Integrand size = 25, antiderivative size = 87 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) f} \]
-1/2*a*arctanh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b)^(3/2 )/f-1/2*cot(f*x+e)*csc(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/(a+b)/f
Time = 0.73 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.61 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x) \sqrt {a+b-a \sin ^2(e+f x)} \left (\frac {(a+b) \csc ^2(e+f x)}{a}+\frac {\text {arctanh}\left (\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}\right )}{\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}\right )}{2 \sqrt {2} (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}} \]
-1/2*(a*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*x]*Sqrt[a + b - a*Sin [e + f*x]^2]*(((a + b)*Csc[e + f*x]^2)/a + ArcTanh[Sqrt[1 - (a*Sin[e + f*x ]^2)/(a + b)]]/Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)]))/(Sqrt[2]*(a + b)^2*f *Sqrt[a + b*Sec[e + f*x]^2])
Time = 0.28 (sec) , antiderivative size = 93, 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, 4622, 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.
| \(\displaystyle \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^3 \sqrt {a+b \sec (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4622 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\int \frac {a}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{2 (a+b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {a \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{2 (a+b)}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {a \int \frac {1}{1-\frac {(a+b) \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}}{2 (a+b)}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)^{3/2}}}{f}\) |
(-1/2*(a*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/( a + b)^(3/2) + (Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*(a + b)*(1 - S ec[e + f*x]^2)))/f
3.1.97.3.1 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[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si mp[1/(f*ff^m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p /x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
Leaf count of result is larger than twice the leaf count of optimal. \(1214\) vs. \(2(75)=150\).
Time = 1.28 (sec) , antiderivative size = 1215, normalized size of antiderivative = 13.97
1/8/f/(a+b)^(5/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc( f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e) ^2+a+b)^(1/2)/((a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x +e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+ a+b)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^2)^(1/2)/((1-cos(f*x+e))^2*csc(f*x+ e)^2-1)/(1-cos(f*x+e))^2*((a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e) )^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*cs c(f*x+e)^2+a+b)^(1/2)*(a+b)^(3/2)*(1-cos(f*x+e))^2+2*ln((a*(1-cos(f*x+e))^ 2*csc(f*x+e)^2+b*(1-cos(f*x+e))^2*csc(f*x+e)^2+(a*(1-cos(f*x+e))^4*csc(f*x +e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2* b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(1/2)-a+b)/(a+b)^(1/2))*a ^2*(1-cos(f*x+e))^2+2*ln((a*(1-cos(f*x+e))^2*csc(f*x+e)^2+b*(1-cos(f*x+e)) ^2*csc(f*x+e)^2+(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f* x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2 +a+b)^(1/2)*(a+b)^(1/2)-a+b)/(a+b)^(1/2))*(1-cos(f*x+e))^2*a*b+2*ln(2/(1-c os(f*x+e))^2*(-a*(1-cos(f*x+e))^2+b*(1-cos(f*x+e))^2+(a*(1-cos(f*x+e))^4*c sc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e )^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(1/2)*sin(f*x+e)^2+ a*sin(f*x+e)^2+b*sin(f*x+e)^2))*a^2*(1-cos(f*x+e))^2+2*ln(2/(1-cos(f*x+e)) ^2*(-a*(1-cos(f*x+e))^2+b*(1-cos(f*x+e))^2+(a*(1-cos(f*x+e))^4*csc(f*x+...
Time = 0.34 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.51 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {2 \, {\left (a + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}, \frac {{\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) + {\left (a + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}\right ] \]
[1/4*(2*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + (a*cos(f*x + e)^2 - a)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b )*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos (f*x + e)^2 - 1)))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f), 1/2*((a*cos(f*x + e)^2 - a)*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt ((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + (a + b)*sq rt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/((a^2 + 2*a*b + b^ 2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f)]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]