Integrand size = 27, antiderivative size = 653 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3} \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{(c-d)^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {3 \sqrt {a} d^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{4 c (c-d) (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\sqrt {a} (2 c-d) d^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^2 (c-d)^2 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 \sqrt {a} d^{3/2} \left (3 c^2-3 c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^3 (c-d)^3 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {d^2 \tan (e+f x)}{2 c \left (c^2-d^2\right ) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {3 d^2 \tan (e+f x)}{4 c (c-d) (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {(2 c-d) d^2 \tan (e+f x)}{c^2 (c-d)^2 (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
1/2*d^2*tan(f*x+e)/c/(c^2-d^2)/f/(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2) +3/4*d^2*tan(f*x+e)/c/(c-d)/(c+d)^2/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1 /2)+(2*c-d)*d^2*tan(f*x+e)/c^2/(c-d)^2/(c+d)/f/(c+d*sec(f*x+e))/(a+a*sec(f *x+e))^(1/2)+2*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*a^(1/2)*tan(f*x+e)/ c^3/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+3/4*d^(3/2)*arctanh(d^ (1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*a^(1/2)*tan(f*x+e)/c/(c- d)/(c+d)^(5/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+(2*c-d)*d^( 3/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*a^(1/2)*t an(f*x+e)/c^2/(c-d)^2/(c+d)^(3/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e) )^(1/2)-arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2 )*tan(f*x+e)/(c-d)^3/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+2*d^( 3/2)*(3*c^2-3*c*d+d^2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d )^(1/2))*a^(1/2)*tan(f*x+e)/c^3/(c-d)^3/f/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/ 2)/(a+a*sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2940\) vs. \(2(653)=1306\).
Time = 22.45 (sec) , antiderivative size = 2940, normalized size of antiderivative = 4.50 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3} \, dx=\text {Result too large to show} \]
(Cos[(e + f*x)/2]*(d + c*Cos[e + f*x])^3*Sec[e + f*x]^4*(-1/2*(d^2*(-13*c^ 2 - c*d + 6*d^2)*Sin[(e + f*x)/2])/(c^3*(-c + d)^2*(c + d)^2) - (d^4*Sin[( e + f*x)/2])/(c^3*(-c + d)*(c + d)*(d + c*Cos[e + f*x])^2) + (-15*c^2*d^3* Sin[(e + f*x)/2] - c*d^4*Sin[(e + f*x)/2] + 8*d^5*Sin[(e + f*x)/2])/(2*c^3 *(-c + d)^2*(c + d)^2*(d + c*Cos[e + f*x]))))/(f*Sqrt[a*(1 + Sec[e + f*x]) ]*(c + d*Sec[e + f*x])^3) - (Cos[(e + f*x)/2]*(d + c*Cos[e + f*x])^3*((Sqr t[2]*d^(3/2)*(35*c^4 + 14*c^3*d - 21*c^2*d^2 - 4*c*d^3 + 8*d^4)*ArcTan[(Sq rt[d]*Tan[(e + f*x)/2])/(Sqrt[-c - d]*Sqrt[-(Cos[e + f*x]/(1 + Cos[e + f*x ]))])])/(Sqrt[-c - d]*(c - d)) - 2*Sqrt[2]*(c^2 - d^2)^2*Log[Sec[(e + f*x) /2]^2*(-1 + 2*Cos[e + f*x] - 2*Sqrt[-(Cos[e + f*x]/(1 + Cos[e + f*x]))]*Si n[e + f*x])] + 2*Sqrt[2]*(c^2 - d^2)^2*Log[Sec[(e + f*x)/2]^2*(-1 + 2*Cos[ e + f*x] + 2*Sqrt[-(Cos[e + f*x]/(1 + Cos[e + f*x]))]*Sin[e + f*x])] + (8* c^3*(c + d)^2*Log[Tan[(e + f*x)/2] + Sqrt[-1 + Tan[(e + f*x)/2]^2]])/(c - d))*((-2*c*d*Sec[(e + f*x)/2])/((-c + d)^2*(c + d)^2*(d + c*Cos[e + f*x])* Sqrt[Sec[e + f*x]]) - (13*d^2*Sec[(e + f*x)/2])/(8*(-c + d)^2*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]]) + (d^3*Sec[(e + f*x)/2])/(8*c*(-c + d)^2*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]]) + (d^4*Sec[(e + f *x)/2])/(2*c^2*(-c + d)^2*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x] ]) + (c^2*Sec[(e + f*x)/2]*Sqrt[Sec[e + f*x]])/(2*(-c + d)^2*(c + d)^2*(d + c*Cos[e + f*x])) + (3*d^2*Sec[(e + f*x)/2]*Sqrt[Sec[e + f*x]])/(8*(-c...
