Integrand size = 23, antiderivative size = 97 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\cos (e+f x)}{a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \sec (e+f x)}{3 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \sec (e+f x)}{3 a^3 f \sqrt {a+b \sec ^2(e+f x)}} \] Output:
-cos(f*x+e)/a/f/(a+b*sec(f*x+e)^2)^(3/2)-4/3*b*sec(f*x+e)/a^2/f/(a+b*sec(f *x+e)^2)^(3/2)-8/3*b*sec(f*x+e)/a^3/f/(a+b*sec(f*x+e)^2)^(1/2)
Time = 0.86 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left ((3 a+8 b)^2+12 a (a+4 b) \cos (2 (e+f x))+3 a^2 \cos (4 (e+f x))\right ) \sec ^5(e+f x)}{48 a^3 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:
Integrate[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]
Output:
-1/48*((a + 2*b + a*Cos[2*(e + f*x)])*((3*a + 8*b)^2 + 12*a*(a + 4*b)*Cos[ 2*(e + f*x)] + 3*a^2*Cos[4*(e + f*x)])*Sec[e + f*x]^5)/(a^3*f*(a + b*Sec[e + f*x]^2)^(5/2))
Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4622, 245, 209, 208}
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.
\(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {-\frac {4 b \int \frac {1}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {-\frac {4 b \left (\frac {2 \int \frac {1}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 a}+\frac {\sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {-\frac {4 b \left (\frac {2 \sec (e+f x)}{3 a^2 \sqrt {a+b \sec ^2(e+f x)}}+\frac {\sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\) |
Input:
Int[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]
Output:
(-(Cos[e + f*x]/(a*(a + b*Sec[e + f*x]^2)^(3/2))) - (4*b*(Sec[e + f*x]/(3* a*(a + b*Sec[e + f*x]^2)^(3/2)) + (2*Sec[e + f*x])/(3*a^2*Sqrt[a + b*Sec[e + f*x]^2])))/a)/f
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si mp[1/(f*ff^m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p /x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {1}{a \sec \left (f x +e \right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\sec \left (f x +e \right )}{3 a \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sec \left (f x +e \right )}{3 a^{2} \sqrt {a +b \sec \left (f x +e \right )^{2}}}\right )}{a}}{f}\) | \(90\) |
default | \(\frac {-\frac {1}{a \sec \left (f x +e \right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\sec \left (f x +e \right )}{3 a \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sec \left (f x +e \right )}{3 a^{2} \sqrt {a +b \sec \left (f x +e \right )^{2}}}\right )}{a}}{f}\) | \(90\) |
Input:
int(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/a/sec(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2)-4*b/a*(1/3*sec(f*x+e)/a/(a+b *sec(f*x+e)^2)^(3/2)+2/3/a^2*sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2)))
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{5} + 12 \, a b \cos \left (f x + e\right )^{3} + 8 \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \] Input:
integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
Output:
-1/3*(3*a^2*cos(f*x + e)^5 + 12*a*b*cos(f*x + e)^3 + 8*b^2*cos(f*x + e))*s qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^5*f*cos(f*x + e)^4 + 2*a^4*b *f*cos(f*x + e)^2 + a^3*b^2*f)
\[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sin {\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(sin(f*x+e)/(a+b*sec(f*x+e)**2)**(5/2),x)
Output:
Integral(sin(e + f*x)/(a + b*sec(e + f*x)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {3 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} + \frac {6 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{3} \cos \left (f x + e\right )^{3}}}{3 \, f} \] Input:
integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
Output:
-1/3*(3*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^3 + (6*(a + b/cos(f*x + e)^2)*b*cos(f*x + e)^2 - b^2)/((a + b/cos(f*x + e)^2)^(3/2)*a^3*cos(f*x + e)^3))/f
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {3 \, \sqrt {a \cos \left (f x + e\right )^{2} + b} + \frac {6 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} b - b^{2}}{{\left (a \cos \left (f x + e\right )^{2} + b\right )}^{\frac {3}{2}}}}{3 \, a^{3} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \] Input:
integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
Output:
-1/3*(3*sqrt(a*cos(f*x + e)^2 + b) + (6*(a*cos(f*x + e)^2 + b)*b - b^2)/(a *cos(f*x + e)^2 + b)^(3/2))/(a^3*f*sgn(cos(f*x + e)))
Time = 25.01 (sec) , antiderivative size = 26927, normalized size of antiderivative = 277.60 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
int(sin(e + f*x)/(a + b/cos(e + f*x)^2)^(5/2),x)
Output:
((a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(exp(e*3i + f*x*3i)*(((2*a + 4*b)*(((2*a + 4*b)*((((32*a*b^2 + 30*a^2*b + 3*a^3)/(4 8*a^3*b*f*(a + b)) - ((2*a + 4*b)*(8*a*b^2 + 8*a^2*b + a^3))/(48*a^4*b*f*( a + b)))*(2*a + 4*b))/a + (8*a*b^2 + 8*a^2*b + a^3)/(48*a^3*b*f*(a + b)) - (16*a*b + a^2 + 20*b^2)/(24*a^2*b*f*(a + b))))/a - (a + 6*b)/(24*a*b*f*(a + b)) - (32*a*b^2 + 30*a^2*b + 3*a^3)/(48*a^3*b*f*(a + b)) + ((2*a + 4*b) *(8*a*b^2 + 8*a^2*b + a^3))/(48*a^4*b*f*(a + b))))/a - (((32*a*b^2 + 30*a^ 2*b + 3*a^3)/(48*a^3*b*f*(a + b)) - ((2*a + 4*b)*(8*a*b^2 + 8*a^2*b + a^3) )/(48*a^4*b*f*(a + b)))*(2*a + 4*b))/a - (8*a*b^2 + 8*a^2*b + a^3)/(48*a^3 *b*f*(a + b)) + (16*a*b + a^2 + 20*b^2)/(24*a^2*b*f*(a + b)) + (32*a*b^2 + 40*a^2*b + 3*a^3)/(48*a^3*b*f*(a + b))) + exp(e*1i + f*x*1i)*(((2*a + 4*b )*((((32*a*b^2 + 30*a^2*b + 3*a^3)/(48*a^3*b*f*(a + b)) - ((2*a + 4*b)*(8* a*b^2 + 8*a^2*b + a^3))/(48*a^4*b*f*(a + b)))*(2*a + 4*b))/a + (8*a*b^2 + 8*a^2*b + a^3)/(48*a^3*b*f*(a + b)) - (16*a*b + a^2 + 20*b^2)/(24*a^2*b*f* (a + b))))/a + (48*a*b^2 + 18*a^2*b + a^3 + 32*b^3)/(48*a^3*b*f*(a + b)) - (a + 6*b)/(24*a*b*f*(a + b)) - (32*a*b^2 + 30*a^2*b + 3*a^3)/(48*a^3*b*f* (a + b)) + ((2*a + 4*b)*(8*a*b^2 + 8*a^2*b + a^3))/(48*a^4*b*f*(a + b))))* (2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i + f*x*2i) + 1) *(a + exp(e*2i + f*x*2i)*(2*a + 4*b) + a*exp(e*4i + f*x*4i))) - ((a + b/(e xp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(exp(e*1i + f*x*...
\[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:
int(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x)
Output:
int((sqrt(sec(e + f*x)**2*b + a)*sin(e + f*x))/(sec(e + f*x)**6*b**3 + 3*s ec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)