3.701 \(\int \frac {(a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=316 \[ -\frac {8 a b \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2 (5 A+7 C)+48 A b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{105 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b^2 \left (5 a^2 (5 A+7 C)+b^2 (87 A-35 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d}+\frac {4 a b \left (a^2 (101 A+175 C)+96 A b^2\right ) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {2 \left (a^4 (5 A+7 C)+42 a^2 b^2 (A+3 C)+7 b^4 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {16 A b \sin (c+d x) (a+b \cos (c+d x))^3}{35 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-8/5*a*b*(5*b^2*(A-C)+a^2*(3*A+5*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*(7*b^4*(3*A+C)+42*a^2*b^2*(A+3*C)+a^4*(5*A+7*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))/d+2/105*(48*A*b^2+5*a^2*(5*A+7*C))*(a+b*cos(d*x+c))^2*sin(d*x+c)/
d/cos(d*x+c)^(3/2)+16/35*A*b*(a+b*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/7*A*(a+b*cos(d*x+c))^4*sin(d*x
+c)/d/cos(d*x+c)^(7/2)+4/105*a*b*(96*A*b^2+a^2*(101*A+175*C))*sin(d*x+c)/d/cos(d*x+c)^(1/2)-2/105*b^2*(b^2*(87
*A-35*C)+5*a^2*(5*A+7*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 1.14, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3048, 3047, 3031, 3023, 2748, 2641, 2639} \[ \frac {2 \left (42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)+7 b^4 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {8 a b \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2 (5 A+7 C)+48 A b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{105 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b^2 \left (5 a^2 (5 A+7 C)+b^2 (87 A-35 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{105 d}+\frac {4 a b \left (a^2 (101 A+175 C)+96 A b^2\right ) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {16 A b \sin (c+d x) (a+b \cos (c+d x))^3}{35 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(-8*a*b*(5*b^2*(A - C) + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(7*b^4*(3*A + C) + 42*a^2*b^2*
(A + 3*C) + a^4*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (4*a*b*(96*A*b^2 + a^2*(101*A + 175*C))*Sin[c
+ d*x])/(105*d*Sqrt[Cos[c + d*x]]) - (2*b^2*(b^2*(87*A - 35*C) + 5*a^2*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(105*d) + (2*(48*A*b^2 + 5*a^2*(5*A + 7*C))*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(105*d*Cos[c + d*x]^(
3/2)) + (16*A*b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)) + (2*A*(a + b*Cos[c + d*x])^4*S
in[c + d*x])/(7*d*Cos[c + d*x]^(7/2))

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 3031

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[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*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] && Ne
Q[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]

