3.12.100 \(\int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1200]

Optimal. Leaf size=215 \[ -\frac {4 a^2 (5 A+4 B+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (14 A+7 B+6 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a^2 (35 A+49 B+33 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (5 A+4 B+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-4/5*a^2*(5*A+4*B+3*C)*(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
+4/21*a^2*(14*A+7*B+6*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+2/105*a^2*(35*A+49*B+33*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*C*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(
7/2)+2/35*(7*B+4*C)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(5/2)+4/5*a^2*(5*A+4*B+3*C)*sin(d*x+c)/d/cos(
d*x+c)^(1/2)

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Rubi [A]
time = 0.41, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {4197, 3122, 3054, 3047, 3100, 2827, 2716, 2719, 2720} \begin {gather*} \frac {4 a^2 (14 A+7 B+6 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^2 (5 A+4 B+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^2 (35 A+49 B+33 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (5 A+4 B+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (7 B+4 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*(5*A + 4*B + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(14*A + 7*B + 6*C)*EllipticF[(c + d*x)/2,
2])/(21*d) + (2*a^2*(35*A + 49*B + 33*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)) + (4*a^2*(5*A + 4*B + 3*C)*S
in[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (
2*(7*B + 4*C)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2))

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 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]

Rule 3122

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/(b*d*(n +
1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C
- B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x]
, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

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*} \int \frac {(a+a \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 {(a+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 (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{2} a (7 B+4 C)+\frac {1}{2} a (7 A+C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (35 A+49 B+33 C)+\frac {1}{4} a^2 (35 A+7 B+9 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} a^3 (35 A+49 B+33 C)+\left (\frac {1}{4} a^3 (35 A+7 B+9 C)+\frac {1}{4} a^3 (35 A+49 B+33 C)\right ) \cos (c+d x)+\frac {1}{4} a^3 (35 A+7 B+9 C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {2 a^2 (35 A+49 B+33 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 \int \frac {\frac {21}{4} a^3 (5 A+4 B+3 C)+\frac {5}{4} a^3 (14 A+7 B+6 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{105 a}\\ &=\frac {2 a^2 (35 A+49 B+33 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (2 a^2 (5 A+4 B+3 C)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{21} \left (2 a^2 (14 A+7 B+6 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^2 (14 A+7 B+6 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a^2 (35 A+49 B+33 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (5 A+4 B+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} \left (2 a^2 (5 A+4 B+3 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {4 a^2 (5 A+4 B+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (14 A+7 B+6 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a^2 (35 A+49 B+33 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (5 A+4 B+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.75, size = 2041, normalized size = 9.49 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/2*I)*A*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c +
d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1
+ E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^
((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeomet
ric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1
+ E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 +
E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]) - (
((2*I)/5)*B*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c +
d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1
+ E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^
((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeomet
ric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1
+ E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 +
E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]) - (
((3*I)/10)*C*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c +
d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1
+ E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E
^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeome
tric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1
+ E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 +
E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]) +
(Cos[c + d*x]^(9/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(5
*A + 4*B + 3*C)*Csc[c]*Sec[c])/(5*d) + (C*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(7*d) + (Sec[c]*Sec[c + d*x]^3*(5*C*
Sin[c] + 7*B*Sin[d*x] + 14*C*Sin[d*x]))/(35*d) + (Sec[c]*Sec[c + d*x]^2*(21*B*Sin[c] + 42*C*Sin[c] + 35*A*Sin[
d*x] + 70*B*Sin[d*x] + 60*C*Sin[d*x]))/(105*d) + (Sec[c]*Sec[c + d*x]*(35*A*Sin[c] + 70*B*Sin[c] + 60*C*Sin[c]
+ 210*A*Sin[d*x] + 168*B*Sin[d*x] + 126*C*Sin[d*x]))/(105*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*
x]) - (4*A*Cos[c + d*x]^4*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] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Si
n[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTa
n[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c + d*x]^4*
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] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*S
qrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C
+ 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*C*Cos[c + d*x]^4*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]
+ C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*S
in[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Co
s[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(904\) vs. \(2(247)=494\).
time = 0.39, size = 905, normalized size = 4.21

method	result	size
default	\(\text {Expression too large to display}\)	\(905\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*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]

-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)^(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))+(1/4*A+1/2*B+1/4*C)*(-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*B+1/2*C)/(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(1/2*A+1/4*B)/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)-(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+1/4*C*(-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))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.40, size = 281, normalized size = 1.31 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} {\left (14 \, A + 7 \, B + 6 \, C\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 (14 \, A + 7 \, B + 6 \, C\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 (5 \, A + 4 \, B + 3 \, C\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 (5 \, A + 4 \, B + 3 \, C\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 (5 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, A + 14 \, B + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 21 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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]

-2/105*(5*I*sqrt(2)*(14*A + 7*B + 6*C)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c)) - 5*I*sqrt(2)*(14*A + 7*B + 6*C)*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x
+ c)) + 21*I*sqrt(2)*(5*A + 4*B + 3*C)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*(5*A + 4*B + 3*C)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weier
strassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (42*(5*A + 4*B + 3*C)*a^2*cos(d*x + c)^3 + 5*(7*A + 14
*B + 12*C)*a^2*cos(d*x + c)^2 + 21*(B + 2*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*
cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \frac {A}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {2 A \sec {\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {2 B \sec ^{2}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {2 C \sec ^{3}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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]

a**2*(Integral(A/sqrt(cos(c + d*x)), x) + Integral(2*A*sec(c + d*x)/sqrt(cos(c + d*x)), x) + Integral(A*sec(c
+ d*x)**2/sqrt(cos(c + d*x)), x) + Integral(B*sec(c + d*x)/sqrt(cos(c + d*x)), x) + Integral(2*B*sec(c + d*x)*
*2/sqrt(cos(c + d*x)), x) + Integral(B*sec(c + d*x)**3/sqrt(cos(c + d*x)), x) + Integral(C*sec(c + d*x)**2/sqr
t(cos(c + d*x)), x) + Integral(2*C*sec(c + d*x)**3/sqrt(cos(c + d*x)), x) + Integral(C*sec(c + d*x)**4/sqrt(co
s(c + d*x)), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 7.46, size = 346, normalized size = 1.61 \begin {gather*} \frac {6\,B\,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\,B\,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\,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 )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {30\,C\,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\,C\,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\,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 )}{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 {4\,A\,a^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 )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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