Integrand size = 27, antiderivative size = 271 \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx=\frac {2 a d (2 c+d) \left (2 c^2+2 c d+d^2\right ) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{3/2} c^4 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 d^2 \left (6 c^2+8 c d+3 d^2\right ) (a-a \sec (e+f x)) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}+\frac {2 d^3 (4 c+3 d) (a-a \sec (e+f x))^2 \tan (e+f x)}{5 a f \sqrt {a+a \sec (e+f x)}}-\frac {2 d^4 (a-a \sec (e+f x))^3 \tan (e+f x)}{7 a^2 f \sqrt {a+a \sec (e+f x)}} \]
2*a*d*(2*c+d)*(2*c^2+2*c*d+d^2)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)-2/3*d^ 2*(6*c^2+8*c*d+3*d^2)*(a-a*sec(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2) +2/5*d^3*(4*c+3*d)*(a-a*sec(f*x+e))^2*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(1/2 )-2/7*d^4*(a-a*sec(f*x+e))^3*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)+2*a^( 3/2)*c^4*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)
Time = 8.41 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx=\frac {2 \sqrt {a (1+\sec (e+f x))} (c+d \sec (e+f x))^4 \left (105 c^4 \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right ) \sqrt {\frac {\sec (e+f x)}{(1+\sec (e+f x))^2}} \sqrt {1+\sec (e+f x)}+d \sqrt {\sec (e+f x)} \left (420 c^3+420 c^2 d+224 c d^2+48 d^3+2 d \left (105 c^2+56 c d+12 d^2\right ) \sec (e+f x)+6 d^2 (14 c+3 d) \sec ^2(e+f x)+15 d^3 \sec ^3(e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right )}{105 f (d+c \cos (e+f x))^4 \sec ^{\frac {9}{2}}(e+f x)} \]
(2*Sqrt[a*(1 + Sec[e + f*x])]*(c + d*Sec[e + f*x])^4*(105*c^4*ArcTan[Tan[( e + f*x)/2]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]]*Sqrt[Sec[e + f*x]/(1 + Sec[e + f*x])^2]*Sqrt[1 + Sec[e + f*x]] + d*Sqrt[Sec[e + f*x]]*(420*c^3 + 420*c^2*d + 224*c*d^2 + 48*d^3 + 2*d*(105*c^2 + 56*c*d + 12*d^2)*Sec[e + f *x] + 6*d^2*(14*c + 3*d)*Sec[e + f*x]^2 + 15*d^3*Sec[e + f*x]^3)*Tan[(e + f*x)/2]))/(105*f*(d + c*Cos[e + f*x])^4*Sec[e + f*x]^(9/2))
Time = 0.41 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.78, 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.
| \(\displaystyle \int \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {\cos (e+f x) (c+d \sec (e+f x))^4}{\sqrt {a-a \sec (e+f x)}}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \left (\frac {\cos (e+f x) c^4}{\sqrt {a-a \sec (e+f x)}}-\frac {d^4 (a-a \sec (e+f x))^{5/2}}{a^3}+\frac {d^3 (4 c+3 d) (a-a \sec (e+f x))^{3/2}}{a^2}-\frac {d^2 \left (6 c^2+8 d c+3 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a}+\frac {d (2 c+d) \left (2 c^2+2 d c+d^2\right )}{\sqrt {a-a \sec (e+f x)}}\right )d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {2 d^4 (a-a \sec (e+f x))^{7/2}}{7 a^4}-\frac {2 d^3 (4 c+3 d) (a-a \sec (e+f x))^{5/2}}{5 a^3}+\frac {2 d^2 \left (6 c^2+8 c d+3 d^2\right ) (a-a \sec (e+f x))^{3/2}}{3 a^2}-\frac {2 c^4 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 d (2 c+d) \left (2 c^2+2 c d+d^2\right ) \sqrt {a-a \sec (e+f x)}}{a}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*((-2*c^4*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/Sqrt[a] - (2*d* (2*c + d)*(2*c^2 + 2*c*d + d^2)*Sqrt[a - a*Sec[e + f*x]])/a + (2*d^2*(6*c^ 2 + 8*c*d + 3*d^2)*(a - a*Sec[e + f*x])^(3/2))/(3*a^2) - (2*d^3*(4*c + 3*d )*(a - a*Sec[e + f*x])^(5/2))/(5*a^3) + (2*d^4*(a - a*Sec[e + f*x])^(7/2)) /(7*a^4))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x ]]))
3.2.47.3.1 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 = 6.83 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.11
| method | result | size |
| parts | \(\frac {2 c^{4} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )}{f}+\frac {2 d^{4} \left (16 \cos \left (f x +e \right )^{3}+8 \cos \left (f x +e \right )^{2}+6 \cos \left (f x +e \right )+5\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{35 f \left (\cos \left (f x +e \right )+1\right )}-\frac {8 c^{3} d \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}{f}+\frac {4 c^{2} d^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (2 \sin \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f \left (\cos \left (f x +e \right )+1\right )}+\frac {8 c \,d^{3} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (8 \sin \left (f x +e \right )+4 \tan \left (f x +e \right )+3 \sec \left (f x +e \right ) \tan \left (f x +e \right )\right )}{15 f \left (\cos \left (f x +e \right )+1\right )}\) | \(300\) |
