Integrand size = 25, antiderivative size = 111 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{f}-\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f} \] Output:
-a^(1/2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f+(a+b*sec(f*x+e)^2)^(1 /2)/f-1/3*(a+2*b)*(a+b*sec(f*x+e)^2)^(3/2)/b^2/f+1/5*(a+b*sec(f*x+e)^2)^(5 /2)/b^2/f
Time = 1.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.25 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=\frac {15 b^2 \left (a+b \sec ^2(e+f x)\right )-5 a \left (a+b \sec ^2(e+f x)\right )^2-10 b \left (a+b \sec ^2(e+f x)\right )^2+3 \left (a+b \sec ^2(e+f x)\right )^3-15 a b^2 \text {arctanh}\left (\sqrt {1+\frac {b \sec ^2(e+f x)}{a}}\right ) \sqrt {1+\frac {b \sec ^2(e+f x)}{a}}}{15 b^2 f \sqrt {a+b \sec ^2(e+f x)}} \] Input:
Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x]^5,x]
Output:
(15*b^2*(a + b*Sec[e + f*x]^2) - 5*a*(a + b*Sec[e + f*x]^2)^2 - 10*b*(a + b*Sec[e + f*x]^2)^2 + 3*(a + b*Sec[e + f*x]^2)^3 - 15*a*b^2*ArcTanh[Sqrt[1 + (b*Sec[e + f*x]^2)/a]]*Sqrt[1 + (b*Sec[e + f*x]^2)/a])/(15*b^2*f*Sqrt[a + b*Sec[e + f*x]^2])
Time = 0.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4627, 354, 99, 2009}
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.
| \(\displaystyle \int \tan ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (e+f x)^5 \sqrt {a+b \sec (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4627 |
| \(\displaystyle \frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a}d\sec ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {\left (b \sec ^2(e+f x)+a\right )^{3/2}}{b}+\cos (e+f x) \sqrt {b \sec ^2(e+f x)+a}+\frac {(-a-2 b) \sqrt {b \sec ^2(e+f x)+a}}{b}\right )d\sec ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )+\frac {2 \left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2}-\frac {2 (a+2 b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2}+2 \sqrt {a+b \sec ^2(e+f x)}}{2 f}\) |
Input:
Int[Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x]^5,x]
Output:
(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + 2*Sqrt[a + b*Sec [e + f*x]^2] - (2*(a + 2*b)*(a + b*Sec[e + f*x]^2)^(3/2))/(3*b^2) + (2*(a + b*Sec[e + f*x]^2)^(5/2))/(5*b^2))/(2*f)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si mp[1/f Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), 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] || IGtQ[p, 0] || Integers Q[2*n, p])
Leaf count of result is larger than twice the leaf count of optimal. \(321\) vs. \(2(95)=190\).
Time = 28.32 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.90
| method | result | size |
| default | \(-\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (15 \sqrt {a}\, \ln \left (4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \cos \left (f x +e \right ) a \right ) b^{2} \cos \left (f x +e \right )+\left (2 \cos \left (f x +e \right )+2\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \left (10 \cos \left (f x +e \right )+10-\sec \left (f x +e \right )-\sec \left (f x +e \right )^{2}\right )+b^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (-15 \cos \left (f x +e \right )-15+10 \sec \left (f x +e \right )+10 \sec \left (f x +e \right )^{2}-3 \sec \left (f x +e \right )^{3}-3 \sec \left (f x +e \right )^{4}\right )\right )}{15 f \,b^{2} \left (1+\cos \left (f x +e \right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\) | \(322\) |
Input:
int((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)
Output:
-1/15/f/b^2*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/(1 +cos(f*x+e))^2)^(1/2)*(15*a^(1/2)*ln(4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos( f*x+e))^2)^(1/2)*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 )^(1/2)+4*cos(f*x+e)*a)*b^2*cos(f*x+e)+(2*cos(f*x+e)+2)*((b+a*cos(f*x+e)^2 )/(1+cos(f*x+e))^2)^(1/2)*a^2+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)* a*b*(10*cos(f*x+e)+10-sec(f*x+e)-sec(f*x+e)^2)+b^2*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*(-15*cos(f*x+e)-15+10*sec(f*x+e)+10*sec(f*x+e)^2-3*se c(f*x+e)^3-3*sec(f*x+e)^4))
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (95) = 190\).
