3.1.13 \(\int (a \sin ^4(x))^{5/2} \, dx\) [13]

3.1.13.1 Optimal result
3.1.13.2 Mathematica [A] (verified)
3.1.13.3 Rubi [A] (verified)
3.1.13.4 Maple [A] (verified)
3.1.13.5 Fricas [A] (verification not implemented)
3.1.13.6 Sympy [F]
3.1.13.7 Maxima [A] (verification not implemented)
3.1.13.8 Giac [A] (verification not implemented)
3.1.13.9 Mupad [F(-1)]

3.1.13.1 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)
 
3.1.13.2 Mathematica [A] (verified)

Time = 0.13 (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
 
3.1.13.3 Rubi [A] (verified)

Time = 0.47 (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)
 

3.1.13.3.1 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]])
 
3.1.13.4 Maple [A] (verified)

Time = 10.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.46

method result size
default \(-\frac {a^{2} \sqrt {a \left (\sin ^{4}\left (x \right )\right )}\, \left (128 \left (\cos ^{8}\left (x \right )\right ) \cot \left (x \right )-656 \left (\cos ^{6}\left (x \right )\right ) \cot \left (x \right )+1368 \left (\cos ^{4}\left (x \right )\right ) \cot \left (x \right )-1490 \left (\cos ^{2}\left (x \right )\right ) \cot \left (x \right )+965 \cot \left (x \right )-315 \left (\csc ^{2}\left (x \right )\right ) x \right ) \sqrt {16}}{5120}\) \(61\)
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^2*(a*sin(x)^4)^(1/2)*(128*cos(x)^8*cot(x)-656*cos(x)^6*cot(x)+13 
68*cos(x)^4*cot(x)-1490*cos(x)^2*cot(x)+965*cot(x)-315*csc(x)^2*x)*16^(1/2 
)
 
3.1.13.5 Fricas [A] (verification not implemented)

Time = 0.28 (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)
 
3.1.13.6 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)
 
3.1.13.7 Maxima [A] (verification not implemented)

Time = 0.31 (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)
 
3.1.13.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.43 \[ \int \left (a \sin ^4(x)\right )^{5/2} \, dx=\frac {1}{10240} \, {\left (2520 \, a^{2} x - 2 \, a^{2} \sin \left (10 \, x\right ) + 25 \, a^{2} \sin \left (8 \, x\right ) - 150 \, a^{2} \sin \left (6 \, x\right ) + 600 \, a^{2} \sin \left (4 \, x\right ) - 2100 \, a^{2} \sin \left (2 \, x\right )\right )} \sqrt {a} \]

input
integrate((a*sin(x)^4)^(5/2),x, algorithm="giac")
 
output
1/10240*(2520*a^2*x - 2*a^2*sin(10*x) + 25*a^2*sin(8*x) - 150*a^2*sin(6*x) 
 + 600*a^2*sin(4*x) - 2100*a^2*sin(2*x))*sqrt(a)
 
3.1.13.9 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)