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

Optimal. Leaf size=170 \[ \frac {2 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (49 A+32 C) \tan (c+d x)}{21 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (7 A+8 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{21 d}+\frac {2 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

2*a^(5/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/7*a*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/
7*C*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/21*a^3*(49*A+32*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/21*a^2*(7*A
+8*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4055, 3917, 3915, 3774, 203, 3792} \[ \frac {2 a^3 (49 A+32 C) \tan (c+d x)}{21 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (7 A+8 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{21 d}+\frac {2 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(49*A + 32*C)*Tan[c + d*x])/(
21*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(7*A + 8*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(21*d) + (2*a*C*(a
+ a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(7*d) + (2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

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 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3915

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4055

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> -Simp
[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(b*(m + 1)), Int[(a + b*Csc[e + f*x])^m*Simp
[A*b*(m + 1) + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ
[m, -2^(-1)]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {2 \int (a+a \sec (c+d x))^{5/2} \left (\frac {7 a A}{2}+\frac {5}{2} a C \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {4 \int (a+a \sec (c+d x))^{3/2} \left (\frac {35 a^2 A}{4}+\frac {5}{4} a^2 (7 A+8 C) \sec (c+d x)\right ) \, dx}{35 a}\\ &=\frac {2 a^2 (7 A+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {8 \int \sqrt {a+a \sec (c+d x)} \left (\frac {105 a^3 A}{8}+\frac {5}{8} a^3 (49 A+32 C) \sec (c+d x)\right ) \, dx}{105 a}\\ &=\frac {2 a^2 (7 A+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\left (a^2 A\right ) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{21} \left (a^2 (49 A+32 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^3 (49 A+32 C) \tan (c+d x)}{21 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (7 A+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}-\frac {\left (2 a^3 A\right ) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (49 A+32 C) \tan (c+d x)}{21 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (7 A+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.99, size = 151, normalized size = 0.89 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt {a (\sec (c+d x)+1)} \left (\sqrt {\sec (c+d x)-1} ((84 A+93 C) \cos (c+d x)+(7 A+23 C) \cos (2 (c+d x))+28 A \cos (3 (c+d x))+7 A+23 C \cos (3 (c+d x))+29 C)+42 A \cos ^3(c+d x) \tan ^{-1}\left (\sqrt {\sec (c+d x)-1}\right )\right )}{21 d \sqrt {\sec (c+d x)-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(42*A*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[c + d*x]^3 + (7*A + 29*C + (84*A + 93*C)*Cos[c + d*x] + (7*A +
23*C)*Cos[2*(c + d*x)] + 28*A*Cos[3*(c + d*x)] + 23*C*Cos[3*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sec[c + d*x]^
3*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(21*d*Sqrt[-1 + Sec[c + d*x]])

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 408, normalized size = 2.40 \[ \left [\frac {21 \, {\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, {\left (28 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (7 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 12 \, C a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, -\frac {2 \, {\left (21 \, {\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, {\left (28 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (7 \, A + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 12 \, C a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{21 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/21*(21*(A*a^2*cos(d*x + c)^4 + A*a^2*cos(d*x + c)^3)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*(28
*A + 23*C)*a^2*cos(d*x + c)^3 + (7*A + 23*C)*a^2*cos(d*x + c)^2 + 12*C*a^2*cos(d*x + c) + 3*C*a^2)*sqrt((a*cos
(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3), -2/21*(21*(A*a^2*cos(d*x + c
)^4 + A*a^2*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d
*x + c))) - (2*(28*A + 23*C)*a^2*cos(d*x + c)^3 + (7*A + 23*C)*a^2*cos(d*x + c)^2 + 12*C*a^2*cos(d*x + c) + 3*
C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]

________________________________________________________________________________________

giac [B]  time = 2.06, size = 355, normalized size = 2.09 \[ -\frac {\frac {21 \, A \sqrt {-a} a^{3} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {2 \, {\left (63 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 84 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (175 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 140 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (161 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 112 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (49 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 32 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{21 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/21*(21*A*sqrt(-a)*a^3*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4
*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(
2)*abs(a) - 6*a))*sgn(cos(d*x + c))/abs(a) + 2*(63*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 84*sqrt(2)*C*a^6*sgn(cos(
d*x + c)) - (175*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 140*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (161*sqrt(2)*A*a^6*sg
n(cos(d*x + c)) + 112*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (49*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 32*sqrt(2)*C*a^6
*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*
c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

________________________________________________________________________________________

maple [B]  time = 1.79, size = 434, normalized size = 2.55 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (21 A \sqrt {2}\, \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}+63 A \sqrt {2}\, \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}+63 A \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}+21 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sin \left (d x +c \right )-896 A \left (\cos ^{4}\left (d x +c \right )\right )-736 C \left (\cos ^{4}\left (d x +c \right )\right )+784 A \left (\cos ^{3}\left (d x +c \right )\right )+368 C \left (\cos ^{3}\left (d x +c \right )\right )+112 A \left (\cos ^{2}\left (d x +c \right )\right )+176 C \left (\cos ^{2}\left (d x +c \right )\right )+144 C \cos \left (d x +c \right )+48 C \right ) a^{2}}{168 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(21*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3*arctanh(1/2*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+63*A*2^(1/2)*sin(d*x+
c)*cos(d*x+c)^2*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(7/2)+63*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin
(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+21*A*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)-896*A*cos(d*x+
c)^4-736*C*cos(d*x+c)^4+784*A*cos(d*x+c)^3+368*C*cos(d*x+c)^3+112*A*cos(d*x+c)^2+176*C*cos(d*x+c)^2+144*C*cos(
d*x+c)+48*C)/sin(d*x+c)/cos(d*x+c)^3*a^2

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + C/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 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________