3.1388 \(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=360 \[ -\frac {2 b^2 \left (3 a^2 (105 A-41 C)-7 b^2 (9 A+7 C)\right ) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {4 a b \left (a^2 (63 A-31 C)-6 b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \sqrt {\sec (c+d x)}}+\frac {8 a b \left (7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}-\frac {2 b (9 A-C) \sin (c+d x) (a+b \cos (c+d x))^3}{9 d \sqrt {\sec (c+d x)}}-\frac {2 a b (21 A-5 C) \sin (c+d x) (a+b \cos (c+d x))^2}{21 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^4}{d} \]

[Out]

-2/315*b^2*(3*a^2*(105*A-41*C)-7*b^2*(9*A+7*C))*sin(d*x+c)/d/sec(d*x+c)^(3/2)-4/63*a*b*(a^2*(63*A-31*C)-6*b^2*
(7*A+5*C))*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/21*a*b*(21*A-5*C)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-
2/9*b*(9*A-C)*(a+b*cos(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2*A*(a+b*cos(d*x+c))^4*sin(d*x+c)*sec(d*x+c)^(1
/2)/d-2/15*(15*a^4*(A-C)-18*a^2*b^2*(5*A+3*C)-b^4*(9*A+7*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+8/21*a*b*(7*a^2*(3*A+C)+b^2*(7*A+5*C)
)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d
*x+c)^(1/2)/d

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Rubi [A]  time = 1.34, antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4221, 3048, 3049, 3033, 3023, 2748, 2641, 2639} \[ -\frac {2 b^2 \left (3 a^2 (105 A-41 C)-7 b^2 (9 A+7 C)\right ) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {4 a b \left (a^2 (63 A-31 C)-6 b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \sqrt {\sec (c+d x)}}+\frac {8 a b \left (7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (-18 a^2 b^2 (5 A+3 C)+15 a^4 (A-C)-b^4 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}-\frac {2 b (9 A-C) \sin (c+d x) (a+b \cos (c+d x))^3}{9 d \sqrt {\sec (c+d x)}}-\frac {2 a b (21 A-5 C) \sin (c+d x) (a+b \cos (c+d x))^2}{21 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^4}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

(-2*(15*a^4*(A - C) - 18*a^2*b^2*(5*A + 3*C) - b^4*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S
qrt[Sec[c + d*x]])/(15*d) + (8*a*b*(7*a^2*(3*A + C) + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/
2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (2*b^2*(3*a^2*(105*A - 41*C) - 7*b^2*(9*A + 7*C))*Sin[c + d*x])/(315*d*Sec[
c + d*x]^(3/2)) - (4*a*b*(a^2*(63*A - 31*C) - 6*b^2*(7*A + 5*C))*Sin[c + d*x])/(63*d*Sqrt[Sec[c + d*x]]) - (2*
a*b*(21*A - 5*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) - (2*b*(9*A - C)*(a + b*Cos[c
+ d*x])^3*Sin[c + d*x])/(9*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Cos[c + d*x])^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x]
)/d

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2748

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 3023

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 3033

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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 3049

