Integrand size = 21, antiderivative size = 99 \[ \int \sqrt {d \cos (a+b x)} \sin ^4(a+b x) \, dx=\frac {8 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{15 b \sqrt {\cos (a+b x)}}-\frac {4 (d \cos (a+b x))^{3/2} \sin (a+b x)}{15 b d}-\frac {2 (d \cos (a+b x))^{3/2} \sin ^3(a+b x)}{9 b d} \]
-4/15*(d*cos(b*x+a))^(3/2)*sin(b*x+a)/b/d-2/9*(d*cos(b*x+a))^(3/2)*sin(b*x +a)^3/b/d+8/15*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(s in(1/2*a+1/2*b*x),2^(1/2))*(d*cos(b*x+a))^(1/2)/b/cos(b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \sqrt {d \cos (a+b x)} \sin ^4(a+b x) \, dx=\frac {d \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \sin ^5(a+b x)}{5 b \sqrt {d \cos (a+b x)}} \]
(d*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 5/2, 7/2, Sin[a + b*x]^2] *Sin[a + b*x]^5)/(5*b*Sqrt[d*Cos[a + b*x]])
Time = 0.44 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3048, 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.
\(\displaystyle \int \sin ^4(a+b x) \sqrt {d \cos (a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^4 \sqrt {d \cos (a+b x)}dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {2}{3} \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x)dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{3/2}}{9 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \int \sqrt {d \cos (a+b x)} \sin (a+b x)^2dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{3/2}}{9 b d}\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {2}{3} \left (\frac {2}{5} \int \sqrt {d \cos (a+b x)}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{3/2}}{9 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (\frac {2}{5} \int \sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{3/2}}{9 b d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2}{3} \left (\frac {2 \sqrt {d \cos (a+b x)} \int \sqrt {\cos (a+b x)}dx}{5 \sqrt {\cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{3/2}}{9 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (\frac {2 \sqrt {d \cos (a+b x)} \int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{3/2}}{9 b d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2}{3} \left (\frac {4 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{5 b \sqrt {\cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{3/2}}{9 b d}\) |
(-2*(d*Cos[a + b*x])^(3/2)*Sin[a + b*x]^3)/(9*b*d) + (2*((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)))/3
3.3.15.3.1 Defintions of rubi rules used
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]
Time = 0.95 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.23
method | result | size |
default | \(-\frac {8 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d \left (40 \left (\cos ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-120 \left (\cos ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+118 \left (\cos ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-36 \left (\cos ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-5 \left (\cos ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {1-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+3 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{45 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\) | \(221\) |
-8/45*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d*(40*cos( 1/2*b*x+1/2*a)^11-120*cos(1/2*b*x+1/2*a)^9+118*cos(1/2*b*x+1/2*a)^7-36*cos (1/2*b*x+1/2*a)^5-5*cos(1/2*b*x+1/2*a)^3-3*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(1 -2*cos(1/2*b*x+1/2*a)^2)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))+3*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 higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01 \[ \int \sqrt {d \cos (a+b x)} \sin ^4(a+b x) \, dx=\frac {2 \, {\left ({\left (5 \, \cos \left (b x + a\right )^{3} - 11 \, \cos \left (b x + a\right )\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right ) + 6 i \, \sqrt {2} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) - 6 i \, \sqrt {2} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )\right )}}{45 \, b} \]
2/45*((5*cos(b*x + a)^3 - 11*cos(b*x + a))*sqrt(d*cos(b*x + a))*sin(b*x + a) + 6*I*sqrt(2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) - 6*I*sqrt(2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))))/b
Timed out. \[ \int \sqrt {d \cos (a+b x)} \sin ^4(a+b x) \, dx=\text {Timed out} \]
\[ \int \sqrt {d \cos (a+b x)} \sin ^4(a+b x) \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )^{4} \,d x } \]
\[ \int \sqrt {d \cos (a+b x)} \sin ^4(a+b x) \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )^{4} \,d x } \]
Timed out. \[ \int \sqrt {d \cos (a+b x)} \sin ^4(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^4\,\sqrt {d\,\cos \left (a+b\,x\right )} \,d x \]