3.12.99 \(\int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1199]

Optimal. Leaf size=174 \[ -\frac {4 a^2 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (3 A+2 B+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^2 (15 A+25 B+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-4/5*a^2*(5*B+4*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+4/3
*a^2*(3*A+2*B+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+2/5*C
*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/15*(5*B+4*C)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^
(3/2)+2/15*a^2*(15*A+25*B+17*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.37, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4197, 3122, 3054, 3047, 3100, 2827, 2720, 2719} \begin {gather*} \frac {4 a^2 (3 A+2 B+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^2 (15 A+25 B+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}-\frac {4 a^2 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (5 B+4 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*(5*B + 4*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(3*A + 2*B + C)*EllipticF[(c + d*x)/2, 2])/(3*d)
+ (2*a^2*(15*A + 25*B + 17*C)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*C*(a + a*Cos[c + d*x])^2*Sin[c + d
*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(5*B + 4*C)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)
)

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 \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, 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 {7}{2}}(c+d x)} \, dx\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{2} a (5 B+4 C)+\frac {1}{2} a (5 A-C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (15 A+25 B+17 C)+\frac {1}{4} a^2 (15 A-5 B-7 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} a^3 (15 A+25 B+17 C)+\left (\frac {1}{4} a^3 (15 A-5 B-7 C)+\frac {1}{4} a^3 (15 A+25 B+17 C)\right ) \cos (c+d x)+\frac {1}{4} a^3 (15 A-5 B-7 C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac {2 a^2 (15 A+25 B+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 \int \frac {\frac {5}{4} a^3 (3 A+2 B+C)-\frac {3}{4} a^3 (5 B+4 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a}\\ &=\frac {2 a^2 (15 A+25 B+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (2 a^2 (3 A+2 B+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (2 a^2 (5 B+4 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {4 a^2 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (3 A+2 B+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^2 (15 A+25 B+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (5 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{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 = 6.67, size = 1599, normalized size = 9.19 \begin {gather*} -\frac {i B \cos ^4(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{2 (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}-\frac {2 i C \cos ^4(c+d x) \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{5 (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))}+\frac {\cos ^{\frac {9}{2}}(c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {(-5 A-20 B-16 C+5 A \cos (2 c)) \csc (c) \sec (c)}{10 d}+\frac {C \sec (c) \sec ^3(c+d x) \sin (d x)}{5 d}+\frac {\sec (c) \sec ^2(c+d x) (3 C \sin (c)+5 B \sin (d x)+10 C \sin (d x))}{15 d}+\frac {\sec (c) \sec (c+d x) (5 B \sin (c)+10 C \sin (c)+15 A \sin (d x)+30 B \sin (d x)+24 C \sin (d x))}{15 d}\right )}{A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)}-\frac {2 A \cos ^4(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {4 B \cos ^4(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}}-\frac {2 C \cos ^4(c+d x) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/2*I)*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]) - (
((2*I)/5)*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*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]) + (
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)*(-1/10*
((-5*A - 20*B - 16*C + 5*A*Cos[2*c])*Csc[c]*Sec[c])/d + (C*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(5*d) + (Sec[c]*Sec
[c + d*x]^2*(3*C*Sin[c] + 5*B*Sin[d*x] + 10*C*Sin[d*x]))/(15*d) + (Sec[c]*Sec[c + d*x]*(5*B*Sin[c] + 10*C*Sin[
c] + 15*A*Sin[d*x] + 30*B*Sin[d*x] + 24*C*Sin[d*x]))/(15*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]
) - (2*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 - 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]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*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]]]]*Sqrt[
-(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]) - (2*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]*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])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(878\) vs. \(2(210)=420\).
time = 0.30, size = 879, normalized size = 5.05

method	result	size
default	\(\text {Expression too large to display}\)	\(879\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+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*B*(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))+(1/4*B+1/2*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)))+1/20*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)+(1/4*A+1/2*B+1/4*C)/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*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*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(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^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 = 0.34, size = 248, normalized size = 1.43 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\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 ) - 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\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 (3 \, {\left (5 \, A + 10 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(5*I*sqrt(2)*(3*A + 2*B + C)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)
) - 5*I*sqrt(2)*(3*A + 2*B + C)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) +
3*I*sqrt(2)*(5*B + 4*C)*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I
*sin(d*x + c))) - 3*I*sqrt(2)*(5*B + 4*C)*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0,
cos(d*x + c) - I*sin(d*x + c))) - (3*(5*A + 10*B + 8*C)*a^2*cos(d*x + c)^2 + 5*(B + 2*C)*a^2*cos(d*x + c) + 3
*C*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*sqrt(cos(c + d*x)), x) + Integral(2*A*sqrt(cos(c + d*x))*sec(c + d*x), x) + Integral(A*sqrt(c
os(c + d*x))*sec(c + d*x)**2, x) + Integral(B*sqrt(cos(c + d*x))*sec(c + d*x), x) + Integral(2*B*sqrt(cos(c +
d*x))*sec(c + d*x)**2, x) + Integral(B*sqrt(cos(c + d*x))*sec(c + d*x)**3, x) + Integral(C*sqrt(cos(c + d*x))*
sec(c + d*x)**2, x) + Integral(2*C*sqrt(cos(c + d*x))*sec(c + d*x)**3, x) + Integral(C*sqrt(cos(c + d*x))*sec(
c + d*x)**4, 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(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^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 = 7.20, size = 313, normalized size = 1.80 \begin {gather*} \frac {6\,C\,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\,C\,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\,C\,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 {2\,A\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,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 {4\,B\,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\,B\,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(cos(c + d*x)^(1/2)*(a + a/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(6*C*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 20*C*a^2*cos(c + d*x)*sin(c + d*x)*hyperg
eom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 30*C*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)) + (2*A*a^2*ellipticE(c/2 + (d*x)/2, 2))/d + (
4*A*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a^2*sin(c + d*x)*hyper
geom([-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)) + (4*B*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*B*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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