\(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [752]

Optimal result

Integrand size = 33, antiderivative size = 559 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {\left (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) b (a+b)^{3/2} d}+\frac {\left (21 a^2 A b^2-5 a A b^3-15 A b^4+a^3 b (3 A-2 C)+6 a^4 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) b (a+b)^{3/2} d}+\frac {5 A b \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^4 d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (5 A b^2-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

A*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3/2)-1/3*(26*A*a^2*b^2-15*A*b^4-a^4*(3*A-8*C))*cot(d*x+c)*EllipticE((a+b*se
c(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*(21*A*a^2*b^2-5*a*A*b^3-15*A*b^4+a^3*b*(3*A-2*C)+6*a^4*C)*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-b)/b/(a+b)^(3/2)/d+5*A*b*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*(5*A*
b^2-a^2*(3*A-2*C))*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-1/3*b*(26*A*a^2*b^2-15*A*b^4-a^4*(3*A-8*C
))*tan(d*x+c)/a^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 559, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4190, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {5 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^4 d}-\frac {b \left (5 A b^2-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\left (-\left (a^4 (3 A-8 C)\right )+26 a^2 A b^2-15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}+\frac {\left (6 a^4 C+a^3 b (3 A-2 C)+21 a^2 A b^2-5 a A b^3-15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}-\frac {b \left (-\left (a^4 (3 A-8 C)\right )+26 a^2 A b^2-15 A b^4\right ) \tan (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}} \]

[In]

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

[Out]

-1/3*((26*a^2*A*b^2 - 15*A*b^4 - a^4*(3*A - 8*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))])/(a^3*(
a - b)*b*(a + b)^(3/2)*d) + ((21*a^2*A*b^2 - 5*a*A*b^3 - 15*A*b^4 + a^3*b*(3*A - 2*C) + 6*a^4*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^3*(a - b)*b*(a + b)^(3/2)*d) + (5*A*b*Sqrt[a + b]*Cot[c + d*x]*E
llipticPi[(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*(5*A*b^2 - a^2*(3*A - 2*C))*Tan[c + d*x])/(3*a^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (b*(26*a^2
*A*b^2 - 15*A*b^4 - a^4*(3*A - 8*C))*Tan[c + d*x])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

Rule 3869

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

Rule 3917

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

Rule 4006

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 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4143

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 4145

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 4190

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*} \text {integral}& = \frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\frac {5 A b}{2}-a C \sec (c+d x)-\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 (5 A b^2-a^2 (3 A-2 C)\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 {15}{4} A b \left (a^2-b^2\right )+\frac {3}{2} a \left (A b^2+a^2 C\right ) \sec (c+d x)-\frac {1}{4} b \left (5 A b^2-a^2 (3 A-2 C)\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 (5 A b^2-a^2 (3 A-2 C)\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 (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\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 {15}{8} A b \left (a^2-b^2\right )^2+\frac {1}{4} a \left (5 A b^4-3 a^4 C-a^2 b^2 (9 A+C)\right ) \sec (c+d x)-\frac {1}{8} b \left (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\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 (5 A b^2-a^2 (3 A-2 C)\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 (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\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 {15}{8} A b \left (a^2-b^2\right )^2+\left (\frac {1}{8} b \left (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\right )+\frac {1}{4} a \left (5 A b^4-3 a^4 C-a^2 b^2 (9 A+C)\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 (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\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 (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\right ) \cot (c+d x) E\left (\arcsin \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 (5 A b^2-a^2 (3 A-2 C)\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 (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\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) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^3}+\frac {\left (21 a^2 A b^2-5 a A b^3-15 A b^4+a^3 b (3 A-2 C)+6 a^4 C\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 (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\right ) \cot (c+d x) E\left (\arcsin \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 (21 a^2 A b^2-5 a A b^3-15 A b^4+a^3 b (3 A-2 C)+6 a^4 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \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 {5 A b \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \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 (5 A b^2-a^2 (3 A-2 C)\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 (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\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*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1702\) vs. \(2(559)=1118\).

