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

Optimal. Leaf size=582 \[ \frac {\sqrt {a+b} (5 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^4 d}+\frac {b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\left (-3 a^3 (A-4 B)-a^2 b (21 A+2 B)+a b^2 (5 A-6 B)+15 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 d \sqrt {a+b} \left (a^2-b^2\right )}+\frac {b \left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {\left (3 a^4 A+14 a^3 b B-26 a^2 A b^2-6 a b^3 B+15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}} \]

[Out]

A*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3/2)+1/3*(3*A*a^4-26*A*a^2*b^2+15*A*b^4+14*B*a^3*b-6*B*a*b^3)*cot(d*x+c)*El
lipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+
c))/(a-b))^(1/2)/a^3/(a-b)/b/(a+b)^(3/2)/d-1/3*(15*A*b^3+a*b^2*(5*A-6*B)-3*a^3*(A-4*B)-a^2*b*(21*A+2*B))*cot(d
*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+
sec(d*x+c))/(a-b))^(1/2)/a^3/(a^2-b^2)/d/(a+b)^(1/2)+(5*A*b-2*B*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2
)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b)
)^(1/2)/a^4/d+1/3*b*(3*A*a^2-5*A*b^2+2*B*a*b)*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+1/3*b*(3*A*a^4
-26*A*a^2*b^2+15*A*b^4+14*B*a^3*b-6*B*a*b^3)*tan(d*x+c)/a^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

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Rubi [A]  time = 1.21, antiderivative size = 582, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {4034, 4061, 4060, 4058, 3921, 3784, 3832, 4004} \[ \frac {b \left (-26 a^2 A b^2+3 a^4 A+14 a^3 b B-6 a b^3 B+15 A b^4\right ) \tan (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {b \left (3 a^2 A+2 a b B-5 A b^2\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\left (-a^2 b (21 A+2 B)-3 a^3 (A-4 B)+a b^2 (5 A-6 B)+15 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 d \sqrt {a+b} \left (a^2-b^2\right )}+\frac {\left (-26 a^2 A b^2+3 a^4 A+14 a^3 b B-6 a b^3 B+15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}+\frac {\sqrt {a+b} (5 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^4 d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14*a^3*b*B - 6*a*b^3*B)*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^3*(a - b)*b*(a + b)^(3/2)*d) - ((15*A*b^3 + a*b^2*(5*A - 6*B) - 3*a^3*(A - 4*B) - a^2*b*(21*A + 2*B
))*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^3*Sqrt[a + b]*(a^2 - b^2)*d) + (Sqrt[a + b]*(5*A
*b - 2*a*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/(a^4*d) + (A*Sin[c + d*x])/(a*d*
(a + b*Sec[c + d*x])^(3/2)) + (b*(3*a^2*A - 5*A*b^2 + 2*a*b*B)*Tan[c + d*x])/(3*a^2*(a^2 - b^2)*d*(a + b*Sec[c
+ d*x])^(3/2)) + (b*(3*a^4*A - 26*a^2*A*b^2 + 15*A*b^4 + 14*a^3*b*B - 6*a*b^3*B)*Tan[c + d*x])/(3*a^3*(a^2 -
b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

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 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 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]

Rule 4034

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*n), x]
+ Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + A*a*(n +
1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -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 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 4061

