\(\int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1304]

Optimal result

Integrand size = 43, antiderivative size = 249 \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (14 a b B+7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

-2/5*(10*A*a*b+5*B*a^2+3*B*b^2+6*C*a*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+2/21*(14*B*a*b+7*a^2*(3*A+C)+b^2*(7*A+5*C))*(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*b*(7*B*b+4*C*a)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*(7*A*b^2+14*B*
a*b+4*C*a^2+5*C*b^2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*C*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/5*
(10*A*a*b+5*B*a^2+3*B*b^2+6*C*a*b)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4197, 3126, 3110, 3100, 2827, 2716, 2719, 2720} \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (7 a^2 (3 A+C)+14 a b B+b^2 (7 A+5 C)\right )}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d}+\frac {2 \sin (c+d x) \left (4 a^2 C+14 a b B+7 A b^2+5 b^2 C\right )}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b (4 a C+7 b B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[In]

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

[Out]

(-2*(10*a*A*b + 5*a^2*B + 3*b^2*B + 6*a*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(14*a*b*B + 7*a^2*(3*A + C)
+ b^2*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*b*(7*b*B + 4*a*C)*Sin[c + d*x])/(35*d*Cos[c + d*x]^
(5/2)) + (2*(7*A*b^2 + 14*a*b*B + 4*a^2*C + 5*b^2*C)*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (2*(10*a*A*b +
5*a^2*B + 3*b^2*B + 6*a*b*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*C*(b + a*Cos[c + d*x])^2*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 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]

Rule 3110

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

Rule 3126

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

Rule 4197

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(b+a \cos (c+d x))^2 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(b+a \cos (c+d x)) \left (\frac {1}{2} (7 b B+4 a C)+\frac {1}{2} (7 A b+7 a B+5 b C) \cos (c+d x)+\frac {1}{2} a (7 A+C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {4}{35} \int \frac {-\frac {5}{4} \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right )-\frac {7}{4} \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)-\frac {5}{4} a^2 (7 A+C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {8}{105} \int \frac {-\frac {21}{8} \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right )-\frac {5}{8} \left (14 a b B+7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (-10 a A b-5 a^2 B-3 b^2 B-6 a b C\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx-\frac {1}{21} \left (-14 a b B-7 a^2 (3 A+C)-b^2 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 \left (14 a b B+7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (14 a b B+7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.61 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (-21 \left (5 a^2 B+3 b^2 B+2 a b (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (14 a b B+7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {15 b^2 C \sin (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {21 b (b B+2 a C) \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {5 \left (7 A b^2+14 a b B+7 a^2 C+5 b^2 C\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {21 \left (5 a^2 B+3 b^2 B+2 a b (5 A+3 C)\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{105 d} \]

[In]

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

[Out]

(2*(-21*(5*a^2*B + 3*b^2*B + 2*a*b*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 5*(14*a*b*B + 7*a^2*(3*A + C) + b^
2*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2] + (15*b^2*C*Sin[c + d*x])/Cos[c + d*x]^(7/2) + (21*b*(b*B + 2*a*C)*Si
n[c + d*x])/Cos[c + d*x]^(5/2) + (5*(7*A*b^2 + 14*a*b*B + 7*a^2*C + 5*b^2*C)*Sin[c + d*x])/Cos[c + d*x]^(3/2)
+ (21*(5*a^2*B + 3*b^2*B + 2*a*b*(5*A + 3*C))*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(105*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(919\) vs. \(2(281)=562\).

Time = 4.82 (sec) , antiderivative size = 920, normalized size of antiderivative = 3.69

method	result	size
default	\(\text {Expression too large to display}\)	\(920\)

[In]

int((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*a^2*(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)
)+2*C*b^2*(-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*a*(2*A*b+B*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*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-Ellipt
icE(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)^(1/2))+2/5*b*(B*b+2*C*
a)/(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)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*
d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*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)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*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)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1
/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*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)*(s
in(1/2*d*x+1/2*c)^2)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(A*b^2+2*B*a*b+C*a^2)*(-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))))/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 = 0.13 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (7 i \, {\left (3 \, A + C\right )} a^{2} + 14 i \, B a b + i \, {\left (7 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, {\left (3 \, A + C\right )} a^{2} - 14 i \, B a b - i \, {\left (7 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (5 i \, B a^{2} + 2 i \, {\left (5 \, A + 3 \, C\right )} a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-5 i \, B a^{2} - 2 i \, {\left (5 \, A + 3 \, C\right )} a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (21 \, {\left (5 \, B a^{2} + 2 \, {\left (5 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, C b^{2} + 5 \, {\left (7 \, C a^{2} + 14 \, B a b + {\left (7 \, A + 5 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 21 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/105*(5*sqrt(2)*(7*I*(3*A + C)*a^2 + 14*I*B*a*b + I*(7*A + 5*C)*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-7*I*(3*A + C)*a^2 - 14*I*B*a*b - I*(7*A + 5*C)*b^2)*cos(d*x +
c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(5*I*B*a^2 + 2*I*(5*A + 3*C)*a*b +
3*I*B*b^2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) +
21*sqrt(2)*(-5*I*B*a^2 - 2*I*(5*A + 3*C)*a*b - 3*I*B*b^2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPI
nverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(21*(5*B*a^2 + 2*(5*A + 3*C)*a*b + 3*B*b^2)*cos(d*x + c)^3 +
15*C*b^2 + 5*(7*C*a^2 + 14*B*a*b + (7*A + 5*C)*b^2)*cos(d*x + c)^2 + 21*(2*C*a*b + B*b^2)*cos(d*x + c))*sqrt(
cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

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

[In]

integrate((a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.02 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {6\,B\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,B\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,B\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {30\,C\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,C\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,C\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,A\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

int(((a + b/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(1/2),x)

[Out]

(6*B*b^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*B*a^2*cos(c + d*x)^2*sin(c + d*x)*hype
rgeom([-1/4, 1/2], 3/4, cos(c + d*x)^2) + 20*B*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c
+ d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (30*C*b^2*sin(c + d*x)*hypergeom([-7/4, 1/2
], -3/4, cos(c + d*x)^2) + 70*C*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) +
84*C*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 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)) + (2*A*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*b^2*sin(c + d*x)*hypergeom([-3/4, 1/
2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (4*A*a*b*sin(c + d*x)*hypergeom([-
1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))