3.13.99 \(\int \cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1299]

Optimal. Leaf size=250 \[ \frac {2 \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (4 A b^2+18 a b B+a^2 (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a (4 A b+9 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{9 d} \]

[Out]

2/15*(18*a*b*B+3*b^2*(3*A+5*C)+a^2*(7*A+9*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+2/21*(10*A*a*b+5*B*a^2+7*B*b^2+14*C*a*b)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*
c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/45*(4*A*b^2+18*a*b*B+a^2*(7*A+9*C))*cos(d*x+c)^(3/2)*sin(d*x+c)/d
+2/63*a*(4*A*b+9*B*a)*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*A*cos(d*x+c)^(3/2)*(b+a*cos(d*x+c))^2*sin(d*x+c)/d+2/2
1*(10*A*a*b+5*B*a^2+7*B*b^2+14*C*a*b)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.42, antiderivative size = 250, 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, 3128, 3112, 3102, 2827, 2719, 2715, 2720} \begin {gather*} \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right )}{21 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (7 A+9 C)+18 a b B+3 b^2 (3 A+5 C)\right )}{15 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 (7 A+9 C)+18 a b B+4 A b^2\right )}{45 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 B+10 a A b+14 a b C+7 b^2 B\right )}{21 d}+\frac {2 a (9 a B+4 A b) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{63 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)^2}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*(18*a*b*B + 3*b^2*(3*A + 5*C) + a^2*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(10*a*A*b + 5*a^2*B
+ 7*b^2*B + 14*a*b*C)*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*(10*a*A*b + 5*a^2*B + 7*b^2*B + 14*a*b*C)*Sqrt[C
os[c + d*x]]*Sin[c + d*x])/(21*d) + (2*(4*A*b^2 + 18*a*b*B + a^2*(7*A + 9*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])
/(45*d) + (2*a*(4*A*b + 9*a*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d) + (2*A*Cos[c + d*x]^(3/2)*(b + a*Cos[c
+ d*x])^2*Sin[c + d*x])/(9*d)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 3102

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

Rule 3112

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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 \cos ^{\frac {9}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2}{9} \int \sqrt {\cos (c+d x)} (b+a \cos (c+d x)) \left (\frac {3}{2} b (A+3 C)+\frac {1}{2} (7 a A+9 b B+9 a C) \cos (c+d x)+\frac {1}{2} (4 A b+9 a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 a (4 A b+9 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {4}{63} \int \sqrt {\cos (c+d x)} \left (\frac {21}{4} b^2 (A+3 C)+\frac {9}{4} \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \cos (c+d x)+\frac {7}{4} \left (4 A b^2+18 a b B+a^2 (7 A+9 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 \left (4 A b^2+18 a b B+a^2 (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a (4 A b+9 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {8}{315} \int \sqrt {\cos (c+d x)} \left (\frac {21}{8} \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right )+\frac {45}{8} \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {2 \left (4 A b^2+18 a b B+a^2 (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a (4 A b+9 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{7} \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{15} \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (4 A b^2+18 a b B+a^2 (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a (4 A b+9 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{21} \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (10 a A b+5 a^2 B+7 b^2 B+14 a b C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 \left (4 A b^2+18 a b B+a^2 (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a (4 A b+9 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]
time = 1.42, size = 194, normalized size = 0.78 \begin {gather*} \frac {84 \left (18 a b B+3 b^2 (3 A+5 C)+a^2 (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+60 \left (5 a^2 B+7 b^2 B+2 a b (5 A+7 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} \left (7 \left (36 A b^2+72 a b B+a^2 (43 A+36 C)\right ) \cos (c+d x)+5 \left (156 a A b+78 a^2 B+84 b^2 B+168 a b C+18 a (2 A b+a B) \cos (2 (c+d x))+7 a^2 A \cos (3 (c+d x))\right )\right ) \sin (c+d x)}{630 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(84*(18*a*b*B + 3*b^2*(3*A + 5*C) + a^2*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2] + 60*(5*a^2*B + 7*b^2*B + 2*a*b
*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(7*(36*A*b^2 + 72*a*b*B + a^2*(43*A + 36*C))*Cos[
c + d*x] + 5*(156*a*A*b + 78*a^2*B + 84*b^2*B + 168*a*b*C + 18*a*(2*A*b + a*B)*Cos[2*(c + d*x)] + 7*a^2*A*Cos[
3*(c + d*x)]))*Sin[c + d*x])/(630*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(783\) vs. \(2(282)=564\).
time = 0.18, size = 784, normalized size = 3.14

