Integrand size = 35, antiderivative size = 165 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {5 (A-15 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(3 A-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:
5/32*(A-15*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2) )*2^(1/2)/a^(5/2)/d-1/4*(A+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^( 5/2)-1/16*(3*A-13*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/4*(A+9*C)*tan (d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
Time = 3.83 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {-5 \sqrt {2} (A-15 C) \arctan \left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \sqrt {-1+\sec (c+d x)} \sin (c+d x)-(A+113 C+10 (A+17 C) \cos (c+d x)+(A+49 C) \cos (2 (c+d x))) \sec (c+d x) \tan ^3\left (\frac {1}{2} (c+d x)\right )}{32 a^2 d (-1+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/ 2),x]
Output:
(-5*Sqrt[2]*(A - 15*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Sqrt[-1 + S ec[c + d*x]]*Sin[c + d*x] - (A + 113*C + 10*(A + 17*C)*Cos[c + d*x] + (A + 49*C)*Cos[2*(c + d*x)])*Sec[c + d*x]*Tan[(c + d*x)/2]^3)/(32*a^2*d*(-1 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.99 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4573, 27, 3042, 4496, 27, 3042, 4489, 3042, 4282, 216}
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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4573 |
\(\displaystyle -\frac {\int -\frac {\sec ^2(c+d x) (4 a (A-C)+a (A+9 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a (A-C)+a (A+9 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a (A-C)+a (A+9 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4496 |
\(\displaystyle \frac {-\frac {\int -\frac {\sec (c+d x) \left (3 (3 A-13 C) a^2+4 (A+9 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (3 A-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (3 (3 A-13 C) a^2+4 (A+9 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (3 A-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 (3 A-13 C) a^2+4 (A+9 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (3 A-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle \frac {\frac {5 a^2 (A-15 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx+\frac {8 a^2 (A+9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (3 A-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5 a^2 (A-15 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {8 a^2 (A+9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (3 A-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {\frac {\frac {8 a^2 (A+9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {10 a^2 (A-15 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}-\frac {a (3 A-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {5 \sqrt {2} a^{3/2} (A-15 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {8 a^2 (A+9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (3 A-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
Input:
Int[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]
Output:
-1/4*((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a*(3*A - 13*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + ((5 *Sqrt[2]*a^(3/2)*(A - 15*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (8*a^2*(A + 9*C)*Tan[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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]^2*(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/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b *B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(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*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.95 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.62
method | result | size |
default | \(\frac {\left (\frac {\left (-2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+83 \csc \left (d x +c \right )-83 \cot \left (d x +c \right )\right ) C}{32}+\frac {\left (-2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+3 \csc \left (d x +c \right )-3 \cot \left (d x +c \right )\right ) A}{32}+\frac {5 A \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}-\frac {75 C \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}\) | \(267\) |
parts | \(-\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\frac {4 \sqrt {2}\, \left (\cos \left (d x +c \right )-1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cot \left (d x +c \right )}{\cos \left (d x +c \right )+1}+5 \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-5 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{32 d \,a^{3}}+\frac {C \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}-\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}-\frac {75 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}+\frac {83 \csc \left (d x +c \right )}{32}-\frac {83 \cot \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}\) | \(307\) |
Input:
int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETUR NVERBOSE)
Output:
1/d*(1/32*(-2*(1-cos(d*x+c))^5*csc(d*x+c)^5-17*(1-cos(d*x+c))^3*csc(d*x+c) ^3+83*csc(d*x+c)-83*cot(d*x+c))*C+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))*A+5/32*A*ln((-2*co s(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/(cos( d*x+c)+1))^(1/2)-75/32*C*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c )+csc(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/a^3*(a*(1+sec(d*x+c))) ^(1/2)
Time = 0.12 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.81 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {5 \, \sqrt {2} {\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + A - 15 \, C\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (A + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 17 \, C\right )} \cos \left (d x + c\right ) + 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {5 \, \sqrt {2} {\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + A - 15 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (A + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 17 \, C\right )} \cos \left (d x + c\right ) + 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \] Input:
integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algori thm="fricas")
Output:
[1/64*(5*sqrt(2)*((A - 15*C)*cos(d*x + c)^3 + 3*(A - 15*C)*cos(d*x + c)^2 + 3*(A - 15*C)*cos(d*x + c) + A - 15*C)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)* sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*co s(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1) ) + 4*((A + 49*C)*cos(d*x + c)^2 + 5*(A + 17*C)*cos(d*x + c) + 32*C)*sqrt( (a*cos(d*x + c) + a)/cos(d*x + c))*sin(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*(5*sqrt(2)*(( A - 15*C)*cos(d*x + c)^3 + 3*(A - 15*C)*cos(d*x + c)^2 + 3*(A - 15*C)*cos( d*x + c) + A - 15*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos( d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*((A + 49*C)*cos(d*x + c )^2 + 5*(A + 17*C)*cos(d*x + c) + 32*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(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)]
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a*(sec(c + d*x) + 1))**( 5/2), x)
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algori thm="maxima")
Output:
Timed out
Time = 1.11 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.55 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} + \frac {\sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 17 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 83 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} - \frac {5 \, {\left (\sqrt {2} A - 15 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \] Input:
integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algori thm="giac")
Output:
1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*((2*(sqrt(2)*A*a^6*sgn(cos(d*x + c)) + sqrt(2)*C*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a^8 + (sqrt (2)*A*a^6*sgn(cos(d*x + c)) + 17*sqrt(2)*C*a^6*sgn(cos(d*x + c)))/a^8)*tan (1/2*d*x + 1/2*c)^2 - (3*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 83*sqrt(2)*C*a^ 6*sgn(cos(d*x + c)))/a^8)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 - a) - 5*(sqrt(2)*A - 15*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)))/(sqrt(-a)*a^2*sgn(cos(d*x + c))))/ d
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/2)),x)
Output:
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/2)), x)
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:
int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1) *sec(c + d*x)**2)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a))/a**3