3.700 \(\int \frac{(a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=321 \[ \frac{8 a b \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (30 a^2 b^2 (A-C)+a^4 (3 A+5 C)-b^4 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 b^2 \left (3 a^2 (3 A+5 C)+b^2 (59 A-3 C)\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 \left (a^2 (3 A+5 C)+16 A b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \sqrt{\cos (c+d x)}}-\frac{4 a b \left (3 a^2 (3 A+5 C)+2 b^2 (33 A-5 C)\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{16 A b \sin (c+d x) (a+b \cos (c+d x))^3}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
[Out]
(-2*(30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a*b*(b^2*(3
*A + C) + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*d) - (4*a*b*(2*b^2*(33*A - 5*C) + 3*a^2*(3*A + 5*C))*Sq
rt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) - (2*b^2*(b^2*(59*A - 3*C) + 3*a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*Sin[c
+ d*x])/(15*d) + (2*(16*A*b^2 + a^2*(3*A + 5*C))*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]
) + (16*A*b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (2*A*(a + b*Cos[c + d*x])^4*Sin[c
+ d*x])/(5*d*Cos[c + d*x]^(5/2))
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Rubi [A] time = 1.22486, antiderivative size = 321, normalized size of antiderivative = 1.,
number of steps used = 8, 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} \[ \frac{8 a b \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (30 a^2 b^2 (A-C)+a^4 (3 A+5 C)-b^4 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 b^2 \left (3 a^2 (3 A+5 C)+b^2 (59 A-3 C)\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 \left (a^2 (3 A+5 C)+16 A b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \sqrt{\cos (c+d x)}}-\frac{4 a b \left (3 a^2 (3 A+5 C)+2 b^2 (33 A-5 C)\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{16 A b \sin (c+d x) (a+b \cos (c+d x))^3}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
[Out]
(-2*(30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a*b*(b^2*(3
*A + C) + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*d) - (4*a*b*(2*b^2*(33*A - 5*C) + 3*a^2*(3*A + 5*C))*Sq
rt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) - (2*b^2*(b^2*(59*A - 3*C) + 3*a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*Sin[c
+ d*x])/(15*d) + (2*(16*A*b^2 + a^2*(3*A + 5*C))*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]
) + (16*A*b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (2*A*(a + b*Cos[c + d*x])^4*Sin[c
+ d*x])/(5*d*Cos[c + d*x]^(5/2))
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))^4 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \cos (c+d x))^3 \left (4 A b+\frac{1}{2} a (3 A+5 C) \cos (c+d x)-\frac{5}{2} b (A-C) \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4}{15} \int \frac{(a+b \cos (c+d x))^2 \left (\frac{3}{4} \left (16 A b^2+a^2 (3 A+5 C)\right )+\frac{1}{2} a b (A+15 C) \cos (c+d x)-\frac{5}{4} b^2 (11 A-3 C) \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8}{15} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{4} b \left (96 A b^2+a^2 (19 A+45 C)\right )-\frac{1}{8} a \left (b^2 (101 A-45 C)+3 a^2 (3 A+5 C)\right ) \cos (c+d x)-\frac{5}{8} b \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{16}{75} \int \frac{\frac{5}{8} a b \left (96 A b^2+a^2 (19 A+45 C)\right )-\frac{15}{16} \left (30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)-\frac{15}{8} a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{32}{225} \int \frac{\frac{75}{8} a b \left (b^2 (3 A+C)+a^2 (A+3 C)\right )-\frac{45}{32} \left (30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{1}{3} \left (4 a b \left (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 (-30 a^2 b^2 (A-C)+b^4 (5 A+3 C)-a^4 (3 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a b \left (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{4 a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}-\frac{2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 1.57635, size = 233, normalized size = 0.73 \[ \frac{40 a b \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 \left (30 a^2 b^2 (A-C)+a^4 (3 A+5 C)-b^4 (5 A+3 C)\right ) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+90 a^2 A b^2 \sin (2 (c+d x))+40 a^3 A b \sin (c+d x)+9 a^4 A \sin (2 (c+d x))+6 a^4 A \tan (c+d x)+15 a^4 C \sin (2 (c+d x))+40 a b^3 C \sin (c+d x) \cos ^2(c+d x)+6 b^4 C \sin (c+d x) \cos ^3(c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
[Out]
(-6*(30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 40
*a*b*(b^2*(3*A + C) + a^2*(A + 3*C))*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 40*a^3*A*b*Sin[c + d*x] +
40*a*b^3*C*Cos[c + d*x]^2*Sin[c + d*x] + 6*b^4*C*Cos[c + d*x]^3*Sin[c + d*x] + 9*a^4*A*Sin[2*(c + d*x)] + 90*a
^2*A*b^2*Sin[2*(c + d*x)] + 15*a^4*C*Sin[2*(c + d*x)] + 6*a^4*A*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/2))
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Maple [B] time = 1.729, size = 1622, normalized size = 5.1 \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))^4*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/5*C*b^4*(-4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2
*c)+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(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*EllipticE(cos(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)-6*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)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)+1/3*(16*C*a*b^3-12*C*b^4)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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))-3*EllipticE(cos(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)-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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(2*A*b^4+12*C*a^2*b^2-16*C*a*b^3+6*C*b^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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+8*a*A*b^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*A*b^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)/(-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))+8*a^3*b*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c
os(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))-12*a^2*b^2*C*(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))+8*C*a*b^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*C*b^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)/(-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*A*a^4/(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*(12*EllipticE(cos(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)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1
/2*c)-12*EllipticE(cos(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)*s
in(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(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)-8*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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+8*A*a^3*b*(-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)/(-1/2+cos(1/2*d*x+1/2*c)^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*a^2*(6*A*b
^2+C*a^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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-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)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/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., 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 )}^{4}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
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)/cos(d*x+c)^(7/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4/cos(d*x + c)^(7/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{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}}{\cos \left (d x + c\right )^{\frac{7}{2}}}, x\right ) \end{align*}
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)/cos(d*x+c)^(7/2),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)/cos(d*x + c
)^(7/2), 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))**4*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),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 )}^{4}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
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)/cos(d*x+c)^(7/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4/cos(d*x + c)^(7/2), x)