Integrand size = 31, antiderivative size = 153 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx=\frac {a^3 \text {arctanh}(\sin (e+f x))}{2 d f}+\frac {a^3 \left (c^2-3 c d+3 d^2\right ) \text {arctanh}(\sin (e+f x))}{d^3 f}-\frac {2 a^3 (c-d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{d^3 \sqrt {c+d} f}-\frac {a^3 (c-3 d) \tan (e+f x)}{d^2 f}+\frac {a^3 \sec (e+f x) \tan (e+f x)}{2 d f} \]
1/2*a^3*arctanh(sin(f*x+e))/d/f+a^3*(c^2-3*c*d+3*d^2)*arctanh(sin(f*x+e))/ d^3/f-2*a^3*(c-d)^(5/2)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2) )/d^3/f/(c+d)^(1/2)-a^3*(c-3*d)*tan(f*x+e)/d^2/f+1/2*a^3*sec(f*x+e)*tan(f* x+e)/d/f
Result contains complex when optimal does not.
Time = 3.38 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.74 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx=\frac {a^3 \cos ^2(e+f x) (d+c \cos (e+f x)) \sec ^6\left (\frac {1}{2} (e+f x)\right ) (1+\sec (e+f x))^3 \left (-2 \left (2 c^2-6 c d+7 d^2\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 \left (2 c^2-6 c d+7 d^2\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {8 (c-d)^3 \arctan \left (\frac {(i \cos (e)+\sin (e)) \left (c \sin (e)+(-d+c \cos (e)) \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right ) (i \cos (e)+\sin (e))}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}+\frac {d^2}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {4 (c-3 d) d \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}-\frac {d^2}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {4 (c-3 d) d \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}\right )}{32 d^3 f (c+d \sec (e+f x))} \]
(a^3*Cos[e + f*x]^2*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^6*(1 + Sec[e + f *x])^3*(-2*(2*c^2 - 6*c*d + 7*d^2)*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2] ] + 2*(2*c^2 - 6*c*d + 7*d^2)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + ( 8*(c - d)^3*ArcTan[((I*Cos[e] + Sin[e])*(c*Sin[e] + (-d + c*Cos[e])*Tan[(f *x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(I*Cos[e] + Sin[e] ))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2]) + d^2/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 - (4*(c - 3*d)*d*Sin[(f*x)/2])/((Cos[e/2] - Sin[e/2]) *(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])) - d^2/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (4*(c - 3*d)*d*Sin[(f*x)/2])/((Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))))/(32*d^3*f*(c + d*Sec[e + f*x]))
Time = 0.49 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.73, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4475, 113, 25, 27, 171, 25, 27, 175, 45, 104, 218}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^3}{c+d \sec (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}{c+d \csc \left (e+f x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4475 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {(\sec (e+f x) a+a)^{5/2}}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 113 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (-\frac {\int -\frac {a^3 \sqrt {\sec (e+f x) a+a} (c+2 d-(2 c-5 d) \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{2 a d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {a^3 \sqrt {\sec (e+f x) a+a} (c+2 d-(2 c-5 d) \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{2 a d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \int \frac {\sqrt {\sec (e+f x) a+a} (c+2 d-(2 c-5 d) \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \left (\frac {(2 c-5 d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}{a d}-\frac {\int -\frac {a^2 \left (d (c+2 d)+\left (2 c^2-6 d c+7 d^2\right ) \sec (e+f x)\right )}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{a d}\right )}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \left (\frac {\int \frac {a^2 \left (d (c+2 d)+\left (2 c^2-6 d c+7 d^2\right ) \sec (e+f x)\right )}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{a d}+\frac {(2 c-5 d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}{a d}\right )}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \left (\frac {a \int \frac {d (c+2 d)+\left (2 c^2-6 d c+7 d^2\right ) \sec (e+f x)}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{d}+\frac {(2 c-5 d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}{a d}\right )}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \left (\frac {a \left (\frac {\left (2 c^2-6 c d+7 d^2\right ) \int \frac {1}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a}}d\sec (e+f x)}{d}-\frac {2 (c-d)^3 \int \frac {1}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{d}\right )}{d}+\frac {(2 c-5 d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}{a d}\right )}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \left (\frac {a \left (\frac {2 \left (2 c^2-6 c d+7 d^2\right ) \int \frac {1}{-\frac {(a-a \sec (e+f x)) a}{\sec (e+f x) a+a}-a}d\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {\sec (e+f x) a+a}}}{d}-\frac {2 (c-d)^3 \int \frac {1}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a} (c+d \sec (e+f x))}d\sec (e+f x)}{d}\right )}{d}+\frac {(2 c-5 d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}{a d}\right )}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \left (\frac {a \left (\frac {2 \left (2 c^2-6 c d+7 d^2\right ) \int \frac {1}{-\frac {(a-a \sec (e+f x)) a}{\sec (e+f x) a+a}-a}d\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {\sec (e+f x) a+a}}}{d}-\frac {4 (c-d)^3 \int \frac {1}{a (c-d)+\frac {a (c+d) (\sec (e+f x) a+a)}{a-a \sec (e+f x)}}d\frac {\sqrt {\sec (e+f x) a+a}}{\sqrt {a-a \sec (e+f x)}}}{d}\right )}{d}+\frac {(2 c-5 d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}{a d}\right )}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {a^2 \left (\frac {a \left (-\frac {2 \left (2 c^2-6 c d+7 d^2\right ) \arctan \left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a \sec (e+f x)+a}}\right )}{a d}-\frac {4 (c-d)^{5/2} \arctan \left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{a d \sqrt {c+d}}\right )}{d}+\frac {(2 c-5 d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}{a d}\right )}{2 d}-\frac {\sqrt {a-a \sec (e+f x)} (a \sec (e+f x)+a)^{3/2}}{2 d}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*(-1/2*(Sqrt[a - a*Sec[e + f*x]]*(a + a*Sec[e + f*x])^(3/2))/d + (a^ 2*((a*((-2*(2*c^2 - 6*c*d + 7*d^2)*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a + a*Sec[e + f*x]]])/(a*d) - (4*(c - d)^(5/2)*ArcTan[(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])/(Sqrt[c - d]*Sqrt[a - a*Sec[e + f*x]])])/(a*d*Sqrt[c + d] )))/d + ((2*c - 5*d)*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])/(a *d)))/(2*d))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.3.5.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[a ^2*g*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(g*x)^(p - 1)*(a + b*x)^(m - 1/2)*((c + d*x)^n/Sqrt[a - b*x]), x ], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[ b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p, 1] || In tegerQ[m - 1/2])
Time = 0.87 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.46
| method | result | size |
| derivativedivides | \(\frac {16 a^{3} \left (-\frac {\left (c^{3}-3 c^{2} d +3 c \,d^{2}-d^{3}\right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{8 d^{3} \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {1}{32 d \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {5 d -2 c}{32 d^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {\left (-2 c^{2}+6 c d -7 d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{32 d^{3}}-\frac {1}{32 d \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {5 d -2 c}{32 d^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {\left (2 c^{2}-6 c d +7 d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{32 d^{3}}\right )}{f}\) | \(224\) |
| default | \(\frac {16 a^{3} \left (-\frac {\left (c^{3}-3 c^{2} d +3 c \,d^{2}-d^{3}\right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{8 d^{3} \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {1}{32 d \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {5 d -2 c}{32 d^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {\left (-2 c^{2}+6 c d -7 d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{32 d^{3}}-\frac {1}{32 d \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {5 d -2 c}{32 d^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {\left (2 c^{2}-6 c d +7 d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{32 d^{3}}\right )}{f}\) | \(224\) |
