3.692 \(\int \frac{(a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=218 \[ \frac{2 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 a b^2 (5 A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}}-\frac{2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 d} \]
[Out]
(-2*b*(15*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(3*b^2*(3*A + C) + a^2*(A + 3
*C))*EllipticF[(c + d*x)/2, 2])/(3*d) - (2*a*b^2*(5*A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/d - (2*b^3*(35*A -
3*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (4*A*b*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*
x]]) + (2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))
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Rubi [A] time = 0.690981, antiderivative size = 218, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3048, 3047, 3033, 3023, 2748, 2641, 2639} \[ \frac{2 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 a b^2 (5 A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}}-\frac{2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 d} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
[Out]
(-2*b*(15*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(3*b^2*(3*A + C) + a^2*(A + 3
*C))*EllipticF[(c + d*x)/2, 2])/(3*d) - (2*a*b^2*(5*A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/d - (2*b^3*(35*A -
3*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (4*A*b*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*
x]]) + (2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))
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 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 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 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 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 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 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]
Rubi steps
\begin{align*} \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 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \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 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4}{3} \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) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{4 A b (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{8}{15} \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 a b^2 (5 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d}-\frac{2 b^3 (35 A-3 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{4 A b (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{16}{45} \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 a b^2 (5 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d}-\frac{2 b^3 (35 A-3 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{4 A b (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{1}{3} \left (a \left (3 b^2 (3 A+C)+a^2 (A+3 C)\right )\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 )\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a \left (3 b^2 (3 A+C)+a^2 (A+3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 a b^2 (5 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d}-\frac{2 b^3 (35 A-3 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{4 A b (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 1.69354, size = 150, normalized size = 0.69 \[ \frac{10 \left (a^3 (A+3 C)+3 a b^2 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \left (3 b^3 (5 A+3 C)-45 a^2 b (A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{5 a \left (2 a^2 A \tan (c+d x)+3 b^2 C \sin (2 (c+d x))\right )+6 \sin (c+d x) \left (15 a^2 A b+b^3 C \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}}}{15 d} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
[Out]
(2*(-45*a^2*b*(A - C) + 3*b^3*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 10*(3*a*b^2*(3*A + C) + a^3*(A + 3*C))*
EllipticF[(c + d*x)/2, 2] + (6*(15*a^2*A*b + b^3*C*Cos[c + d*x]^2)*Sin[c + d*x] + 5*a*(3*b^2*C*Sin[2*(c + d*x)
] + 2*a^2*A*Tan[c + d*x]))/Sqrt[Cos[c + d*x]])/(15*d)
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Maple [B] time = 1.385, size = 1267, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}
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)/cos(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*(30*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*a^3*sin(1/2*d*x+1/2*c)^2-18*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^
2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*sin(1/2*d*x+1/2*c)^2+10*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*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))+45*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*b-15*C*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))+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+72*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/
2*d*x+1/2*c)^6-36*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-48*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^8-30*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^
3*sin(1/2*d*x+1/2*c)^2-180*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)*sin(
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+120*C*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+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))+9*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^3-15*a^3*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))+30*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2*sin(1/2*d*x+1/2*c)^2-90*C*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b*sin(1/2*d*x+1/2*c)^2+90*A*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2*sin(1/2*d*
x+1/2*c)^2+90*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*a^2*b*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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
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)/cos(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{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}}{\cos \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}
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)/cos(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)/cos(d*x + c)^(5/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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)/cos(d*x+c)**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
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)/cos(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)