Integrand size = 31, antiderivative size = 228 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx=\frac {d^2 (3 b c-2 a d) \text {arctanh}(\sin (e+f x))}{b^3 f}+\frac {2 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a (a-b)^{3/2} b (a+b)^{3/2} f}+\frac {2 (b c-a d)^2 (b c+2 a d) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b^3 \sqrt {a+b} f}-\frac {(b c-a d)^3 \sin (e+f x)}{b^2 \left (a^2-b^2\right ) f (b+a \cos (e+f x))}+\frac {d^3 \tan (e+f x)}{b^2 f} \]
d^2*(-2*a*d+3*b*c)*arctanh(sin(f*x+e))/b^3/f+2*(-a*d+b*c)^3*arctanh((a-b)^ (1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a/(a-b)^(3/2)/b/(a+b)^(3/2)/f-(-a*d+ b*c)^3*sin(f*x+e)/b^2/(a^2-b^2)/f/(b+a*cos(f*x+e))+2*(-a*d+b*c)^2*(2*a*d+b *c)*arctanh((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a/b^3/f/(a-b)^(1/2 )/(a+b)^(1/2)+d^3*tan(f*x+e)/b^2/f
Time = 3.41 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.59 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx=\frac {\cos (e+f x) (b+a \cos (e+f x)) (c+d \sec (e+f x))^3 \left (-\frac {2 (b c-a d)^2 \left (a b c+2 a^2 d-3 b^2 d\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (e+f x))}{\left (a^2-b^2\right )^{3/2}}+d^2 (-3 b c+2 a d) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+d^2 (3 b c-2 a d) (b+a \cos (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {b d^3 (b+a \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {b d^3 (b+a \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {b (b c-a d)^3 \sin (e+f x)}{(-a+b) (a+b)}\right )}{b^3 f (d+c \cos (e+f x))^3 (a+b \sec (e+f x))^2} \]
(Cos[e + f*x]*(b + a*Cos[e + f*x])*(c + d*Sec[e + f*x])^3*((-2*(b*c - a*d) ^2*(a*b*c + 2*a^2*d - 3*b^2*d)*ArcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqrt[a^ 2 - b^2]]*(b + a*Cos[e + f*x]))/(a^2 - b^2)^(3/2) + d^2*(-3*b*c + 2*a*d)*( b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + d^2*(3*b*c - 2*a*d)*(b + a*Cos[e + f*x])*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + ( b*d^3*(b + a*Cos[e + f*x])*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) + (b*d^3*(b + a*Cos[e + f*x])*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (b*(b*c - a*d)^3*Sin[e + f*x])/((-a + b)*(a + b)))) /(b^3*f*(d + c*Cos[e + f*x])^3*(a + b*Sec[e + f*x])^2)
Time = 0.72 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 4476, 3042, 3431, 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.
\(\displaystyle \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}{\left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4476 |
\(\displaystyle \int \frac {\sec ^2(e+f x) (c \cos (e+f x)+d)^3}{(a \cos (e+f x)+b)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (c \sin \left (e+f x+\frac {\pi }{2}\right )+d\right )^3}{\sin \left (e+f x+\frac {\pi }{2}\right )^2 \left (a \sin \left (e+f x+\frac {\pi }{2}\right )+b\right )^2}dx\) |
\(\Big \downarrow \) 3431 |