Time = 1.68 (sec) , antiderivative size = 456, normalized size of antiderivative = 4.11 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=\left [\frac {15 \, \sqrt {a} b^{2} \cos \left (f x + e\right )^{4} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) - 8 \, {\left ({\left (2 \, a^{2} + 10 \, a b - 15 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, b^{2} f \cos \left (f x + e\right )^{4}}, \frac {15 \, \sqrt {-a} b^{2} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{4} - 4 \, {\left ({\left (2 \, a^{2} + 10 \, a b - 15 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, b^{2} f \cos \left (f x + e\right )^{4}}\right ] \] Input:
integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="fricas")
Output:
[1/120*(15*sqrt(a)*b^2*cos(f*x + e)^4*log(128*a^4*cos(f*x + e)^8 + 256*a^3 *b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 - 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f *x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f* x + e)^2)) - 8*((2*a^2 + 10*a*b - 15*b^2)*cos(f*x + e)^4 - (a*b - 10*b^2)* cos(f*x + e)^2 - 3*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b^2* f*cos(f*x + e)^4), 1/60*(15*sqrt(-a)*b^2*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2))*cos(f*x + e)^4 - 4*((2*a^2 + 10*a*b - 15*b^2)*cos(f*x + e)^4 - (a*b - 10*b^2)*cos( f*x + e)^2 - 3*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b^2*f*co s(f*x + e)^4)]
\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \tan ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate((a+b*sec(f*x+e)**2)**(1/2)*tan(f*x+e)**5,x)
Output:
Integral(sqrt(a + b*sec(e + f*x)**2)*tan(e + f*x)**5, x)
\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{5} \,d x } \] Input:
integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="maxima")
Output:
integrate(sqrt(b*sec(f*x + e)^2 + a)*tan(f*x + e)^5, x)
Leaf count of result is larger than twice the leaf count of optimal. 1280 vs. \(2 (95) = 190\).
Time = 0.63 (sec) , antiderivative size = 1280, normalized size of antiderivative = 11.53 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=\text {Too large to display} \] Input:
integrate((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="giac")
Output:
2/15*(15*a*arctan(-1/2*(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) + sqrt(a + b))/sqrt(-a))/sqrt(-a) - 2 *(15*(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))^9*a - 165*(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))^8*sqrt(a + b)*a + 20*(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*(27*a^2 - 5*a*b - 16*b^2) - 20*(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*(33*a^2 - 83*a*b + 32*b^2)*sqrt(a + b) - 2*(15*a^3 + 1230*a^2*b - 625*a*b^2 - 416*b^3)*(sqr t(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 + 10*(81*a^3 + 90*a^2*b - 391*a*b^2 + 256*b^3)*(sqrt(a + b)*t an(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*sqrt(a + b) - 20*(33*a^4 - 45*a^3*b - 157*a^2*b^2 + 161*a*b^3 - 16*b...
Timed out. \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^5\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \] Input:
int(tan(e + f*x)^5*(a + b/cos(e + f*x)^2)^(1/2),x)
Output:
int(tan(e + f*x)^5*(a + b/cos(e + f*x)^2)^(1/2), x)
\[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx=\frac {3 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{4}-4 \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2}+8 \sqrt {\sec \left (f x +e \right )^{2} b +a}-3 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{5}}{\sec \left (f x +e \right )^{2} b +a}d x \right ) b f -4 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a f +8 \left (\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{2} b +a}d x \right ) a f}{12 f} \] Input:
int((a+b*sec(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x)
Output:
(3*sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**4 - 4*sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**2 + 8*sqrt(sec(e + f*x)**2*b + a) - 3*int((sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*tan(e + f*x)**5)/(sec(e + f*x)**2*b + a),x )*b*f - 4*int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**3)/(sec(e + f*x)* *2*b + a),x)*a*f + 8*int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x))/(sec(e + f*x)**2*b + a),x)*a*f)/(12*f)