Integrand size = 21, antiderivative size = 98 \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\frac {4 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{21 b \sqrt {d \cos (a+b x)}}+\frac {4 d \sqrt {d \cos (a+b x)} \sin (a+b x)}{21 b}-\frac {2 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b d} \] Output:
4/21*d^2*cos(b*x+a)^(1/2)*InverseJacobiAM(1/2*a+1/2*b*x,2^(1/2))/b/(d*cos( b*x+a))^(1/2)+4/21*d*(d*cos(b*x+a))^(1/2)*sin(b*x+a)/b-2/7*(d*cos(b*x+a))^ (5/2)*sin(b*x+a)/b/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\frac {(d \cos (a+b x))^{3/2} \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {5}{2},\sin ^2(a+b x)\right ) \tan ^3(a+b x)}{3 b} \] Input:
Integrate[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x]^2,x]
Output:
((d*Cos[a + b*x])^(3/2)*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-1/4, 3/2 , 5/2, Sin[a + b*x]^2]*Tan[a + b*x]^3)/(3*b)
Time = 0.74 (sec) , antiderivative size = 103, 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, 3115, 3042, 3121, 3042, 3120}
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.
\(\displaystyle \int \sin ^2(a+b x) (d \cos (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^2 (d \cos (a+b x))^{3/2}dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {2}{7} \int (d \cos (a+b x))^{3/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{7} \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{3/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {2}{7} \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{7} \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2}{7} \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{7} \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2}{7} \left (\frac {2 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}\) |
Input:
Int[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x]^2,x]
Output:
(-2*(d*Cos[a + b*x])^(5/2)*Sin[a + b*x])/(7*b*d) + (2*((2*d^2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*Sqrt[d*Cos[a + b*x]]) + (2*d*Sqrt[d *Cos[a + b*x]]*Sin[a + b*x])/(3*b)))/7
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[((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[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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. \(207\) vs. \(2(85)=170\).
Time = 5.29 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.12
method | result | size |
default | \(\frac {4 \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, d^{2} \left (24 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{9}-60 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{7}+50 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{5}-15 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}-\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {-2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+\cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{21 \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\) | \(208\) |
Input:
int((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
4/21*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^2*(24*cos (1/2*b*x+1/2*a)^9-60*cos(1/2*b*x+1/2*a)^7+50*cos(1/2*b*x+1/2*a)^5-15*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)*EllipticF(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 = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=-\frac {2 \, {\left (2 i \, \sqrt {\frac {1}{2}} d^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 2 i \, \sqrt {\frac {1}{2}} d^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (3 \, d \cos \left (b x + a\right )^{2} - 2 \, d\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{21 \, b} \] Input:
integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x, algorithm="fricas")
Output:
-2/21*(2*I*sqrt(1/2)*d^(3/2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*s in(b*x + a)) - 2*I*sqrt(1/2)*d^(3/2)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + (3*d*cos(b*x + a)^2 - 2*d)*sqrt(d*cos(b*x + a))*sin (b*x + a))/b
Timed out. \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\text {Timed out} \] Input:
integrate((d*cos(b*x+a))**(3/2)*sin(b*x+a)**2,x)
Output:
Timed out
\[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x, algorithm="maxima")
Output:
integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a)^2, x)
\[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x, algorithm="giac")
Output:
integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a)^2, x)
Timed out. \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2} \,d x \] Input:
int(sin(a + b*x)^2*(d*cos(a + b*x))^(3/2),x)
Output:
int(sin(a + b*x)^2*(d*cos(a + b*x))^(3/2), x)
\[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}d x \right ) d \] Input:
int((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x)
Output:
sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)*sin(a + b*x)**2,x)*d