3.193 \(\int \frac{\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=214 \[ -\frac{(7 A+15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(5 A+13 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{10 a^2 d}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(5 A+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}+\frac{(15 A+31 C) \tan (c+d x)}{5 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((7*A + 15*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A
+ C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((15*A + 31*C)*Tan[c + d*x])/(5*a*d*Sqrt[
a + a*Sec[c + d*x]]) + ((5*A + 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) - ((5*A + 1
3*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(10*a^2*d)

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Rubi [A]  time = 0.6272, antiderivative size = 214, normalized size of antiderivative = 1., 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 = {4085, 4021, 4010, 4001, 3795, 203} \[ -\frac{(7 A+15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(5 A+13 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{10 a^2 d}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(5 A+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}+\frac{(15 A+31 C) \tan (c+d x)}{5 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*A + 15*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A
+ C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((15*A + 31*C)*Tan[c + d*x])/(5*a*d*Sqrt[
a + a*Sec[c + d*x]]) + ((5*A + 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) - ((5*A + 1
3*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(10*a^2*d)

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)]

Rule 4021

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

Rule 4010

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)), I
nt[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] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

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 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 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])

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\sec ^3(c+d x) \left (a (A+3 C)-\frac{1}{2} a (5 A+9 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{\int \frac{\sec ^2(c+d x) \left (-a^2 (5 A+9 C)+\frac{3}{4} a^2 (5 A+13 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(5 A+13 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac{2 \int \frac{\sec (c+d x) \left (\frac{3}{8} a^3 (5 A+13 C)-\frac{3}{4} a^3 (15 A+31 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(15 A+31 C) \tan (c+d x)}{5 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(5 A+13 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac{(7 A+15 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(15 A+31 C) \tan (c+d x)}{5 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(5 A+13 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}+\frac{(7 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 )}{2 a d}\\ &=-\frac{(7 A+15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(15 A+31 C) \tan (c+d x)}{5 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(5 A+13 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}\\ \end{align*}

Mathematica [A]  time = 4.90577, size = 189, normalized size = 0.88 \[ \frac{\sin (c+d x) \cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (\sec ^3(c+d x) ((75 A+131 C) \cos (c+d x)+8 (5 A+9 C) \cos (2 (c+d x))+25 A \cos (3 (c+d x))+40 A+49 C \cos (3 (c+d x))+88 C)-10 \sqrt{2} (7 A+15 C) \cot ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right )\right )}{20 d (a (\sec (c+d x)+1))^{3/2} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*(-10*Sqrt[2]*(7*A + 15*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cot[(c
+ d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] + (40*A + 88*C + (75*A + 131*C)*Cos[c + d*x] + 8*(5*A + 9*C)*Cos[2*(c + d*
x)] + 25*A*Cos[3*(c + d*x)] + 49*C*Cos[3*(c + d*x)])*Sec[c + d*x]^3)*Sin[c + d*x])/(20*d*(A + 2*C + A*Cos[2*(c
+ d*x)])*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.339, size = 784, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80/d/a^2*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)
+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)+75*C*cos(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)+105*A*ln(-
(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/
2)*sin(d*x+c)*cos(d*x+c)^2+225*C*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c
))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)*cos(d*x+c)^2+105*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)*cos(d*x+c)+225*C*ln(-(
-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2
)*sin(d*x+c)*cos(d*x+c)+35*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+75*C*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(
d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+200*A*cos(d*x+c)^4+392*C*cos(d*x+c)^4-40
*A*cos(d*x+c)^3-104*C*cos(d*x+c)^3-160*A*cos(d*x+c)^2-320*C*cos(d*x+c)^2+64*C*cos(d*x+c)-32*C)*(a*(cos(d*x+c)+
1)/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 0.626887, size = 1268, normalized size = 5.93 \begin{align*} \left [-\frac{5 \, \sqrt{2}{\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2}\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 (25 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{40 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac{5 \, \sqrt{2}{\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2}\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 (25 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{20 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/40*(5*sqrt(2)*((7*A + 15*C)*cos(d*x + c)^4 + 2*(7*A + 15*C)*cos(d*x + c)^3 + (7*A + 15*C)*cos(d*x + c)^2)*
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*((25*A + 49*C)*cos(d*x + c)^3 +
4*(5*A + 9*C)*cos(d*x + c)^2 - 4*C*cos(d*x + c) + 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(
a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2), 1/20*(5*sqrt(2)*((7*A + 15*C)*cos(d*x +
c)^4 + 2*(7*A + 15*C)*cos(d*x + c)^3 + (7*A + 15*C)*cos(d*x + c)^2)*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*((25*A + 49*C)*cos(d*x + c)^3 + 4*(5*A + 9*C)*c
os(d*x + c)^2 - 4*C*cos(d*x + c) + 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x +
c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 9.49401, size = 420, normalized size = 1.96 \begin{align*} -\frac{\frac{5 \, \sqrt{2}{\left (7 \, A + 15 \, 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 \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{{\left ({\left ({\left (\frac{5 \, \sqrt{2}{\left (A a^{3} + C a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (55 \, A a^{3} + 127 \, C a^{3}\right )}}{a^{2} \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} + \frac{5 \, \sqrt{2}{\left (19 \, A a^{3} + 35 \, C a^{3}\right )}}{a^{2} \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} - \frac{5 \, \sqrt{2}{\left (9 \, A a^{3} + 17 \, C a^{3}\right )}}{a^{2} \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 )}{{\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}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(5*sqrt(2)*(7*A + 15*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*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - (((5*sqrt(2)*(A*a^3 + C*a^3)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(tan
(1/2*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(55*A*a^3 + 127*C*a^3)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x
+ 1/2*c)^2 + 5*sqrt(2)*(19*A*a^3 + 35*C*a^3)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 - 5
*sqrt(2)*(9*A*a^3 + 17*C*a^3)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*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