\(\displaystyle \int \left (\frac {d^2 (3 b c-2 a d) \sec (e+f x)}{b^3}+\frac {(a d-b c)^2 (2 a d+b c)}{a b^3 (a \cos (e+f x)+b)}+\frac {(a d-b c)^3}{a b^2 (a \cos (e+f x)+b)^2}+\frac {d^3 \sec ^2(e+f x)}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(b c-a d)^3 \sin (e+f x)}{b^2 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac {d^2 (3 b c-2 a d) \text {arctanh}(\sin (e+f x))}{b^3 f}+\frac {2 (b c-a d)^2 (2 a d+b c) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b^3 f \sqrt {a-b} \sqrt {a+b}}+\frac {2 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a b f (a-b)^{3/2} (a+b)^{3/2}}+\frac {d^3 \tan (e+f x)}{b^2 f}\) |
(d^2*(3*b*c - 2*a*d)*ArcTanh[Sin[e + f*x]])/(b^3*f) + (2*(b*c - a*d)^3*Arc Tanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a*(a - b)^(3/2)*b*(a + b)^(3/2)*f) + (2*(b*c - a*d)^2*(b*c + 2*a*d)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*b^3*Sqrt[a + b]*f) - ((b*c - a*d)^3 *Sin[e + f*x])/(b^2*(a^2 - b^2)*f*(b + a*Cos[e + f*x])) + (d^3*Tan[e + f*x ])/(b^2*f)
3.3.60.3.1 Defintions of rubi rules used
Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[Exp andTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])^n, x ], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (Int egersQ[m, n] || IntegersQ[m, p] || IntegersQ[n, p]) && NeQ[p, 2]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[1 /g^(m + n) Int[(g*Csc[e + f*x])^(m + n + p)*(b + a*Sin[e + f*x])^m*(d + c *Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && IntegerQ[m] && IntegerQ[n]
Time = 1.47 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.38
method | result | size |
derivativedivides | \(\frac {-\frac {d^{3}}{b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {d^{2} \left (2 a d -3 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{b^{3}}-\frac {2 \left (\frac {b \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 a^{4} d^{3}-3 a^{3} b c \,d^{2}-3 a^{2} b^{2} d^{3}+b^{3} c^{3} a +6 a \,b^{3} c \,d^{2}-3 c^{2} d \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}-\frac {d^{3}}{b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {d^{2} \left (2 a d -3 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{b^{3}}}{f}\) | \(314\) |
default | \(\frac {-\frac {d^{3}}{b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {d^{2} \left (2 a d -3 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{b^{3}}-\frac {2 \left (\frac {b \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 a^{4} d^{3}-3 a^{3} b c \,d^{2}-3 a^{2} b^{2} d^{3}+b^{3} c^{3} a +6 a \,b^{3} c \,d^{2}-3 c^{2} d \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}-\frac {d^{3}}{b^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {d^{2} \left (2 a d -3 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{b^{3}}}{f}\) | \(314\) |
risch | \(\text {Expression too large to display}\) | \(1533\) |
1/f*(-d^3/b^2/(tan(1/2*f*x+1/2*e)-1)+d^2*(2*a*d-3*b*c)/b^3*ln(tan(1/2*f*x+ 1/2*e)-1)-2/b^3*(b*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/(a^2-b^2) *tan(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)-(2 *a^4*d^3-3*a^3*b*c*d^2-3*a^2*b^2*d^3+a*b^3*c^3+6*a*b^3*c*d^2-3*b^4*c^2*d)/ (a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a-b)*(a +b))^(1/2)))-d^3/b^2/(tan(1/2*f*x+1/2*e)+1)-d^2*(2*a*d-3*b*c)/b^3*ln(tan(1 /2*f*x+1/2*e)+1))
Leaf count of result is larger than twice the leaf count of optimal. 635 vs. \(2 (210) = 420\).
