3.2.57 \(\int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [157]
Optimal. Leaf size=137 \[ \frac {2 a (35 A+27 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+18 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d} \]
[Out]
2/35*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/105*a*(35*A+27*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/105*(35
*A+18*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/7*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A]
time = 0.25, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {4174, 4095,
4086, 3877} \begin {gather*} \frac {2 (35 A+18 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (35 A+27 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2),x]
[Out]
(2*a*(35*A + 27*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A + 18*C)*Sqrt[a + a*Sec[c + d*x]]*
Tan[c + d*x])/(105*d) + (2*C*Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(7*d) + (2*C*(a + a*Sec[c +
d*x])^(3/2)*Tan[c + d*x])/(35*a*d)
Rule 3877
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 4086
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 4095
Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && !LtQ[m, -1]
Rule 4174
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(m + n + 1
))), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n +
a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(
-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Rubi steps
\begin {align*} \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac {2 \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (7 A+4 C)+\frac {1}{2} a C \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac {2 C \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac {4 \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3 a^2 C}{4}+\frac {1}{4} a^2 (35 A+18 C) \sec (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac {2 (35 A+18 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac {1}{105} (35 A+27 C) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (35 A+27 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+18 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}\\ \end {align*}
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Mathematica [A]
time = 0.87, size = 99, normalized size = 0.72 \begin {gather*} \frac {(35 A+54 C+3 (35 A+36 C) \cos (c+d x)+(35 A+24 C) \cos (2 (c+d x))+35 A \cos (3 (c+d x))+24 C \cos (3 (c+d x))) \sec ^3(c+d x) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{105 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2),x]
[Out]
((35*A + 54*C + 3*(35*A + 36*C)*Cos[c + d*x] + (35*A + 24*C)*Cos[2*(c + d*x)] + 35*A*Cos[3*(c + d*x)] + 24*C*C
os[3*(c + d*x)])*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(105*d)
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Maple [A]
time = 10.86, size = 107, normalized size = 0.78
| | |
method |
result |
size |
| | |
default |
\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (70 A \left (\cos ^{3}\left (d x +c \right )\right )+48 C \left (\cos ^{3}\left (d x +c \right )\right )+35 A \left (\cos ^{2}\left (d x +c \right )\right )+24 C \left (\cos ^{2}\left (d x +c \right )\right )+18 C \cos \left (d x +c \right )+15 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{105 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )}\) |
\(107\) |
| | |
|
|
|
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))^(1/2),x,method=_RETURNVERBOSE)
[Out]
-2/105/d*(-1+cos(d*x+c))*(70*A*cos(d*x+c)^3+48*C*cos(d*x+c)^3+35*A*cos(d*x+c)^2+24*C*cos(d*x+c)^2+18*C*cos(d*x
+c)+15*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3/sin(d*x+c)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
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))^(1/2),x, algorithm="maxima")
[Out]
-4/105*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(7*(5*A*sin(4*d*x + 4*c) + 2*
(5*A + 6*C)*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (35*A*cos(4*d*x + 4*c
) + 14*(5*A + 6*C)*cos(2*d*x + 2*c) + 35*A + 24*C)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*s
qrt(a) - 105*((A*d*cos(2*d*x + 2*c)^4 + A*d*sin(2*d*x + 2*c)^4 + 4*A*d*cos(2*d*x + 2*c)^3 + 6*A*d*cos(2*d*x +
2*c)^2 + 4*A*d*cos(2*d*x + 2*c) + 2*(A*d*cos(2*d*x + 2*c)^2 + 2*A*d*cos(2*d*x + 2*c) + A*d)*sin(2*d*x + 2*c)^2
+ A*d)*integrate((((cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x +
6*c)*cos(2*d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(10*d*x + 10*c)*sin(2*d
*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin
(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*s
in(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x +
2*c)*sin(4*d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 6*cos(6*d*
x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 4*
cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x + 4*c)
- cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2
*c) - 4*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (cos(10*d*x
+ 10*c)*cos(2*d*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 4*cos(4
*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(10*d*x + 10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*s
in(2*d*x + 2*c) + 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)
^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)))/(((2*(4*cos(8*d*x + 8*c) + 6*cos(6*d*x + 6*c) + 4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(10*d*x + 10*c
) + cos(10*d*x + 10*c)^2 + 8*(6*cos(6*d*x + 6*c) + 4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(8*d*x + 8*c) + 1
6*cos(8*d*x + 8*c)^2 + 12*(4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + 36*cos(6*d*x + 6*c)^2 + 1
6*cos(4*d*x + 4*c)^2 + 8*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + 2*(4*sin(8*d*x + 8*c) + 6*si