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 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^3 \left (4 A b-\frac {1}{2} a (A-C) \cos (c+d x)-\frac {1}{2} b (9 A-C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b (9 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{9} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{4} a b (63 A+C)-\frac {1}{4} \left (9 a^2 (A-C)-b^2 (9 A+7 C)\right ) \cos (c+d x)-\frac {3}{4} a b (21 A-5 C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 a b (21 A-5 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (9 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{63} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} a^2 b (189 A+11 C)-\frac {1}{8} a \left (63 a^2 (A-C)-b^2 (189 A+131 C)\right ) \cos (c+d x)-\frac {1}{8} b \left (a^2 (315 A-123 C)-7 b^2 (9 A+7 C)\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 \left (3 a^2 (105 A-41 C)-7 b^2 (9 A+7 C)\right ) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 a b (21 A-5 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (9 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{315} \left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} a^3 b (189 A+11 C)-\frac {21}{16} \left (15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \cos (c+d x)-\frac {15}{8} a b \left (a^2 (63 A-31 C)-6 b^2 (7 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 \left (3 a^2 (105 A-41 C)-7 b^2 (9 A+7 C)\right ) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {4 a b \left (a^2 (63 A-31 C)-6 b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \sqrt {\sec (c+d x)}}-\frac {2 a b (21 A-5 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (9 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{945} \left (32 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {45}{8} a b \left (7 a^2 (3 A+C)+b^2 (7 A+5 C)\right )-\frac {63}{32} \left (15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 \left (3 a^2 (105 A-41 C)-7 b^2 (9 A+7 C)\right ) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {4 a b \left (a^2 (63 A-31 C)-6 b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \sqrt {\sec (c+d x)}}-\frac {2 a b (21 A-5 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (9 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{21} \left (4 a b \left (7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{15} \left (\left (15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {8 a b \left (7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}-\frac {2 b^2 \left (3 a^2 (105 A-41 C)-7 b^2 (9 A+7 C)\right ) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {4 a b \left (a^2 (63 A-31 C)-6 b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \sqrt {\sec (c+d x)}}-\frac {2 a b (21 A-5 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (9 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{9 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 1.77, size = 252, normalized size = 0.70 \[ \frac {\sqrt {\sec (c+d x)} \left (960 a b \left (7 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-336 \left (15 a^4 (A-C)-18 a^2 b^2 (5 A+3 C)-b^4 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) \left (2520 a^4 A+120 a b \left (28 a^2 C+28 A b^2+29 b^2 C\right ) \cos (c+d x)+84 \left (18 a^2 b^2 C+3 A b^4+4 b^4 C\right ) \cos (2 (c+d x))+1512 a^2 b^2 C+360 a b^3 C \cos (3 (c+d x))+252 A b^4+35 b^4 C \cos (4 (c+d x))+301 b^4 C\right )\right )}{2520 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

(Sqrt[Sec[c + d*x]]*(-336*(15*a^4*(A - C) - 18*a^2*b^2*(5*A + 3*C) - b^4*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*Ellip
ticE[(c + d*x)/2, 2] + 960*a*b*(7*a^2*(3*A + C) + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2
] + 2*(2520*a^4*A + 252*A*b^4 + 1512*a^2*b^2*C + 301*b^4*C + 120*a*b*(28*A*b^2 + 28*a^2*C + 29*b^2*C)*Cos[c +
d*x] + 84*(3*A*b^4 + 18*a^2*b^2*C + 4*b^4*C)*Cos[2*(c + d*x)] + 360*a*b^3*C*Cos[3*(c + d*x)] + 35*b^4*C*Cos[4*
(c + d*x)])*Sin[c + d*x]))/(2520*d)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{4} \cos \left (d x + c\right )^{6} + 4 \, C a b^{3} \cos \left (d x + c\right )^{5} + 4 \, A a^{3} b \cos \left (d x + c\right ) + A a^{4} + {\left (6 \, C a^{2} b^{2} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (C a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{4} + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sec \left (d x + c\right )^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((C*b^4*cos(d*x + c)^6 + 4*C*a*b^3*cos(d*x + c)^5 + 4*A*a^3*b*cos(d*x + c) + A*a^4 + (6*C*a^2*b^2 + A*
b^4)*cos(d*x + c)^4 + 4*(C*a^3*b + A*a*b^3)*cos(d*x + c)^3 + (C*a^4 + 6*A*a^2*b^2)*cos(d*x + c)^2)*sec(d*x + c
)^(3/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4*sec(d*x + c)^(3/2), x)

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maple [B]  time = 3.89, size = 1209, normalized size = 3.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x)

[Out]

-2/315*(-1120*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^10+320*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(9*a+7*b)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+
1/2*c)-8*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(63*A*b^2+378*C*a^2+540*C*a*b+259*C*b^2)*sin
(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+56*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(30*A*a*b^2+9*A
*b^3+30*C*a^3+54*C*a^2*b+60*C*a*b^2+17*C*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-6*(-2*sin(1/2*d*x+1/2*c)
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(105*A*a^4+140*A*a*b^3+21*A*b^4+140*C*a^3*b+126*C*a^2*b^2+160*C*a*b^3+28*C*b^4)
*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+315*A*(-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)^(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^4-1890*A*(-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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2-189*A*(-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)^(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^4+1260*A*a^3*b*(s
in(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))*(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+420*a*A*b^3*(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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-315*C*(-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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4-1134*C*(-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)^(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*b^2-147*C*(-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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^4+420*a^3*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+300*C*a*b^3*(si
n(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))*(-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)^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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4*sec(d*x + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^4,x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2),x)

[Out]

Timed out

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