )^{5} - 1490 \, a^{2} \cos \left (x\right )^{3} + 965 \, a^{2} \cos \left (x\right )\right )} \sin \left (x\right )\right )}}{1280 \, {\left (\cos \left (x\right )^{2} - 1\right )}} \] Input:

integrate((a*sin(x)^4)^(5/2),x, algorithm="fricas")
 

Output:

-1/1280*sqrt(a*cos(x)^4 - 2*a*cos(x)^2 + a)*(315*a^2*x - (128*a^2*cos(x)^9 
 - 656*a^2*cos(x)^7 + 1368*a^2*cos(x)^5 - 1490*a^2*cos(x)^3 + 965*a^2*cos( 
x))*sin(x))/(cos(x)^2 - 1)
 

Sympy [F]

\[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=\int \left (a \sin ^{4}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \] Input:

integrate((a*sin(x)**4)**(5/2),x)
 

Output:

Integral((a*sin(x)**4)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.64 \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=\frac {63}{256} \, a^{\frac {5}{2}} x - \frac {965 \, a^{\frac {5}{2}} \tan \left (x\right )^{9} + 2370 \, a^{\frac {5}{2}} \tan \left (x\right )^{7} + 2688 \, a^{\frac {5}{2}} \tan \left (x\right )^{5} + 1470 \, a^{\frac {5}{2}} \tan \left (x\right )^{3} + 315 \, a^{\frac {5}{2}} \tan \left (x\right )}{1280 \, {\left (\tan \left (x\right )^{10} + 5 \, \tan \left (x\right )^{8} + 10 \, \tan \left (x\right )^{6} + 10 \, \tan \left (x\right )^{4} + 5 \, \tan \left (x\right )^{2} + 1\right )}} \] Input:

integrate((a*sin(x)^4)^(5/2),x, algorithm="maxima")
 

Output:

63/256*a^(5/2)*x - 1/1280*(965*a^(5/2)*tan(x)^9 + 2370*a^(5/2)*tan(x)^7 + 
2688*a^(5/2)*tan(x)^5 + 1470*a^(5/2)*tan(x)^3 + 315*a^(5/2)*tan(x))/(tan(x 
)^10 + 5*tan(x)^8 + 10*tan(x)^6 + 10*tan(x)^4 + 5*tan(x)^2 + 1)
 

Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.30 \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=\frac {1}{10240} \, a^{\frac {5}{2}} {\left (2520 \, x - 2 \, \sin \left (10 \, x\right ) + 25 \, \sin \left (8 \, x\right ) - 150 \, \sin \left (6 \, x\right ) + 600 \, \sin \left (4 \, x\right ) - 2100 \, \sin \left (2 \, x\right )\right )} \] Input:

integrate((a*sin(x)^4)^(5/2),x, algorithm="giac")
 

Output:

1/10240*a^(5/2)*(2520*x - 2*sin(10*x) + 25*sin(8*x) - 150*sin(6*x) + 600*s 
in(4*x) - 2100*sin(2*x))
 

Mupad [F(-1)]

Timed out. \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=\int {\left (a\,{\sin \left (x\right )}^4\right )}^{5/2} \,d x \] Input:

int((a*sin(x)^4)^(5/2),x)
 

Output:

int((a*sin(x)^4)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.37 \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=\frac {\sqrt {a}\, a^{2} \left (-128 \cos \left (x \right ) \sin \left (x \right )^{9}-144 \cos \left (x \right ) \sin \left (x \right )^{7}-168 \cos \left (x \right ) \sin \left (x \right )^{5}-210 \cos \left (x \right ) \sin \left (x \right )^{3}-315 \cos \left (x \right ) \sin \left (x \right )+315 x \right )}{1280} \] Input:

int((a*sin(x)^4)^(5/2),x)
 

Output:

(sqrt(a)*a**2*( - 128*cos(x)*sin(x)**9 - 144*cos(x)*sin(x)**7 - 168*cos(x) 
*sin(x)**5 - 210*cos(x)*sin(x)**3 - 315*cos(x)*sin(x) + 315*x))/1280