Integrand size = 25, antiderivative size = 134 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{5/2} f}-\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f} \]
1/8*(8*a^2+8*a*b+3*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b )^(5/2)/f-1/8*(8*a+5*b)*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)^2/f+1/ 4*sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)/f
Time = 0.50 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b} \sec ^2(e+f x) \left (-8 a-5 b+2 (a+b) \sec ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^{5/2} f} \]
((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] + Sqrt[a + b]*Sec[e + f*x]^2*(-8*a - 5*b + 2*(a + b)*Sec[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^(5/2)*f)
Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 100, 27, 87, 73, 221}
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.
| \(\displaystyle \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^5}{\sqrt {a+b \sin (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 3673 |
| \(\displaystyle \frac {\int \frac {\sin ^4(e+f x)}{\left (1-\sin ^2(e+f x)\right )^3 \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {\frac {\sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\int \frac {4 (a+b) \sin ^2(e+f x)+4 a+b}{2 \left (1-\sin ^2(e+f x)\right )^2 \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 (a+b)}}{2 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\int \frac {4 (a+b) \sin ^2(e+f x)+4 a+b}{\left (1-\sin ^2(e+f x)\right )^2 \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{4 (a+b)}}{2 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {\frac {\sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 (a+b)}}{4 (a+b)}}{2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b (a+b)}}{4 (a+b)}}{2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}}{4 (a+b)}}{2 f}\) |
(Sqrt[a + b*Sin[e + f*x]^2]/(2*(a + b)*(1 - Sin[e + f*x]^2)^2) - (-(((8*a^ 2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(a + b )^(3/2)) + ((8*a + 5*b)*Sqrt[a + b*Sin[e + f*x]^2])/((a + b)*(1 - Sin[e + f*x]^2)))/(4*(a + b)))/(2*f)
3.6.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(643\) vs. \(2(118)=236\).
Time = 1.58 (sec) , antiderivative size = 644, normalized size of antiderivative = 4.81
| method | result | size |
| default | \(\frac {\left (8 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{4}+24 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3} b +27 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b^{2}+14 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{3}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{4}+8 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{4}+24 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3} b +27 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b^{2}+14 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{3}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{4}\right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \left (a +b \right )^{\frac {5}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (8 a +5 b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+4 a \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}}+4 b \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}}}{16 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f}\) | \(644\) |
1/16*((8*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin (f*x+e)+a))*a^4+24*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^( 1/2)-b*sin(f*x+e)+a))*a^3*b+27*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos (f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2+14*ln(2/(1+sin(f*x+e))*((a+b)^(1 /2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^3+3*ln(2/(1+sin(f*x+e) )*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^4+8*ln(2/(sin (f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^4+24 *ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+ a))*a^3*b+27*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b *sin(f*x+e)+a))*a^2*b^2+14*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x +e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^3+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+ b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^4)*cos(f*x+e)^4-2*(a+b)^(5/2)*( a+b-b*cos(f*x+e)^2)^(1/2)*(8*a+5*b)*cos(f*x+e)^2+4*a*(a+b-b*cos(f*x+e)^2)^ (1/2)*(a+b)^(5/2)+4*b*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2))/(a+b)^(5/2)/ cos(f*x+e)^4/(a^2+2*a*b+b^2)/f
Time = 0.39 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.45 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}, -\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{4} + {\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}\right ] \]
[1/16*((8*a^2 + 8*a*b + 3*b^2)*sqrt(a + b)*cos(f*x + e)^4*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 2*((8*a^2 + 13*a*b + 5*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b - 2* b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*c os(f*x + e)^4), -1/8*((8*a^2 + 8*a*b + 3*b^2)*sqrt(-a - b)*arctan(sqrt(-b* cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^4 + ((8*a^2 + 1 3*a*b + 5*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b - 2*b^2)*sqrt(-b*cos(f*x + e )^2 + a + b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4)]
\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (118) = 236\).
Time = 0.30 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.85 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {{\left (8 \, a^{2} b^{3} + 8 \, a b^{4} + 3 \, b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} - \frac {2 \, {\left ({\left (8 \, a b^{4} + 5 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - {\left (8 \, a^{2} b^{4} + 11 \, a b^{5} + 3 \, b^{6}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}\right )}}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}}}{16 \, b^{3} f} \]
-1/16*((8*a^2*b^3 + 8*a*b^4 + 3*b^5)*log((sqrt(b*sin(f*x + e)^2 + a) - sqr t(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/((a^2 + 2*a*b + b^2) *sqrt(a + b)) - 2*((8*a*b^4 + 5*b^5)*(b*sin(f*x + e)^2 + a)^(3/2) - (8*a^2 *b^4 + 11*a*b^5 + 3*b^6)*sqrt(b*sin(f*x + e)^2 + a))/(a^4 + 4*a^3*b + 6*a^ 2*b^2 + 4*a*b^3 + b^4 + (b*sin(f*x + e)^2 + a)^2*(a^2 + 2*a*b + b^2) - 2*( a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(b*sin(f*x + e)^2 + a)))/(b^3*f)
\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^5}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]