Integrand size = 21, antiderivative size = 69 \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\frac {4 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)}}-\frac {2 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d} \] Output:
4/5*(d*cos(b*x+a))^(1/2)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))/b/cos(b*x+a )^(1/2)-2/5*(d*cos(b*x+a))^(3/2)*sin(b*x+a)/b/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\frac {d \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{2},\frac {5}{2},\sin ^2(a+b x)\right ) \sin ^3(a+b x)}{3 b \sqrt {d \cos (a+b x)}} \] Input:
Integrate[Sqrt[d*Cos[a + b*x]]*Sin[a + b*x]^2,x]
Output:
(d*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 3/2, 5/2, Sin[a + b*x]^2] *Sin[a + b*x]^3)/(3*b*Sqrt[d*Cos[a + b*x]])
Time = 0.57 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3048, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \sin ^2(a+b x) \sqrt {d \cos (a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^2 \sqrt {d \cos (a+b x)}dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {2}{5} \int \sqrt {d \cos (a+b x)}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{5} \int \sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 \sqrt {d \cos (a+b x)} \int \sqrt {\cos (a+b x)}dx}{5 \sqrt {\cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sqrt {d \cos (a+b x)} \int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {4 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{5 b \sqrt {\cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\) |
Input:
Int[Sqrt[d*Cos[a + b*x]]*Sin[a + b*x]^2,x]
Output:
(4*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]] ) - (2*(d*Cos[a + b*x])^(3/2)*Sin[a + b*x])/(5*b*d)
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Leaf count of result is larger than twice the leaf count of optimal. \(193\) vs. \(2(61)=122\).
Time = 4.57 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.81
method | result | size |
default | \(\frac {4 \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, d \left (4 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{7}-8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{5}+5 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {-2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-\cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\) | \(194\) |
Input:
int((d*cos(b*x+a))^(1/2)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
4/5*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d*(4*cos(1/2 *b*x+1/2*a)^7-8*cos(1/2*b*x+1/2*a)^5+5*cos(1/2*b*x+1/2*a)^3+(sin(1/2*b*x+1 /2*a)^2)^(1/2)*(-2*cos(1/2*b*x+1/2*a)^2+1)^(1/2)*EllipticE(cos(1/2*b*x+1/2 *a),2^(1/2))-cos(1/2*b*x+1/2*a))/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1 /2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=-\frac {2 \, {\left (\sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 i \, \sqrt {\frac {1}{2}} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 2 i \, \sqrt {\frac {1}{2}} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )\right )}}{5 \, b} \] Input:
integrate((d*cos(b*x+a))^(1/2)*sin(b*x+a)^2,x, algorithm="fricas")
Output:
-2/5*(sqrt(d*cos(b*x + a))*cos(b*x + a)*sin(b*x + a) - 2*I*sqrt(1/2)*sqrt( d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin( b*x + a))) + 2*I*sqrt(1/2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInve rse(-4, 0, cos(b*x + a) - I*sin(b*x + a))))/b
\[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\int \sqrt {d \cos {\left (a + b x \right )}} \sin ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate((d*cos(b*x+a))**(1/2)*sin(b*x+a)**2,x)
Output:
Integral(sqrt(d*cos(a + b*x))*sin(a + b*x)**2, x)
\[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*cos(b*x+a))^(1/2)*sin(b*x+a)^2,x, algorithm="maxima")
Output:
integrate(sqrt(d*cos(b*x + a))*sin(b*x + a)^2, x)
\[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*cos(b*x+a))^(1/2)*sin(b*x+a)^2,x, algorithm="giac")
Output:
integrate(sqrt(d*cos(b*x + a))*sin(b*x + a)^2, x)
Timed out. \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,\sqrt {d\,\cos \left (a+b\,x\right )} \,d x \] Input:
int(sin(a + b*x)^2*(d*cos(a + b*x))^(1/2),x)
Output:
int(sin(a + b*x)^2*(d*cos(a + b*x))^(1/2), x)
\[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}d x \right ) \] Input:
int((d*cos(b*x+a))^(1/2)*sin(b*x+a)^2,x)
Output:
sqrt(d)*int(sqrt(cos(a + b*x))*sin(a + b*x)**2,x)