Integrand size = 27, antiderivative size = 802 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^3} \, dx=-\frac {\tan (e+f x)}{2 a (c-d)^3 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} (c-4 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} (c-d)^4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} \sqrt {a} (c-d)^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {3 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{4 \sqrt {a} c (c-d)^2 (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {(3 c-d) d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c^2 (c-d)^3 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 d^{5/2} \left (6 c^2-4 c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c^3 (c-d)^4 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {d^3 \tan (e+f x)}{2 a c (c-d)^2 (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {(3 c-d) d^3 \tan (e+f x)}{a c^2 (c-d)^3 (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {3 d^3 \tan (e+f x)}{4 a c \left (c^2-d^2\right )^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
-1/2*tan(f*x+e)/a/(c-d)^3/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)-1/2*d^3* tan(f*x+e)/a/c/(c-d)^2/(c+d)/f/(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/2)-( 3*c-d)*d^3*tan(f*x+e)/a/c^2/(c-d)^3/(c+d)/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+ e))^(1/2)-3/4*d^3*tan(f*x+e)/a/c/(c^2-d^2)^2/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))*tan(f*x+e)/c^3/f/a^ (1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-3/4*d^(5/2)*arctanh(d^ (1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/c/(c-d)^2/(c+ d)^(5/2)/f/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-(3*c-d)*d ^(5/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x +e)/c^2/(c-d)^3/(c+d)^(3/2)/f/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+ e))^(1/2)-1/4*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*tan(f*x+ e)/(c-d)^3/f*2^(1/2)/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2) -(c-4*d)*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*tan(f *x+e)/(c-d)^4/f/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-2*d^ (5/2)*(6*c^2-4*c*d+d^2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+ d)^(1/2))*tan(f*x+e)/c^3/(c-d)^4/f/a^(1/2)/(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. \(2632\) vs. \(2(802)=1604\).
Time = 18.76 (sec) , antiderivative size = 2632, normalized size of antiderivative = 3.28 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^3} \, dx=\text {Result too large to show} \]
(Cos[(e + f*x)/2]^3*(d + c*Cos[e + f*x])^3*Sec[e + f*x]^5*(-(((-2*c^5 - 4* c^4*d - 2*c^3*d^2 - 17*c^2*d^3 - 5*c*d^4 + 6*d^5)*Sin[(e + f*x)/2])/(c^3*( -c + d)^3*(c + d)^2)) - (2*d^5*Sin[(e + f*x)/2])/(c^3*(-c + d)^2*(c + d)*( d + c*Cos[e + f*x])^2) + (-19*c^2*d^4*Sin[(e + f*x)/2] - 5*c*d^5*Sin[(e + f*x)/2] + 8*d^6*Sin[(e + f*x)/2])/(c^3*(-c + d)^3*(c + d)^2*(d + c*Cos[e + f*x])) - (Sec[(e + f*x)/2]*Tan[(e + f*x)/2])/(-c + d)^3))/(f*(a*(1 + Sec[ e + f*x]))^(3/2)*(c + d*Sec[e + f*x])^3) - ((2*c^3*(5*c - 17*d)*(c + d)^2* ArcSin[Tan[(e + f*x)/2]] - 8*Sqrt[2]*(c - d)^4*(c + d)^2*ArcTan[Tan[(e + f *x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]] - (Sqrt[2]*d^(5/2)*(63*c^4 + 54*c^3*d - 17*c^2*d^2 - 12*c*d^3 + 8*d^4)*ArcTanh[(Sqrt[d]*Tan[(e + f*x)/ 2])/(Sqrt[-c - d]*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])])])/Sqrt[-c - d])*C os[(e + f*x)/2]^3*(d + c*Cos[e + f*x])^3*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2 ]^2]*((c^3*Sec[(e + f*x)/2])/(2*(-c + d)^3*(c + d)^2*(d + c*Cos[e + f*x])* Sqrt[Sec[e + f*x]]) - (c^2*d*Sec[(e + f*x)/2])/((-c + d)^3*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]]) - (19*c*d^2*Sec[(e + f*x)/2])/(2*(-c + d)^3*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]]) - (33*d^3*Sec[(e + f*x)/2])/(4*(-c + d)^3*(c + d)^2*(d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]] ) - (3*d^4*Sec[(e + f*x)/2])/(4*c*(-c + d)^3*(c + d)^2*(d + c*Cos[e + f*x] )*Sqrt[Sec[e + f*x]]) + (d^5*Sec[(e + f*x)/2])/(c^2*(-c + d)^3*(c + d)^2*( d + c*Cos[e + f*x])*Sqrt[Sec[e + f*x]]) - (c^3*Sec[(e + f*x)/2]*Sqrt[Se...
