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

Optimal. Leaf size=645 \[ -\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}-\frac {\sqrt {a+b} \left (4 a^2 (A+2 C)+35 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^5 d}-\frac {b^2 \left (a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)+105 A b^4\right ) \tan (c+d x)}{12 a^4 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {\left (a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)+105 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 d \sqrt {a+b} \left (a^2-b^2\right )}+\frac {b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \tan (c+d x)}{12 a^3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\left (6 a^4 (A-8 C)-a^3 (27 A b-8 b C)-3 a^2 b^2 (45 A-8 C)+35 a A b^3+105 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 d \sqrt {a+b} \left (a^2-b^2\right )}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}} \]

[Out]

-7/4*A*b*sin(d*x+c)/a^2/d/(a+b*sec(d*x+c))^(3/2)+1/2*A*cos(d*x+c)*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3/2)-1/12*(
105*A*b^4+a^4*(33*A-56*C)-2*a^2*b^2*(85*A-12*C))*cot(d*x+c)*EllipticE((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^4/(a^2-b^2)/d/(a+b)^(1/2)+1/1
2*(35*a*A*b^3+105*A*b^4+6*a^4*(A-8*C)-3*a^2*b^2*(45*A-8*C)-a^3*(27*A*b-8*C*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
^4/(a^2-b^2)/d/(a+b)^(1/2)-1/4*(35*A*b^2+4*a^2*(A+2*C))*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^5
/d+1/12*b^2*(35*A*b^2-a^2*(27*A-8*C))*tan(d*x+c)/a^3/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-1/12*b^2*(105*A*b^4+a^
4*(33*A-56*C)-2*a^2*b^2*(85*A-12*C))*tan(d*x+c)/a^4/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.56, antiderivative size = 645, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4105, 4104, 4060, 4058, 3921, 3784, 3832, 4004} \[ -\frac {b^2 \left (-2 a^2 b^2 (85 A-12 C)+a^4 (33 A-56 C)+105 A b^4\right ) \tan (c+d x)}{12 a^4 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \tan (c+d x)}{12 a^3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\left (-3 a^2 b^2 (45 A-8 C)-a^3 (27 A b-8 b C)+6 a^4 (A-8 C)+35 a A b^3+105 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\left (-2 a^2 b^2 (85 A-12 C)+a^4 (33 A-56 C)+105 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\sqrt {a+b} \left (4 a^2 (A+2 C)+35 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^5 d}-\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((105*A*b^4 + a^4*(33*A - 56*C) - 2*a^2*b^2*(85*A - 12*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)
)])/(12*a^4*Sqrt[a + b]*(a^2 - b^2)*d) + ((35*a*A*b^3 + 105*A*b^4 + 6*a^4*(A - 8*C) - 3*a^2*b^2*(45*A - 8*C) -
a^3*(27*A*b - 8*b*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(12*a^4*Sqrt[a + b]*(a^2 - b^2)*d
) - (Sqrt[a + b]*(35*A*b^2 + 4*a^2*(A + 2*C))*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))]
)/(4*a^5*d) - (7*A*b*Sin[c + d*x])/(4*a^2*d*(a + b*Sec[c + d*x])^(3/2)) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d
*(a + b*Sec[c + d*x])^(3/2)) + (b^2*(35*A*b^2 - a^2*(27*A - 8*C))*Tan[c + d*x])/(12*a^3*(a^2 - b^2)*d*(a + b*S
ec[c + d*x])^(3/2)) - (b^2*(105*A*b^4 + a^4*(33*A - 56*C) - 2*a^2*b^2*(85*A - 12*C))*Tan[c + d*x])/(12*a^4*(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 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 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 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 + A*n + C*n)*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, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4105

