3.853 \(\int \frac{\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=495 \[ -\frac{2 \left (6 a^2 b B+a^2 b C-3 a^3 C-a b^2 B-3 b^3 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 a^2 b d (a-b) (a+b)^{3/2}}+\frac{2 b \left (7 a^2 b B-4 a^3 C-3 b^3 B\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 b (b B-a C) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (7 a^2 b B-4 a^3 C-3 b^3 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b d (a-b) (a+b)^{3/2}}-\frac{2 B \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^3 d} \]

[Out]

(2*(7*a^2*b*B - 3*b^3*B - 4*a^3*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))])/(3*a^2*(a - b)*b*(a +
b)^(3/2)*d) - (2*(6*a^2*b*B - a*b^2*B - 3*b^3*B - 3*a^3*C + a^2*b*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))])/(3*a^2*(a - b)*b*(a + b)^(3/2)*d) - (2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[S
qrt[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))])/(a^3*d) + (2*b*(b*B - a*C)*Tan[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2))
+ (2*b*(7*a^2*b*B - 3*b^3*B - 4*a^3*C)*Tan[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.856809, antiderivative size = 495, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4072, 3923, 4060, 4058, 3921, 3784, 3832, 4004} \[ \frac{2 b \left (7 a^2 b B-4 a^3 C-3 b^3 B\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 b (b B-a C) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (6 a^2 b B+a^2 b C-3 a^3 C-a b^2 B-3 b^3 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b d (a-b) (a+b)^{3/2}}+\frac{2 \left (7 a^2 b B-4 a^3 C-3 b^3 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b d (a-b) (a+b)^{3/2}}-\frac{2 B \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*a^2*b*B - 3*b^3*B - 4*a^3*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))])/(3*a^2*(a - b)*b*(a +
b)^(3/2)*d) - (2*(6*a^2*b*B - a*b^2*B - 3*b^3*B - 3*a^3*C + a^2*b*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))])/(3*a^2*(a - b)*b*(a + b)^(3/2)*d) - (2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[S
qrt[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))])/(a^3*d) + (2*b*(b*B - a*C)*Tan[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2))
+ (2*b*(7*a^2*b*B - 3*b^3*B - 4*a^3*C)*Tan[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

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 3923

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*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[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

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 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

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{\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=\int \frac{B+C \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\\ &=\frac{2 b (b B-a C) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{-\frac{3}{2} \left (a^2-b^2\right ) B+\frac{3}{2} a (b B-a C) \sec (c+d x)-\frac{1}{2} b (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{2 b (b B-a C) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 b B-3 b^3 B-4 a^3 C\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} \left (a^2-b^2\right )^2 B-\frac{1}{4} a \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \sec (c+d x)-\frac{1}{4} b \left (7 a^2 b B-3 b^3 B-4 a^3 C\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 b (b B-a C) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 b B-3 b^3 B-4 a^3 C\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} \left (a^2-b^2\right )^2 B+\left (\frac{1}{4} b \left (7 a^2 b B-3 b^3 B-4 a^3 C\right )-\frac{1}{4} a \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}-\frac{\left (b \left (7 a^2 b B-3 b^3 B-4 a^3 C\right )\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (7 a^2 b B-3 b^3 B-4 a^3 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}}}{3 a^2 (a-b) b (a+b)^{3/2} d}+\frac{2 b (b B-a C) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 b B-3 b^3 B-4 a^3 C\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{B \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{a^2}+\frac{\left (a b^2 B+3 b^3 B+3 a^3 C-a^2 b (6 B+C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 (a-b) (a+b)^2}\\ &=\frac{2 \left (7 a^2 b B-3 b^3 B-4 a^3 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}}}{3 a^2 (a-b) b (a+b)^{3/2} d}+\frac{2 \left (a b^2 B+3 b^3 B+3 a^3 C-a^2 b (6 B+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}}}{3 a^2 (a-b) b (a+b)^{3/2} d}-\frac{2 \sqrt{a+b} B \cot (c+d x) \Pi \left (\frac{a+b}{a};\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}}}{a^3 d}+\frac{2 b (b B-a C) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 b B-3 b^3 B-4 a^3 C\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 16.2126, size = 2039, normalized size = 4.12 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*(-7*a^2*b*B + 3*b^3*B + 4*a^3*C)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2
) - (2*(b^3*B*Sin[c + d*x] - a*b^2*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) - (2*(-8*a^2*b^
2*B*Sin[c + d*x] + 4*b^4*B*Sin[c + d*x] + 5*a^3*b*C*Sin[c + d*x] - a*b^3*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2
*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)*Sqr
t[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(7*a^3*b*Sqrt[(-a + b)/(a +
b)]*B*Tan[(c + d*x)/2] + 7*a^2*b^2*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2] - 3*a*b^3*Sqrt[(-a + b)/(a + b)]*
B*Tan[(c + d*x)/2] - 3*b^4*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2] - 4*a^4*Sqrt[(-a + b)/(a + b)]*C*Tan[(c +
d*x)/2] - 4*a^3*b*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/2] - 14*a^3*b*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)
/2]^3 + 6*a*b^3*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^3 + 8*a^4*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/2]^
3 + 7*a^3*b*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5 - 7*a^2*b^2*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^
5 - 3*a*b^3*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5 + 3*b^4*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5 -
4*a^4*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/2]^5 + 4*a^3*b*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/2]^5 - (6*I
)*a^4*B*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sq
rt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (12*I)*a^2*b^
2*B*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1
- Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (6*I)*b^4*B*Ellip
ticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c
+ d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (6*I)*a^4*B*EllipticPi[-((
a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[
1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (12*I)*a^2*b^2*B
*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c +
d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (
6*I)*b^4*B*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]
*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a
+ b)] + I*(a - b)*(-7*a^2*b*B + 3*b^3*B + 4*a^3*C)*EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2
]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2
+ b*Tan[(c + d*x)/2]^2)/(a + b)] + I*(a - b)*(-4*a*b^2*B - 6*b^3*B + 3*a^3*(B - C) + a^2*b*(9*B + C))*Elliptic
F[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(
c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(3*a^2*Sqrt[(-a + b)/(a +
b)]*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^(5/2)*(-1 + Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - T
an[(c + d*x)/2]^2)]*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2)))

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Maple [B]  time = 0.426, size = 5710, normalized size = 11.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)*sec(d*x + c)^2 + B*cos(d*x + c)*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^3*sec(d*x +
c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*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(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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