| risch | \(-\frac {i a^{3} \left (d \,{\mathrm e}^{3 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )} c -6 d \,{\mathrm e}^{2 i \left (f x +e \right )}-d \,{\mathrm e}^{i \left (f x +e \right )}+2 c -6 d \right )}{f \,d^{2} \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{2}}+\frac {\sqrt {\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i \sqrt {\left (c +d \right ) \left (c -d \right )}-d}{c}\right ) c^{2}}{\left (c +d \right ) f \,d^{3}}-\frac {2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i \sqrt {\left (c +d \right ) \left (c -d \right )}-d}{c}\right ) c}{\left (c +d \right ) f \,d^{2}}+\frac {\sqrt {\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i \sqrt {\left (c +d \right ) \left (c -d \right )}-d}{c}\right )}{\left (c +d \right ) f d}-\frac {\sqrt {\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {\left (c +d \right ) \left (c -d \right )}+d}{c}\right ) c^{2}}{\left (c +d \right ) f \,d^{3}}+\frac {2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {\left (c +d \right ) \left (c -d \right )}+d}{c}\right ) c}{\left (c +d \right ) f \,d^{2}}-\frac {\sqrt {\left (c +d \right ) \left (c -d \right )}\, a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {\left (c +d \right ) \left (c -d \right )}+d}{c}\right )}{\left (c +d \right ) f d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c^{2}}{f \,d^{3}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c}{f \,d^{2}}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 f d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c^{2}}{f \,d^{3}}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c}{f \,d^{2}}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 f d}\) | \(595\) |
16/f*a^3*(-1/8*(c^3-3*c^2*d+3*c*d^2-d^3)/d^3/((c+d)*(c-d))^(1/2)*arctanh(( c-d)*tan(1/2*f*x+1/2*e)/((c+d)*(c-d))^(1/2))+1/32/d/(tan(1/2*f*x+1/2*e)-1) ^2-1/32*(5*d-2*c)/d^2/(tan(1/2*f*x+1/2*e)-1)+1/32/d^3*(-2*c^2+6*c*d-7*d^2) *ln(tan(1/2*f*x+1/2*e)-1)-1/32/d/(tan(1/2*f*x+1/2*e)+1)^2-1/32*(5*d-2*c)/d ^2/(tan(1/2*f*x+1/2*e)+1)+1/32*(2*c^2-6*c*d+7*d^2)/d^3*ln(tan(1/2*f*x+1/2* e)+1))
Time = 0.53 (sec) , antiderivative size = 532, normalized size of antiderivative = 3.48 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx=\left [\frac {2 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt {\frac {c - d}{c + d}} \cos \left (f x + e\right )^{2} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{2} + c d + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {c - d}{c + d}} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (a^{3} d^{2} - 2 \, {\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, d^{3} f \cos \left (f x + e\right )^{2}}, -\frac {4 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} \sqrt {-\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (d \cos \left (f x + e\right ) + c\right )} \sqrt {-\frac {c - d}{c + d}}}{{\left (c - d\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{2} - {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) + {\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (a^{3} d^{2} - 2 \, {\left (a^{3} c d - 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, d^{3} f \cos \left (f x + e\right )^{2}}\right ] \]
[1/4*(2*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*sqrt((c - d)/(c + d))*cos(f*x + e) ^2*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 - 2*(c^2 + c*d + (c*d + d^2)*cos(f*x + e))*sqrt((c - d)/(c + d))*sin(f*x + e) + 2*c^2 - d^ 2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2)) + (2*a^3*c^2 - 6*a^3*c *d + 7*a^3*d^2)*cos(f*x + e)^2*log(sin(f*x + e) + 1) - (2*a^3*c^2 - 6*a^3* c*d + 7*a^3*d^2)*cos(f*x + e)^2*log(-sin(f*x + e) + 1) + 2*(a^3*d^2 - 2*(a ^3*c*d - 3*a^3*d^2)*cos(f*x + e))*sin(f*x + e))/(d^3*f*cos(f*x + e)^2), -1 /4*(4*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*sqrt(-(c - d)/(c + d))*arctan(-(d*co s(f*x + e) + c)*sqrt(-(c - d)/(c + d))/((c - d)*sin(f*x + e)))*cos(f*x + e )^2 - (2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*cos(f*x + e)^2*log(sin(f*x + e) + 1) + (2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*cos(f*x + e)^2*log(-sin(f*x + e ) + 1) - 2*(a^3*d^2 - 2*(a^3*c*d - 3*a^3*d^2)*cos(f*x + e))*sin(f*x + e))/ (d^3*f*cos(f*x + e)^2)]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx=a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{c + d \sec {\left (e + f x \right )}}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{c + d \sec {\left (e + f x \right )}}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{c + d \sec {\left (e + f x \right )}}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{c + d \sec {\left (e + f x \right )}}\, dx\right ) \]
a**3*(Integral(sec(e + f*x)/(c + d*sec(e + f*x)), x) + Integral(3*sec(e + f*x)**2/(c + d*sec(e + f*x)), x) + Integral(3*sec(e + f*x)**3/(c + d*sec(e + f*x)), x) + Integral(sec(e + f*x)**4/(c + d*sec(e + f*x)), x))
Exception generated. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx=\text {Exception raised: ValueError} \]
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*c^2-4*d^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (140) = 280\).
