3.14.11 \(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1311]

Optimal. Leaf size=357 \[ -\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]

[Out]

-2/15*(15*a^3*B+27*a*b^2*B+9*a^2*b*(5*A+3*C)+b^3*(9*A+7*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*(21*a^2*b*B+5*b^3*B+7*a^3*(3*A+C)+3*a*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+2/315*b*(63*A*b^2+99*B*a*b+24*C*a^2+
49*C*b^2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/63*(54*a^2*b*B+15*b^3*B+8*a^3*C+9*a*b^2*(7*A+5*C))*sin(d*x+c)/d/cos(
d*x+c)^(3/2)+2/21*(3*B*b+2*C*a)*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/9*C*(b+a*cos(d*x+c))^3*sin(
d*x+c)/d/cos(d*x+c)^(9/2)+2/15*(15*a^3*B+27*a*b^2*B+9*a^2*b*(5*A+3*C)+b^3*(9*A+7*C))*sin(d*x+c)/d/cos(d*x+c)^(
1/2)

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Rubi [A]
time = 0.66, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4197, 3126, 3110, 3100, 2827, 2716, 2719, 2720} \begin {gather*} \frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right )}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right )}{15 d}+\frac {2 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{63 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*a^3*B + 27*a*b^2*B + 9*a^2*b*(5*A + 3*C) + b^3*(9*A + 7*C))*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(21
*a^2*b*B + 5*b^3*B + 7*a^3*(3*A + C) + 3*a*b^2*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*b*(63*A*b^2
+ 99*a*b*B + 24*a^2*C + 49*b^2*C)*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)) + (2*(54*a^2*b*B + 15*b^3*B + 8*a^
3*C + 9*a*b^2*(7*A + 5*C))*Sin[c + d*x])/(63*d*Cos[c + d*x]^(3/2)) + (2*(15*a^3*B + 27*a*b^2*B + 9*a^2*b*(5*A
+ 3*C) + b^3*(9*A + 7*C))*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(3*b*B + 2*a*C)*(b + a*Cos[c + d*x])^2*
Sin[c + d*x])/(21*d*Cos[c + d*x]^(7/2)) + (2*C*(b + a*Cos[c + d*x])^3*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2))

