Integrand size = 45, antiderivative size = 211 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(75 A-19 B-5 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:
-1/32*(75*A-19*B-5*C)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1 /2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A-B+C)*sec(d*x+c)^(1/2) *sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(13*A-5*B-3*C)*sec(d*x+c)^(1/2)* sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/16*(49*A-9*B+C)*sec(d*x+c)^(1/2)*s in(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(653\) vs. \(2(211)=422\).
Time = 7.29 (sec) , antiderivative size = 653, normalized size of antiderivative = 3.09 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {A \sqrt {1+\sec (c+d x)} \left (\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}-\frac {98 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}-\frac {75 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}}-\frac {B \sqrt {1+\sec (c+d x)} \left (\frac {8 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+\frac {18 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}+\frac {19 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}}-\frac {C \sqrt {1+\sec (c+d x)} \left (\frac {8 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+5 \left (\frac {2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}-\frac {\left (2 \arcsin \left (\sqrt {\sec (c+d x)}\right )-\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \left (\arcsin \left (\sqrt {1-\sec (c+d x)}\right )+\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)),x]
Output:
-1/32*(A*Sqrt[1 + Sec[c + d*x]]*((8*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(5/2)) + (26*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 + Sec [c + d*x])^(3/2)) - (98*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]) - (75*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])) )/(a^2*Sqrt[a*(1 + Sec[c + d*x])]) - (B*Sqrt[1 + Sec[c + d*x]]*((8*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(5/2)) + (18*Sec[c + d*x]^ (3/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) + (19*Sqrt[2]*ArcTan[(Sqr t[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])))/(32*a^2*Sqrt[a*(1 + Sec[c + d*x]) ]) - (C*Sqrt[1 + Sec[c + d*x]]*((8*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(5/2)) + 5*((2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Se c[c + d*x])^(3/2)) - ((2*ArcSin[Sqrt[Sec[c + d*x]]] - Sqrt[2]*ArcTan[(Sqrt [2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) - (2*(ArcSin[Sqrt[1 - Sec[c + d*x]] ] + Sqrt[1 - Sec[c + d*x]]*Sqrt[Sec[c + d*x]])*Tan[c + d*x])/(d*Sqrt[1 - S ec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]))))/(32*a^2*Sqrt[a*(1 + Sec[c + d*x])] )
Time = 1.19 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4572, 27, 3042, 4508, 27, 3042, 4501, 3042, 4295, 219}
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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (75 A-19 B-5 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} a^{3/2} (75 A-19 B-5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
Input:
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Sec[c + d*x]]*(a + a*Sec [c + d*x])^(5/2)),x]
Output:
-1/4*((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) ^(5/2)) + (-1/2*(a*(13*A - 5*B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d* (a + a*Sec[c + d*x])^(3/2)) + (-((Sqrt[2]*a^(3/2)*(75*A - 19*B - 5*C)*ArcT anh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d) + (2*a^2*(49*A - 9*B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d* Sqrt[a + a*Sec[c + d*x]]))/(4*a^2))/(8*a^2)
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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f)) Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(379\) vs. \(2(180)=360\).
Time = 4.12 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.80
| method | result | size |
| default | \(\frac {\left (-\frac {\left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-3 \csc \left (d x +c \right )+3 \cot \left (d x +c \right )\right ) C}{32}-\frac {\left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-83 \csc \left (d x +c \right )+83 \cot \left (d x +c \right )\right ) A}{32}-\frac {\left (-2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+9 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+11 \csc \left (d x +c \right )-11 \cot \left (d x +c \right )\right ) B}{32}+\frac {75 A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{32}-\frac {19 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, B}{32}-\frac {5 C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )}{32}\right ) \sqrt {-a \left (-1-\sec \left (d x +c \right )\right )}}{d \,a^{3} \sqrt {\sec \left (d x +c \right )}}\) | \(380\) |
| parts | \(\frac {A \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}+\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}+\frac {75 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{32}+\frac {83 \csc \left (d x +c \right )}{32}-\frac {83 \cot \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3} \sqrt {\sec \left (d x +c \right )}}+\frac {B \left (\left (19 \cos \left (d x +c \right )^{2}+38 \cos \left (d x +c \right )+19\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (-13 \cos \left (d x +c \right )-9\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {C \left (\left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (5 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )^{2} \sqrt {2}\, \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(484\) |
Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2 ),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/32*(2*(1-cos(d*x+c))^5*csc(d*x+c)^5-(1-cos(d*x+c))^3*csc(d*x+c)^3- 3*csc(d*x+c)+3*cot(d*x+c))*C-1/32*(2*(1-cos(d*x+c))^5*csc(d*x+c)^5-17*(1-c os(d*x+c))^3*csc(d*x+c)^3-83*csc(d*x+c)+83*cot(d*x+c))*A-1/32*(-2*(1-cos(d *x+c))^5*csc(d*x+c)^5+9*(1-cos(d*x+c))^3*csc(d*x+c)^3+11*csc(d*x+c)-11*cot (d*x+c))*B+75/32*A*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+ c)+cot(d*x+c)))*(-2/(cos(d*x+c)+1))^(1/2)-19/32*arctan(1/2*2^(1/2)/(-1/(co s(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))*(-2/(cos(d*x+c)+1))^(1/2)*B-5 /32*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/ 2)*(-csc(d*x+c)+cot(d*x+c))))/a^3*(-a*(-1-sec(d*x+c)))^(1/2)/sec(d*x+c)^(1 /2)
Time = 0.10 (sec) , antiderivative size = 556, normalized size of antiderivative = 2.64 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) )^(5/2),x, algorithm="fricas")
Output:
[-1/64*(sqrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C )*cos(d*x + c)^2 + 3*(75*A - 19*B - 5*C)*cos(d*x + c) + 75*A - 19*B - 5*C) *sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a )/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(32*A*cos(d*x + c)^3 + (85*A - 13*B + 5*C)*cos(d*x + c)^2 + (49*A - 9*B + C)*cos(d*x + c))*sqrt((a*cos( d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d* x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(s qrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C)*cos(d*x + c)^2 + 3*(75*A - 19*B - 5*C)*cos(d*x + c) + 75*A - 19*B - 5*C)*sqrt(-a) *arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos( d*x + c))/(a*sin(d*x + c))) + 2*(32*A*cos(d*x + c)^3 + (85*A - 13*B + 5*C) *cos(d*x + c)^2 + (49*A - 9*B + C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a) /cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3* a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2)/(a+a*sec(d*x+ c))**(5/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 264379 vs. \(2 (180) = 360\).
Time = 3.52 (sec) , antiderivative size = 264379, normalized size of antiderivative = 1252.98 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) )^(5/2),x, algorithm="maxima")
Output:
-1/32*((576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si n(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/ 2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*cos(5/2*d* x + 5/2*c)^6 + 14400*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2 *d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*c os(3/2*d*x + 3/2*c)^6 + 187500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin (1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^6 + 576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2* d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/ 2*c)^6 + 5184*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2* sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(3/2* d*x + 3/2*c)^6 + 262500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d* x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^4*sin(1/2 *d*x + 1/2*c)^2 + 77700*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2...
Time = 4.02 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.05 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} - \sqrt {2} B a^{6} + \sqrt {2} C a^{6}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} - \frac {17 \, \sqrt {2} A a^{6} - 9 \, \sqrt {2} B a^{6} + \sqrt {2} C a^{6}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {83 \, \sqrt {2} A a^{6} - 11 \, \sqrt {2} B a^{6} + 3 \, \sqrt {2} C a^{6}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {{\left (75 \, \sqrt {2} A - 19 \, \sqrt {2} B - 5 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}}}{32 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) )^(5/2),x, algorithm="giac")
Output:
-1/32*(((2*(sqrt(2)*A*a^6 - sqrt(2)*B*a^6 + sqrt(2)*C*a^6)*tan(1/2*d*x + 1 /2*c)^2/a^8 - (17*sqrt(2)*A*a^6 - 9*sqrt(2)*B*a^6 + sqrt(2)*C*a^6)/a^8)*ta n(1/2*d*x + 1/2*c)^2 - (83*sqrt(2)*A*a^6 - 11*sqrt(2)*B*a^6 + 3*sqrt(2)*C* a^6)/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) - (75*sq rt(2)*A - 19*sqrt(2)*B - 5*sqrt(2)*C)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c ) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/a^(5/2))/(d*sgn(cos(d*x + c)))
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^(5/2)*(1 /cos(c + d*x))^(1/2)),x)
Output:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^(5/2)*(1 /cos(c + d*x))^(1/2)), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}+3 \sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b \right )}{a^{3}} \] Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2 ),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x))/(se c(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(s ec(c + d*x))*sqrt(sec(c + d*x) + 1))/(sec(c + d*x)**4 + 3*sec(c + d*x)**3 + 3*sec(c + d*x)**2 + sec(c + d*x)),x)*a + int((sqrt(sec(c + d*x))*sqrt(se c(c + d*x) + 1))/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1 ),x)*b))/a**3