Integrand size = 25, antiderivative size = 97 \[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{3 b d^2 \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \] Output:
2/3*(c*sin(b*x+a))^(1/2)/b/c/d/(d*cos(b*x+a))^(3/2)+2/3*InverseJacobiAM(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 = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=\frac {2 \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{4},\frac {5}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)}}{b c d (d \cos (a+b x))^{3/2}} \] Input:
Integrate[1/((d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]]),x]
Output:
(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[1/4, 7/4, 5/4, Sin[a + b*x]^2] *Sqrt[c*Sin[a + b*x]])/(b*c*d*(d*Cos[a + b*x])^(3/2))
Time = 0.41 (sec) , antiderivative size = 97, 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, 3051, 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.
\(\displaystyle \int \frac {1}{\sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3051 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx}{3 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx}{3 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\) |
Input:
Int[1/((d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]]),x]
Output:
(2*Sqrt[c*Sin[a + b*x]])/(3*b*c*d*(d*Cos[a + b*x])^(3/2)) + (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]]*Sq rt[c*Sin[a + b*x]])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-(b*Sin[e + f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1) /(a*b*f*(m + 1))), x] + Simp[(m + n + 2)/(a^2*(m + 1)) Int[(b*Sin[e + f*x ])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m , -1] && IntegersQ[2*m, 2*n]
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 = 4.56 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.28
method | result | size |
default | \(\frac {\frac {2 \left (\cos \left (b x +a \right )+1\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \tan \left (b x +a \right )}{3}}{b \sqrt {c \sin \left (b x +a \right )}\, \sqrt {d \cos \left (b x +a \right )}\, d^{2}}\) | \(124\) |
Input:
int(1/(d*cos(b*x+a))^(5/2)/(c*sin(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3/b/(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(1/2)/d^2*((cos(b*x+a)+1)*(-cot( b*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a) -csc(b*x+a))^(1/2)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2)) +tan(b*x+a))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=-\frac {2 \, {\left (\sqrt {i \, c d} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-i \, c d} \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )}\right )}}{3 \, b c d^{3} \cos \left (b x + a\right )^{2}} \] Input:
integrate(1/(d*cos(b*x+a))^(5/2)/(c*sin(b*x+a))^(1/2),x, algorithm="fricas ")
Output:
-2/3*(sqrt(I*c*d)*cos(b*x + a)^2*elliptic_f(arcsin(cos(b*x + a) + I*sin(b* x + a)), -1) + sqrt(-I*c*d)*cos(b*x + a)^2*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)))/(b*c*d ^3*cos(b*x + a)^2)
Timed out. \[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=\text {Timed out} \] Input:
integrate(1/(d*cos(b*x+a))**(5/2)/(c*sin(b*x+a))**(1/2),x)
Output:
Timed out
\[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=\int { \frac {1}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \sqrt {c \sin \left (b x + a\right )}} \,d x } \] Input:
integrate(1/(d*cos(b*x+a))^(5/2)/(c*sin(b*x+a))^(1/2),x, algorithm="maxima ")
Output:
integrate(1/((d*cos(b*x + a))^(5/2)*sqrt(c*sin(b*x + a))), x)
\[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=\int { \frac {1}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \sqrt {c \sin \left (b x + a\right )}} \,d x } \] Input:
integrate(1/(d*cos(b*x+a))^(5/2)/(c*sin(b*x+a))^(1/2),x, algorithm="giac")
Output:
integrate(1/((d*cos(b*x + a))^(5/2)*sqrt(c*sin(b*x + a))), x)
Timed out. \[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=\int \frac {1}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}\,\sqrt {c\,\sin \left (a+b\,x\right )}} \,d x \] Input:
int(1/((d*cos(a + b*x))^(5/2)*(c*sin(a + b*x))^(1/2)),x)
Output:
int(1/((d*cos(a + b*x))^(5/2)*(c*sin(a + b*x))^(1/2)), x)
\[ \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right )^{3} \sin \left (b x +a \right )}d x \right )}{c \,d^{3}} \] Input:
int(1/(d*cos(b*x+a))^(5/2)/(c*sin(b*x+a))^(1/2),x)
Output:
(sqrt(d)*sqrt(c)*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x)))/(cos(a + b*x) **3*sin(a + b*x)),x))/(c*d**3)