Time = 0.86 (sec) , antiderivative size = 579, normalized size of antiderivative = 0.72, 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.
\(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^{3/2} (c+d \sec (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4428 |
\(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x)}{a^2 (\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 198 |
\(\displaystyle -\frac {\tan (e+f x) \int \left (-\frac {\left (6 c^2-4 d c+d^2\right ) d^3}{c^3 (c-d)^4 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {(3 c-d) d^3}{c^2 (c-d)^3 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}-\frac {d^3}{c (c-d)^2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}+\frac {\cos (e+f x)}{c^3 \sqrt {a-a \sec (e+f x)}}+\frac {4 d-c}{(c-d)^4 (\sec (e+f x)+1) \sqrt {a-a \sec (e+f x)}}-\frac {1}{(c-d)^3 (\sec (e+f x)+1)^2 \sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\tan (e+f x) \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c^3}+\frac {d^{5/2} (3 c-d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^2 (c-d)^3 (c+d)^{3/2}}+\frac {2 d^{5/2} \left (6 c^2-4 c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c^3 (c-d)^4 \sqrt {c+d}}+\frac {3 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{4 \sqrt {a} c (c-d)^2 (c+d)^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} \sqrt {a} (c-d)^3}+\frac {\sqrt {2} (c-4 d) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a} (c-d)^4}+\frac {d^3 (3 c-d) \sqrt {a-a \sec (e+f x)}}{a c^2 (c-d)^3 (c+d) (c+d \sec (e+f x))}+\frac {3 d^3 \sqrt {a-a \sec (e+f x)}}{4 a c \left (c^2-d^2\right )^2 (c+d \sec (e+f x))}+\frac {d^3 \sqrt {a-a \sec (e+f x)}}{2 a c (c-d)^2 (c+d) (c+d \sec (e+f x))^2}+\frac {\sqrt {a-a \sec (e+f x)}}{2 a (c-d)^3 (\sec (e+f x)+1)}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((((-2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/(Sqrt[a]*c^3) + (Sqrt[2 ]*(c - 4*d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[a]* (c - d)^4) + ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2 ]*Sqrt[a]*(c - d)^3) + (3*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x] ])/(Sqrt[a]*Sqrt[c + d])])/(4*Sqrt[a]*c*(c - d)^2*(c + d)^(5/2)) + ((3*c - d)*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d ])])/(Sqrt[a]*c^2*(c - d)^3*(c + d)^(3/2)) + (2*d^(5/2)*(6*c^2 - 4*c*d + d ^2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sq rt[a]*c^3*(c - d)^4*Sqrt[c + d]) + Sqrt[a - a*Sec[e + f*x]]/(2*a*(c - d)^3 *(1 + Sec[e + f*x])) + (d^3*Sqrt[a - a*Sec[e + f*x]])/(2*a*c*(c - d)^2*(c + d)*(c + d*Sec[e + f*x])^2) + ((3*c - d)*d^3*Sqrt[a - a*Sec[e + f*x]])/(a *c^2*(c - d)^3*(c + d)*(c + d*Sec[e + f*x])) + (3*d^3*Sqrt[a - a*Sec[e + f *x]])/(4*a*c*(c^2 - d^2)^2*(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]]))
3.2.77.3.1 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. \(117746\) vs. \(2(694)=1388\).
Time = 21.10 (sec) , antiderivative size = 117747, normalized size of antiderivative = 146.82
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^3} \, dx=\int \frac {1}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^3} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^3} \, dx=\text {Hanged} \]