Integrand size = 27, antiderivative size = 144 \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2 \, dx=\frac {2 a d (2 c+d) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{3/2} c^2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 d^2 (a-a \sec (e+f x)) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}} \]
2*a*d*(2*c+d)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)-2/3*d^2*(a-a*sec(f*x+e)) *tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2*a^(3/2)*c^2*arctanh((a-a*sec(f*x+e) )^(1/2)/a^(1/2))*tan(f*x+e)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2 )
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.98 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.08 \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2 \, dx=\frac {\csc ^3\left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (1+\sec (e+f x))} (c+d \sec (e+f x))^2 \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )}} \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )} \left (256 \, _3F_2\left (\frac {3}{2},2,\frac {7}{2};1,\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin ^6\left (\frac {1}{2} (e+f x)\right ) \left (c+d-2 c \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )^2+1024 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin ^6\left (\frac {1}{2} (e+f x)\right ) \left (d^2+c d \left (2-3 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )+c^2 \left (1-3 \sin ^2\left (\frac {1}{2} (e+f x)\right )+2 \sin ^4\left (\frac {1}{2} (e+f x)\right )\right )\right )-\frac {7 \sqrt {2} \left (-3 \arcsin \left (\sqrt {2} \sqrt {\sin ^2\left (\frac {1}{2} (e+f x)\right )}\right )+\sqrt {2} \sqrt {\sin ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )} \left (3+4 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \left (15 d^2+10 c d \left (3-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )+c^2 \left (15-20 \sin ^2\left (\frac {1}{2} (e+f x)\right )+12 \sin ^4\left (\frac {1}{2} (e+f x)\right )\right )\right )}{\sqrt {\sin ^2\left (\frac {1}{2} (e+f x)\right )}}\right )}{672 f (d+c \cos (e+f x))^2 \sec ^{\frac {5}{2}}(e+f x)} \]
(Csc[(e + f*x)/2]^3*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])]*(c + d*Sec [e + f*x])^2*Sqrt[(1 - 2*Sin[(e + f*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[(e + f*x )/2]^2]*(256*HypergeometricPFQ[{3/2, 2, 7/2}, {1, 9/2}, 2*Sin[(e + f*x)/2] ^2]*Sin[(e + f*x)/2]^6*(c + d - 2*c*Sin[(e + f*x)/2]^2)^2 + 1024*Hypergeom etric2F1[3/2, 7/2, 9/2, 2*Sin[(e + f*x)/2]^2]*Sin[(e + f*x)/2]^6*(d^2 + c* d*(2 - 3*Sin[(e + f*x)/2]^2) + c^2*(1 - 3*Sin[(e + f*x)/2]^2 + 2*Sin[(e + f*x)/2]^4)) - (7*Sqrt[2]*(-3*ArcSin[Sqrt[2]*Sqrt[Sin[(e + f*x)/2]^2]] + Sq rt[2]*Sqrt[Sin[(e + f*x)/2]^2]*Sqrt[1 - 2*Sin[(e + f*x)/2]^2]*(3 + 4*Sin[( e + f*x)/2]^2))*(15*d^2 + 10*c*d*(3 - 2*Sin[(e + f*x)/2]^2) + c^2*(15 - 20 *Sin[(e + f*x)/2]^2 + 12*Sin[(e + f*x)/2]^4)))/Sqrt[Sin[(e + f*x)/2]^2]))/ (672*f*(d + c*Cos[e + f*x])^2*Sec[e + f*x]^(5/2))
Time = 0.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 4428, 99, 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.
| \(\displaystyle \int \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x) (c+d \sec (e+f x))^2}{\sqrt {a-a \sec (e+f x)}}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \left (\frac {\cos (e+f x) c^2}{\sqrt {a-a \sec (e+f x)}}-\frac {d^2 \sqrt {a-a \sec (e+f x)}}{a}+\frac {d (2 c+d)}{\sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {2 d^2 (a-a \sec (e+f x))^{3/2}}{3 a^2}-\frac {2 c^2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 d (2 c+d) \sqrt {a-a \sec (e+f x)}}{a}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*((-2*c^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/Sqrt[a] - (2*d* (2*c + d)*Sqrt[a - a*Sec[e + f*x]])/a + (2*d^2*(a - a*Sec[e + f*x])^(3/2)) /(3*a^2))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x ]]))
3.2.49.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]
Time = 4.58 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.10
| method | result | size |
| parts | \(\frac {2 c^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )}{f}+\frac {2 d^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (2 \sin \left (f x +e \right )+\tan \left (f x +e \right )\right )}{3 f \left (\cos \left (f x +e \right )+1\right )}-\frac {4 c d \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}{f}\) | \(159\) |
| default | \(\frac {2 \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (3 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) c^{2} \cos \left (f x +e \right )+3 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) c^{2}+6 \sin \left (f x +e \right ) c d +2 \sin \left (f x +e \right ) d^{2}+d^{2} \tan \left (f x +e \right )\right )}{3 f \left (\cos \left (f x +e \right )+1\right )}\) | \(192\) |
2*c^2/f*(a*(sec(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arctan h(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))+2/3*d^2/f* (a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)+1)*(2*sin(f*x+e)+tan(f*x+e))-4*c*d/f* (a*(sec(f*x+e)+1))^(1/2)*(cot(f*x+e)-csc(f*x+e))
Time = 0.28 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.22 \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2 \, dx=\left [\frac {3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (d^{2} + 2 \, {\left (3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (d^{2} + 2 \, {\left (3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}\right ] \]
[1/3*(3*(c^2*cos(f*x + e)^2 + c^2*cos(f*x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*s in(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 2*(d^2 + 2*(3*c*d + d^2)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e)) /(f*cos(f*x + e)^2 + f*cos(f*x + e)), -2/3*(3*(c^2*cos(f*x + e)^2 + c^2*co s(f*x + e))*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (d^2 + 2*(3*c*d + d^2)*cos(f*x + e))*sqrt( (a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^2 + f*cos (f*x + e))]
\[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2 \, dx=\int \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )^{2}\, dx \]
\[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2 \, dx=\int { \sqrt {a \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}^{2} \,d x } \]
-1/6*(8*(3*c*d*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*si n(2*f*x + 2*e) - (3*c*d*cos(2*f*x + 2*e) + 3*c*d + d^2)*sin(3/2*arctan2(si n(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sqrt(a) + 3*((c^2*cos(2*f*x + 2*e)^ 2 + c^2*sin(2*f*x + 2*e)^2 + 2*c^2*cos(2*f*x + 2*e) + c^2)*arctan2((cos(2* f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2* arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + si n(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1) - (c^2*cos(2*f*x + 2*e)^2 + c^2*sin(2 *f*x + 2*e)^2 + 2*c^2*cos(2*f*x + 2*e) + c^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(sin(2 *f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e )^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos( 2*f*x + 2*e) + 1)) - 1) - 2*(c^2*f*cos(2*f*x + 2*e)^2 + c^2*f*sin(2*f*x + 2*e)^2 + 2*c^2*f*cos(2*f*x + 2*e) + c^2*f)*integrate((((cos(6*f*x + 6*e)*c os(2*f*x + 2*e) + 2*cos(4*f*x + 4*e)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + sin(6*f*x + 6*e)*sin(2*f*x + 2*e) + 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + sin(2*f*x + 2*e)^2)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)) ) + (cos(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4*f*x + 4* e) - cos(6*f*x + 6*e)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e)*sin(2*f*x +...
\[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2 \, dx=\int { \sqrt {a \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}^{2} \,d x } \]
Timed out. \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2 \, dx=\int \sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \]