3.216 \(\int \frac{\sec ^{\frac{9}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=292 \[ -\frac{(13 A-33 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}+\frac{7 (7 A-17 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{30 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(13 A-33 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 a^3 d}+\frac{7 (7 A-17 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{7 (7 A-17 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A-B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-2 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{3 a d (a \sec (c+d x)+a)^2} \]

[Out]

(-7*(7*A - 17*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*A - 33*B)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + (7*(7*A - 17*B)*Sqrt[Sec[c + d*x]
]*Sin[c + d*x])/(10*a^3*d) - ((13*A - 33*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*a^3*d) + ((A - B)*Sec[c + d*x]
^(9/2)*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((A - 2*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*a*d*(a + a*
Sec[c + d*x])^2) + (7*(7*A - 17*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(30*d*(a^3 + a^3*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.559542, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4019, 3787, 3768, 3771, 2639, 2641} \[ \frac{7 (7 A-17 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{30 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(13 A-33 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 a^3 d}+\frac{7 (7 A-17 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-33 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{7 (7 A-17 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A-B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-2 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{3 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^(9/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(-7*(7*A - 17*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*A - 33*B)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + (7*(7*A - 17*B)*Sqrt[Sec[c + d*x]
]*Sin[c + d*x])/(10*a^3*d) - ((13*A - 33*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*a^3*d) + ((A - B)*Sec[c + d*x]
^(9/2)*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((A - 2*B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*a*d*(a + a*
Sec[c + d*x])^2) + (7*(7*A - 17*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(30*d*(a^3 + a^3*Sec[c + d*x]))

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 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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{\sec ^{\frac{9}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac{(A-B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^{\frac{7}{2}}(c+d x) \left (\frac{7}{2} a (A-B)-\frac{1}{2} a (3 A-13 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (\frac{25}{2} a^2 (A-2 B)-\frac{3}{2} a^2 (8 A-23 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \sec ^{\frac{3}{2}}(c+d x) \left (\frac{21}{4} a^3 (7 A-17 B)-\frac{15}{4} a^3 (13 A-33 B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(A-B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(13 A-33 B) \int \sec ^{\frac{5}{2}}(c+d x) \, dx}{4 a^3}+\frac{(7 (7 A-17 B)) \int \sec ^{\frac{3}{2}}(c+d x) \, dx}{20 a^3}\\ &=\frac{7 (7 A-17 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac{(13 A-33 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 a^3 d}+\frac{(A-B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(13 A-33 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}-\frac{(7 (7 A-17 B)) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=\frac{7 (7 A-17 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac{(13 A-33 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 a^3 d}+\frac{(A-B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left ((13 A-33 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}-\frac{\left (7 (7 A-17 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac{7 (7 A-17 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-33 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}+\frac{7 (7 A-17 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac{(13 A-33 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 a^3 d}+\frac{(A-B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-2 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac{7 (7 A-17 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 7.96317, size = 953, normalized size = 3.26 \[ \frac{49 \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 ^2(c+d x) (A+B \sec (c+d x)) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}-\frac{119 \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 ^2(c+d x) (A+B \sec (c+d x)) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}-\frac{26 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{5}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}+\frac{22 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{5}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^3}+\frac{\sec ^{\frac{5}{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 ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}+\frac{2 (B-A) \tan \left (\frac{c}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \left (13 B \sin \left (\frac{d x}{2}\right )-8 A \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d}+\frac{4 (13 B-8 A) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \left (29 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{14 (17 B-7 A) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{5 d}+\frac{16 B \sec (c) \sec (c+d x) \sin (d x)}{3 d}+\frac{4 (33 \cos (c) B+4 B-13 A \cos (c)) \sec (c) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{(B+A \cos (c+d x)) (\sec (c+d x) a+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^(9/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(49*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]
^6*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]^2*(A + B*Sec[c + d*x]))/(15*d*E^(I*d*x)*(B + A*Cos[c + d*x])*
(a + a*Sec[c + d*x])^3) - (119*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]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hy
pergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(15*d*E^(I
*d*x)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) - (26*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*El
lipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d*(B + A*Cos[c + d*x])*(a
+ a*Sec[c + d*x])^3) + (22*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/
2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*Sin[c])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (Cos[c/2
+ (d*x)/2]^6*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*((-14*(-7*A + 17*B)*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) + (
2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3
*(-8*A*Sin[(d*x)/2] + 13*B*Sin[(d*x)/2]))/(15*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-13*A*Sin[(d*x)/2] + 29*B*S
in[(d*x)/2]))/(3*d) + (16*B*Sec[c]*Sec[c + d*x]*Sin[d*x])/(3*d) + (4*(4*B - 13*A*Cos[c] + 33*B*Cos[c])*Sec[c]*
Tan[c/2])/(3*d) + (4*(-8*A + 13*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(-A + B)*Sec[c/2 + (d*x)/2]^4*Ta
n[c/2])/(5*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3)

________________________________________________________________________________________

Maple [B]  time = 2.949, size = 876, normalized size = 3. \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)^(9/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x)

[Out]

-1/60*(4*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-165*B*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+357*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^6-10*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-165
*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+357*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^4*cos
(1/2*d*x+1/2*c)+8*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
-165*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+357*B*EllipticE(cos(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*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))-165*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+357*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*
c)-168*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(7*A-17*B)*sin(1/2*d*x+1/2*c)^10+8*(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(482*A-1167*B)*sin(1/2*d*x+1/2*c)^8-10*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)*(461*A-1111*B)*sin(1/2*d*x+1/2*c)^6+14*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)*(169*A-404*B)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(439*A-1029*B)*sin(1
/2*d*x+1/2*c)^2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/a^3/cos(1/2*d*x+1/2*c)^5/(-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)/d

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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)^(9/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(9/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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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(sec(d*x+c)**(9/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{9}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(9/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^(9/2)/(a*sec(d*x + c) + a)^3, x)