3.202 \(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=165 \[ \frac {5 (A-15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {(3 A-13 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

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))/a^(5/2)/d*2^(1/2)-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)*t
an(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4085, 4008, 4001, 3795, 203} \[ \frac {5 (A-15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {(3 A-13 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(5*(A - 15*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A
+ C)*Sec[c + d*x]^2*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((3*A - 13*C)*Tan[c + d*x])/(16*a*d*(a +
a*Sec[c + d*x])^(3/2)) + ((A + 9*C)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-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
]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(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)]

Rule 4008

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] + Dist[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] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4085

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] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[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)]

Rubi steps

\begin {align*} \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 {(A+C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {\int \frac {\sec ^2(c+d x) \left (-2 a (A-C)-\frac {1}{2} a (A+9 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\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 {\int \frac {\sec (c+d x) \left (\frac {3}{4} a^2 (3 A-13 C)+a^2 (A+9 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\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)}}+\frac {(5 (A-15 C)) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\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)}}-\frac {(5 (A-15 C)) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {5 (A-15 C) \tan ^{-1}\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)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.89, size = 136, normalized size = 0.82 \[ \frac {-\left (\tan ^3\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (10 (A+17 C) \cos (c+d x)+(A+49 C) \cos (2 (c+d x))+A+113 C)\right )-5 \sqrt {2} (A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)-1} \tan ^{-1}\left (\frac {\sqrt {\sec (c+d x)-1}}{\sqrt {2}}\right )}{32 a^2 d (\cos (c+d x)-1) \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(-5*Sqrt[2]*(A - 15*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*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[(c + d*x)/2]^3)/(32*a^2*d*(-1
+ Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 464, normalized size = 2.81 \[ \left [\frac {5 \, \sqrt {2} {\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + A - 15 \, C\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (A + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 17 \, C\right )} \cos \left (d x + c\right ) + 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {5 \, \sqrt {2} {\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + A - 15 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (A + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 17 \, C\right )} \cos \left (d x + c\right ) + 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[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*
cos(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)]

________________________________________________________________________________________

giac [B]  time = 6.65, size = 286, normalized size = 1.73 \[ \frac {\frac {{\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + \sqrt {2} C a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} + \frac {\sqrt {2} A a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 17 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 83 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + \frac {5 \, {\left (\sqrt {2} A - 15 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/32*(((2*(sqrt(2)*A*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + sqrt(2)*C*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(
1/2*d*x + 1/2*c)^2/a^8 + (sqrt(2)*A*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 17*sqrt(2)*C*a^6*sgn(tan(1/2*d*x + 1
/2*c)^2 - 1))/a^8)*tan(1/2*d*x + 1/2*c)^2 - (3*sqrt(2)*A*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 83*sqrt(2)*C*a^
6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(-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(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

________________________________________________________________________________________

maple [B]  time = 1.84, size = 597, normalized size = 3.62 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (-5 A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+75 C \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-10 A \sin \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+150 C \sin \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+2 A \left (\cos ^{3}\left (d x +c \right )\right )-5 A \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+98 C \left (\cos ^{3}\left (d x +c \right )\right )+75 C \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+8 A \left (\cos ^{2}\left (d x +c \right )\right )+72 C \left (\cos ^{2}\left (d x +c \right )\right )-10 A \cos \left (d x +c \right )-106 C \cos \left (d x +c \right )-64 C \right )}{32 d \sin \left (d x +c \right )^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/32/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-5*A*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+75*C*cos(d*x+
c)^2*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*sin(d*x+c)-10*A*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x
+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+150*C*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+2*A*cos(d*x+c)^3-5*A*ln(
((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*sin(d*x+c)+98*C*cos(d*x+c)^3+75*C*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+
c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+8*A*cos(d*x+c)^2+72*C*cos(d*x+c)^2-10*A*cos(d*x+c)-106*C*c
os(d*x+c)-64*C)/sin(d*x+c)^5/a^3

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/2)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a*(sec(c + d*x) + 1))**(5/2), x)

________________________________________________________________________________________