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*} \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx &=\int \frac {(b+a \cos (c+d x))^3 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\\ &=\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {2}{9} \int \frac {(b+a \cos (c+d x))^2 \left (\frac {3}{2} (3 b B+2 a C)+\frac {1}{2} (9 A b+9 a B+7 b C) \cos (c+d x)+\frac {1}{2} a (9 A+C) \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {4}{63} \int \frac {(b+a \cos (c+d x)) \left (\frac {1}{4} \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right )+\frac {1}{4} \left (126 a A b+63 a^2 B+45 b^2 B+86 a b C\right ) \cos (c+d x)+\frac {1}{4} a (63 a A+9 b B+13 a C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}-\frac {8}{315} \int \frac {-\frac {15}{8} \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right )-\frac {21}{8} \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \cos (c+d x)-\frac {5}{8} a^2 (63 a A+9 b B+13 a C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}-\frac {16}{945} \int \frac {-\frac {63}{16} \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right )-\frac {45}{16} \left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (7 A+5 C)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}-\frac {1}{21} \left (-21 a^2 b B-5 b^3 B-7 a^3 (3 A+C)-3 a b^2 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{15} \left (-15 a^3 B-27 a b^2 B-9 a^2 b (5 A+3 C)-b^3 (9 A+7 C)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}-\frac {1}{15} \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{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 = 8.45, size = 3345, normalized size = 9.37 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]^(11/2)*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(45*a^2*A*b + 9*A*b^3
+ 15*a^3*B + 27*a*b^2*B + 27*a^2*b*C + 7*b^3*C)*Csc[c]*Sec[c])/(15*d) + (4*b^3*C*Sec[c]*Sec[c + d*x]^5*Sin[d*x
])/(9*d) + (4*Sec[c]*Sec[c + d*x]^4*(7*b^3*C*Sin[c] + 9*b^3*B*Sin[d*x] + 27*a*b^2*C*Sin[d*x]))/(63*d) + (4*Sec
[c]*Sec[c + d*x]^2*(63*A*b^3*Sin[c] + 189*a*b^2*B*Sin[c] + 189*a^2*b*C*Sin[c] + 49*b^3*C*Sin[c] + 315*a*A*b^2*
Sin[d*x] + 315*a^2*b*B*Sin[d*x] + 75*b^3*B*Sin[d*x] + 105*a^3*C*Sin[d*x] + 225*a*b^2*C*Sin[d*x]))/(315*d) + (4
*Sec[c]*Sec[c + d*x]^3*(45*b^3*B*Sin[c] + 135*a*b^2*C*Sin[c] + 63*A*b^3*Sin[d*x] + 189*a*b^2*B*Sin[d*x] + 189*
a^2*b*C*Sin[d*x] + 49*b^3*C*Sin[d*x]))/(315*d) + (4*Sec[c]*Sec[c + d*x]*(105*a*A*b^2*Sin[c] + 105*a^2*b*B*Sin[
c] + 25*b^3*B*Sin[c] + 35*a^3*C*Sin[c] + 75*a*b^2*C*Sin[c] + 315*a^2*A*b*Sin[d*x] + 63*A*b^3*Sin[d*x] + 105*a^
3*B*Sin[d*x] + 189*a*b^2*B*Sin[d*x] + 189*a^2*b*C*Sin[d*x] + 49*b^3*C*Sin[d*x]))/(105*d)))/((b + a*Cos[c + d*x
])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) - (4*a^3*A*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Se
c[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*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[
2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*a*A*b^2*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[
d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^3*(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*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1
+ Cot[c]^2]) - (4*a^2*b*B*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]
]^2]*(a + b*Sec[c + d*x])^3*(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*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (20*
b^3*B*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d
*x])^3*(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]]]])/(21*d*(b + a*
Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*a^3*C*Cos[c + d*x]^
5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^3*(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*(b + a*Cos[c + d*x])^3*(A +
2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (20*a*b^2*C*Cos[c + d*x]^5*Csc[c]*Hypergeo
metricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^3*(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]]]])/(7*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c
+ d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) + (6*a^2*A*b*Cos[c + d*x]^5*Csc[c]*(a + b*Sec[c + d*x])^3*(A
+ B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((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[C
os[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]]))/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*
Cos[2*c + 2*d*x])) + (6*A*b^3*Cos[c + d*x]^5*Csc[c]*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x
]^2)*((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]]]*Sq
rt[1 + Tan[c]^2]]))/(5*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (2*a^3*B*
Cos[c + d*x]^5*Csc[c]*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2
, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Si...

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1264\) vs. \(2(385)=770\).
time = 0.67, size = 1265, normalized size = 3.54