n(6*d*x + 6*c) + 4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(10*d*x + 10*c) + sin(10*d*x + 10*c)^2 + 8*(6*sin(6
*d*x + 6*c) + 4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*sin(8*d*x + 8*c)^2 + 12*(4*sin(4*d*
x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 36*sin(6*d*x + 6*c)^2 + 16*sin(4*d*x + 4*c)^2 + 8*sin(4*d*x +
4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (2*(4
*cos(8*d*x + 8*c) + 6*cos(6*d*x + 6*c) + 4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(10*d*x + 10*c) + cos(10*d*
x + 10*c)^2 + 8*(6*cos(6*d*x + 6*c) + 4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(8*d*x + 8*c) + 16*cos(8*d*x +
8*c)^2 + 12*(4*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + 36*cos(6*d*x + 6*c)^2 + 16*cos(4*d*x +
4*c)^2 + 8*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + 2*(4*sin(8*d*x + 8*c) + 6*sin(6*d*x + 6*c
) + 4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(10*d*x + 10*c) + sin(10*d*x + 10*c)^2 + 8*(6*sin(6*d*x + 6*c) +
4*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*sin(8*d*x + 8*c)^2 + 12*(4*sin(4*d*x + 4*c) + si
n(2*d*x + 2*c))*sin(6*d*x + 6*c) + 36*sin(6*d*x + 6*c)^2 + 16*sin(4*d*x + 4*c)^2 + 8*sin(4*d*x + 4*c)*sin(2*d*
x + 2*c) + sin(2*d*x + 2*c)^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2)*(cos(2*d*x + 2*c)^2
+ sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)), x) + 2*((A + 2*C)*d*cos(2*d*x + 2*c)^4 + (A + 2*C)*d*s
in(2*d*x + 2*c)^4 + 4*(A + 2*C)*d*cos(2*d*x + 2*c)^3 + 6*(A + 2*C)*d*cos(2*d*x + 2*c)^2 + 4*(A + 2*C)*d*cos(2*
d*x + 2*c) + 2*((A + 2*C)*d*cos(2*d*x + 2*c)^2 + 2*(A + 2*C)*d*cos(2*d*x + 2*c) + (A + 2*C)*d)*sin(2*d*x + 2*c
)^2 + (A + 2*C)*d)*integrate((((cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*
cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(10*d*x + 10
*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x
+ 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d
*x + 2*c)*sin(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*c
os(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) -
6*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*...
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Fricas [A]
time = 2.41, size = 98, normalized size = 0.72 \begin {gather*} \frac {2 \, {\left (2 \, {\left (35 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (35 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 18 \, C \cos \left (d x + c\right ) + 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}
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))^(1/2),x, algorithm="fricas")
[Out]
2/105*(2*(35*A + 24*C)*cos(d*x + c)^3 + (35*A + 24*C)*cos(d*x + c)^2 + 18*C*cos(d*x + c) + 15*C)*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)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \end {gather*}
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))**(1/2),x)
[Out]
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + C*sec(c + d*x)**2)*sec(c + d*x)**2, x)
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Giac [A]
time = 1.04, size = 222, normalized size = 1.62 \begin {gather*} -\frac {2 \, {\left (105 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (245 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (175 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (35 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 27 \, \sqrt {2} C a^{4} \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 )}{105 \, {\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} d} \end {gather*}
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))^(1/2),x, algorithm="giac")
[Out]
-2/105*(105*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (245*sqrt(2)*A*a^4*sgn(cos
(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (175*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 147*sqrt(2)*C*a^4*sg
n(cos(d*x + c)) - (35*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 27*sqrt(2)*C*a^4*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*s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)
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Mupad [B]
time = 6.21, size = 423, normalized size = 3.09 \begin {gather*} \frac {\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,4{}\mathrm {i}}{3\,d}-\frac {C\,16{}\mathrm {i}}{35\,d}\right )+\frac {A\,4{}\mathrm {i}}{3\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {A\,8{}\mathrm {i}}{7\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,8{}\mathrm {i}}{7\,d}-\frac {\left (8\,A+16\,C\right )\,1{}\mathrm {i}}{7\,d}\right )-\frac {\left (8\,A+16\,C\right )\,1{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {A\,4{}\mathrm {i}}{5\,d}+\frac {C\,16{}\mathrm {i}}{35\,d}+\frac {\left (28\,A+112\,C\right )\,1{}\mathrm {i}}{35\,d}\right )-\frac {A\,4{}\mathrm {i}}{5\,d}+\frac {\left (28\,A+112\,C\right )\,1{}\mathrm {i}}{35\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (140\,A+96\,C\right )\,1{}\mathrm {i}}{105\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \end {gather*}
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))^(1/2))/cos(c + d*x)^2,x)
[Out]
((exp(c*1i + d*x*1i)*((A*4i)/(3*d) - (C*16i)/(35*d)) + (A*4i)/(3*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + ((a + a/(exp(- c*1i - d*x*1i)/2 +
exp(c*1i + d*x*1i)/2))^(1/2)*((A*8i)/(7*d) + exp(c*1i + d*x*1i)*((A*8i)/(7*d) - ((8*A + 16*C)*1i)/(7*d)) - ((8
*A + 16*C)*1i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2
+ exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*16i)/(35*d) - (A*4i)/(5*d) + ((28*A + 112*C)*1i)/(35*d
)) - (A*4i)/(5*d) + ((28*A + 112*C)*1i)/(35*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (exp(
c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(140*A + 96*C)*1i)/(105*d*(exp(c*
1i + d*x*1i) + 1))
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