Integrand size = 25, antiderivative size = 141 \[ \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a f} \]
arctanh(b^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-cot(f*x+e)* (a+b*tan(f*x+e)^2)^(1/2)/f-2/15*(5*a-b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3 /2)/a^2/f-1/5*cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2)/a/f
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 3.82 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.04 \[ \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\left (\left (80 a^3+198 a^2 b+98 a b^2-20 b^3+\left (40 a^3-241 a^2 b-149 a b^2+30 b^3\right ) \cos (2 (e+f x))+\left (-32 a^3+42 a^2 b+62 a b^2-12 b^3\right ) \cos (4 (e+f x))+8 a^3 \cos (6 (e+f x))+a^2 b \cos (6 (e+f x))-11 a b^2 \cos (6 (e+f x))+2 b^3 \cos (6 (e+f x))\right ) \csc ^6(e+f x)-240 \sqrt {2} a^2 b \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \tan (e+f x)}{240 \sqrt {2} a^2 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
-1/240*(((80*a^3 + 198*a^2*b + 98*a*b^2 - 20*b^3 + (40*a^3 - 241*a^2*b - 1 49*a*b^2 + 30*b^3)*Cos[2*(e + f*x)] + (-32*a^3 + 42*a^2*b + 62*a*b^2 - 12* b^3)*Cos[4*(e + f*x)] + 8*a^3*Cos[6*(e + f*x)] + a^2*b*Cos[6*(e + f*x)] - 11*a*b^2*Cos[6*(e + f*x)] + 2*b^3*Cos[6*(e + f*x)])*Csc[e + f*x]^6 - 240*S qrt[2]*a^2*b*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*E llipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b ]/Sqrt[2]], 1])*Tan[e + f*x])/(Sqrt[2]*a^2*f*Sqrt[(a + b + (a - b)*Cos[2*( e + f*x)])*Sec[e + f*x]^2])
Time = 0.33 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4146, 365, 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.
| \(\displaystyle \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (e+f x)^2}}{\sin (e+f x)^6}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int \cot ^4(e+f x) \left (5 a \tan ^2(e+f x)+2 (5 a-b)\right ) \sqrt {b \tan ^2(e+f x)+a}d\tan (e+f x)}{5 a}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a}}{f}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {\frac {5 a \int \cot ^2(e+f x) \sqrt {b \tan ^2(e+f x)+a}d\tan (e+f x)-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}}{5 a}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {\frac {5 a \left (b \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}}{5 a}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {5 a \left (b \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}}{5 a}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {5 a \left (\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {2 (5 a-b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}}{5 a}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{5 a}}{f}\) |
(-1/5*(Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(3/2))/a + ((-2*(5*a - b)*Cot [e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2))/(3*a) + 5*a*(Sqrt[b]*ArcTanh[(Sq rt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]] - Cot[e + f*x]*Sqrt[a + b* Tan[e + f*x]^2]))/(5*a))/f
3.2.3.3.1 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_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Leaf count of result is larger than twice the leaf count of optimal. \(507\) vs. \(2(125)=250\).
Time = 5.04 (sec) , antiderivative size = 508, normalized size of antiderivative = 3.60
| method | result | size |
| default | \(\frac {\left (-15 \sin \left (f x +e \right )^{3} a^{2} \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {b}}\right ) \cos \left (f x +e \right )+2 \sin \left (f x +e \right )^{4} \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b^{2}+15 \sin \left (f x +e \right )^{3} a^{2} \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {b}}\right )+9 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{2} \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a b -8 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{4} a^{2}-10 \sin \left (f x +e \right )^{2} \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a b +20 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{2} a^{2}-15 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{2}\right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}\, \cot \left (f x +e \right ) \csc \left (f x +e \right )^{4}}{15 f \,a^{2} \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(508\) |
1/15/f/a^2*(-15*sin(f*x+e)^3*a^2*b^(1/2)*arctanh(1/b^(1/2)*((a*cos(f*x+e)^ 2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))*cos(f*x +e)+2*sin(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*sin(f*x+e)^3*a^2*b^(1/2)*arctanh(1/b^(1/2)*((a*cos(f*x+e)^2+b*s in(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))+9*sin(f*x+e) ^2*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) *a*b-8*((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^2-10*sin(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)*a*b+20*((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^2-15*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2) ^(1/2)*a^2)*(a+b*tan(f*x+e)^2)^(1/2)/((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(c os(f*x+e)+1)^2)^(1/2)*cot(f*x+e)*csc(f*x+e)^4
Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (125) = 250\).
Time = 0.87 (sec) , antiderivative size = 587, normalized size of antiderivative = 4.16 \[ \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (8 \, a^{2} + 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 19 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b - 2 \, 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}}}}{60 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )} \sin \left (f x + e\right )}, -\frac {15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, {\left ({\left (8 \, a^{2} + 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (20 \, a^{2} + 19 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b - 2 \, 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} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )} \sin \left (f x + e\right )}\right ] \]
[1/60*(15*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^2)*sqrt(b)*log((( a^2 - 8*a*b + 8*b^2)*cos(f*x + e)^4 + 8*(a*b - 2*b^2)*cos(f*x + e)^2 + 4*( (a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 4*((8*a^2 + 9*a*b - 2*b^2)*cos(f*x + e)^5 - (20*a^2 + 19*a*b - 4*b ^2)*cos(f*x + e)^3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*x + e))*sqrt(((a - b) *cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos (f*x + e)^2 + a^2*f)*sin(f*x + e)), -1/30*(15*(a^2*cos(f*x + e)^4 - 2*a^2* cos(f*x + e)^2 + a^2)*sqrt(-b)*arctan(1/2*((a - 2*b)*cos(f*x + e)^3 + 2*b* cos(f*x + e))*sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/( ((a*b - b^2)*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*sin(f*x + e) + 2*((8*a^2 + 9*a*b - 2*b^2)*cos(f*x + e)^5 - (20*a^2 + 19*a*b - 4*b^2)*cos(f*x + e)^ 3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2 *f)*sin(f*x + e))]
\[ \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \csc ^{6}{\left (e + f x \right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93 \[ \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {15 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) - \frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{\tan \left (f x + e\right )} - \frac {10 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \tan \left (f x + e\right )^{3}} + \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a^{2} \tan \left (f x + e\right )^{3}} - \frac {3 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \tan \left (f x + e\right )^{5}}}{15 \, f} \]
1/15*(15*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt(a*b)) - 15*sqrt(b*tan(f*x + e )^2 + a)/tan(f*x + e) - 10*(b*tan(f*x + e)^2 + a)^(3/2)/(a*tan(f*x + e)^3) + 2*(b*tan(f*x + e)^2 + a)^(3/2)*b/(a^2*tan(f*x + e)^3) - 3*(b*tan(f*x + e)^2 + a)^(3/2)/(a*tan(f*x + e)^5))/f
\[ \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \csc ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^6} \,d x \]