method	result	size
default	\(\text {Expression too large to display}\)	\(1265\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^3*(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^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)
)+2*b^2*(B*b+3*C*a)*(-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*C*b^3*(-1/144*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)^5-7/180*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)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(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))
-7/15*(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))))+2*a^2*(3*A*b+B*a)/s
in(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)))+2*a*(3*A*b^2+3*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)))+2/5*b*(A*b^2
+3*B*a*b+3*C*a^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)/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))/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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.66, size = 467, normalized size = 1.31 \begin {gather*} -\frac {15 \, \sqrt {2} {\left (7 i \, {\left (3 \, A + C\right )} a^{3} + 21 i \, B a^{2} b + 3 i \, {\left (7 \, A + 5 \, C\right )} a b^{2} + 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, \sqrt {2} {\left (-7 i \, {\left (3 \, A + C\right )} a^{3} - 21 i \, B a^{2} b - 3 i \, {\left (7 \, A + 5 \, C\right )} a b^{2} - 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (15 i \, B a^{3} + 9 i \, {\left (5 \, A + 3 \, C\right )} a^{2} b + 27 i \, B a b^{2} + i \, {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} {\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 (-15 i \, B a^{3} - 9 i \, {\left (5 \, A + 3 \, C\right )} a^{2} b - 27 i \, B a b^{2} - i \, {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} {\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 (15 \, B a^{3} + 9 \, {\left (5 \, A + 3 \, C\right )} a^{2} b + 27 \, B a b^{2} + {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} + 35 \, C b^{3} + 15 \, {\left (7 \, C a^{3} + 21 \, B a^{2} b + 3 \, {\left (7 \, A + 5 \, C\right )} a b^{2} + 5 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 7 \, {\left (27 \, C a^{2} b + 27 \, B a b^{2} + {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 45 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/315*(15*sqrt(2)*(7*I*(3*A + C)*a^3 + 21*I*B*a^2*b + 3*I*(7*A + 5*C)*a*b^2 + 5*I*B*b^3)*cos(d*x + c)^5*weier
strassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*sqrt(2)*(-7*I*(3*A + C)*a^3 - 21*I*B*a^2*b - 3*I*(7*
A + 5*C)*a*b^2 - 5*I*B*b^3)*cos(d*x + c)^5*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt
(2)*(15*I*B*a^3 + 9*I*(5*A + 3*C)*a^2*b + 27*I*B*a*b^2 + I*(9*A + 7*C)*b^3)*cos(d*x + c)^5*weierstrassZeta(-4,
0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(-15*I*B*a^3 - 9*I*(5*A + 3*C)*a^2
*b - 27*I*B*a*b^2 - I*(9*A + 7*C)*b^3)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*
x + c) - I*sin(d*x + c))) - 2*(21*(15*B*a^3 + 9*(5*A + 3*C)*a^2*b + 27*B*a*b^2 + (9*A + 7*C)*b^3)*cos(d*x + c)
^4 + 35*C*b^3 + 15*(7*C*a^3 + 21*B*a^2*b + 3*(7*A + 5*C)*a*b^2 + 5*B*b^3)*cos(d*x + c)^3 + 7*(27*C*a^2*b + 27*
B*a*b^2 + (9*A + 7*C)*b^3)*cos(d*x + c)^2 + 45*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{3} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 10.70, size = 463, normalized size = 1.30 \begin {gather*} \frac {\frac {2\,B\,b^3\,\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 )}{7}+2\,B\,a^3\,{\cos \left (c+d\,x\right )}^3\,\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 )+2\,B\,a^2\,b\,{\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 )+\frac {6\,B\,a\,b^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 )}{5}}{d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {70\,C\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {9}{4},\frac {1}{2};\ -\frac {5}{4};\ {\cos \left (c+d\,x\right )}^2\right )+210\,C\,a^3\,{\cos \left (c+d\,x\right )}^3\,\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 )+378\,C\,a^2\,b\,{\cos \left (c+d\,x\right )}^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 )+270\,C\,a\,b^2\,\cos \left (c+d\,x\right )\,\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 )}{315\,d\,{\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^3\,\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 )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,A\,a^2\,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}}+\frac {2\,A\,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 )}{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 + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(1/2),x)

[Out]

((2*B*b^3*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + 2*B*a^3*cos(c + d*x)^3*sin(c + d*x)*h
ypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2) + 2*B*a^2*b*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4,
cos(c + d*x)^2) + (6*B*a*b^2*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/5)/(d*co
s(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2)) + (70*C*b^3*sin(c + d*x)*hypergeom([-9/4, 1/2], -5/4, cos(c + d*x
)^2) + 210*C*a^3*cos(c + d*x)^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 378*C*a^2*b*cos(c +
d*x)^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 270*C*a*b^2*cos(c + d*x)*sin(c + d*x)*hype
rgeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/(315*d*cos(c + d*x)^(9/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*A*a^3*ell
ipticF(c/2 + (d*x)/2, 2))/d + (2*A*b^3*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c +
d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)) + (6*A*a^2*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)) + (2*A*a*b^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^
2))/(d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))

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