Integrand size = 21, antiderivative size = 66 \[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=-\frac {d (d \cos (a+b x))^{3/2} \csc (a+b x)}{b}-\frac {3 d^2 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b \sqrt {\cos (a+b x)}} \] Output:
-d*(d*cos(b*x+a))^(3/2)*csc(b*x+a)/b-3*d^2*(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)
Time = 0.49 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=-\frac {(d \cos (a+b x))^{5/2} \left (\cos ^{\frac {3}{2}}(a+b x) \csc (a+b x)+3 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )\right )}{b \cos ^{\frac {5}{2}}(a+b x)} \] Input:
Integrate[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x]^2,x]
Output:
-(((d*Cos[a + b*x])^(5/2)*(Cos[a + b*x]^(3/2)*Csc[a + b*x] + 3*EllipticE[( a + b*x)/2, 2]))/(b*Cos[a + b*x]^(5/2)))
Time = 0.58 (sec) , antiderivative size = 66, 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, 3047, 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.
\(\displaystyle \int \csc ^2(a+b x) (d \cos (a+b x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (a+b x))^{5/2}}{\sin (a+b x)^2}dx\) |
\(\Big \downarrow \) 3047 |
\(\displaystyle -\frac {3}{2} d^2 \int \sqrt {d \cos (a+b x)}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{3/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3}{2} d^2 \int \sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{3/2}}{b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\cos (a+b x)}dx}{2 \sqrt {\cos (a+b x)}}-\frac {d \csc (a+b x) (d \cos (a+b x))^{3/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx}{2 \sqrt {\cos (a+b x)}}-\frac {d \csc (a+b x) (d \cos (a+b x))^{3/2}}{b}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {3 d^2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{b \sqrt {\cos (a+b x)}}-\frac {d \csc (a+b x) (d \cos (a+b x))^{3/2}}{b}\) |
Input:
Int[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x]^2,x]
Output:
-((d*(d*Cos[a + b*x])^(3/2)*Csc[a + b*x])/b) - (3*d^2*Sqrt[d*Cos[a + b*x]] *EllipticE[(a + b*x)/2, 2])/(b*Sqrt[Cos[a + b*x]])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ (b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Cos[e + f*x] )^(m - 2)*(b*Sin[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. \(202\) vs. \(2(62)=124\).
Time = 6.53 (sec) , antiderivative size = 203, normalized size of antiderivative = 3.08
method | result | size |
default | \(\frac {\sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, d^{4} \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (6 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{\frac {3}{2}} \operatorname {EllipticE}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}+8 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}-12 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+6 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}{2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} d +\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d \right )^{\frac {3}{2}} \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\) | \(203\) |
Input:
int((d*cos(b*x+a))^(5/2)*csc(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/2*(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)/(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2*d)^(3/2)*sin(1/ 2*b*x+1/2*a)*(6*cos(1/2*b*x+1/2*a)*(2*sin(1/2*b*x+1/2*a)^2-1)^(3/2)*Ellipt icE(cos(1/2*b*x+1/2*a),2^(1/2))*(sin(1/2*b*x+1/2*a)^2)^(1/2)+8*sin(1/2*b*x +1/2*a)^6-12*sin(1/2*b*x+1/2*a)^4+6*sin(1/2*b*x+1/2*a)^2-1)/(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 = 104, normalized size of antiderivative = 1.58 \[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=\frac {-3 i \, \sqrt {\frac {1}{2}} d^{\frac {5}{2}} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 i \, \sqrt {\frac {1}{2}} d^{\frac {5}{2}} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) - \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right )}{b \sin \left (b x + a\right )} \] Input:
integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a)^2,x, algorithm="fricas")
Output:
(-3*I*sqrt(1/2)*d^(5/2)*sin(b*x + a)*weierstrassZeta(-4, 0, weierstrassPIn verse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*I*sqrt(1/2)*d^(5/2)*sin(b *x + a)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I *sin(b*x + a))) - sqrt(d*cos(b*x + a))*d^2*cos(b*x + a))/(b*sin(b*x + a))
Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=\text {Timed out} \] Input:
integrate((d*cos(b*x+a))**(5/2)*csc(b*x+a)**2,x)
Output:
Timed out
\[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a)^2,x, algorithm="maxima")
Output:
integrate((d*cos(b*x + a))^(5/2)*csc(b*x + a)^2, x)
\[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )^{2} \,d x } \] Input:
integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a)^2,x, algorithm="giac")
Output:
integrate((d*cos(b*x + a))^(5/2)*csc(b*x + a)^2, x)
Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \] Input:
int((d*cos(a + b*x))^(5/2)/sin(a + b*x)^2,x)
Output:
int((d*cos(a + b*x))^(5/2)/sin(a + b*x)^2, x)
\[ \int (d \cos (a+b x))^{5/2} \csc ^2(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right )^{2} \csc \left (b x +a \right )^{2}d x \right ) d^{2} \] Input:
int((d*cos(b*x+a))^(5/2)*csc(b*x+a)^2,x)
Output:
sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)**2*csc(a + b*x)**2,x)*d**2