Integrand size = 25, antiderivative size = 161 \[ \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f} \]
2/15*(5*a-4*b)*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)^2/f-1/5*cos(f *x+e)^5*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)/f+arctanh(sec(f*x+e)*b^(1/2)/(a-b +b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f
Time = 3.92 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.29 \[ \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\cos (e+f x) \left (120 \sqrt {2} (a-b)^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right )+\sqrt {a+b+(a-b) \cos (2 (e+f x))} \left (-89 a^2+254 a b-149 b^2+4 \left (7 a^2-15 a b+8 b^2\right ) \cos (2 (e+f x))-3 (a-b)^2 \cos (4 (e+f x))\right )\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{120 \sqrt {2} (a-b)^2 f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \]
(Cos[e + f*x]*(120*Sqrt[2]*(a - b)^2*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)* Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])] + Sqrt[a + b + (a - b)*Cos[2*(e + f*x )]]*(-89*a^2 + 254*a*b - 149*b^2 + 4*(7*a^2 - 15*a*b + 8*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)^2*f*Sqrt[a + b + (a - b)*Cos[2*( e + f*x)]])
Time = 0.35 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4147, 365, 25, 358, 247, 224, 219}
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.
| \(\displaystyle \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^5 \sqrt {a+b \tan (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int -\cos ^4(e+f x) \left (2 (5 a-4 b)-5 (a-b) \sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 (a-b)}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \cos ^4(e+f x) \left (2 (5 a-4 b)-5 (a-b) \sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 (a-b)}}{f}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {-\frac {-5 (a-b) \int \cos ^2(e+f x) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)-\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}}{5 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 (a-b)}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {-\frac {-5 (a-b) \left (b \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}}{5 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 (a-b)}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {-\frac {-5 (a-b) \left (b \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}}{5 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 (a-b)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {-5 (a-b) \left (\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}\right )-\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}}{5 (a-b)}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 (a-b)}}{f}\) |
(-1/5*(Cos[e + f*x]^5*(a - b + b*Sec[e + f*x]^2)^(3/2))/(a - b) - ((-2*(5* a - 4*b)*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(3/2))/(3*(a - b)) - 5* (a - b)*(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x ]^2]] - Cos[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2]))/(5*(a - b)))/f
3.1.92.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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] + S imp[d/e^2 Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, -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]
Leaf count of result is larger than twice the leaf count of optimal. \(1241\) vs. \(2(145)=290\).
Time = 2.16 (sec) , antiderivative size = 1242, normalized size of antiderivative = 7.71
1/15/f/(a-b)^2*(-3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1 /2)*cos(f*x+e)^5*a^2+6*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2 )^(1/2)*cos(f*x+e)^5*a*b-3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+ 1)^2)^(1/2)*b^2*cos(f*x+e)^5-3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x +e)+1)^2)^(1/2)*cos(f*x+e)^4*a^2+6*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos (f*x+e)+1)^2)^(1/2)*cos(f*x+e)^4*a*b-3*cos(f*x+e)^4*((a*cos(f*x+e)^2-b*cos (f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b^2+15*ln(-4*b^(1/2)*((a*cos(f*x+e)^2 -b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2-b*co s(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*b^(5/2)+1 0*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^3* a^2-21*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+ e)^3*a*b+11*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b^2 *cos(f*x+e)^3-30*ln(-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x +e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1 )^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*b^(3/2)*a+10*((a*cos(f*x+e)^2-b*cos( f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2*a^2-21*((a*cos(f*x+e)^2-b *cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2*a*b+11*cos(f*x+e)^2* ((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b^2+15*ln(-4*b^ (1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2) *((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)-...
Time = 0.41 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.35 \[ \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - 21 \, a b + 11 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}, -\frac {15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) + {\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - 21 \, a b + 11 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}\right ] \]
[1/30*(15*(a^2 - 2*a*b + b^2)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqr t(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b) /cos(f*x + e)^2) - 2*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - (10*a^2 - 21* a*b + 11*b^2)*cos(f*x + e)^3 + (15*a^2 - 40*a*b + 23*b^2)*cos(f*x + e))*sq rt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2 - 2*a*b + b^2)*f), -1/15*(15*(a^2 - 2*a*b + b^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f *x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + (3*(a^2 - 2*a*b + b^2)*co s(f*x + e)^5 - (10*a^2 - 21*a*b + 11*b^2)*cos(f*x + e)^3 + (15*a^2 - 40*a* b + 23*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2 ))/((a^2 - 2*a*b + b^2)*f)]
Timed out. \[ \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.26 \[ \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\frac {20 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a - b} - 30 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 15 \, \sqrt {b} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) - \frac {2 \, {\left (3 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 5 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3}\right )}}{a^{2} - 2 \, a b + b^{2}}}{30 \, f} \]
1/30*(20*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3/(a - b) - 30*sqrt (a - b + b/cos(f*x + e)^2)*cos(f*x + e) - 15*sqrt(b)*log((sqrt(a - b + b/c os(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos (f*x + e) + sqrt(b))) - 2*(3*(a - b + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e) ^5 - 5*(a - b + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3)/(a^2 - 2*a*b + b ^2))/f
Leaf count of result is larger than twice the leaf count of optimal. 2554 vs. \(2 (145) = 290\).
Time = 1.03 (sec) , antiderivative size = 2554, normalized size of antiderivative = 15.86 \[ \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \]
2/15*(15*b*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f* x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a ) - sqrt(a))/sqrt(-b))/sqrt(-b) - 2*(15*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f *x + 1/2*e)^2 + a))^9*b + 165*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan (1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e )^2 + a))^8*sqrt(a)*b - 320*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1 /2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^ 2 + a))^7*a^2 + 540*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^ 7*a*b + 320*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^ 4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^7*b^2 + 640*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a* tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^6*a^(5/2) - 2940 *(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan (1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^6*a^(3/2)*b + 2960* (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan( 1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^6*sqrt(a)*b^2 + 832* (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan( 1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*a^3 - 1246*(sqr...
Timed out. \[ \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int {\sin \left (e+f\,x\right )}^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]