Time = 0.76 (sec) , antiderivative size = 490, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 4428, 27, 198, 2009}
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.
\(\displaystyle \int \frac {1}{\sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4428 |
\(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x)}{a (\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \tan (e+f x) \int \frac {\cos (e+f x)}{(\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 198 |
\(\displaystyle -\frac {a \tan (e+f x) \int \left (\frac {\left (3 c^2-3 d c+d^2\right ) d^2}{c^3 (c-d)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {(2 c-d) d^2}{c^2 (c-d)^2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {d^2}{c (c-d) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}+\frac {\cos (e+f x)}{c^3 \sqrt {a-a \sec (e+f x)}}-\frac {1}{(c-d)^3 (\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \tan (e+f x) \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c^3}-\frac {d^{3/2} (2 c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^2 (c-d)^2 (c+d)^{3/2}}-\frac {2 d^{3/2} \left (3 c^2-3 c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^3 (c-d)^3 \sqrt {c+d}}-\frac {3 d^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{4 \sqrt {a} c (c-d) (c+d)^{5/2}}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a} (c-d)^3}-\frac {d^2 (2 c-d) \sqrt {a-a \sec (e+f x)}}{a c^2 (c-d)^2 (c+d) (c+d \sec (e+f x))}-\frac {d^2 \sqrt {a-a \sec (e+f x)}}{2 a c \left (c^2-d^2\right ) (c+d \sec (e+f x))^2}-\frac {3 d^2 \sqrt {a-a \sec (e+f x)}}{4 a c (c-d) (c+d)^2 (c+d \sec (e+f x))}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a*((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c^3) + (Sqrt [2]*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[a]*(c - d)^ 3) - (3*d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(4*Sqrt[a]*c*(c - d)*(c + d)^(5/2)) - ((2*c - d)*d^(3/2)*ArcTanh[ (Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^2*(c - d)^2*(c + d)^(3/2)) - (2*d^(3/2)*(3*c^2 - 3*c*d + d^2)*ArcTanh[(Sqrt[d] *Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c^3*(c - d)^3* Sqrt[c + d]) - (d^2*Sqrt[a - a*Sec[e + f*x]])/(2*a*c*(c^2 - d^2)*(c + d*Se c[e + f*x])^2) - (3*d^2*Sqrt[a - a*Sec[e + f*x]])/(4*a*c*(c - d)*(c + d)^2 *(c + d*Sec[e + f*x])) - ((2*c - d)*d^2*Sqrt[a - a*Sec[e + f*x]])/(a*c^2*( c - d)^2*(c + d)*(c + d*Sec[e + f*x])))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.2.71.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_))^(q_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && IntegersQ[p, q]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[a^2*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d *x)^n/(x*Sqrt[a - b*x])), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0 ] && IntegerQ[m - 1/2]
Leaf count of result is larger than twice the leaf count of optimal. \(97276\) vs. \(2(564)=1128\).
Time = 18.67 (sec) , antiderivative size = 97277, normalized size of antiderivative = 148.97
Timed out. \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3} \, dx=\int \frac {1}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]
\[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3} \, dx=\int { \frac {1}{\sqrt {a \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}^{3}} \,d x } \]
Exception generated. \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^3} \, dx=\int \frac {1}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^3} \,d x \]