3.2.64 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\) [164]

Optimal. Leaf size=225 \[ \frac {2 a^2 (143 A+112 C) \tan (c+d x)}{165 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (143 A+112 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac {2 (143 A+112 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{385 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d} \]

[Out]

2/385*(143*A+112*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/11*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/
d+2/165*a^2*(143*A+112*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/231*a^2*(33*A+28*C)*sec(d*x+c)^3*tan(d*x+c)/d/
(a+a*sec(d*x+c))^(1/2)-4/1155*a*(143*A+112*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/33*a*C*sec(d*x+c)^3*(a+a*s
ec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.44, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4174, 4103, 4101, 3885, 4086, 3877} \begin {gather*} \frac {2 a^2 (33 A+28 C) \tan (c+d x) \sec ^3(c+d x)}{231 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (143 A+112 C) \tan (c+d x)}{165 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (143 A+112 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{385 d}-\frac {4 a (143 A+112 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{1155 d}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac {2 a C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{33 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(2*a^2*(143*A + 112*C)*Tan[c + d*x])/(165*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(33*A + 28*C)*Sec[c + d*x]^3*Ta
n[c + d*x])/(231*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*(143*A + 112*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11
55*d) + (2*a*C*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(33*d) + (2*(143*A + 112*C)*(a + a*Sec[c
+ d*x])^(3/2)*Tan[c + d*x])/(385*d) + (2*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(11*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 3885

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-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*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

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 4101

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n
, 0] &&  !LtQ[n, 0]

Rule 4103

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m +
n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d
*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] &&
NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -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 ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {2 \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (11 A+6 C)+\frac {3}{2} a C \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 a C \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {4 \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {9}{4} a^2 (11 A+8 C)+\frac {3}{4} a^2 (33 A+28 C) \sec (c+d x)\right ) \, dx}{99 a}\\ &=\frac {2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{77} (a (143 A+112 C)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac {2 (143 A+112 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{385 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{385} (2 (143 A+112 C)) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (143 A+112 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac {2 (143 A+112 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{385 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{165} (a (143 A+112 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (143 A+112 C) \tan (c+d x)}{165 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (143 A+112 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{1155 d}+\frac {2 a C \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac {2 (143 A+112 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{385 d}+\frac {2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]
time = 1.43, size = 144, normalized size = 0.64 \begin {gather*} \frac {a (1188 A+1652 C+(4147 A+4228 C) \cos (c+d x)+2 (737 A+728 C) \cos (2 (c+d x))+1859 A \cos (3 (c+d x))+1456 C \cos (3 (c+d x))+286 A \cos (4 (c+d x))+224 C \cos (4 (c+d x))+286 A \cos (5 (c+d x))+224 C \cos (5 (c+d x))) \sec ^5(c+d x) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{2310 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a*(1188*A + 1652*C + (4147*A + 4228*C)*Cos[c + d*x] + 2*(737*A + 728*C)*Cos[2*(c + d*x)] + 1859*A*Cos[3*(c +
d*x)] + 1456*C*Cos[3*(c + d*x)] + 286*A*Cos[4*(c + d*x)] + 224*C*Cos[4*(c + d*x)] + 286*A*Cos[5*(c + d*x)] + 2
24*C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(2310*d)

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Maple [A]
time = 10.51, size = 152, normalized size = 0.68

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (1144 A \left (\cos ^{5}\left (d x +c \right )\right )+896 C \left (\cos ^{5}\left (d x +c \right )\right )+572 A \left (\cos ^{4}\left (d x +c \right )\right )+448 C \left (\cos ^{4}\left (d x +c \right )\right )+429 A \left (\cos ^{3}\left (d x +c \right )\right )+336 C \left (\cos ^{3}\left (d x +c \right )\right )+165 A \left (\cos ^{2}\left (d x +c \right )\right )+280 C \left (\cos ^{2}\left (d x +c \right )\right )+245 C \cos \left (d x +c \right )+105 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{1155 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-2/1155/d*(-1+cos(d*x+c))*(1144*A*cos(d*x+c)^5+896*C*cos(d*x+c)^5+572*A*cos(d*x+c)^4+448*C*cos(d*x+c)^4+429*A*
cos(d*x+c)^3+336*C*cos(d*x+c)^3+165*A*cos(d*x+c)^2+280*C*cos(d*x+c)^2+245*C*cos(d*x+c)+105*C)*(a*(1+cos(d*x+c)
)/cos(d*x+c))^(1/2)/cos(d*x+c)^5/sin(d*x+c)*a

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 2.48, size = 139, normalized size = 0.62 \begin {gather*} \frac {2 \, {\left (8 \, {\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{5} + 4 \, {\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{4} + 3 \, {\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (33 \, A + 56 \, C\right )} a \cos \left (d x + c\right )^{2} + 245 \, C a \cos \left (d x + c\right ) + 105 \, 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 )}{1155 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/1155*(8*(143*A + 112*C)*a*cos(d*x + c)^5 + 4*(143*A + 112*C)*a*cos(d*x + c)^4 + 3*(143*A + 112*C)*a*cos(d*x
+ c)^3 + 5*(33*A + 56*C)*a*cos(d*x + c)^2 + 245*C*a*cos(d*x + c) + 105*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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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 (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**(3/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(A + C*sec(c + d*x)**2)*sec(c + d*x)**3, x)