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), 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*(m + n + 1)) + a*(A + A*n
+ C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && NeQ[
a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\cos (c+d x) \left (\frac {7 A b}{2}-a (A+2 C) \sec (c+d x)-\frac {5}{2} A b \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx}{2 a}\\ &=-\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{4} \left (35 A b^2+4 a^2 (A+2 C)\right )+\frac {5}{2} a A b \sec (c+d x)-\frac {21}{4} A b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{2 a^2}\\ &=-\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {-\frac {3}{8} \left (a^2-b^2\right ) \left (35 A b^2+4 a^2 (A+2 C)\right )+\frac {3}{4} a b \left (7 A b^2-a^2 (3 A-4 C)\right ) \sec (c+d x)-\frac {1}{8} b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a^3 \left (a^2-b^2\right )}\\ &=-\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \int \frac {\frac {3}{16} \left (a^2-b^2\right )^2 \left (35 A b^2+4 a^2 (A+2 C)\right )+\frac {1}{8} a b \left (35 A b^4+3 a^4 (A-8 C)-2 a^2 b^2 (27 A-4 C)\right ) \sec (c+d x)+\frac {1}{16} b^2 \left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \int \frac {\frac {3}{16} \left (a^2-b^2\right )^2 \left (35 A b^2+4 a^2 (A+2 C)\right )+\left (-\frac {1}{16} b^2 \left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)\right )+\frac {1}{8} a b \left (35 A b^4+3 a^4 (A-8 C)-2 a^2 b^2 (27 A-4 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2}+\frac {\left (b^2 \left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 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}}}{12 a^4 (a-b) (a+b)^{3/2} d}-\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (35 A b^2+4 a^2 (A+2 C)\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 a^4}+\frac {\left (b \left (35 a A b^3+105 A b^4+6 a^4 (A-8 C)-3 a^2 b^2 (45 A-8 C)-a^3 (27 A b-8 b C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a^4 (a-b) (a+b)^2}\\ &=-\frac {\left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 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}}}{12 a^4 (a-b) (a+b)^{3/2} d}+\frac {\left (35 a A b^3+105 A b^4+6 a^4 (A-8 C)-3 a^2 b^2 (45 A-8 C)-a^3 (27 A b-8 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}}}{12 a^4 (a-b) (a+b)^{3/2} d}-\frac {\sqrt {a+b} \left (35 A b^2+4 a^2 (A+2 C)\right ) \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}}}{4 a^5 d}-\frac {7 A b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {b^2 \left (35 A b^2-a^2 (27 A-8 C)\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (105 A b^4+a^4 (33 A-56 C)-2 a^2 b^2 (85 A-12 C)\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 17.46, size = 977, normalized size = 1.51 \[ \frac {1}{2} \left (\frac {(b+a \cos (c+d x))^3 \left (\frac {4 b \left (-7 C a^4-13 A b^2 a^2+3 b^2 C a^2+9 A b^4\right ) \sin (c+d x)}{3 a^4 \left (b^2-a^2\right )^2}-\frac {4 \left (A \sin (c+d x) b^5+a^2 C \sin (c+d x) b^3\right )}{3 a^4 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}-\frac {8 \left (5 A \sin (c+d x) b^6-7 a^2 A \sin (c+d x) b^4+2 a^2 C \sin (c+d x) b^4-4 a^4 C \sin (c+d x) b^2\right )}{3 a^4 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}+\frac {A \sin (2 (c+d x))}{2 a^3}\right ) \sec ^3(c+d x)}{d (a+b \sec (c+d x))^{5/2}}+\frac {(b+a \cos (c+d x))^{5/2} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}} \left (3 (a-b)^2 \left (4 (A+2 C) a^2+35 A b^2\right ) \left ((a-b) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 a \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} (a+b)^2+a b \left ((33 A-56 C) a^4+2 b^2 (12 C-85 A) a^2+105 A b^4\right ) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} (a+b)-b \left (-6 (A+12 C) a^5+b (39 A+16 C) a^4+12 b^2 (4 C-19 A) a^3+2 b^3 (29 A-12 C) a^2+210 A b^4 a-105 A b^5\right ) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} (a+b)+a b \left ((33 A-56 C) a^4+2 b^2 (12 C-85 A) a^2+105 A b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b\right )\right ) \sec ^{\frac {5}{2}}(c+d x)}{6 a^5 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{5/2} \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (a \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right )-b \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )\right )}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^2*(A + 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*((4*b*(-13*a^2*A*b^2 + 9*A*b^4 - 7*a^4*C + 3*a^2*b^2*C)*Sin[c + d*x])/
(3*a^4*(-a^2 + b^2)^2) - (4*(A*b^5*Sin[c + d*x] + a^2*b^3*C*Sin[c + d*x]))/(3*a^4*(a^2 - b^2)*(b + a*Cos[c + d
*x])^2) - (8*(-7*a^2*A*b^4*Sin[c + d*x] + 5*A*b^6*Sin[c + d*x] - 4*a^4*b^2*C*Sin[c + d*x] + 2*a^2*b^4*C*Sin[c
+ d*x]))/(3*a^4*(a^2 - b^2)^2*(b + a*Cos[c + d*x])) + (A*Sin[2*(c + d*x)])/(2*a^3)))/(d*(a + b*Sec[c + d*x])^(
5/2)) + ((b + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]
^2)/(1 + Tan[(c + d*x)/2]^2)]*(a*b*(105*A*b^4 + a^4*(33*A - 56*C) + 2*a^2*b^2*(-85*A + 12*C))*Tan[(c + d*x)/2]
*Sqrt[1 - Tan[(c + d*x)/2]^2]*(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2) + a*b*(a + b)*(105*A*b^4 +
a^4*(33*A - 56*C) + 2*a^2*b^2*(-85*A + 12*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(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)] - b*(a + b)*(210*a*A*b^4 -
105*A*b^5 + 2*a^2*b^3*(29*A - 12*C) + 12*a^3*b^2*(-19*A + 4*C) - 6*a^5*(A + 12*C) + a^4*b*(39*A + 16*C))*Ellip
ticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(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 - b)^2*(a + b)^2*(35*A*b^2 + 4*a^2*(A + 2*C))*((a - b)*EllipticF[ArcSin[
Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*a*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(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*a^5*(a^2 - b^2)^2*d*(
a + b*Sec[c + d*x])^(5/2)*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x)/2]^2)]
*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2))))/2

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fricas [F]  time = 3.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + A*cos(d*x + c)^2)*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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 2.32, size = 9631, normalized size = 14.93 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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