Integrand size = 23, antiderivative size = 85 \[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}+\frac {E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (e+f x)}}{2 f \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}} \] Output:
-1/3*b*sin(f*x+e)^(3/2)/f/(b*sec(f*x+e))^(3/2)-1/2*EllipticE(cos(e+1/4*Pi+ f*x),2^(1/2))*sin(f*x+e)^(1/2)/f/(b*sec(f*x+e))^(1/2)/sin(2*f*x+2*e)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.64 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.87 \[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {b \left (-1+\cos (2 (e+f x))-3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\sec ^2(e+f x)\right ) \sqrt [4]{-\tan ^2(e+f x)}\right )}{6 f (b \sec (e+f x))^{3/2} \sqrt {\sin (e+f x)}} \] Input:
Integrate[Sin[e + f*x]^(5/2)/Sqrt[b*Sec[e + f*x]],x]
Output:
(b*(-1 + Cos[2*(e + f*x)] - 3*Hypergeometric2F1[-1/2, 1/4, 1/2, Sec[e + f* x]^2]*(-Tan[e + f*x]^2)^(1/4)))/(6*f*(b*Sec[e + f*x])^(3/2)*Sqrt[Sin[e + f *x]])
Time = 0.77 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3063, 3042, 3065, 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.
\(\displaystyle \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^{5/2}}{\sqrt {b \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3063 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {\sin (e+f x)}}{\sqrt {b \sec (e+f x)}}dx-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {\sin (e+f x)}}{\sqrt {b \sec (e+f x)}}dx-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3065 |
\(\displaystyle \frac {\int \sqrt {b \cos (e+f x)} \sqrt {\sin (e+f x)}dx}{2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {b \cos (e+f x)} \sqrt {\sin (e+f x)}dx}{2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle \frac {\sqrt {\sin (e+f x)} \int \sqrt {\sin (2 e+2 f x)}dx}{2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (e+f x)} \int \sqrt {\sin (2 e+2 f x)}dx}{2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\sqrt {\sin (e+f x)} E\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{2 f \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\) |
Input:
Int[Sin[e + f*x]^(5/2)/Sqrt[b*Sec[e + f*x]],x]
Output:
-1/3*(b*Sin[e + f*x]^(3/2))/(f*(b*Sec[e + f*x])^(3/2)) + (EllipticE[e - Pi /4 + f*x, 2]*Sqrt[Sin[e + f*x]])/(2*f*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])
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[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*b*(a*Sin[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(m - n))), x] + Simp[a^2*((m - 1)/(m - n)) Int[(a*Sin[e + f*x])^( m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m - n, 0] && IntegersQ[2*m, 2*n]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(b*Cos[e + f*x])^n*(b*Sec[e + f*x])^n Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && Int egerQ[m - 1/2] && IntegerQ[n - 1/2]
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. \(223\) vs. \(2(72)=144\).
Time = 0.37 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.64
method | result | size |
default | \(\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (3+3 \sec \left (f x +e \right )\right )+\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-6-6 \sec \left (f x +e \right )\right )+4 \cos \left (f x +e \right )^{3}-10 \cos \left (f x +e \right )+6}{12 f \sqrt {\sin \left (f x +e \right )}\, \sqrt {b \sec \left (f x +e \right )}}\) | \(224\) |
Input:
int(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/12/f/sin(f*x+e)^(1/2)/(b*sec(f*x+e))^(1/2)*((csc(f*x+e)-cot(f*x+e)+1)^(1 /2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*El lipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))*(3+3*sec(f*x+e))+(csc (f*x+e)-cot(f*x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x +e)+cot(f*x+e))^(1/2)*EllipticE((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2 ))*(-6-6*sec(f*x+e))+4*cos(f*x+e)^3-10*cos(f*x+e)+6)
\[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (f x + e\right )}} \,d x } \] Input:
integrate(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")
Output:
integral(-(cos(f*x + e)^2 - 1)*sqrt(b*sec(f*x + e))*sqrt(sin(f*x + e))/(b* sec(f*x + e)), x)
Timed out. \[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**(5/2)/(b*sec(f*x+e))**(1/2),x)
Output:
Timed out
\[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (f x + e\right )}} \,d x } \] Input:
integrate(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate(sin(f*x + e)^(5/2)/sqrt(b*sec(f*x + e)), x)
\[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (f x + e\right )}} \,d x } \] Input:
integrate(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate(sin(f*x + e)^(5/2)/sqrt(b*sec(f*x + e)), x)
Timed out. \[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^{5/2}}{\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \] Input:
int(sin(e + f*x)^(5/2)/(b/cos(e + f*x))^(1/2),x)
Output:
int(sin(e + f*x)^(5/2)/(b/cos(e + f*x))^(1/2), x)
\[ \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sin \left (f x +e \right )^{2}}{\sec \left (f x +e \right )}d x \right )}{b} \] Input:
int(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x)
Output:
(sqrt(b)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x))*sin(e + f*x)**2)/sec(e + f*x),x))/b