Integrand size = 21, antiderivative size = 72 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=-\frac {4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}+\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}} \] Output:
-4/3*cos(b*x+a)^(1/2)*InverseJacobiAM(1/2*a+1/2*b*x,2^(1/2))/b/d^2/(d*cos( b*x+a))^(1/2)+2/3*sin(b*x+a)/b/d/(d*cos(b*x+a))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{4},\frac {5}{2},\sin ^2(a+b x)\right ) \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}} \] Input:
Integrate[Sin[a + b*x]^2/(d*Cos[a + b*x])^(5/2),x]
Output:
((Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[3/2, 7/4, 5/2, Sin[a + b*x]^2]*S in[a + b*x]^3)/(3*b*d*(d*Cos[a + b*x])^(3/2))
Time = 0.60 (sec) , antiderivative size = 72, 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, 3046, 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.
\(\displaystyle \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^2}{(d \cos (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \int \frac {1}{\sqrt {d \cos (a+b x)}}dx}{3 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 d^2}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 d^2 \sqrt {d \cos (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 d^2 \sqrt {d \cos (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}\) |
Input:
Int[Sin[a + b*x]^2/(d*Cos[a + b*x])^(5/2),x]
Output:
(-4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]) + (2*Sin[a + b*x])/(3*b*d*(d*Cos[a + b*x])^(3/2))
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(a*Sin[e + f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Sin[e + f *x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
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. \(241\) vs. \(2(63)=126\).
Time = 4.30 (sec) , antiderivative size = 242, normalized size of antiderivative = 3.36
method | result | size |
default | \(-\frac {4 \left (2 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}}{3 d^{2} \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\) | \(242\) |
Input:
int(sin(b*x+a)^2/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
Output:
-4/3*(2*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*Elli pticF(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+1/2*a)^2-cos(1/2*b*x+1/2*a)* sin(1/2*b*x+1/2*a)^2-(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2- 1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2)))/d^2*(d*(2*cos(1/2*b*x+1/2* a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(2*cos(1/2*b*x+1/2*a)^2-1)/(-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 = 103, normalized size of antiderivative = 1.43 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=-\frac {2 \, {\left (-2 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \] Input:
integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")
Output:
-2/3*(-2*I*sqrt(1/2)*sqrt(d)*cos(b*x + a)^2*weierstrassPInverse(-4, 0, cos (b*x + a) + I*sin(b*x + a)) + 2*I*sqrt(1/2)*sqrt(d)*cos(b*x + a)^2*weierst rassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) - sqrt(d*cos(b*x + a))* sin(b*x + a))/(b*d^3*cos(b*x + a)^2)
Timed out. \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)**2/(d*cos(b*x+a))**(5/2),x)
Output:
Timed out
\[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(5/2), x)
\[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(5/2),x, algorithm="giac")
Output:
integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(5/2), x)
Timed out. \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \] Input:
int(sin(a + b*x)^2/(d*cos(a + b*x))^(5/2),x)
Output:
int(sin(a + b*x)^2/(d*cos(a + b*x))^(5/2), x)
\[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{3}}d x \right )}{d^{3}} \] Input:
int(sin(b*x+a)^2/(d*cos(b*x+a))^(5/2),x)
Output:
(sqrt(d)*int((sqrt(cos(a + b*x))*sin(a + b*x)**2)/cos(a + b*x)**3,x))/d**3