Integrand size = 25, antiderivative size = 98 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{3 b d^2 \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \] Output:
2/3*c*(c*sin(b*x+a))^(1/2)/b/d/(d*cos(b*x+a))^(3/2)-1/3*c^2*InverseJacobiA M(a-1/4*Pi+b*x,2^(1/2))*sin(2*b*x+2*a)^(1/2)/b/d^2/(d*cos(b*x+a))^(1/2)/(c *sin(b*x+a))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {9}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{5/2}}{5 b c d (d \cos (a+b x))^{3/2}} \] Input:
Integrate[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(5/2),x]
Output:
(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[5/4, 7/4, 9/4, Sin[a + b*x]^2] *(c*Sin[a + b*x])^(5/2))/(5*b*c*d*(d*Cos[a + b*x])^(3/2))
Time = 0.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3046, 3042, 3053, 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.
\(\displaystyle \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx}{3 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx}{3 d^2}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}\) |
Input:
Int[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(5/2),x]
Output:
(2*c*Sqrt[c*Sin[a + b*x]])/(3*b*d*(d*Cos[a + b*x])^(3/2)) - (c^2*EllipticF [a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]* Sqrt[c*Sin[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[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, x]
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]
Time = 11.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.39
method | result | size |
default | \(\frac {\sqrt {c \sin \left (b x +a \right )}\, c \left (\operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \left (-\cot \left (b x +a \right )-\csc \left (b x +a \right )\right )+2 \sec \left (b x +a \right )\right )}{3 b \sqrt {d \cos \left (b x +a \right )}\, d^{2}}\) | \(136\) |
Input:
int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3/b*(c*sin(b*x+a))^(1/2)*c/(d*cos(b*x+a))^(1/2)/d^2*(EllipticF((-cot(b*x +a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(2*c ot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-cot(b*x+a) -csc(b*x+a))+2*sec(b*x+a))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt {i \, c d} c \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-i \, c d} c \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} c}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \] Input:
integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")
Output:
1/3*(sqrt(I*c*d)*c*cos(b*x + a)^2*elliptic_f(arcsin(cos(b*x + a) + I*sin(b *x + a)), -1) + sqrt(-I*c*d)*c*cos(b*x + a)^2*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) + 2*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*c) /(b*d^3*cos(b*x + a)^2)
Timed out. \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((c*sin(b*x+a))**(3/2)/(d*cos(b*x+a))**(5/2),x)
Output:
Timed out
\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(5/2), x)
\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x, algorithm="giac")
Output:
integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(5/2), x)
Timed out. \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^{3/2}}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \] Input:
int((c*sin(a + b*x))^(3/2)/(d*cos(a + b*x))^(5/2),x)
Output:
int((c*sin(a + b*x))^(3/2)/(d*cos(a + b*x))^(5/2), x)
\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, c \left (-\cos \left (b x +a \right )^{2} \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right ) \sin \left (b x +a \right )}d x \right ) b +2 \sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\right )}{3 \cos \left (b x +a \right )^{2} b \,d^{3}} \] Input:
int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x)
Output:
(sqrt(d)*sqrt(c)*c*( - cos(a + b*x)**2*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x)))/(cos(a + b*x)*sin(a + b*x)),x)*b + 2*sqrt(sin(a + b*x))*sqrt(cos( a + b*x))))/(3*cos(a + b*x)**2*b*d**3)