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

Optimal. Leaf size=216 \[ \frac{(11 B-15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(35 B-39 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{30 a^2 d}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(5 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(65 B-93 C) \tan (c+d x)}{15 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

((11*B - 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) + ((B
- C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((65*B - 93*C)*Tan[c + d*x])/(15*a*d*Sqrt
[a + a*Sec[c + d*x]]) - ((5*B - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((35*B -
39*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(30*a^2*d)

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Rubi [A]  time = 0.725143, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4072, 4019, 4021, 4010, 4001, 3795, 203} \[ \frac{(11 B-15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(35 B-39 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{30 a^2 d}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(5 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(65 B-93 C) \tan (c+d x)}{15 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*B - 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) + ((B
- C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((65*B - 93*C)*Tan[c + d*x])/(15*a*d*Sqrt
[a + a*Sec[c + d*x]]) - ((5*B - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((35*B -
39*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(30*a^2*d)

Rule 4072

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

Rule 4019

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

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 (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=\int \frac{\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=\frac{(B-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 (3 a (B-C)-\frac{1}{2} a (5 B-9 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(5 B-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 B-9 C)+\frac{1}{4} a^2 (35 B-39 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(35 B-39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac{2 \int \frac{\sec (c+d x) \left (\frac{1}{8} a^3 (35 B-39 C)-\frac{1}{4} a^3 (65 B-93 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(65 B-93 C) \tan (c+d x)}{15 a d \sqrt{a+a \sec (c+d x)}}-\frac{(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(35 B-39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac{(11 B-15 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(65 B-93 C) \tan (c+d x)}{15 a d \sqrt{a+a \sec (c+d x)}}-\frac{(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(35 B-39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}-\frac{(11 B-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{(11 B-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{(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(65 B-93 C) \tan (c+d x)}{15 a d \sqrt{a+a \sec (c+d x)}}-\frac{(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(35 B-39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}\\ \end{align*}

Mathematica [A]  time = 2.34895, size = 160, normalized size = 0.74 \[ \frac{\tan (c+d x) \left (\sqrt{1-\sec (c+d x)} \left (4 (5 B-3 C) \sec ^2(c+d x)-12 (5 B-9 C) \sec (c+d x)-95 B+12 C \sec ^3(c+d x)+147 C\right )+15 \sqrt{2} (11 B-15 C) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{30 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*Sqrt[2]*(11*B - 15*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sec[c + d*x] + Sqrt[1 -
Sec[c + d*x]]*(-95*B + 147*C - 12*(5*B - 9*C)*Sec[c + d*x] + 4*(5*B - 3*C)*Sec[c + d*x]^2 + 12*C*Sec[c + d*x]^
3))*Tan[c + d*x])/(30*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.335, size = 793, 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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/240/d/a^2*(-1+cos(d*x+c))*(165*B*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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))-225*C*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(5/2)*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))+495*B*cos(
d*x+c)^2*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)-675*C*cos(d*x+c)^2*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)+495*B*cos(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)*sin(d*x+c)-675*C*cos(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)*sin(d*x+c)+165*B*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)-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)+760*B*cos(d*x+c)^4-1176*C*cos(d*x+c)^4-280*B*cos(d
*x+c)^3+312*C*cos(d*x+c)^3-640*B*cos(d*x+c)^2+960*C*cos(d*x+c)^2+160*B*cos(d*x+c)-192*C*cos(d*x+c)+96*C)*(a*(c
os(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*(B*sec(d*x+c)+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 = 0.649612, size = 1315, normalized size = 6.09 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left ({\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (11 \, B - 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 (95 \, B - 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (5 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 12 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{120 \,{\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{15 \, \sqrt{2}{\left ({\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (11 \, B - 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 (95 \, B - 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (5 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 12 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{60 \,{\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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*sqrt(2)*((11*B - 15*C)*cos(d*x + c)^4 + 2*(11*B - 15*C)*cos(d*x + c)^3 + (11*B - 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*((95*B - 147*C)*cos(d*x + c)
^3 + 12*(5*B - 9*C)*cos(d*x + c)^2 - 4*(5*B - 3*C)*cos(d*x + c) - 12*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), -1/60*(15*sqrt(2)*((11
*B - 15*C)*cos(d*x + c)^4 + 2*(11*B - 15*C)*cos(d*x + c)^3 + (11*B - 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*((95*B - 147*C)*cos(d*x +
c)^3 + 12*(5*B - 9*C)*cos(d*x + c)^2 - 4*(5*B - 3*C)*cos(d*x + c) - 12*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 (B + C \sec{\left (c + d x \right )}\right ) \sec ^{4}{\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*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [A]  time = 9.14024, size = 421, normalized size = 1.95 \begin{align*} \frac{\frac{15 \, \sqrt{2}{\left (11 \, B - 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{15 \, \sqrt{2}{\left (B 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 (245 \, B a^{3} - 381 \, 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 (73 \, B a^{3} - 105 \, 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{15 \, \sqrt{2}{\left (9 \, B 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}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*sqrt(2)*(11*B - 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)) - (((15*sqrt(2)*(B*a^3 - C*a^3)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(t
an(1/2*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(245*B*a^3 - 381*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)*(73*B*a^3 - 105*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
- 15*sqrt(2)*(9*B*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