Integrand size = 25, antiderivative size = 100 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f} \]
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-1/3*cot(f*x+e)^3*(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 = 4.58 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.04 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\left (\left (6 a^2+11 a b+3 b^2+4 \left (a^2-3 a b-b^2\right ) \cos (2 (e+f x))+\left (-2 a^2+a b+b^2\right ) \cos (4 (e+f x))\right ) \csc ^4(e+f x)-12 \sqrt {2} a b \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \tan (e+f x)}{12 \sqrt {2} a f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
-1/12*(((6*a^2 + 11*a*b + 3*b^2 + 4*(a^2 - 3*a*b - b^2)*Cos[2*(e + f*x)] + (-2*a^2 + a*b + b^2)*Cos[4*(e + f*x)])*Csc[e + f*x]^4 - 12*Sqrt[2]*a*b*Sq rt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[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*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4146, 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.
| \(\displaystyle \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (e+f x)^2}}{\sin (e+f x)^4}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \cot ^4(e+f x) \left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {\int \cot ^2(e+f x) \sqrt {b \tan ^2(e+f x)+a}d\tan (e+f x)-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}}{f}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {b \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a}-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
(Sqrt[b]*ArcTanh[(Sqrt[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] - (Cot[e + f*x]^3*(a + b*Tan[e + f*x]^ 2)^(3/2))/(3*a))/f
3.2.2.3.1 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[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. \(339\) vs. \(2(88)=176\).
Time = 3.98 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.40
| method | result | size |
| default | \(-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}\, \left (3 a \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {b}}\right ) \cot \left (f x +e \right )^{2}+\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \cot \left (f x +e \right )-3 a \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {b}}\right ) \cot \left (f x +e \right ) \csc \left (f x +e \right )-2 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a \cot \left (f x +e \right )^{3}+3 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a \cot \left (f x +e \right ) \csc \left (f x +e \right )^{2}\right )}{3 f a \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(340\) |
-1/3/f/a*(a+b*tan(f*x+e)^2)^(1/2)/((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos( f*x+e)+1)^2)^(1/2)*(3*a*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)))*cot(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*cot(f*x+e)-3*a*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)))*cot(f*x+e)*csc(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*cot(f*x+e)^3+3*((a*cos(f*x+e)^ 2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a*cot(f*x+e)*csc(f*x+e)^2)
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (88) = 176\).
Time = 0.45 (sec) , antiderivative size = 435, normalized size of antiderivative = 4.35 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )} \sin \left (f x + e\right )}, -\frac {3 \, {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, {\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )} \sin \left (f x + e\right )}\right ] \]
[1/12*(3*(a*cos(f*x + e)^2 - a)*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*((2*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f* x + e)^2))/((a*f*cos(f*x + e)^2 - a*f)*sin(f*x + e)), -1/6*(3*(a*cos(f*x + e)^2 - a)*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*((2*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f* x + e)^2))/((a*f*cos(f*x + e)^2 - a*f)*sin(f*x + e))]
\[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \csc ^{4}{\left (e + f x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {3 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) - \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{\tan \left (f x + e\right )} - \frac {{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \tan \left (f x + e\right )^{3}}}{3 \, f} \]
1/3*(3*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt(a*b)) - 3*sqrt(b*tan(f*x + e)^2 + a)/tan(f*x + e) - (b*tan(f*x + e)^2 + a)^(3/2)/(a*tan(f*x + e)^3))/f
\[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^4} \,d x \]