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

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\frac {1}{2} (5 A b-2 a B)-\frac {3}{2} A b \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{a}\\ &=\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2 A-5 A b^2+2 a b B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \int \frac {-\frac {3}{4} \left (a^2-b^2\right ) (5 A b-2 a B)+\frac {3}{2} a b (A b-a B) \sec (c+d x)+\frac {1}{4} b \left (3 a^2 A-5 A b^2+2 a b B\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2 A-5 A b^2+2 a b B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\frac {3}{8} \left (a^2-b^2\right )^2 (5 A b-2 a B)-\frac {1}{4} a b \left (9 a^2 A b-5 A b^3-6 a^3 B+2 a b^2 B\right ) \sec (c+d x)+\frac {1}{8} b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2 A-5 A b^2+2 a b B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\frac {3}{8} \left (a^2-b^2\right )^2 (5 A b-2 a B)+\left (-\frac {1}{4} a b \left (9 a^2 A b-5 A b^3-6 a^3 B+2 a b^2 B\right )-\frac {1}{8} b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}-\frac {\left (b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\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^3 (a-b) b (a+b)^{3/2} d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2 A-5 A b^2+2 a b B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {(5 A b-2 a B) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^3}-\frac {\left (b \left (15 A b^3+a b^2 (5 A-6 B)-3 a^3 (A-4 B)-a^2 b (21 A+2 B)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^3 (a-b) (a+b)^2}\\ &=\frac {\left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\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^3 (a-b) b (a+b)^{3/2} d}-\frac {\left (15 A b^3+a b^2 (5 A-6 B)-3 a^3 (A-4 B)-a^2 b (21 A+2 B)\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^3 (a-b) (a+b)^{3/2} d}+\frac {\sqrt {a+b} (5 A b-2 a 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^4 d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2 A-5 A b^2+2 a b B\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^4 A-26 a^2 A b^2+15 A b^4+14 a^3 b B-6 a b^3 B\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [B]  time = 21.99, size = 2366, normalized size = 4.07 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-2*b*(-10*a^2*A*b + 6*A*b^3 + 7*a^3*B - 3*a*b^2*B)*Sin[c + d*x])/(3*a
^3*(-a^2 + b^2)^2) + (2*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])^2
) + (2*(-11*a^2*A*b^3*Sin[c + d*x] + 7*A*b^5*Sin[c + d*x] + 8*a^3*b^2*B*Sin[c + d*x] - 4*a*b^4*B*Sin[c + d*x])
)/(3*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) - ((b + a*Cos[c + d*x])^(5/2)*Se
c[c + d*x]^(5/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2
)/(1 + Tan[(c + d*x)/2]^2)]*(3*a^5*A*Tan[(c + d*x)/2] + 3*a^4*A*b*Tan[(c + d*x)/2] - 26*a^3*A*b^2*Tan[(c + d*x
)/2] - 26*a^2*A*b^3*Tan[(c + d*x)/2] + 15*a*A*b^4*Tan[(c + d*x)/2] + 15*A*b^5*Tan[(c + d*x)/2] + 14*a^4*b*B*Ta
n[(c + d*x)/2] + 14*a^3*b^2*B*Tan[(c + d*x)/2] - 6*a^2*b^3*B*Tan[(c + d*x)/2] - 6*a*b^4*B*Tan[(c + d*x)/2] - 6
*a^5*A*Tan[(c + d*x)/2]^3 + 52*a^3*A*b^2*Tan[(c + d*x)/2]^3 - 30*a*A*b^4*Tan[(c + d*x)/2]^3 - 28*a^4*b*B*Tan[(
c + d*x)/2]^3 + 12*a^2*b^3*B*Tan[(c + d*x)/2]^3 + 3*a^5*A*Tan[(c + d*x)/2]^5 - 3*a^4*A*b*Tan[(c + d*x)/2]^5 -
26*a^3*A*b^2*Tan[(c + d*x)/2]^5 + 26*a^2*A*b^3*Tan[(c + d*x)/2]^5 + 15*a*A*b^4*Tan[(c + d*x)/2]^5 - 15*A*b^5*T
an[(c + d*x)/2]^5 + 14*a^4*b*B*Tan[(c + d*x)/2]^5 - 14*a^3*b^2*B*Tan[(c + d*x)/2]^5 - 6*a^2*b^3*B*Tan[(c + d*x
)/2]^5 + 6*a*b^4*B*Tan[(c + d*x)/2]^5 - 30*a^4*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*S
qrt[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)] + 60*a^2*A*b^3
*EllipticPi[-1, ArcSin[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)] - 30*A*b^5*EllipticPi[-1, ArcSin[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)] + 12*a^5
*B*EllipticPi[-1, ArcSin[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)] - 24*a^3*b^2*B*EllipticPi[-1, ArcSin[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)] +
12*a*b^4*B*EllipticPi[-1, ArcSin[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)] - 30*a^4*A*b*EllipticPi[-1, ArcSin[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)] + 60*a^2*A*b^3*EllipticPi[-1, ArcSin[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)] - 30*A*b
^5*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*S
qrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 12*a^5*B*EllipticPi[-1, ArcSin[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)] - 24*a^3*b^2*B*EllipticPi[-1, ArcSin[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*a*b^4*B*EllipticPi[-1, ArcSin[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)] + (a + b)*(3*a^4*A - 26*a^2*A*b^
2 + 15*A*b^4 + 14*a^3*b*B - 6*a*b^3*B)*EllipticE[ArcSin[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)] - 2*a*
(a + b)*(5*A*b^3 + 3*a^3*B + 3*a^2*b*(-2*A + B) - a*b^2*(3*A + 2*B))*EllipticF[ArcSin[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*(a^3 - a*b^2)^2*d*(a + b*Sec[c + d*x])^(5/2)*Sqrt[1 + Tan[(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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fricas [F]  time = 1.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A \cos \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.01, size = 8545, normalized size = 14.68 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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