Integrand size = 25, antiderivative size = 123 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\left (15 a^2+20 a b+8 b^2\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{15 a^3 f}+\frac {2 (5 a+2 b) \cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{15 a^2 f}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{5 a f} \] Output:
-1/15*(15*a^2+20*a*b+8*b^2)*cos(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/a^3/f+2/15 *(5*a+2*b)*cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2)/a^2/f-1/5*cos(f*x+e)^5*(a +b*sec(f*x+e)^2)^(1/2)/a/f
Time = 0.78 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.76 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (89 a^2+144 a b+64 b^2-4 a (7 a+4 b) \cos (2 (e+f x))+3 a^2 \cos (4 (e+f x))\right ) \sec (e+f x)}{240 a^3 f \sqrt {a+b \sec ^2(e+f x)}} \] Input:
Integrate[Sin[e + f*x]^5/Sqrt[a + b*Sec[e + f*x]^2],x]
Output:
-1/240*((a + 2*b + a*Cos[2*(e + f*x)])*(89*a^2 + 144*a*b + 64*b^2 - 4*a*(7 *a + 4*b)*Cos[2*(e + f*x)] + 3*a^2*Cos[4*(e + f*x)])*Sec[e + f*x])/(a^3*f* Sqrt[a + b*Sec[e + f*x]^2])
Time = 0.31 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4622, 365, 25, 359, 242}
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.
\(\displaystyle \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^5}{\sqrt {a+b \sec (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int \frac {\cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 365 |
\(\displaystyle \frac {\frac {\int -\frac {\cos ^4(e+f x) \left (2 (5 a+2 b)-5 a \sec ^2(e+f x)\right )}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{5 a}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{5 a}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {\int \frac {\cos ^4(e+f x) \left (2 (5 a+2 b)-5 a \sec ^2(e+f x)\right )}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{5 a}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{5 a}}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {-\frac {-\frac {\left (15 a^2+20 a b+8 b^2\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{3 a}-\frac {2 (5 a+2 b) \cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{3 a}}{5 a}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{5 a}}{f}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {-\frac {\frac {\left (15 a^2+20 a b+8 b^2\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{3 a^2}-\frac {2 (5 a+2 b) \cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{3 a}}{5 a}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{5 a}}{f}\) |
Input:
Int[Sin[e + f*x]^5/Sqrt[a + b*Sec[e + f*x]^2],x]
Output:
(-1/5*(Cos[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2])/a - (((15*a^2 + 20*a*b + 8*b^2)*Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(3*a^2) - (2*(5*a + 2*b)* Cos[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2])/(3*a))/(5*a))/f
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[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 = 2.41 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.72
method | result | size |
default | \(-\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (-4 \cos \left (f x +e \right )^{2} a b +20 a b +\left (3 \cos \left (f x +e \right )^{4}-10 \cos \left (f x +e \right )^{2}+15\right ) a^{2}+8 b^{2}\right ) \sec \left (f x +e \right )}{15 f \,a^{3} \sqrt {a +b \sec \left (f x +e \right )^{2}}}\) | \(89\) |
Input:
int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/15/f/a^3*(b+a*cos(f*x+e)^2)*(-4*cos(f*x+e)^2*a*b+20*a*b+(3*cos(f*x+e)^4 -10*cos(f*x+e)^2+15)*a^2+8*b^2)/(a+b*sec(f*x+e)^2)^(1/2)*sec(f*x+e)
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 2 \, {\left (5 \, a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, a^{3} f} \] Input:
integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
-1/15*(3*a^2*cos(f*x + e)^5 - 2*(5*a^2 + 2*a*b)*cos(f*x + e)^3 + (15*a^2 + 20*a*b + 8*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2) /(a^3*f)
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**5/(a+b*sec(f*x+e)**2)**(1/2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.32 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a} - \frac {10 \, {\left ({\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{2}} + \frac {3 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 10 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{3}}}{15 \, f} \] Input:
integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
-1/15*(15*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a - 10*((a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 3*sqrt(a + b/cos(f*x + e)^2)*b*cos(f*x + e)) /a^2 + (3*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 10*(a + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 15*sqrt(a + b/cos(f*x + e)^2)*b^2*cos(f*x + e))/a^3)/f
Leaf count of result is larger than twice the leaf count of optimal. 928 vs. \(2 (111) = 222\).
Time = 0.67 (sec) , antiderivative size = 928, normalized size of antiderivative = 7.54 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:
integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
-256/15*(5*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2* e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2 *f*x + 1/2*e)^2 + a + b))^7*(a + b) - 5*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^ 2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2 *f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^6*(2*a - b)*sqrt(a + b) - (sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^5*(13*a^2 + 40*a*b + 15*b^2) + 5*(sqrt(a + b)*tan(1/ 2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^ 4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^4*(8 *a^2 + 5*a*b - 3*b^2)*sqrt(a + b) - 5*(a^3 - 9*a^2*b - 13*a*b^2 - 3*b^3)*( sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan (1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e )^2 + a + b))^3 - 5*(10*a^3 + 17*a^2*b + 4*a*b^2 - 3*b^3)*(sqrt(a + b)*tan (1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2* e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 *sqrt(a + b) + 5*(9*a^4 + 14*a^3*b - 6*a*b^3 - b^4)*(sqrt(a + b)*tan(1/2*f *x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)) - (12*a ^4 + 9*a^3*b - 13*a^2*b^2 - 5*a*b^3 + 5*b^4)*sqrt(a + b))/(((sqrt(a + b...
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^5}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \] Input:
int(sin(e + f*x)^5/(a + b/cos(e + f*x)^2)^(1/2),x)
Output:
int(sin(e + f*x)^5/(a + b/cos(e + f*x)^2)^(1/2), x)
\[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{5}}{\sec \left (f x +e \right )^{2} b +a}d x \] Input:
int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x)
Output:
int((sqrt(sec(e + f*x)**2*b + a)*sin(e + f*x)**5)/(sec(e + f*x)**2*b + a), x)