Time = 39.31 (sec) , antiderivative size = 1326, normalized size of antiderivative = 5.82 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \]
[1/2*(((a^2*b^3*c^3 - 3*a*b^4*c^2*d - 3*(a^4*b - 2*a^2*b^3)*c*d^2 + (2*a^5 - 3*a^3*b^2)*d^3)*cos(f*x + e)^2 + (a*b^4*c^3 - 3*b^5*c^2*d - 3*(a^3*b^2 - 2*a*b^4)*c*d^2 + (2*a^4*b - 3*a^2*b^3)*d^3)*cos(f*x + e))*sqrt(a^2 - b^2 )*log((2*a*b*cos(f*x + e) - (a^2 - 2*b^2)*cos(f*x + e)^2 + 2*sqrt(a^2 - b^ 2)*(b*cos(f*x + e) + a)*sin(f*x + e) + 2*a^2 - b^2)/(a^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + b^2)) + ((3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2*(a^ 6 - 2*a^4*b^2 + a^2*b^4)*d^3)*cos(f*x + e)^2 + (3*(a^4*b^2 - 2*a^2*b^4 + b ^6)*c*d^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d^3)*cos(f*x + e))*log(sin(f*x + e) + 1) - ((3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2*(a^6 - 2*a^4*b^2 + a^ 2*b^4)*d^3)*cos(f*x + e)^2 + (3*(a^4*b^2 - 2*a^2*b^4 + b^6)*c*d^2 - 2*(a^5 *b - 2*a^3*b^3 + a*b^5)*d^3)*cos(f*x + e))*log(-sin(f*x + e) + 1) + 2*((a^ 4*b^2 - 2*a^2*b^4 + b^6)*d^3 - ((a^2*b^4 - b^6)*c^3 - 3*(a^3*b^3 - a*b^5)* c^2*d + 3*(a^4*b^2 - a^2*b^4)*c*d^2 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*d^3)*c os(f*x + e))*sin(f*x + e))/((a^5*b^3 - 2*a^3*b^5 + a*b^7)*f*cos(f*x + e)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*f*cos(f*x + e)), 1/2*(2*((a^2*b^3*c^3 - 3*a *b^4*c^2*d - 3*(a^4*b - 2*a^2*b^3)*c*d^2 + (2*a^5 - 3*a^3*b^2)*d^3)*cos(f* x + e)^2 + (a*b^4*c^3 - 3*b^5*c^2*d - 3*(a^3*b^2 - 2*a*b^4)*c*d^2 + (2*a^4 *b - 3*a^2*b^3)*d^3)*cos(f*x + e))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^ 2)*(b*cos(f*x + e) + a)/((a^2 - b^2)*sin(f*x + e))) + ((3*(a^5*b - 2*a^3*b ^3 + a*b^5)*c*d^2 - 2*(a^6 - 2*a^4*b^2 + a^2*b^4)*d^3)*cos(f*x + e)^2 +...
\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx=\int \frac {\left (c + d \sec {\left (e + f x \right )}\right )^{3} \sec {\left (e + f x \right )}}{\left (a + b \sec {\left (e + f x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 539 vs. \(2 (210) = 420\).
Time = 0.38 (sec) , antiderivative size = 539, normalized size of antiderivative = 2.36 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx=-\frac {\frac {2 \, {\left (a b^{3} c^{3} - 3 \, b^{4} c^{2} d - 3 \, a^{3} b c d^{2} + 6 \, a b^{3} c d^{2} + 2 \, a^{4} d^{3} - 3 \, a^{2} b^{2} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, {\left (b^{3} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a b^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a^{2} b c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{2} b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a b^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - b^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - b^{3} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a b^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, a^{2} b c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{2} b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a b^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {{\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{b^{3}} + \frac {{\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{b^{3}}}{f} \]
