Integrand size = 10, antiderivative size = 78 \[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=-\frac {5}{16} a \cot (x) \sqrt {a \sin ^4(x)}+\frac {5}{16} a x \csc ^2(x) \sqrt {a \sin ^4(x)}-\frac {5}{24} a \cos (x) \sin (x) \sqrt {a \sin ^4(x)}-\frac {1}{6} a \cos (x) \sin ^3(x) \sqrt {a \sin ^4(x)} \] Output:
-5/16*a*cot(x)*(a*sin(x)^4)^(1/2)+5/16*a*x*csc(x)^2*(a*sin(x)^4)^(1/2)-5/2 4*a*cos(x)*sin(x)*(a*sin(x)^4)^(1/2)-1/6*a*cos(x)*sin(x)^3*(a*sin(x)^4)^(1 /2)
Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.49 \[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=-\frac {1}{192} \csc ^6(x) \left (a \sin ^4(x)\right )^{3/2} (-60 x+45 \sin (2 x)-9 \sin (4 x)+\sin (6 x)) \] Input:
Integrate[(a*Sin[x]^4)^(3/2),x]
Output:
-1/192*(Csc[x]^6*(a*Sin[x]^4)^(3/2)*(-60*x + 45*Sin[2*x] - 9*Sin[4*x] + Si n[6*x]))
Time = 0.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {3042, 3686, 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.
\(\displaystyle \int \left (a \sin ^4(x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \sin (x)^4\right )^{3/2}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \int \sin ^6(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \int \sin (x)^6dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {5}{6} \int \sin ^4(x)dx-\frac {1}{6} \sin ^5(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {5}{6} \int \sin (x)^4dx-\frac {1}{6} \sin ^5(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {5}{6} \left (\frac {3}{4} \int \sin ^2(x)dx-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {5}{6} \left (\frac {3}{4} \int \sin (x)^2dx-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \csc ^2(x) \sqrt {a \sin ^4(x)} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {x}{2}-\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin ^3(x) \cos (x)\right )-\frac {1}{6} \sin ^5(x) \cos (x)\right )\) |
Input:
Int[(a*Sin[x]^4)^(3/2),x]
Output:
a*Csc[x]^2*Sqrt[a*Sin[x]^4]*(-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)
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]
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]])
Time = 0.42 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.51
method | result | size |
default | \(-\frac {\sqrt {a \sin \left (x \right )^{4}}\, a \left (\cot \left (x \right ) \left (8 \cos \left (x \right )^{4}-26 \cos \left (x \right )^{2}+33\right )-15 \csc \left (x \right )^{2} x \right ) \sqrt {16}}{192}\) | \(40\) |
risch | \(-\frac {5 a \,{\mathrm e}^{2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}\, x}{16 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {i a \,{\mathrm e}^{8 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{384 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {3 i a \,{\mathrm e}^{6 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{128 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {15 i a \,{\mathrm e}^{4 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{128 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {15 i a \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{128 \left ({\mathrm e}^{2 i x}-1\right )^{2}}-\frac {3 i a \,{\mathrm e}^{-2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{128 \left ({\mathrm e}^{2 i x}-1\right )^{2}}+\frac {i a \,{\mathrm e}^{-4 i x} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}{384 \left ({\mathrm e}^{2 i x}-1\right )^{2}}\) | \(249\) |
Input:
int((a*sin(x)^4)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/192*(a*sin(x)^4)^(1/2)*a*(cot(x)*(8*cos(x)^4-26*cos(x)^2+33)-15*csc(x)^ 2*x)*16^(1/2)
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=-\frac {\sqrt {a \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a} {\left (15 \, a x - {\left (8 \, a \cos \left (x\right )^{5} - 26 \, a \cos \left (x\right )^{3} + 33 \, a \cos \left (x\right )\right )} \sin \left (x\right )\right )}}{48 \, {\left (\cos \left (x\right )^{2} - 1\right )}} \] Input:
integrate((a*sin(x)^4)^(3/2),x, algorithm="fricas")
Output:
-1/48*sqrt(a*cos(x)^4 - 2*a*cos(x)^2 + a)*(15*a*x - (8*a*cos(x)^5 - 26*a*c os(x)^3 + 33*a*cos(x))*sin(x))/(cos(x)^2 - 1)
\[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=\int \left (a \sin ^{4}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a*sin(x)**4)**(3/2),x)
Output:
Integral((a*sin(x)**4)**(3/2), x)
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71 \[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=\frac {5}{16} \, a^{\frac {3}{2}} x - \frac {33 \, a^{\frac {3}{2}} \tan \left (x\right )^{5} + 40 \, a^{\frac {3}{2}} \tan \left (x\right )^{3} + 15 \, a^{\frac {3}{2}} \tan \left (x\right )}{48 \, {\left (\tan \left (x\right )^{6} + 3 \, \tan \left (x\right )^{4} + 3 \, \tan \left (x\right )^{2} + 1\right )}} \] Input:
integrate((a*sin(x)^4)^(3/2),x, algorithm="maxima")
Output:
5/16*a^(3/2)*x - 1/48*(33*a^(3/2)*tan(x)^5 + 40*a^(3/2)*tan(x)^3 + 15*a^(3 /2)*tan(x))/(tan(x)^6 + 3*tan(x)^4 + 3*tan(x)^2 + 1)
Time = 0.41 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.35 \[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=\frac {1}{192} \, a^{\frac {3}{2}} {\left (60 \, x - \sin \left (6 \, x\right ) + 9 \, \sin \left (4 \, x\right ) - 45 \, \sin \left (2 \, x\right )\right )} \] Input:
integrate((a*sin(x)^4)^(3/2),x, algorithm="giac")
Output:
1/192*a^(3/2)*(60*x - sin(6*x) + 9*sin(4*x) - 45*sin(2*x))
Timed out. \[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=\int {\left (a\,{\sin \left (x\right )}^4\right )}^{3/2} \,d x \] Input:
int((a*sin(x)^4)^(3/2),x)
Output:
int((a*sin(x)^4)^(3/2), x)
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.40 \[ \int \left (a \sin ^4(x)\right )^{3/2} \, dx=\frac {\sqrt {a}\, a \left (-8 \cos \left (x \right ) \sin \left (x \right )^{5}-10 \cos \left (x \right ) \sin \left (x \right )^{3}-15 \cos \left (x \right ) \sin \left (x \right )+15 x \right )}{48} \] Input:
int((a*sin(x)^4)^(3/2),x)
Output:
(sqrt(a)*a*( - 8*cos(x)*sin(x)**5 - 10*cos(x)*sin(x)**3 - 15*cos(x)*sin(x) + 15*x))/48