Integrand size = 25, antiderivative size = 144 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{15 (a-b)^3 f}+\frac {2 (5 a-3 b) \cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{5 (a-b) f} \]
-1/15*(15*a^2-10*a*b+3*b^2)*cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/(a-b)^3/ f+2/15*(5*a-3*b)*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(1/2)/(a-b)^2/f-1/5*cos (f*x+e)^5*(a-b+b*sec(f*x+e)^2)^(1/2)/(a-b)/f
Time = 1.64 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\cos (e+f x) \left (-89 a^2+34 a b-9 b^2+4 \left (7 a^2-10 a b+3 b^2\right ) \cos (2 (e+f x))-3 (a-b)^2 \cos (4 (e+f x))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{120 \sqrt {2} (a-b)^3 f} \]
(Cos[e + f*x]*(-89*a^2 + 34*a*b - 9*b^2 + 4*(7*a^2 - 10*a*b + 3*b^2)*Cos[2 *(e + f*x)] - 3*(a - b)^2*Cos[4*(e + f*x)])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(120*Sqrt[2]*(a - b)^3*f)
Time = 0.33 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4147, 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.
\(\displaystyle \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^5}{\sqrt {a+b \tan (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4147 |
\(\displaystyle \frac {\int \frac {\cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 365 |
\(\displaystyle \frac {\frac {\int -\frac {\cos ^4(e+f x) \left (2 (5 a-3 b)-5 (a-b) \sec ^2(e+f x)\right )}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{5 (a-b)}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {\int \frac {\cos ^4(e+f x) \left (2 (5 a-3 b)-5 (a-b) \sec ^2(e+f x)\right )}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{5 (a-b)}}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {-\frac {-\frac {\left (15 a^2-10 a b+3 b^2\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{3 (a-b)}-\frac {2 (5 a-3 b) \cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{3 (a-b)}}{5 (a-b)}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{5 (a-b)}}{f}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {-\frac {\frac {\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{3 (a-b)^2}-\frac {2 (5 a-3 b) \cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{3 (a-b)}}{5 (a-b)}-\frac {\cos ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{5 (a-b)}}{f}\) |
(-1/5*(Cos[e + f*x]^5*Sqrt[a - b + b*Sec[e + f*x]^2])/(a - b) - (((15*a^2 - 10*a*b + 3*b^2)*Cos[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(3*(a - b)^ 2) - (2*(5*a - 3*b)*Cos[e + f*x]^3*Sqrt[a - b + b*Sec[e + f*x]^2])/(3*(a - b)))/(5*(a - b)))/f
3.2.16.3.1 Defintions of rubi rules used
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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Time = 0.89 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91
method | result | size |
default | \(-\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (3 \sin \left (f x +e \right )^{4} b^{2}+6 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a b +3 a^{2} \cos \left (f x +e \right )^{4}-10 a b \sin \left (f x +e \right )^{2}-10 \cos \left (f x +e \right )^{2} a^{2}+15 a^{2}\right ) \sec \left (f x +e \right )}{15 f \left (a -b \right )^{3} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\) | \(131\) |
-1/15/f/(a-b)^3*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*(3*sin(f*x+e)^4*b^2+6*cos( f*x+e)^2*sin(f*x+e)^2*a*b+3*a^2*cos(f*x+e)^4-10*a*b*sin(f*x+e)^2-10*cos(f* x+e)^2*a^2+15*a^2)/(a+b*tan(f*x+e)^2)^(1/2)*sec(f*x+e)
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {{\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (5 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 10 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f} \]
-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 2*(5*a^2 - 8*a*b + 3*b^2)*co s(f*x + e)^3 + (15*a^2 - 10*a*b + 3*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f *x + e)^2 + b)/cos(f*x + e)^2)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \]
Time = 0.25 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.49 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a - b} + \frac {3 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 10 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {10 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{2} - 2 \, a b + b^{2}}}{15 \, f} \]
-1/15*(15*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a - b) + (3*(a - b + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 10*(a - b + b/cos(f*x + e)^2)^( 3/2)*b*cos(f*x + e)^3 + 15*sqrt(a - b + b/cos(f*x + e)^2)*b^2*cos(f*x + e) )/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 10*((a - b + b/cos(f*x + e)^2)^(3/2)*c os(f*x + e)^3 - 3*sqrt(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e))/(a^2 - 2* a*b + b^2))/f
\[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{5}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^5}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]