-(2*(a*b^3*c^3 - 3*b^4*c^2*d - 3*a^3*b*c*d^2 + 6*a*b^3*c*d^2 + 2*a^4*d^3 - 3*a^2*b^2*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*a - 2*b) + arctan( (a*tan(1/2*f*x + 1/2*e) - b*tan(1/2*f*x + 1/2*e))/sqrt(-a^2 + b^2)))/((a^2 *b^3 - b^5)*sqrt(-a^2 + b^2)) - 2*(b^3*c^3*tan(1/2*f*x + 1/2*e)^3 - 3*a*b^ 2*c^2*d*tan(1/2*f*x + 1/2*e)^3 + 3*a^2*b*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 2* a^3*d^3*tan(1/2*f*x + 1/2*e)^3 + a^2*b*d^3*tan(1/2*f*x + 1/2*e)^3 + a*b^2* d^3*tan(1/2*f*x + 1/2*e)^3 - b^3*d^3*tan(1/2*f*x + 1/2*e)^3 - b^3*c^3*tan( 1/2*f*x + 1/2*e) + 3*a*b^2*c^2*d*tan(1/2*f*x + 1/2*e) - 3*a^2*b*c*d^2*tan( 1/2*f*x + 1/2*e) + 2*a^3*d^3*tan(1/2*f*x + 1/2*e) + a^2*b*d^3*tan(1/2*f*x + 1/2*e) - a*b^2*d^3*tan(1/2*f*x + 1/2*e) - b^3*d^3*tan(1/2*f*x + 1/2*e))/ ((a*tan(1/2*f*x + 1/2*e)^4 - b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + a + b)*(a^2*b^2 - b^4)) - (3*b*c*d^2 - 2*a*d^3)*log(abs(tan(1/2 *f*x + 1/2*e) + 1))/b^3 + (3*b*c*d^2 - 2*a*d^3)*log(abs(tan(1/2*f*x + 1/2* e) - 1))/b^3)/f
Time = 22.92 (sec) , antiderivative size = 7958, normalized size of antiderivative = 34.90 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \]
(d^2*atan(((d^2*((32*tan(e/2 + (f*x)/2)*(8*a^8*d^6 - 8*a^7*b*d^6 + a^2*b^6 *c^6 + 4*a^2*b^6*d^6 - 8*a^3*b^5*d^6 + 5*a^4*b^4*d^6 + 16*a^5*b^3*d^6 - 16 *a^6*b^2*d^6 + 9*b^8*c^2*d^4 + 9*b^8*c^4*d^2 - 18*a*b^7*c^2*d^4 - 36*a*b^7 *c^3*d^3 + 24*a^2*b^6*c*d^5 - 24*a^3*b^5*c*d^5 - 48*a^4*b^4*c*d^5 + 54*a^5 *b^3*c*d^5 + 24*a^6*b^2*c*d^5 + 45*a^2*b^6*c^2*d^4 + 12*a^2*b^6*c^4*d^2 + 36*a^3*b^5*c^2*d^4 + 12*a^3*b^5*c^3*d^3 - 57*a^4*b^4*c^2*d^4 - 6*a^4*b^4*c ^4*d^2 - 18*a^5*b^3*c^2*d^4 + 4*a^5*b^3*c^3*d^3 + 18*a^6*b^2*c^2*d^4 - 12* a*b^7*c*d^5 - 6*a*b^7*c^5*d - 24*a^7*b*c*d^5))/(a*b^6 + b^7 - a^2*b^5 - a^ 3*b^4) + (d^2*((32*(a*b^11*c^3 + 2*a*b^11*d^3 - 3*b^12*c*d^2 - 3*b^12*c^2* d - a^2*b^10*c^3 - a^3*b^9*c^3 + a^4*b^8*c^3 - 3*a^2*b^10*d^3 - 3*a^3*b^9* d^3 + 5*a^4*b^8*d^3 + a^5*b^7*d^3 - 2*a^6*b^6*d^3 + 3*a^2*b^10*c*d^2 + 3*a ^2*b^10*c^2*d - 9*a^3*b^9*c*d^2 - 3*a^3*b^9*c^2*d + 3*a^5*b^7*c*d^2 + 6*a* b^11*c*d^2 + 3*a*b^11*c^2*d))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*d^2* tan(e/2 + (f*x)/2)*(2*a*d - 3*b*c)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4* a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4))) *(2*a*d - 3*b*c))/b^3)*(2*a*d - 3*b*c)*1i)/b^3 + (d^2*((32*tan(e/2 + (f*x) /2)*(8*a^8*d^6 - 8*a^7*b*d^6 + a^2*b^6*c^6 + 4*a^2*b^6*d^6 - 8*a^3*b^5*d^6 + 5*a^4*b^4*d^6 + 16*a^5*b^3*d^6 - 16*a^6*b^2*d^6 + 9*b^8*c^2*d^4 + 9*b^8 *c^4*d^2 - 18*a*b^7*c^2*d^4 - 36*a*b^7*c^3*d^3 + 24*a^2*b^6*c*d^5 - 24*a^3 *b^5*c*d^5 - 48*a^4*b^4*c*d^5 + 54*a^5*b^3*c*d^5 + 24*a^6*b^2*c*d^5 + 4...