Integrand size = 35, antiderivative size = 203 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(11 A-7 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(7 A-3 B) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(19 A-15 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}} \] Output:
1/4*(11*A-7*B)*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^(3/2)/d-1/2*(A-B)*sin(d*x+c)/d/sec(d*x+c)^( 1/2)/(a+a*sec(d*x+c))^(3/2)+1/6*(7*A-3*B)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/ (a+a*sec(d*x+c))^(1/2)-1/6*(19*A-15*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+ a*sec(d*x+c))^(1/2)
Time = 1.19 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.85 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {-6 \sqrt {2} (11 A-7 B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+\sqrt {1-\sec (c+d x)} (12 (-A+B)+(-17 A+15 B+2 A \cos (2 (c+d x))) \sec (c+d x)) \tan (c+d x)}{6 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (a (1+\sec (c+d x)))^{3/2}} \] Input:
Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3 /2)),x]
Output:
(-6*Sqrt[2]*(11*A - 7*B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[ c + d*x]]]*Cos[(c + d*x)/2]^3*Sec[c + d*x]^(5/2)*Sin[(c + d*x)/2] + Sqrt[1 - Sec[c + d*x]]*(12*(-A + B) + (-17*A + 15*B + 2*A*Cos[2*(c + d*x)])*Sec[ c + d*x])*Tan[c + d*x])/(6*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*(a* (1 + Sec[c + d*x]))^(3/2))
Time = 1.10 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4508, 27, 3042, 4510, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {a (7 A-3 B)-4 a (A-B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (7 A-3 B)-4 a (A-B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 A-3 B)-4 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {2 \int -\frac {a^2 (19 A-15 B)-2 a^2 (7 A-3 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A-15 B)-2 a^2 (7 A-3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A-15 B)-2 a^2 (7 A-3 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A-7 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A-7 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {6 a^2 (11 A-7 B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (19 A-15 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {2 a (7 A-3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} a^{3/2} (11 A-7 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\) |
Input:
Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)),x ]
Output:
-1/2*((A - B)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/ 2)) + ((2*a*(7*A - 3*B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*S ec[c + d*x]]) - ((-3*Sqrt[2]*a^(3/2)*(11*A - 7*B)*ArcTanh[(Sqrt[a]*Sqrt[Se c[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^2* (19*A - 15*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]] ))/(3*a))/(4*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[(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[1/(b*d *n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
Time = 2.34 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.16
| method | result | size |
| default | \(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (33 \cos \left (d x +c \right )+66+33 \sec \left (d x +c \right )\right )+B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-21 \cos \left (d x +c \right )-42-21 \sec \left (d x +c \right )\right )+\left (-8 \cos \left (d x +c \right )^{2}+24 \cos \left (d x +c \right )+38\right ) A \tan \left (d x +c \right )+B \left (-24 \sin \left (d x +c \right )-30 \tan \left (d x +c \right )\right )\right )}{12 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(235\) |
| parts | \(-\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-8 \cos \left (d x +c \right )^{2}+24 \cos \left (d x +c \right )+38\right ) \tan \left (d x +c \right )+\arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (33 \cos \left (d x +c \right )+66+33 \sec \left (d x +c \right )\right )\right )}{12 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \left (\frac {\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{4}+\frac {7 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{4}-\frac {9 \cot \left (d x +c \right )}{4}+\frac {9 \csc \left (d x +c \right )}{4}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{2} \sqrt {\sec \left (d x +c \right )}}\) | \(265\) |
Input:
int((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x,method=_RET URNVERBOSE)
Output:
-1/12/d/a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)/sec(d*x +c)^(3/2)*((-2/(1+cos(d*x+c)))^(1/2)*A*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot (d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(33*cos(d*x+c)+66+33*sec(d*x+c))+B*(-2 /(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+ cos(d*x+c)))^(1/2))*(-21*cos(d*x+c)-42-21*sec(d*x+c))+(-8*cos(d*x+c)^2+24* cos(d*x+c)+38)*A*tan(d*x+c)+B*(-24*sin(d*x+c)-30*tan(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.29 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left ({\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right ) + 11 \, A - 7 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \frac {4 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 12 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} - {\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right ) + 11 \, A - 7 \, B\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 12 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} - {\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algo rithm="fricas")
Output:
[-1/24*(3*sqrt(2)*((11*A - 7*B)*cos(d*x + c)^2 + 2*(11*A - 7*B)*cos(d*x + c) + 11*A - 7*B)*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*co s(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(4*A*cos(d*x + c)^3 - 12*(A - B)*cos(d*x + c)^2 - (19*A - 15*B)*cos(d*x + c))*sqrt((a*c os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos (d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/12*(3*sqrt(2)*((11*A - 7*B )*cos(d*x + c)^2 + 2*(11*A - 7*B)*cos(d*x + c) + 11*A - 7*B)*sqrt(-a)*arct an(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*(4*A*cos(d*x + c)^3 - 12*(A - B)*cos(d*x + c)^2 - (19*A - 15*B)*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^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c))**(3/2),x)
Output:
Integral((A + B*sec(c + d*x))/((a*(sec(c + d*x) + 1))**(3/2)*sec(c + d*x)* *(3/2)), x)
Leaf count of result is larger than twice the leaf count of optimal. 41138 vs. \(2 (172) = 344\).
Time = 0.71 (sec) , antiderivative size = 41138, normalized size of antiderivative = 202.65 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algo rithm="maxima")
Output:
-1/12*(3*(4*(7*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) - 7*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) - 8*sin(1/2*d*x + 1/2*c))*cos(3/2*d*x + 3/2*c)^4 + 63*(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) - 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 + 4*(7*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) - 7*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) - 8*sin(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c)^4 + 70*(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)^2*sin(1/2*d*x + 1/2*c)^2 + 7*(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) - lo g(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))*sin(1/2*d*x + 1/2*c)^4 - 8*sin(1/2*d*x + 1/2*c)^5 + 28*(7*(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) - 8*cos(1/2*d*x + 1/2*c)*sin(1/2*d*x + 1/2 *c))*cos(3/2*d*x + 3/2*c)^3 + 4*(21*(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)...
Time = 2.17 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.87 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {3 \, {\left (11 \, \sqrt {2} A - 7 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}} + \frac {{\left ({\left (\frac {3 \, {\left (\sqrt {2} A a - \sqrt {2} B a\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a} + \frac {2 \, {\left (23 \, \sqrt {2} A a - 15 \, \sqrt {2} B a\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {27 \, {\left (\sqrt {2} A a - \sqrt {2} B a\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}}}{12 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algo rithm="giac")
Output:
-1/12*(3*(11*sqrt(2)*A - 7*sqrt(2)*B)*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^(3/2) + ((3*(sqrt(2)*A*a - sqrt (2)*B*a)*tan(1/2*d*x + 1/2*c)^2/a + 2*(23*sqrt(2)*A*a - 15*sqrt(2)*B*a)/a) *tan(1/2*d*x + 1/2*c)^2 + 27*(sqrt(2)*A*a - sqrt(2)*B*a)/a)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2))/(d*sgn(cos(d*x + c)))
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(3/2 )),x)
Output:
int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(3/2 )), x)
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}+2 \sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}+2 \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )}d x \right ) b \right )}{a^{2}} \] Input:
int((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/(sec(c + d*x)**4 + 2*sec(c + d*x)**3 + sec(c + d*x)**2),x)*a + int((sqrt(sec(c + d*x))*sqr t(sec(c + d*x) + 1))/(sec(c + d*x)**3 + 2*sec(c + d*x)**2 + sec(c + d*x)), x)*b))/a**2