3.152 \(\int \frac{\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.632908, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4019, 4021, 4010, 4001, 3795, 203} \[ \frac{(11 A-15 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(35 A-39 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{30 a^2 d}+\frac{(A-B) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(5 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(65 A-93 B) \tan (c+d x)}{15 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 2.37172, size = 160, normalized size = 0.74 \[ \frac{\tan (c+d x) \left (\sqrt{1-\sec (c+d x)} \left (4 (5 A-3 B) \sec ^2(c+d x)-12 (5 A-9 B) \sec (c+d x)-95 A+12 B \sec ^3(c+d x)+147 B\right )+15 \sqrt{2} (11 A-15 B) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{30 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((15*Sqrt[2]*(11*A - 15*B)*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*A + 147*B - 12*(5*A - 9*B)*Sec[c + d*x] + 4*(5*A - 3*B)*Sec[c + d*x]^2 + 12*B*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 = 0.299, 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)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/240/d/a^2*(-1+cos(d*x+c))*(165*A*sin(d*x+c)*cos(d*x+c)^3*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)-225*B*sin(d*x+c)*cos(d*x+c)^3*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)+495*A*
sin(d*x+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*co
s(d*x+c)/(cos(d*x+c)+1))^(5/2)-675*B*sin(d*x+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)+495*A*sin(d*x+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)-675*B*s
in(d*x+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)+165*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)-225*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)+760*A*cos(d*x+c)^4-1176*B*cos(d*x+
c)^4-280*A*cos(d*x+c)^3+312*B*cos(d*x+c)^3-640*A*cos(d*x+c)^2+960*B*cos(d*x+c)^2+160*A*cos(d*x+c)-192*B*cos(d*
x+c)+96*B)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)^2

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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)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.610318, size = 1315, normalized size = 6.09 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left ({\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (11 \, A - 15 \, B\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 \, A - 147 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 12 \, B\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 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (11 \, A - 15 \, B\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 \, A - 147 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 12 \, B\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)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*sqrt(2)*((11*A - 15*B)*cos(d*x + c)^4 + 2*(11*A - 15*B)*cos(d*x + c)^3 + (11*A - 15*B)*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*A - 147*B)*cos(d*x + c)
^3 + 12*(5*A - 9*B)*cos(d*x + c)^2 - 4*(5*A - 3*B)*cos(d*x + c) - 12*B)*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
*A - 15*B)*cos(d*x + c)^4 + 2*(11*A - 15*B)*cos(d*x + c)^3 + (11*A - 15*B)*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*A - 147*B)*cos(d*x +
c)^3 + 12*(5*A - 9*B)*cos(d*x + c)^2 - 4*(5*A - 3*B)*cos(d*x + c) - 12*B)*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 + B \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)**4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + B*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 = 9.68822, size = 421, normalized size = 1.95 \begin{align*} \frac{\frac{15 \, \sqrt{2}{\left (11 \, A - 15 \, B\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 (A a^{3} - B 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 \, A a^{3} - 381 \, B 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 \, A a^{3} - 105 \, B 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 \, A a^{3} - 17 \, B 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)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/60*(15*sqrt(2)*(11*A - 15*B)*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)*(A*a^3 - B*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*A*a^3 - 381*B*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*A*a^3 - 105*B*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*A*a^3 - 17*B*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