method	result	size
default	\(\text {Expression too large to display}\)	\(784\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(9/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
10*a^2+(2240*A*a^2+1440*A*a*b+720*B*a^2)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-2072*A*a^2-2160*A*a*b-504*A
*b^2-1080*B*a^2-1008*B*a*b-504*C*a^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(952*A*a^2+1680*A*a*b+504*A*b^2+
840*B*a^2+1008*B*a*b+420*B*b^2+504*C*a^2+840*C*a*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-168*A*a^2-480*A*
a*b-126*A*b^2-240*B*a^2-252*B*a*b-210*B*b^2-126*C*a^2-420*C*a*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+150*a
*A*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-189*A
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+75*a^
2*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+105*b^
2*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-378*B*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+210*a*
b*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-189*C*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-315*C*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2)/(-2*s
in(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(cos(d*x+c)^(9/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^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 = 1.50, size = 318, normalized size = 1.27 \begin {gather*} \frac {2 \, {\left (35 \, A a^{2} \cos \left (d x + c\right )^{3} + 75 \, B a^{2} + 30 \, {\left (5 \, A + 7 \, C\right )} a b + 105 \, B b^{2} + 45 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{2} + 7 \, {\left ({\left (7 \, A + 9 \, C\right )} a^{2} + 18 \, B a b + 9 \, A b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \, \sqrt {2} {\left (5 i \, B a^{2} + 2 i \, {\left (5 \, A + 7 \, C\right )} a b + 7 i \, B b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 \, \sqrt {2} {\left (-5 i \, B a^{2} - 2 i \, {\left (5 \, A + 7 \, C\right )} a b - 7 i \, B b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 \, \sqrt {2} {\left (-i \, {\left (7 \, A + 9 \, C\right )} a^{2} - 18 i \, B a b - 3 i \, {\left (3 \, A + 5 \, C\right )} b^{2}\right )} {\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 (i \, {\left (7 \, A + 9 \, C\right )} a^{2} + 18 i \, B a b + 3 i \, {\left (3 \, A + 5 \, C\right )} b^{2}\right )} {\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 )}{315 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/315*(2*(35*A*a^2*cos(d*x + c)^3 + 75*B*a^2 + 30*(5*A + 7*C)*a*b + 105*B*b^2 + 45*(B*a^2 + 2*A*a*b)*cos(d*x +
c)^2 + 7*((7*A + 9*C)*a^2 + 18*B*a*b + 9*A*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 15*sqrt(2)*(5
*I*B*a^2 + 2*I*(5*A + 7*C)*a*b + 7*I*B*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*sqr
t(2)*(-5*I*B*a^2 - 2*I*(5*A + 7*C)*a*b - 7*I*B*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))
- 21*sqrt(2)*(-I*(7*A + 9*C)*a^2 - 18*I*B*a*b - 3*I*(3*A + 5*C)*b^2)*weierstrassZeta(-4, 0, weierstrassPInvers
e(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*sqrt(2)*(I*(7*A + 9*C)*a^2 + 18*I*B*a*b + 3*I*(3*A + 5*C)*b^2)*w
eierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

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Sympy [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(cos(d*x+c)**(9/2)*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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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)^(9/2)*(a+b*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)*(b*sec(d*x + c) + a)^2*cos(d*x + c)^(9/2), x)

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Mupad [B]
time = 5.88, size = 366, normalized size = 1.46 \begin {gather*} \frac {2\,B\,b^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,C\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,A\,a\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,B\,a\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\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)^(9/2)*(a + b/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*B*b^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*C*b^2*ellipticE(c/2 + (d*
x)/2, 2))/d + (2*C*a*b*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d - (2*A*a
^2*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)
) - (2*A*b^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2
)^(1/2)) - (2*B*a^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c +
d*x)^2)^(1/2)) - (2*C*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(
sin(c + d*x)^2)^(1/2)) - (4*A*a*b*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))
/(9*d*(sin(c + d*x)^2)^(1/2)) - (4*B*a*b*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d
*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))

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