Integrand size = 21, antiderivative size = 68 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=-\frac {4 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b d^2 \sqrt {\cos (a+b x)}}+\frac {2 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}} \] Output:
-4*(d*cos(b*x+a))^(1/2)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))/b/d^2/cos(b* x+a)^(1/2)+2*sin(b*x+a)/b/d/(d*cos(b*x+a))^(1/2)
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 = 0.88 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {3}{2},\frac {5}{2},\sin ^2(a+b x)\right ) \sin ^3(a+b x)}{3 b d \sqrt {d \cos (a+b x)}} \] Input:
Integrate[Sin[a + b*x]^2/(d*Cos[a + b*x])^(3/2),x]
Output:
((Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[5/4, 3/2, 5/2, Sin[a + b*x]^2]*S in[a + b*x]^3)/(3*b*d*Sqrt[d*Cos[a + b*x]])
Time = 0.60 (sec) , antiderivative size = 68, 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, 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.
\(\displaystyle \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^2}{(d \cos (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {2 \int \sqrt {d \cos (a+b x)}dx}{d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {2 \int \sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}dx}{d^2}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {2 \sqrt {d \cos (a+b x)} \int \sqrt {\cos (a+b x)}dx}{d^2 \sqrt {\cos (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {2 \sqrt {d \cos (a+b x)} \int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx}{d^2 \sqrt {\cos (a+b x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {4 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\cos (a+b x)}}\) |
Input:
Int[Sin[a + b*x]^2/(d*Cos[a + b*x])^(3/2),x]
Output:
(-4*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(b*d^2*Sqrt[Cos[a + b* x]]) + (2*Sin[a + b*x])/(b*d*Sqrt[d*Cos[a + b*x]])
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[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. \(197\) vs. \(2(64)=128\).
Time = 4.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.91
method | result | size |
default | \(-\frac {4 \left (-\cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} d +\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} d +\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d}\, \operatorname {EllipticE}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{d \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\) | \(198\) |
Input:
int(sin(b*x+a)^2/(d*cos(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
-4/d*(-cos(1/2*b*x+1/2*a)*(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2* d)^(1/2)*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)*(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2*d)^(1/2) *EllipticE(cos(1/2*b*x+1/2*a),2^(1/2)))/(-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 = 105, normalized size of antiderivative = 1.54 \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=-\frac {2 \, {\left (2 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) - 2 i \, \sqrt {\frac {1}{2}} \sqrt {d} \cos \left (b x + a\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) - \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{b d^{2} \cos \left (b x + a\right )} \] Input:
integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(3/2),x, algorithm="fricas")
Output:
-2*(2*I*sqrt(1/2)*sqrt(d)*cos(b*x + a)*weierstrassZeta(-4, 0, weierstrassP Inverse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) - 2*I*sqrt(1/2)*sqrt(d)*cos (b*x + a)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))) - sqrt(d*cos(b*x + a))*sin(b*x + a))/(b*d^2*cos(b*x + a) )
Timed out. \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)**2/(d*cos(b*x+a))**(3/2),x)
Output:
Timed out
\[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(3/2),x, algorithm="maxima")
Output:
integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(3/2), x)
\[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(3/2),x, algorithm="giac")
Output:
integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(3/2), x)
Timed out. \[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}} \,d x \] 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)
\[ \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{2}}d x \right )}{d^{2}} \] Input:
int(sin(b*x+a)^2/(d*cos(b*x+a))^(3/2),x)
Output:
(sqrt(d)*int((sqrt(cos(a + b*x))*sin(a + b*x)**2)/cos(a + b*x)**2,x))/d**2