3.854 \(\int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=446 \[ \frac{2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b d \sqrt{a+b} \left (a^2-b^2\right )^2}-\frac{2 \left (23 a^2 b B-3 a^3 C-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}-\frac{2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (23 a^2 b B-3 a^3 C-29 a b^2 C+9 b^3 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^2 d (a-b)^2 (a+b)^{5/2}} \]

[Out]

(-2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[
a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15*(a
- b)^2*b^2*(a + b)^(5/2)*d) + (2*(3*a^2*(5*B + C) - 8*a*b*(B + 3*C) + b^2*(9*B + 5*C))*Cot[c + d*x]*EllipticF
[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b
*(1 + Sec[c + d*x]))/(a - b))])/(15*b*Sqrt[a + b]*(a^2 - b^2)^2*d) - (2*(b*B - a*C)*Tan[c + d*x])/(5*(a^2 - b^
2)*d*(a + b*Sec[c + d*x])^(5/2)) - (2*(8*a*b*B - 3*a^2*C - 5*b^2*C)*Tan[c + d*x])/(15*(a^2 - b^2)^2*d*(a + b*S
ec[c + d*x])^(3/2)) - (2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*Tan[c + d*x])/(15*(a^2 - b^2)^3*d*Sqrt[
a + b*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.828185, antiderivative size = 446, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {4060, 4058, 12, 3832, 4004} \[ -\frac{2 \left (23 a^2 b B-3 a^3 C-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}-\frac{2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}+\frac{2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b d \sqrt{a+b} \left (a^2-b^2\right )^2}-\frac{2 \left (23 a^2 b B-3 a^3 C-29 a b^2 C+9 b^3 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^2 d (a-b)^2 (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

(-2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[
a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15*(a
- b)^2*b^2*(a + b)^(5/2)*d) + (2*(3*a^2*(5*B + C) - 8*a*b*(B + 3*C) + b^2*(9*B + 5*C))*Cot[c + d*x]*EllipticF
[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b
*(1 + Sec[c + d*x]))/(a - b))])/(15*b*Sqrt[a + b]*(a^2 - b^2)^2*d) - (2*(b*B - a*C)*Tan[c + d*x])/(5*(a^2 - b^
2)*d*(a + b*Sec[c + d*x])^(5/2)) - (2*(8*a*b*B - 3*a^2*C - 5*b^2*C)*Tan[c + d*x])/(15*(a^2 - b^2)^2*d*(a + b*S
ec[c + d*x])^(3/2)) - (2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*Tan[c + d*x])/(15*(a^2 - b^2)^3*d*Sqrt[
a + b*Sec[c + d*x]])

Rule 4060

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

Rule 4058

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[(Csc[e + f*
x]*(1 + Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx &=-\frac{2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \int \frac{-\frac{5}{2} a (a B-b C) \sec (c+d x)+\frac{3}{2} a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{5 a \left (a^2-b^2\right )}\\ &=-\frac{2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac{4 \int \frac{\frac{3}{4} a^2 \left (5 a^2 B+3 b^2 B-8 a b C\right ) \sec (c+d x)-\frac{1}{4} a^2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{15 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}-\frac{8 \int \frac{-\frac{1}{8} a^3 \left (15 a^3 B+17 a b^2 B-27 a^2 b C-5 b^3 C\right ) \sec (c+d x)-\frac{1}{8} a^3 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}\\ &=-\frac{2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}-\frac{8 \int \frac{\left (\frac{1}{8} a^3 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right )-\frac{1}{8} a^3 \left (15 a^3 B+17 a b^2 B-27 a^2 b C-5 b^3 C\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}+\frac{\left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=-\frac{2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}-\frac{2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 (a-b)^2 (a+b)^3}\\ &=-\frac{2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac{2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b (a+b)^{5/2} d}-\frac{2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 24.7313, size = 3729, normalized size = 8.36 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