Time = 22.25 (sec) , antiderivative size = 1702, normalized size of antiderivative = 3.04 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {8 \left (-5 a^2 A b^2+3 A b^4-2 a^4 C\right ) \sin (c+d x)}{3 a^3 \left (-a^2+b^2\right )^2}+\frac {4 \left (A b^4 \sin (c+d x)+a^2 b^2 C \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {4 \left (-11 a^2 A b^3 \sin (c+d x)+7 A b^5 \sin (c+d x)-5 a^4 b C \sin (c+d x)+a^2 b^3 C \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{5/2}}-\frac {2 (b+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (3 a^5 A \tan \left (\frac {1}{2} (c+d x)\right )+3 a^4 A b \tan \left (\frac {1}{2} (c+d x)\right )-26 a^3 A b^2 \tan \left (\frac {1}{2} (c+d x)\right )-26 a^2 A b^3 \tan \left (\frac {1}{2} (c+d x)\right )+15 a A b^4 \tan \left (\frac {1}{2} (c+d x)\right )+15 A b^5 \tan \left (\frac {1}{2} (c+d x)\right )-8 a^5 C \tan \left (\frac {1}{2} (c+d x)\right )-8 a^4 b C \tan \left (\frac {1}{2} (c+d x)\right )-6 a^5 A \tan ^3\left (\frac {1}{2} (c+d x)\right )+52 a^3 A b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )-30 a A b^4 \tan ^3\left (\frac {1}{2} (c+d x)\right )+16 a^5 C \tan ^3\left (\frac {1}{2} (c+d x)\right )+3 a^5 A \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a^4 A b \tan ^5\left (\frac {1}{2} (c+d x)\right )-26 a^3 A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+26 a^2 A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )+15 a A b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )-15 A b^5 \tan ^5\left (\frac {1}{2} (c+d x)\right )-8 a^5 C \tan ^5\left (\frac {1}{2} (c+d x)\right )+8 a^4 b C \tan ^5\left (\frac {1}{2} (c+d x)\right )-30 a^4 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+60 a^2 A b^3 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-30 A b^5 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-30 a^4 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+60 a^2 A b^3 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-30 A b^5 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-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) \left (-26 a^2 A b^2+15 A b^4+a^4 (3 A-8 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+2 a (a+b) \left (3 a A b^2-5 A b^3+3 a^3 C+a^2 b (6 A+C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-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 )}{3 a \left (a^3-a b^2\right )^2 d (A+2 C+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{5/2} \sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]

[In]

Integrate[(Cos[c + d*x]*(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]*(A + C*Sec[c + d*x]^2)*((-8*(-5*a^2*A*b^2 + 3*A*b^4 - 2*a^4*C)*Sin[c + d*
x])/(3*a^3*(-a^2 + b^2)^2) + (4*(A*b^4*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)*(b + a*Cos[c
+ d*x])^2) + (4*(-11*a^2*A*b^3*Sin[c + d*x] + 7*A*b^5*Sin[c + d*x] - 5*a^4*b*C*Sin[c + d*x] + a^2*b^3*C*Sin[c
+ d*x]))/(3*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^
(5/2)) - (2*(b + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^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] - 8*a^5*C*Tan[(c + d*x)/2] - 8*a^4*b*C*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 + 16*a^5*C*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*Tan[(c + d*x)/2]^5 - 8*a^5*C*Tan[(c + d
*x)/2]^5 + 8*a^4*b*C*Tan[(c + d*x)/2]^5 - 30*a^4*A*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)] + 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)] - 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[Ta
n[(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)]*T
an[(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)*(-26*a^2*A*b^2 + 15*A*b^4 + a^4*(3*A - 8*C))*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)*(3*a*A*b^2 - 5*A*b^3 + 3*a^3*C + a^2*b*(6*A + C))*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 + 2*C + A*Cos[2*c + 2*d*x])*(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)))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(8178\) vs. \(2(518)=1036\).

Time = 7.15 (sec) , antiderivative size = 8179, normalized size of antiderivative = 14.63

method	result	size
default	\(\text {Expression too large to display}\)	\(8179\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

integrate(cos(d*x+c)*(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)*sec(d*x + c)^2 + 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)

Sympy [F]

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

[In]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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