Integrand size = 25, antiderivative size = 91 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 a f} \]
-1/2*(a-b)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(3/2)/ f-1/2*cot(f*x+e)*csc(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/a/f
Leaf count is larger than twice the leaf count of optimal. \(367\) vs. \(2(91)=182\).
Time = 3.79 (sec) , antiderivative size = 367, normalized size of antiderivative = 4.03 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\cot ^2(e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (2 (-a+b) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )-\left (-2 (a-b) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+\sqrt {2} \sqrt {a} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}\right ) \sec (e+f x)+8 (a-b) \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right ) \sec (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )}{4 a^{3/2} f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}} \]
(Cot[e + f*x]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(2 *(-a + b)*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]] - (-2*(a - 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]] + Sqrt[2]*Sqrt[a]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])* Sec[(e + f*x)/2]^4])*Sec[e + f*x] + 8*(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]]*Sec [e + f*x]*Sin[(e + f*x)/2]^2))/(4*a^(3/2)*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4147, 373, 27, 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.
| \(\displaystyle \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^3 \sqrt {a+b \tan (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {\int \frac {a-b}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{2 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {(a-b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{2 a}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {(a-b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{2 a}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a^{3/2}}}{f}\) |
(-1/2*((a - b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^ 2]])/a^(3/2) + (Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*a*(1 - Sec [e + f*x]^2)))/f
3.2.20.3.1 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_)^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]*((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 + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, 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. \(954\) vs. \(2(79)=158\).
Time = 0.94 (sec) , antiderivative size = 955, normalized size of antiderivative = 10.49
1/8/f/a^(5/2)*(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*(-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)/(-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^(3/2)*(-cos(f*x+e)+1)^2+2*ln((a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+(a*(-co s(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+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) )*a^2*(-cos(f*x+e)+1)^2-2*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*(-cos(f*x+e)+1)^2*b- 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))*a* (-cos(f*x+e)+1)^2*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)*a^(3/2)*sin(...
Time = 0.39 (sec) , antiderivative size = 284, normalized size of antiderivative = 3.12 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {2 \, a \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}, \frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + a \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{2 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}\right ] \]
[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)))/(a^2*f*cos(f*x + e)^2 - a^2*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)*cos(f*x + e )^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f)]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]