Integrand size = 25, antiderivative size = 127 \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 \sqrt {a} f}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f} \]
-1/2*(a+b)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/f/a^(1/2 )+arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-1/2*cot (f*x+e)*csc(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f
Leaf count is larger than twice the leaf count of optimal. \(586\) vs. \(2(127)=254\).
Time = 5.75 (sec) , antiderivative size = 586, normalized size of antiderivative = 4.61 \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-a \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )-b \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+a \cos (e+f x) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+b \cos (e+f x) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+\frac {\sqrt {a} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}{\sqrt {2}}-16 \sqrt {a} \sqrt {b} \text {arctanh}\left (\frac {-\sqrt {a} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{2 \sqrt {b}}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )-4 (a+b) \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )}{2 \sqrt {a} f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}} \]
-1/2*(Cot[e + f*x]*Csc[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Se c[e + f*x]^2]*(-(a*Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*T an[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]]) - b*Log[a - 2*b - a*T an[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] + a*Cos[e + f*x]*Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt [a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] + b*Cos[ e + f*x]*Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f* x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] + (Sqrt[a]*Sqrt[(a + b + (a - b) *Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])/Sqrt[2] - 16*Sqrt[a]*Sqrt[b]*ArcTa nh[(-(Sqrt[a]*(-1 + Tan[(e + f*x)/2]^2)) + Sqrt[4*b*Tan[(e + f*x)/2]^2 + a *(-1 + Tan[(e + f*x)/2]^2)^2])/(2*Sqrt[b])]*Sin[(e + f*x)/2]^2 - 4*(a + b) *ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]]*Sin[(e + f*x)/2]^2))/(Sqrt[a]*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])
Time = 0.34 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4147, 369, 398, 224, 219, 291, 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.
| \(\displaystyle \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (e+f x)^2}}{\sin (e+f x)^3}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x) \sqrt {b \sec ^2(e+f x)+a-b}}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 369 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 \left (1-\sec ^2(e+f x)\right )}-\frac {1}{2} \int \frac {2 b \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 b \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)-(a+b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)\right )+\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 b \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}-(a+b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)\right )+\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )-(a+b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)\right )+\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )-(a+b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}\right )+\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{2} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )-\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{\sqrt {a}}\right )+\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 \left (1-\sec ^2(e+f x)\right )}}{f}\) |
((-(((a + b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2] ])/Sqrt[a]) + 2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[ e + f*x]^2]])/2 + (Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*(1 - Se c[e + f*x]^2)))/f
3.1.96.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2* b*(p + 1))), x] - Simp[e^2/(2*b*(p + 1)) Int[(e*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(m - 1) + d*(m + 2*q - 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 0 ] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
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. \(1344\) vs. \(2(109)=218\).
Time = 0.61 (sec) , antiderivative size = 1345, normalized size of antiderivative = 10.59
1/8/f/a^(3/2)*((a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc (f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)/((-cos(f*x+e)+1)^2*csc(f*x +e)^2-1)^2)^(1/2)*((-cos(f*x+e)+1)^2*csc(f*x+e)^2-1)*(-(a*(-cos(f*x+e)+1)^ 4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*cs c(f*x+e)^2+a)^(1/2)*a^(3/2)*(-cos(f*x+e)+1)^4*csc(f*x+e)^4+2*ln((a*(-cos(f *x+e)+1)^2*csc(f*x+e)^2+(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e) +1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)*a^(1/2)-a+2 *b)/a^(1/2))*a^2*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+2*a^2*ln(2/(-cos(f*x+e)+1) ^2*(-a*(-cos(f*x+e)+1)^2+2*b*(-cos(f*x+e)+1)^2+(a*(-cos(f*x+e)+1)^4*csc(f* x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e) ^2+a)^(1/2)*a^(1/2)*sin(f*x+e)^2+a*sin(f*x+e)^2))*(-cos(f*x+e)+1)^2*csc(f* x+e)^2-8*b^(1/2)*ln(4*(b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+b^(1/2)*(a*(-cos(f *x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e )+1)^2*csc(f*x+e)^2+a)^(1/2)+b)/((-cos(f*x+e)+1)^2*csc(f*x+e)^2-1))*a^(3/2 )*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+3*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*( -cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)* a^(3/2)*(-cos(f*x+e)+1)^2*csc(f*x+e)^2-4*b*(a*(-cos(f*x+e)+1)^4*csc(f*x+e) ^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a )^(1/2)*(-cos(f*x+e)+1)^2*a^(1/2)*csc(f*x+e)^2+2*ln((a*(-cos(f*x+e)+1)^2*c sc(f*x+e)^2+(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc...
Time = 0.79 (sec) , antiderivative size = 849, normalized size of antiderivative = 6.69 \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \]
[1/4*(2*a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + ((a + b)*cos(f*x + e)^2 - a - b)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + 2*(a*cos(f*x + e)^2 - a)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(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))/(a*f*cos(f*x + e)^2 - a*f), 1/2*(((a + b)*cos(f*x + e)^2 - a - b)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + a*sqrt(((a - b)*c os(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + (a*cos(f*x + e)^2 - a)*s qrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(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))/(a*f*cos(f*x + e)^2 - a*f), -1/4*(4*(a*cos(f*x + e)^2 - a)*sqrt(-b)*arctan(sqrt(-b)*sqr t(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - 2*a*sqrt( ((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - ((a + b)*cos(f *x + e)^2 - a - b)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt (((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f *x + e)^2 - 1)))/(a*f*cos(f*x + e)^2 - a*f), 1/2*(((a + b)*cos(f*x + e)^2 - a - b)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f* x + e)^2)*cos(f*x + e)/a) - 2*(a*cos(f*x + e)^2 - a)*sqrt(-b)*arctan(sqrt( -b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) +...
\[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \csc ^{3}{\left (e + f x \right )}\, dx \]
\[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{3} \,d x } \]
Exception generated. \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^3} \,d x \]