\(\int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [495]

Optimal result

Integrand size = 33, antiderivative size = 194 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {4 a^2 (4 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (7 A+6 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+3 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

-4/5*a^2*(4*A+3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/2
1*a^2*(7*A+6*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/35*a
^2*(7*A+9*B)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+4/21*a^2*(7*A+6*B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*B*(a^2+a^2*cos
(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(7/2)+4/5*a^2*(4*A+3*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3033, 3054, 3047, 3100, 2827, 2716, 2720, 2719} \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^2 (7 A+6 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {4 a^2 (4 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (7 A+6 B) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (4 A+3 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(-4*a^2*(4*A + 3*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(7*A + 6*B)*EllipticF[(c + d*x)/2, 2])/(21*d) +
(2*a^2*(7*A + 9*B)*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)) + (4*a^2*(7*A + 6*B)*Sin[c + d*x])/(21*d*Cos[c + d*
x]^(3/2)) + (4*a^2*(4*A + 3*B)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*B*(a^2 + a^2*Cos[c + d*x])*Sin[c +
d*x])/(7*d*Cos[c + d*x]^(7/2))

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3054

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d
*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*
n] || EqQ[c, 0])

Rule 3100

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+a \cos (c+d x))^2 (B+A \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{2} a (7 A+9 B)+\frac {1}{2} a (7 A+3 B) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {\frac {1}{2} a^2 (7 A+9 B)+\left (\frac {1}{2} a^2 (7 A+3 B)+\frac {1}{2} a^2 (7 A+9 B)\right ) \cos (c+d x)+\frac {1}{2} a^2 (7 A+3 B) \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (7 A+9 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4}{35} \int \frac {\frac {5}{2} a^2 (7 A+6 B)+\frac {7}{2} a^2 (4 A+3 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (7 A+9 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{5} \left (2 a^2 (4 A+3 B)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{7} \left (2 a^2 (7 A+6 B)\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (7 A+9 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+3 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (2 a^2 (4 A+3 B)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (2 a^2 (7 A+6 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {4 a^2 (4 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (7 A+6 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^2 (7 A+9 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (4 A+3 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.82 (sec) , antiderivative size = 1067, normalized size of antiderivative = 5.50 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\cos ^{\frac {7}{2}}(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \left (\frac {(4 A+3 B) \csc (c) \sec (c)}{5 d}+\frac {B \sec (c) \sec ^4(c+d x) \sin (d x)}{14 d}+\frac {\sec (c) \sec ^3(c+d x) (5 B \sin (c)+7 A \sin (d x)+14 B \sin (d x))}{70 d}+\frac {\sec (c) \sec ^2(c+d x) (21 A \sin (c)+42 B \sin (c)+70 A \sin (d x)+60 B \sin (d x))}{210 d}+\frac {\sec (c) \sec (c+d x) (35 A \sin (c)+30 B \sin (c)+84 A \sin (d x)+63 B \sin (d x))}{105 d}\right )}{B+A \cos (c+d x)}-\frac {A \cos ^3(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)}}-\frac {2 B \cos ^3(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{7 d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)}}+\frac {2 A \cos ^3(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (B+A \cos (c+d x))}+\frac {3 B \cos ^3(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{10 d (B+A \cos (c+d x))} \]

[In]

Integrate[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(Cos[c + d*x]^(7/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*(((4*A + 3*B)*Csc[c]*Sec[
c])/(5*d) + (B*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(14*d) + (Sec[c]*Sec[c + d*x]^3*(5*B*Sin[c] + 7*A*Sin[d*x] + 14
*B*Sin[d*x]))/(70*d) + (Sec[c]*Sec[c + d*x]^2*(21*A*Sin[c] + 42*B*Sin[c] + 70*A*Sin[d*x] + 60*B*Sin[d*x]))/(21
0*d) + (Sec[c]*Sec[c + d*x]*(35*A*Sin[c] + 30*B*Sin[c] + 84*A*Sin[d*x] + 63*B*Sin[d*x]))/(105*d)))/(B + A*Cos[
c + d*x]) - (A*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2
+ (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[
Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3
*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4},
Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan
[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqr
t[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (2*A*Cos[c + d*x]^3*Csc[c]*S
ec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d
*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*
x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x
+ ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos
[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(B + A*Cos[c + d*x])) + (3
*B*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*((HypergeometricPFQ[
{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan
[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[
1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]
]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(10*d
*(B + A*Cos[c + d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(823\) vs. \(2(226)=452\).

Time = 23.11 (sec) , antiderivative size = 824, normalized size of antiderivative = 4.25

method	result	size
default	\(\text {Expression too large to display}\)	\(824\)

[In]

int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/4*A/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/
2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(sin
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))+1/4*B*(-1/56*
cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(
1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(1/2*A+1/4*B)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d
*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(
1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+1/5*(1/4*A+1/
2*B)/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*sin(1/
2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin
(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2))/sin(1/2*d*x+1/2*c)/(2*cos(
1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.36 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (42 \, {\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 21 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, B a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-2/105*(5*I*sqrt(2)*(7*A + 6*B)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) -
5*I*sqrt(2)*(7*A + 6*B)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I*s
qrt(2)*(4*A + 3*B)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c))) - 21*I*sqrt(2)*(4*A + 3*B)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(
d*x + c) - I*sin(d*x + c))) - (42*(4*A + 3*B)*a^2*cos(d*x + c)^3 + 10*(7*A + 6*B)*a^2*cos(d*x + c)^2 + 21*(A +
2*B)*a^2*cos(d*x + c) + 15*B*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=a^{2} \left (\int \frac {A}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A \sec {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 B \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))**2*(A+B*sec(d*x+c))/cos(d*x+c)**(3/2),x)

[Out]

a**2*(Integral(A/cos(c + d*x)**(3/2), x) + Integral(2*A*sec(c + d*x)/cos(c + d*x)**(3/2), x) + Integral(A*sec(
c + d*x)**2/cos(c + d*x)**(3/2), x) + Integral(B*sec(c + d*x)/cos(c + d*x)**(3/2), x) + Integral(2*B*sec(c + d
*x)**2/cos(c + d*x)**(3/2), x) + Integral(B*sec(c + d*x)**3/cos(c + d*x)**(3/2), x))

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/cos(d*x + c)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 16.88 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {30\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,B\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,B\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2)/cos(c + d*x)^(3/2),x)

[Out]

(6*A*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 20*A*a^2*cos(c + d*x)*sin(c + d*x)*hyperg
eom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 30*A*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c
+ d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (30*B*a^2*sin(c + d*x)*hypergeom([-7/4, 1/2
], -3/4, cos(c + d*x)^2) + 84*B*a^2*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 7
0*B*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(105*d*cos(c + d*x)^(7/2)*(1
- cos(c + d*x)^2)^(1/2))