3.4.68 \(\int (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [368]
Optimal. Leaf size=101 \[ \frac {8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
[Out]
2/5*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+8/15*a^2*(5*B+3*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a*(5*B+3
*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A]
time = 0.09, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4139, 12, 3878,
3877} \begin {gather*} \frac {8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 B+3 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(8*a^2*(5*B + 3*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*B + 3*C)*Sqrt[a + a*Sec[c + d*x]]*T
an[c + d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 3878
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b)*Cot[e + f*x]*(
(a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Dist[a*((2*m - 1)/m), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1),
x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && IntegerQ[2*m]
Rule 4139
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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 + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b
, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2 \int \frac {1}{2} a (5 B+3 C) \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx}{5 a}\\ &=\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{5} (5 B+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{15} (4 a (5 B+3 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {8 a^2 (5 B+3 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A]
time = 0.29, size = 62, normalized size = 0.61 \begin {gather*} \frac {2 a^2 \left (25 B+18 C+(5 B+9 C) \sec (c+d x)+3 C \sec ^2(c+d x)\right ) \tan (c+d x)}{15 d \sqrt {a (1+\sec (c+d x))}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(2*a^2*(25*B + 18*C + (5*B + 9*C)*Sec[c + d*x] + 3*C*Sec[c + d*x]^2)*Tan[c + d*x])/(15*d*Sqrt[a*(1 + Sec[c + d
*x])])
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Maple [A]
time = 12.33, size = 95, normalized size = 0.94
| | |
| method |
result |
size |
| | |
| default |
\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (25 B \left (\cos ^{2}\left (d x +c \right )\right )+18 C \left (\cos ^{2}\left (d x +c \right )\right )+5 B \cos \left (d x +c \right )+9 C \cos \left (d x +c \right )+3 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{15 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}\) |
\(95\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
[Out]
-2/15/d*(-1+cos(d*x+c))*(25*B*cos(d*x+c)^2+18*C*cos(d*x+c)^2+5*B*cos(d*x+c)+9*C*cos(d*x+c)+3*C)*(a*(1+cos(d*x+
c))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)*a
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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((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
2/15*(15*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((3*B + 2*C)*a*d*cos(2*d*x
+ 2*c)^2 + (3*B + 2*C)*a*d*sin(2*d*x + 2*c)^2 + 2*(3*B + 2*C)*a*d*cos(2*d*x + 2*c) + (3*B + 2*C)*a*d)*integrat
e((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(6*d*x + 6*c)*cos(2*d*x + 2*c
) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*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(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4
*d*x + 4*c)*sin(2*d*x + 2*c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - c
os(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c))) - (cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2
+ sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*sin(5/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))/((cos(2*d*x
+ 2*c)^4 + sin(2*d*x + 2*c)^4 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(6*d*x
+ 6*c)^2 + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c)^2 + 2*cos(2*d
*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c)^2 + 4*(cos(2
*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c)^2 + (2*cos(2*d*x + 2*c)^2 + 2*co
s(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 2*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*(cos(
2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c)^2 + cos(2*
d*x + 2*c))*cos(6*d*x + 6*c) + 4*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c
)^2 + cos(2*d*x + 2*c))*cos(4*d*x + 4*c) + cos(2*d*x + 2*c)^2 + 2*(sin(2*d*x + 2*c)^3 + 2*(cos(2*d*x + 2*c)^2
+ sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1
)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 4*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*s
in(2*d*x + 2*c))*sin(4*d*x + 4*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (cos(2*d*x + 2
*c)^4 + sin(2*d*x + 2*c)^4 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*
c)^2 + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c)^2 + 2*cos(2*d*x +
2*c)^3 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c)^2 + 4*(cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c)^2 + (2*cos(2*d*x + 2*c)^2 + 2*cos(2*
d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 2*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*(cos(2*d*
x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x
+ 2*c))*cos(6*d*x + 6*c) + 4*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2
+ cos(2*d*x + 2*c))*cos(4*d*x + 4*c) + cos(2*d*x + 2*c)^2 + 2*(sin(2*d*x + 2*c)^3 + 2*(cos(2*d*x + 2*c)^2 + si
n(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*si
n(2*d*x + 2*c))*sin(6*d*x + 6*c) + 4*(sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2
*d*x + 2*c))*sin(4*d*x + 4*c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2), x) + 2*((2*B + 3*C
)*a*d*cos(2*d*x + 2*c)^2 + (2*B + 3*C)*a*d*sin(2*d*x + 2*c)^2 + 2*(2*B + 3*C)*a*d*cos(2*d*x + 2*c) + (2*B + 3*
C)*a*d)*integrate((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(6*d*x + 6*c)
*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*
c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x
+ 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(
4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c))) - (cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + co
s(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^
2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)))/((cos(2*d*x + 2*c)^4 + sin(2*d*x + 2*c)^4 ...
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Fricas [A]
time = 3.25, size = 89, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left ({\left (25 \, B + 18 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (5 \, B + 9 \, C\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
2/15*((25*B + 18*C)*a*cos(d*x + c)^2 + (5*B + 9*C)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Integral((a*(sec(c + d*x) + 1))**(3/2)*(B + C*sec(c + d*x))*sec(c + d*x), x)
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Giac [A]
time = 1.14, size = 170, normalized size = 1.68 \begin {gather*} \frac {4 \, {\left ({\left (2 \, \sqrt {2} {\left (5 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, 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} - 5 \, \sqrt {2} {\left (5 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \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((a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
4/15*((2*sqrt(2)*(5*B*a^4*sgn(cos(d*x + c)) + 3*C*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 5*sqrt(2)*(5
*B*a^4*sgn(cos(d*x + c)) + 3*C*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(B*a^4*sgn(cos(d*x
+ c)) + C*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a)*d)
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Mupad [B]
time = 7.58, size = 213, normalized size = 2.11 \begin {gather*} -\frac {2\,a\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\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 (B\,25{}\mathrm {i}+C\,18{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,50{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,25{}\mathrm {i}+C\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,18{}\mathrm {i}+C\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,48{}\mathrm {i}+C\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,18{}\mathrm {i}+C\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}\right )}{15\,d\,\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2),x)
[Out]
-(2*a*(exp(c*1i + d*x*1i) - 1)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(B*25i + C*18i +
B*exp(c*1i + d*x*1i)*10i + B*exp(c*2i + d*x*2i)*50i + B*exp(c*3i + d*x*3i)*10i + B*exp(c*4i + d*x*4i)*25i + C*
exp(c*1i + d*x*1i)*18i + C*exp(c*2i + d*x*2i)*48i + C*exp(c*3i + d*x*3i)*18i + C*exp(c*4i + d*x*4i)*18i))/(15*
d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)
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