3.1378 \(\int (a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\)

Optimal. Leaf size=258 \[ \frac {2 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 a b^2 (5 A-C) \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3}{3 d}+\frac {4 A b \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^2}{d}-\frac {2 b^3 (35 A-3 C) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

-2/15*b^3*(35*A-3*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/3*A*(a+b*cos(d*x+c))^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d-2*a*
b^2*(5*A-C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+4*A*b*(a+b*cos(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*b*(15*a^2
*(A-C)-b^2*(5*A+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))*co
s(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/3*a*(3*b^2*(3*A+C)+a^2*(A+3*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 = 0.83, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4221, 3048, 3047, 3033, 3023, 2748, 2641, 2639} \[ \frac {2 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 a b^2 (5 A-C) \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3}{3 d}+\frac {4 A b \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^2}{d}-\frac {2 b^3 (35 A-3 C) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]

[Out]

(-2*b*(15*a^2*(A - C) - b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d
) + (2*a*(3*b^2*(3*A + C) + a^2*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3
*d) - (2*b^3*(35*A - 3*C)*Sin[c + d*x])/(15*d*Sec[c + d*x]^(3/2)) - (2*a*b^2*(5*A - C)*Sin[c + d*x])/(d*Sqrt[S
ec[c + d*x]]) + (4*A*b*(a + b*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d + (2*A*(a + b*Cos[c + d*x])^3
*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*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 3047

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 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 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))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{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))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^2 \left (3 A b+\frac {1}{2} a (A+3 C) \cos (c+d x)-\frac {1}{2} b (5 A-3 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 A b (a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} \left (24 A b^2+a^2 (A+3 C)\right )-\frac {1}{2} a b (5 A-3 C) \cos (c+d x)-\frac {1}{4} b^2 (35 A-3 C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^3 (35 A-3 C) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 A b (a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{15} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} a \left (24 A b^2+a^2 (A+3 C)\right )-\frac {3}{8} b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)-\frac {45}{8} a b^2 (5 A-C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^3 (35 A-3 C) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 a b^2 (5 A-C) \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {4 A b (a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{45} \left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {15}{16} a \left (3 b^2 (3 A+C)+a^2 (A+3 C)\right )-\frac {9}{16} b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^3 (35 A-3 C) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 a b^2 (5 A-C) \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {4 A b (a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (a \left (3 b^2 (3 A+C)+a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a \left (3 b^2 (3 A+C)+a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b^3 (35 A-3 C) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 a b^2 (5 A-C) \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {4 A b (a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 A (a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 2.66, size = 179, normalized size = 0.69 \[ \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (20 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+12 b \left (b^2 (5 A+3 C)-15 a^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {\sin (c+d x) \left (20 a^3 A+9 b \left (20 a^2 A+b^2 C\right ) \cos (c+d x)+30 a b^2 C \cos (2 (c+d x))+30 a b^2 C+3 b^3 C \cos (3 (c+d x))\right )}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(12*b*(-15*a^2*(A - C) + b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 2
0*a*(3*b^2*(3*A + C) + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2] + ((20*a^3*A + 30*a*b^2*C + 9*b*(20*a^2*A + b^
2*C)*Cos[c + d*x] + 30*a*b^2*C*Cos[2*(c + d*x)] + 3*b^3*C*Cos[3*(c + d*x)])*Sin[c + d*x])/Cos[c + d*x]^(3/2)))
/(30*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 8.97, size = 1267, normalized size = 4.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+
1)/sin(1/2*d*x+1/2*c)^3*(-180*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-120*C*a*b^2*cos(1/2*d*x+1/2*c)*s
in(1/2*d*x+1/2*c)^4+90*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+30*C*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^2+9*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))*b^3-15*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))*a^3+120*C*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+72*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-
48*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-36*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+10*A*a^3*cos
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+6*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-45*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*b-45*A*a*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+90*A*(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))*a^2*b*sin(1/2*d*x+1/2*c)^2+9
0*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2*
sin(1/2*d*x+1/2*c)^2-90*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))*a^2*b*sin(1/2*d*x+1/2*c)^2+30*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)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2*sin(1/2*d*x+1/2*c)^2+15*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^3-5*A*a^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))+45*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))*a^2*b-15*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-30*A*(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))*b^3*sin(1/2*d*x+1/2*c)^2+10*A*(2*sin(1/2*d*x+1/2*c)
^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*sin(1/2*d*x+1/2*c)^2-18*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))*b^3*sin(1/2
*d*x+1/2*c)^2+30*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)*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2))*a^3*sin(1/2*d*x+1/2*c)^2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^3,x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^3, 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))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2),x)

[Out]

Timed out

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