3.215 \(\int \frac{A+B \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=244 \[ -\frac{5 (3 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}-\frac{(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac{7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{7 (8 A-5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

[Out]

(7*(8*A - 5*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A - 2*B)*Sqr
t[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (7*(8*A - 5*B)*Sin[c + d*x])/(15*a^2
*d*Sec[c + d*x]^(3/2)) - (5*(3*A - 2*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3*A - 2*B)*Sin[c + d*x]
)/(a^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c
+ d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.381588, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4020, 3787, 3769, 3771, 2639, 2641} \[ -\frac{(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac{7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{5 (3 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{7 (8 A-5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(7*(8*A - 5*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A - 2*B)*Sqr
t[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (7*(8*A - 5*B)*Sin[c + d*x])/(15*a^2
*d*Sec[c + d*x]^(3/2)) - (5*(3*A - 2*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3*A - 2*B)*Sin[c + d*x]
)/(a^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c
+ d*x])^2)

Rule 4020

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a (11 A-5 B)-\frac{7}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{7}{2} a^2 (8 A-5 B)-\frac{15}{2} a^2 (3 A-2 B) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac{(7 (8 A-5 B)) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{6 a^2}-\frac{(5 (3 A-2 B)) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}\\ &=\frac{7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac{(7 (8 A-5 B)) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{10 a^2}-\frac{(5 (3 A-2 B)) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}\\ &=\frac{7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac{\left (7 (8 A-5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{10 a^2}-\frac{\left (5 (3 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{7 (8 A-5 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 a^2 d}-\frac{5 (3 A-2 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 6.90719, size = 946, normalized size = 3.88 \[ -\frac{56 \sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{7 \sqrt{2} B e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}-\frac{10 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{20 B \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{\sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \left (\frac{2 \sec \left (\frac{c}{2}\right ) \left (B \sin \left (\frac{d x}{2}\right )-A \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{2 (B-A) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{4 \sec \left (\frac{c}{2}\right ) \left (10 B \sin \left (\frac{d x}{2}\right )-13 A \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{(-73 \cos (2 c) A-151 A+100 B+40 B \cos (2 c)) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{10 d}+\frac{4 (B-2 A) \cos (2 d x) \sin (2 c)}{3 d}+\frac{2 A \cos (3 d x) \sin (3 c)}{5 d}-\frac{2 (40 B-73 A) \cos (c) \sin (d x)}{5 d}+\frac{4 (B-2 A) \cos (2 c) \sin (2 d x)}{3 d}+\frac{2 A \cos (3 c) \sin (3 d x)}{5 d}-\frac{4 (10 B-13 A) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(B+A \cos (c+d x)) (\sec (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(-56*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2
]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4,
7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(15*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(
a + a*Sec[c + d*x])^2) + (7*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d
*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hyper
geometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(3*d*E^(I*d*x)*
(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) - (10*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*Elliptic
F[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[
c + d*x])^2) + (20*B*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c
+ d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x
)/2]^4*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*(((-151*A + 100*B - 73*A*Cos[2*c] + 40*B*Cos[2*c])*Cos[d*x]*Csc
[c/2]*Sec[c/2])/(10*d) + (4*(-2*A + B)*Cos[2*d*x]*Sin[2*c])/(3*d) + (2*A*Cos[3*d*x]*Sin[3*c])/(5*d) + (2*Sec[c
/2]*Sec[c/2 + (d*x)/2]^3*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(3*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-13*A*S
in[(d*x)/2] + 10*B*Sin[(d*x)/2]))/(3*d) - (2*(-73*A + 40*B)*Cos[c]*Sin[d*x])/(5*d) + (4*(-2*A + B)*Cos[2*c]*Si
n[2*d*x])/(3*d) + (2*A*Cos[3*c]*Sin[3*d*x])/(5*d) - (4*(-13*A + 10*B)*Tan[c/2])/(3*d) + (2*(-A + B)*Sec[c/2 +
(d*x)/2]^2*Tan[c/2])/(3*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2)

________________________________________________________________________________________

Maple [A]  time = 2.023, size = 465, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/30/a^2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*cos(1/2*d*x+1/2*c)^10-352*A*cos(1/2*d*
x+1/2*c)^8+80*B*cos(1/2*d*x+1/2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6-150*A*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-336*A*cos(1/2*d*x+1/2*c)^3*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+60*B*cos(1
/2*d*x+1/2*c)^6+100*B*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+210*B*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*
c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+266*A*cos(1/2*d*x+1/2*c)^4-240*B*cos(1/2*d*x+1/2*c)^4-135*
A*cos(1/2*d*x+1/2*c)^2+105*B*cos(1/2*d*x+1/2*c)^2+5*A-5*B)/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{5} + 2 \, a^{2} \sec \left (d x + c\right )^{4} + a^{2} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a^2*sec(d*x + c)^5 + 2*a^2*sec(d*x + c)^4 + a^2*sec(d*x + c)
^3), x)

________________________________________________________________________________________

Sympy [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((A+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)), x)