((b + a*Cos[c + d*x])^4*Sec[c + d*x]^4*((-2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*Sin[c + d*x])/(15*b*
(-a^2 + b^2)^3) - (2*(b^3*B*Sin[c + d*x] - a*b^2*C*Sin[c + d*x]))/(5*a^2*(a^2 - b^2)*(b + a*Cos[c + d*x])^3) -
(2*(-14*a^2*b^2*B*Sin[c + d*x] + 6*b^4*B*Sin[c + d*x] + 9*a^3*b*C*Sin[c + d*x] - a*b^3*C*Sin[c + d*x]))/(15*a
^2*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (2*(-34*a^4*b*B*Sin[c + d*x] + 5*a^2*b^3*B*Sin[c + d*x] - 3*b^5*B*S
in[c + d*x] + 9*a^5*C*Sin[c + d*x] + 25*a^3*b^2*C*Sin[c + d*x] - 2*a*b^4*C*Sin[c + d*x]))/(15*a^2*(a^2 - b^2)^
3*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(7/2)) - (2*(b + a*Cos[c + d*x])^3*((23*a^2*b*B)/(15*(-a^2 +
b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (3*b^3*B)/(5*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sq
rt[Sec[c + d*x]]) - (a^3*C)/(5*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (29*a*b^2*C)/(15*
(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (8*a^3*B*Sqrt[Sec[c + d*x]])/(15*(-a^2 + b^2)^3*
Sqrt[b + a*Cos[c + d*x]]) - (8*a*b^2*B*Sqrt[Sec[c + d*x]])/(15*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (a^4
*C*Sqrt[Sec[c + d*x]])/(5*b*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (2*a^2*b*C*Sqrt[Sec[c + d*x]])/(15*(-a^
2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) + (b^3*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]])
+ (23*a^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) + (3*a*b^2*B*Cos
[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (a^4*C*Cos[2*(c + d*x)]*Sqrt[S
ec[c + d*x]])/(5*b*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (29*a^2*b*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])
/(15*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]))*Sec[c + d*x]^(7/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a
+ b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c +
d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(b^2*(9
*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])
/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-23*a^2*b*B - 9*b^3*B +
3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(15*b*(-a^2 + b
^2)^3*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(7/2)*(-(a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c +
d*x]*(2*(a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b
+ a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a
+ b)*(b^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*C
os[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-23*a^2*b*B
- 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(15
*b*(-a^2 + b^2)^3*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) + (Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]
]*Tan[(c + d*x)/2]*(2*(a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*
x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a +
b)] + 2*b*(a + b)*(b^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*
Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] +
(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c +
d*x)/2]))/(15*b*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) - (2*Sqrt[Cos[(c + d*x)/2]^2
*Sec[c + d*x]]*(((-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x
)/2]^4)/2 + ((a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Co
s[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d
*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + (b*(a + b)*(b^2*(9*B - 5*C)
+ 8*a*b*(B - 3*C) + 3*a^2*(5*B - C))*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin
[Tan[(c + d*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos
[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + ((a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*S
qrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/
((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b +
a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + (b*(a + b)*(b^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B -
C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d
*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt
[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] - a*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c +
d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] - (-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*(b + a
*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] + (-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*
C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2 + (b*(a + b)*(b^2*(9*B - 5*C) + 8*a
*b*(B - 3*C) + 3*a^2*(5*B - C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 +
Cos[c + d*x]))]*Sec[(c + d*x)/2]^2)/(Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b
)]) + ((a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b +
a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Sec[(c + d*x)/2]^2*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)
])/Sqrt[1 - Tan[(c + d*x)/2]^2]))/(15*b*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) - ((
2*(a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos
[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(b
^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c +
d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-23*a^2*b*B - 9*b^
3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])*(-(Cos[(c +
d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(15*b*(-a^2 + b^2)^3*
Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]])))

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Maple [B]  time = 0.577, size = 7695, normalized size = 17.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x)

[Out]

result too large to display

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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((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),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 (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{b^{4} \sec \left (d x + c\right )^{4} + 4 \, a b^{3} \sec \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sec \left (d x + c\right )^{2} + 4 \, a^{3} b \sec \left (d x + c\right ) + a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^4*sec(d*x + c)^4 + 4*a*b^3*sec(d*x +
c)^3 + 6*a^2*b^2*sec(d*x + c)^2 + 4*a^3*b*sec(d*x + c) + a^4), 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((B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))/(b*sec(d*x + c) + a)^(7/2), x)