| default | \(\frac {2 \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (105 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, c^{4} \cos \left (f x +e \right )+105 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, c^{4}+420 \sin \left (f x +e \right ) c^{3} d +420 \sin \left (f x +e \right ) c^{2} d^{2}+224 c \,d^{3} \sin \left (f x +e \right )+48 \sin \left (f x +e \right ) d^{4}+210 c^{2} d^{2} \tan \left (f x +e \right )+112 c \,d^{3} \tan \left (f x +e \right )+24 d^{4} \tan \left (f x +e \right )+84 c \,d^{3} \tan \left (f x +e \right ) \sec \left (f x +e \right )+18 d^{4} \tan \left (f x +e \right ) \sec \left (f x +e \right )+15 d^{4} \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}\right )}{105 f \left (\cos \left (f x +e \right )+1\right )}\) | \(301\) |
2*c^4/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))+2/35*d^4/f *(16*cos(f*x+e)^3+8*cos(f*x+e)^2+6*cos(f*x+e)+5)*(a*(sec(f*x+e)+1))^(1/2)/ (cos(f*x+e)+1)*tan(f*x+e)*sec(f*x+e)^2-8*c^3*d/f*(a*(sec(f*x+e)+1))^(1/2)* (cot(f*x+e)-csc(f*x+e))+4*c^2*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))+8/15*c*d^3/f*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x +e)+1)*(8*sin(f*x+e)+4*tan(f*x+e)+3*sec(f*x+e)*tan(f*x+e))
Time = 0.30 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.74 \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx=\left [\frac {105 \, {\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (15 \, d^{4} + 4 \, {\left (105 \, c^{3} d + 105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (14 \, c d^{3} + 3 \, d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{105 \, {\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}, -\frac {2 \, {\left (105 \, {\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (15 \, d^{4} + 4 \, {\left (105 \, c^{3} d + 105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (105 \, c^{2} d^{2} + 56 \, c d^{3} + 12 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (14 \, c d^{3} + 3 \, d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{105 \, {\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}\right ] \]
[1/105*(105*(c^4*cos(f*x + e)^4 + c^4*cos(f*x + e)^3)*sqrt(-a)*log((2*a*co s(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 2*(15*d^4 + 4*(105*c^3*d + 105*c^2*d^2 + 56*c*d^3 + 12*d^4)*cos(f*x + e)^3 + 2*(105*c^ 2*d^2 + 56*c*d^3 + 12*d^4)*cos(f*x + e)^2 + 6*(14*c*d^3 + 3*d^4)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e) ^4 + f*cos(f*x + e)^3), -2/105*(105*(c^4*cos(f*x + e)^4 + c^4*cos(f*x + e) ^3)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(s qrt(a)*sin(f*x + e))) - (15*d^4 + 4*(105*c^3*d + 105*c^2*d^2 + 56*c*d^3 + 12*d^4)*cos(f*x + e)^3 + 2*(105*c^2*d^2 + 56*c*d^3 + 12*d^4)*cos(f*x + e)^ 2 + 6*(14*c*d^3 + 3*d^4)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 + f*cos(f*x + e)^3)]
\[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx=\int \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )^{4}\, dx \]
\[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx=\int { \sqrt {a \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}^{4} \,d x } \]
-1/210*(16*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(7*(15*c^3*d*sin(6*f*x + 6*e) + 5*(9*c^3*d + 3*c^2*d^2 + 4*c*d^3 )*sin(4*f*x + 4*e) + (45*c^3*d + 30*c^2*d^2 + 28*c*d^3 + 6*d^4)*sin(2*f*x + 2*e))*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - (105*c^ 3*d*cos(6*f*x + 6*e) + 105*c^3*d + 105*c^2*d^2 + 56*c*d^3 + 12*d^4 + 35*(9 *c^3*d + 3*c^2*d^2 + 4*c*d^3)*cos(4*f*x + 4*e) + 7*(45*c^3*d + 30*c^2*d^2 + 28*c*d^3 + 6*d^4)*cos(2*f*x + 2*e))*sin(7/2*arctan2(sin(2*f*x + 2*e), co s(2*f*x + 2*e) + 1)))*sqrt(a) + 105*((c^4*cos(2*f*x + 2*e)^4 + c^4*sin(2*f *x + 2*e)^4 + 4*c^4*cos(2*f*x + 2*e)^3 + 6*c^4*cos(2*f*x + 2*e)^2 + 4*c^4* cos(2*f*x + 2*e) + c^4 + 2*(c^4*cos(2*f*x + 2*e)^2 + 2*c^4*cos(2*f*x + 2*e ) + c^4)*sin(2*f*x + 2*e)^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) - (c^4*cos(2*f*x + 2*e)^4 + c^4*sin(2*f*x + 2*e)^4 + 4*c^4*cos(2*f* x + 2*e)^3 + 6*c^4*cos(2*f*x + 2*e)^2 + 4*c^4*cos(2*f*x + 2*e) + c^4 + 2*( c^4*cos(2*f*x + 2*e)^2 + 2*c^4*cos(2*f*x + 2*e) + c^4)*sin(2*f*x + 2*e)^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/...
\[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx=\int { \sqrt {a \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}^{4} \,d x } \]
Timed out. \[ \int \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^4 \, dx=\int \sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^4 \,d x \]