Time = 0.37 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.86 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx=\frac {\frac {{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{d^{3}} - \frac {{\left (2 \, a^{3} c^{2} - 6 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{d^{3}} + \frac {4 \, {\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )}}{\sqrt {-c^{2} + d^{2}} d^{3}} + \frac {2 \, {\left (2 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} d^{2}}}{2 \, f} \]
1/2*((2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*log(abs(tan(1/2*f*x + 1/2*e) + 1) )/d^3 - (2*a^3*c^2 - 6*a^3*c*d + 7*a^3*d^2)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/d^3 + 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*(pi*floor(1/2 *(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan(1/2*f*x + 1/2*e) - d*t an(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))/(sqrt(-c^2 + d^2)*d^3) + 2*(2*a^3* c*tan(1/2*f*x + 1/2*e)^3 - 5*a^3*d*tan(1/2*f*x + 1/2*e)^3 - 2*a^3*c*tan(1/ 2*f*x + 1/2*e) + 7*a^3*d*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*d^2))/f
Time = 14.36 (sec) , antiderivative size = 1902, normalized size of antiderivative = 12.43 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c+d \sec (e+f x)} \, dx=\text {Too large to display} \]
(atanh((18824*a^9*c^2*tan(e/2 + (f*x)/2))/(18824*a^9*c^2 + 2968*a^9*d^2 - (16680*a^9*c^3)/d + (8608*a^9*c^4)/d^2 - (2480*a^9*c^5)/d^3 + (320*a^9*c^6 )/d^4 - 11560*a^9*c*d) - (16680*a^9*c^3*tan(e/2 + (f*x)/2))/(2968*a^9*d^3 - 16680*a^9*c^3 - 11560*a^9*c*d^2 + 18824*a^9*c^2*d + (8608*a^9*c^4)/d - ( 2480*a^9*c^5)/d^2 + (320*a^9*c^6)/d^3) + (8608*a^9*c^4*tan(e/2 + (f*x)/2)) /(8608*a^9*c^4 + 2968*a^9*d^4 - 11560*a^9*c*d^3 - 16680*a^9*c^3*d + 18824* a^9*c^2*d^2 - (2480*a^9*c^5)/d + (320*a^9*c^6)/d^2) - (2480*a^9*c^5*tan(e/ 2 + (f*x)/2))/(2968*a^9*d^5 - 2480*a^9*c^5 - 11560*a^9*c*d^4 + 8608*a^9*c^ 4*d + 18824*a^9*c^2*d^3 - 16680*a^9*c^3*d^2 + (320*a^9*c^6)/d) + (320*a^9* c^6*tan(e/2 + (f*x)/2))/(320*a^9*c^6 + 2968*a^9*d^6 - 11560*a^9*c*d^5 - 24 80*a^9*c^5*d + 18824*a^9*c^2*d^4 - 16680*a^9*c^3*d^3 + 8608*a^9*c^4*d^2) + (2968*a^9*d^2*tan(e/2 + (f*x)/2))/(18824*a^9*c^2 + 2968*a^9*d^2 - (16680* a^9*c^3)/d + (8608*a^9*c^4)/d^2 - (2480*a^9*c^5)/d^3 + (320*a^9*c^6)/d^4 - 11560*a^9*c*d) - (11560*a^9*c*d*tan(e/2 + (f*x)/2))/(18824*a^9*c^2 + 2968 *a^9*d^2 - (16680*a^9*c^3)/d + (8608*a^9*c^4)/d^2 - (2480*a^9*c^5)/d^3 + ( 320*a^9*c^6)/d^4 - 11560*a^9*c*d))*(2*a^3*c^2 + 7*a^3*d^2 - 6*a^3*c*d))/(d ^3*f) - ((tan(e/2 + (f*x)/2)*(2*a^3*c - 7*a^3*d))/d^2 - (a^3*tan(e/2 + (f* x)/2)^3*(2*c - 5*d))/d^2)/(f*(tan(e/2 + (f*x)/2)^4 - 2*tan(e/2 + (f*x)/2)^ 2 + 1)) - (a^3*atan(((a^3*((c + d)*(c - d)^5)^(1/2)*((8*tan(e/2 + (f*x)/2) *(8*a^6*c^7 - 53*a^6*d^7 + 259*a^6*c*d^6 - 64*a^6*c^6*d - 547*a^6*c^2*d...