Integrand size = 25, antiderivative size = 68 \[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=-\frac {8 a^2 b \sqrt {a \sin (e+f x)}}{5 f \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}} \] Output:
-8/5*a^2*b*(a*sin(f*x+e))^(1/2)/f/(b*tan(f*x+e))^(1/2)-2/5*b*(a*sin(f*x+e) )^(5/2)/f/(b*tan(f*x+e))^(1/2)
Time = 0.99 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=-\frac {a^2 \sqrt {a \sin (e+f x)} (8 \cot (e+f x)+\sin (2 (e+f x))) \sqrt {b \tan (e+f x)}}{5 f} \] Input:
Integrate[(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]
Output:
-1/5*(a^2*Sqrt[a*Sin[e + f*x]]*(8*Cot[e + f*x] + Sin[2*(e + f*x)])*Sqrt[b* Tan[e + f*x]])/f
Time = 0.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3078, 3042, 3069}
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.
\(\displaystyle \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}dx\) |
\(\Big \downarrow \) 3078 |
\(\displaystyle \frac {4}{5} a^2 \int \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}dx-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a^2 \int \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}dx-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3069 |
\(\displaystyle -\frac {8 a^2 b \sqrt {a \sin (e+f x)}}{5 f \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt {b \tan (e+f x)}}\) |
Input:
Int[(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]
Output:
(-8*a^2*b*Sqrt[a*Sin[e + f*x]])/(5*f*Sqrt[b*Tan[e + f*x]]) - (2*b*(a*Sin[e + f*x])^(5/2))/(5*f*Sqrt[b*Tan[e + f*x]])
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f* m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/( f*m)), x] + Simp[a^2*((m + n - 1)/m) Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[ e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1 ] && EqQ[n, 1/2])) && IntegersQ[2*m, 2*n]
Leaf count of result is larger than twice the leaf count of optimal. \(339\) vs. \(2(56)=112\).
Time = 1.76 (sec) , antiderivative size = 340, normalized size of antiderivative = 5.00
method | result | size |
default | \(-\frac {\sec \left (\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+5 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\cos \left (f x +e \right )+1}{1+\cos \left (f x +e \right )}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right )+2}{1+\cos \left (f x +e \right )}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+24 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-8\right ) \sqrt {\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) a}\, a^{2} \sqrt {\frac {\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) b}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}}{5 f}\) | \(340\) |
Input:
int((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/5/f*sec(1/2*f*x+1/2*e)*csc(1/2*f*x+1/2*e)*(-16*cos(1/2*f*x+1/2*e)^6+5*( -cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*ln((2*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f *x+e))^2)^(1/2)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)+1)/(1+co s(f*x+e)))*cos(1/2*f*x+1/2*e)^2-5*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*ln( 2*(2*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x+e)/(1+cos (f*x+e))^2)^(1/2)-cos(f*x+e)+1)/(1+cos(f*x+e)))*cos(1/2*f*x+1/2*e)^2+24*co s(1/2*f*x+1/2*e)^4+8*cos(1/2*f*x+1/2*e)^2-8)*(sin(1/2*f*x+1/2*e)*cos(1/2*f *x+1/2*e)*a)^(1/2)*a^2*(cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)/(2*cos(1/2*f *x+1/2*e)^2-1)*b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\frac {2 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{5 \, f \sin \left (f x + e\right )} \] Input:
integrate((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
2/5*(a^2*cos(f*x + e)^3 - 5*a^2*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b* sin(f*x + e)/cos(f*x + e))/(f*sin(f*x + e))
Timed out. \[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\text {Timed out} \] Input:
integrate((a*sin(f*x+e))**(5/2)*(b*tan(f*x+e))**(1/2),x)
Output:
Timed out
\[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {b \tan \left (f x + e\right )} \,d x } \] Input:
integrate((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)
\[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}} \sqrt {b \tan \left (f x + e\right )} \,d x } \] Input:
integrate((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate((a*sin(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)
Time = 2.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18 \[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\frac {a^2\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (18\,\sin \left (2\,e+2\,f\,x\right )-\sin \left (4\,e+4\,f\,x\right )\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{10\,f\,\left (\cos \left (2\,e+2\,f\,x\right )-1\right )} \] Input:
int((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(1/2),x)
Output:
(a^2*(a*sin(e + f*x))^(1/2)*(18*sin(2*e + 2*f*x) - sin(4*e + 4*f*x))*((b*s in(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2))/(10*f*(cos(2*e + 2*f*x) - 1))
\[ \int (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)} \, dx=\sqrt {b}\, \sqrt {a}\, \left (\int \sqrt {\tan \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right )}\, \sin \left (f x +e \right )^{2}d x \right ) a^{2} \] Input:
int((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x)
Output:
sqrt(b)*sqrt(a)*int(sqrt(tan(e + f*x))*sqrt(sin(e + f*x))*sin(e + f*x)**2, x)*a**2