Integrand size = 25, antiderivative size = 134 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}+\frac {4 (c \sin (a+b x))^{3/2}}{5 b c d^3 \sqrt {d \cos (a+b x)}}-\frac {4 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b d^4 \sqrt {\sin (2 a+2 b x)}} \] Output:
2/5*(c*sin(b*x+a))^(3/2)/b/c/d/(d*cos(b*x+a))^(5/2)+4/5*(c*sin(b*x+a))^(3/ 2)/b/c/d^3/(d*cos(b*x+a))^(1/2)+4/5*(d*cos(b*x+a))^(1/2)*EllipticE(cos(a+1 /4*Pi+b*x),2^(1/2))*(c*sin(b*x+a))^(1/2)/b/d^4/sin(2*b*x+2*a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {7}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)} \tan (a+b x)}{3 b d^4} \] Input:
Integrate[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(7/2),x]
Output:
(2*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[3/4, 9/4, 7/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a + b*x]]*Tan[a + b*x])/(3*b*d^4)
Time = 0.57 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3051, 3042, 3051, 3042, 3052, 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.
| \(\displaystyle \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3051 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{3/2}}dx}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{3/2}}dx}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\) |
| \(\Big \downarrow \) 3051 |
| \(\displaystyle \frac {2 \left (\frac {2 (c \sin (a+b x))^{3/2}}{b c d \sqrt {d \cos (a+b x)}}-\frac {2 \int \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}dx}{d^2}\right )}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {2 (c \sin (a+b x))^{3/2}}{b c d \sqrt {d \cos (a+b x)}}-\frac {2 \int \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}dx}{d^2}\right )}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\) |
| \(\Big \downarrow \) 3052 |
| \(\displaystyle \frac {2 \left (\frac {2 (c \sin (a+b x))^{3/2}}{b c d \sqrt {d \cos (a+b x)}}-\frac {2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{d^2 \sqrt {\sin (2 a+2 b x)}}\right )}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {2 (c \sin (a+b x))^{3/2}}{b c d \sqrt {d \cos (a+b x)}}-\frac {2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{d^2 \sqrt {\sin (2 a+2 b x)}}\right )}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 \left (\frac {2 (c \sin (a+b x))^{3/2}}{b c d \sqrt {d \cos (a+b x)}}-\frac {2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\sin (2 a+2 b x)}}\right )}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\) |
Input:
Int[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(7/2),x]
Output:
(2*(c*Sin[a + b*x])^(3/2))/(5*b*c*d*(d*Cos[a + b*x])^(5/2)) + (2*((2*(c*Si n[a + b*x])^(3/2))/(b*c*d*Sqrt[d*Cos[a + b*x]]) - (2*Sqrt[d*Cos[a + b*x]]* EllipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(b*d^2*Sqrt[Sin[2*a + 2 *b*x]])))/(5*d^2)
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[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(252\) vs. \(2(115)=230\).
Time = 11.50 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.89
| method | result | size |
| default | \(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, \left (\operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \left (2 \cot \left (b x +a \right )+2 \csc \left (b x +a \right )\right )+\operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \left (-\cot \left (b x +a \right )-\csc \left (b x +a \right )\right )-2 \cot \left (b x +a \right )+\csc \left (b x +a \right )+\csc \left (b x +a \right ) \sec \left (b x +a \right )^{2}\right )}{5 b \sqrt {d \cos \left (b x +a \right )}\, d^{3}}\) | \(253\) |
Input:
int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x,method=_RETURNVERBOSE)
Output:
2/5/b*(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(1/2)/d^3*(EllipticE((-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*cot (b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*(2*cot(b*x+a)+ 2*csc(b*x+a))+EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*(-co t(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)*(-cot(b*x+a)-csc(b*x+a))-2*cot(b*x+a)+csc(b*x+a)+csc( b*x+a)*sec(b*x+a)^2)
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=-\frac {2 \, {\left (i \, \sqrt {i \, c d} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - i \, \sqrt {-i \, c d} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - i \, \sqrt {i \, c d} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + i \, \sqrt {-i \, c d} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {d \cos \left (b x + a\right )} {\left (2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{5 \, b d^{4} \cos \left (b x + a\right )^{3}} \] Input:
integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")
Output:
-2/5*(I*sqrt(I*c*d)*cos(b*x + a)^3*elliptic_e(arcsin(cos(b*x + a) + I*sin( b*x + a)), -1) - I*sqrt(-I*c*d)*cos(b*x + a)^3*elliptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - I*sqrt(I*c*d)*cos(b*x + a)^3*elliptic_f(arcsi n(cos(b*x + a) + I*sin(b*x + a)), -1) + I*sqrt(-I*c*d)*cos(b*x + a)^3*elli ptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - sqrt(d*cos(b*x + a))*( 2*cos(b*x + a)^2 + 1)*sqrt(c*sin(b*x + a))*sin(b*x + a))/(b*d^4*cos(b*x + a)^3)
Timed out. \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((c*sin(b*x+a))**(1/2)/(d*cos(b*x+a))**(7/2),x)
Output:
Timed out
\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")
Output:
integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(7/2), x)
\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x, algorithm="giac")
Output:
integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(7/2), x)
Timed out. \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\int \frac {\sqrt {c\,\sin \left (a+b\,x\right )}}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \] Input:
int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(7/2),x)
Output:
int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(7/2), x)
\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}}{\cos \left (b x +a \right )^{4}}d x \right )}{d^{4}} \] Input:
int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(7/2),x)
Output:
(sqrt(d)*sqrt(c)*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x)))/cos(a + b*x)* *4,x))/d**4