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Giac [A]
time = 1.39, size = 305, normalized size = 1.36 \begin {gather*} \frac {4 \, {\left ({\left ({\left ({\left ({\left (2 \, \sqrt {2} {\left (209 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 161 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} {\left (209 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 161 \, C a^{7} \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} + 132 \, \sqrt {2} {\left (37 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 28 \, C a^{7} \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} - 154 \, \sqrt {2} {\left (37 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 33 \, C a^{7} \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} + 770 \, \sqrt {2} {\left (5 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{7} \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} - 1155 \, \sqrt {2} {\left (A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{1155 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \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)^3*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

4/1155*(((((2*sqrt(2)*(209*A*a^7*sgn(cos(d*x + c)) + 161*C*a^7*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 11*
sqrt(2)*(209*A*a^7*sgn(cos(d*x + c)) + 161*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 132*sqrt(2)*(37*
A*a^7*sgn(cos(d*x + c)) + 28*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 154*sqrt(2)*(37*A*a^7*sgn(cos(
d*x + c)) + 33*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 770*sqrt(2)*(5*A*a^7*sgn(cos(d*x + c)) + 3*C
*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 1155*sqrt(2)*(A*a^7*sgn(cos(d*x + c)) + C*a^7*sgn(cos(d*x +
c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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Mupad [B]
time = 13.15, size = 751, normalized size = 3.34 \begin {gather*} \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\,a\,24{}\mathrm {i}}{7\,d}+\frac {a\,\left (A+12\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {C\,a\,416{}\mathrm {i}}{231\,d}\right )-\frac {a\,\left (3\,A+4\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {A\,a\,8{}\mathrm {i}}{7\,d}+\frac {C\,a\,32{}\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\,\left (5\,A+4\,C\right )\,8{}\mathrm {i}}{11\,d}-\frac {a\,\left (7\,A+12\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {A\,a\,16{}\mathrm {i}}{11\,d}\right )+\frac {a\,\left (5\,A+4\,C\right )\,8{}\mathrm {i}}{11\,d}-\frac {a\,\left (7\,A+12\,C\right )\,8{}\mathrm {i}}{11\,d}+\frac {A\,a\,16{}\mathrm {i}}{11\,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 )}^5}+\frac {\left (\frac {A\,a\,8{}\mathrm {i}}{3\,d}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (143\,A+112\,C\right )\,8{}\mathrm {i}}{1155\,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 (-{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a\,8{}\mathrm {i}}{3\,d}-\frac {a\,\left (A+3\,C\right )\,32{}\mathrm {i}}{9\,d}+\frac {a\,\left (A-8\,C\right )\,8{}\mathrm {i}}{9\,d}-\frac {C\,a\,64{}\mathrm {i}}{99\,d}\right )+\frac {a\,\left (3\,A+8\,C\right )\,8{}\mathrm {i}}{9\,d}-\frac {a\,\left (A+C\right )\,32{}\mathrm {i}}{9\,d}+\frac {A\,a\,8{}\mathrm {i}}{9\,d}+\frac {C\,a\,64{}\mathrm {i}}{9\,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 )}^4}+\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\,\left (11\,A-14\,C\right )\,16{}\mathrm {i}}{385\,d}+\frac {A\,a\,24{}\mathrm {i}}{5\,d}\right )-\frac {A\,a\,8{}\mathrm {i}}{5\,d}+\frac {a\,\left (A+2\,C\right )\,16{}\mathrm {i}}{5\,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 {a\,{\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 (143\,A+112\,C\right )\,16{}\mathrm {i}}{1155\,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))^(3/2))/cos(c + d*x)^3,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(A + 12*C)*8i)/(7*d) -
(A*a*24i)/(7*d) + (C*a*416i)/(231*d)) - (a*(3*A + 4*C)*8i)/(7*d) + (A*a*8i)/(7*d) + (C*a*32i)/(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)*((a*(5*A + 4*C)*8i)/(11*d) - (a*(7*A + 12*C)*8i)/(11*d) + (A*a*16i)/(11*d)) + (a*(5*A + 4
*C)*8i)/(11*d) - (a*(7*A + 12*C)*8i)/(11*d) + (A*a*16i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i)
+ 1)^5) + (((A*a*8i)/(3*d) - (a*exp(c*1i + d*x*1i)*(143*A + 112*C)*8i)/(1155*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*(3*A + 8*C)*8i)/(9*d) - exp(c*1i + d*x*1i)*((A*a*8i)/(3*d) - (a*
(A + 3*C)*32i)/(9*d) + (a*(A - 8*C)*8i)/(9*d) - (C*a*64i)/(99*d)) - (a*(A + C)*32i)/(9*d) + (A*a*8i)/(9*d) + (
C*a*64i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp
(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(11*A - 14*C)*16i)/(385*d) + (A*a*24i)/(5*d)) - (A*a*8i)/(5*
d) + (a*(A + 2*C)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (a*exp(c*1i + d*x*1i)*(
a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(143*A + 112*C)*16i)/(1155*d*(exp(c*1i + d*x*1i)
+ 1))

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