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+b \cos (c+d x))^3 \left (4 A b+\frac {1}{2} a (5 A+7 C) \cos (c+d x)-\frac {1}{2} b (3 A-7 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4}{35} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{4} \left (48 A b^2+5 a^2 (5 A+7 C)\right )+\frac {1}{2} a b (17 A+35 C) \cos (c+d x)-\frac {1}{4} b^2 (39 A-35 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (48 A b^2+5 a^2 (5 A+7 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {8}{105} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} b \left (96 A b^2+a^2 (101 A+175 C)\right )+\frac {1}{8} a \left (5 a^2 (5 A+7 C)+3 b^2 (11 A+105 C)\right ) \cos (c+d x)-\frac {3}{8} b \left (b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 a b \left (96 A b^2+a^2 (101 A+175 C)\right ) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {2 \left (48 A b^2+5 a^2 (5 A+7 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {16}{105} \int \frac {\frac {1}{16} \left (-192 A b^4-5 a^4 (5 A+7 C)-5 a^2 b^2 (47 A+133 C)\right )+\frac {21}{4} a b \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos (c+d x)+\frac {3}{16} b^2 \left (b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a b \left (96 A b^2+a^2 (101 A+175 C)\right ) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 \left (b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 \left (48 A b^2+5 a^2 (5 A+7 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {32}{315} \int \frac {-\frac {15}{32} \left (7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right )+\frac {63}{8} a b \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a b \left (96 A b^2+a^2 (101 A+175 C)\right ) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 \left (b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 \left (48 A b^2+5 a^2 (5 A+7 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (4 a b \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right )\right ) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{21} \left (-7 b^4 (3 A+C)-42 a^2 b^2 (A+3 C)-a^4 (5 A+7 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {8 a b \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 b^4 (3 A+C)+42 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a b \left (96 A b^2+a^2 (101 A+175 C)\right ) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 \left (b^2 (87 A-35 C)+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 \left (48 A b^2+5 a^2 (5 A+7 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 2.61, size = 276, normalized size = 0.87 \[ \frac {25 a^4 A \sin (2 (c+d x))+30 a^4 A \tan (c+d x)+35 a^4 C \sin (2 (c+d x))+168 a^3 A b \sin (c+d x)+504 a^3 A b \sin (c+d x) \cos ^2(c+d x)+840 a^3 b C \sin (c+d x) \cos ^2(c+d x)-168 a b \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+210 a^2 A b^2 \sin (2 (c+d x))+10 \left (a^4 (5 A+7 C)+42 a^2 b^2 (A+3 C)+7 b^4 (3 A+C)\right ) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+840 a A b^3 \sin (c+d x) \cos ^2(c+d x)+70 b^4 C \sin (c+d x) \cos ^3(c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(-168*a*b*(5*b^2*(A - C) + a^2*(3*A + 5*C))*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 10*(7*b^4*(3*A + C)
+ 42*a^2*b^2*(A + 3*C) + a^4*(5*A + 7*C))*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 168*a^3*A*b*Sin[c +
d*x] + 504*a^3*A*b*Cos[c + d*x]^2*Sin[c + d*x] + 840*a*A*b^3*Cos[c + d*x]^2*Sin[c + d*x] + 840*a^3*b*C*Cos[c +
d*x]^2*Sin[c + d*x] + 70*b^4*C*Cos[c + d*x]^3*Sin[c + d*x] + 25*a^4*A*Sin[2*(c + d*x)] + 210*a^2*A*b^2*Sin[2*
(c + d*x)] + 35*a^4*C*Sin[2*(c + d*x)] + 30*a^4*A*Tan[c + d*x])/(105*d*Cos[c + d*x]^(5/2))

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{4} \cos \left (d x + c\right )^{6} + 4 \, C a b^{3} \cos \left (d x + c\right )^{5} + 4 \, A a^{3} b \cos \left (d x + c\right ) + A a^{4} + {\left (6 \, C a^{2} b^{2} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (C a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{4} + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}}{\cos \left (d x + c\right )^{\frac {9}{2}}}, x\right ) \]

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)/cos(d*x+c)^(9/2),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
)^(9/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

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)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4/cos(d*x + c)^(9/2), x)

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maple [B]  time = 9.36, size = 1531, normalized size = 4.84 \[ \text {result too large to display} \]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/3*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*(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)*EllipticE(cos(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*sin(1/2*d*x+1/2*c)^4+sin(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(co
s(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-4*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))-EllipticE(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))+12*C*a^2*b^
2*(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*(s
in(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*a^4*(-1/56*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)^4-5/42*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+5/21*(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*b*(A*b^2+C
*a^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)^(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))+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)+2*a^2*(6*A*b^2+C*a^2)*(-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)))-8/5*A*a^3*b/(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))*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*Elliptic
E(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)*sin(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*(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))-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))/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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

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)/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^4/cos(d*x + c)^(9/2), x)

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mupad [B]  time = 5.53, size = 378, normalized size = 1.20 \[ \frac {2\,\left (C\,b^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+12\,C\,a\,b^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+C\,b^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+18\,C\,a^2\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,b^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,A\,a\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,A\,a^3\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,C\,a^3\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,A\,a^2\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^4)/cos(c + d*x)^(9/2),x)

[Out]

(2*(C*b^4*ellipticF(c/2 + (d*x)/2, 2) + 12*C*a*b^3*ellipticE(c/2 + (d*x)/2, 2) + C*b^4*cos(c + d*x)^(1/2)*sin(
c + d*x) + 18*C*a^2*b^2*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*A*b^4*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a
^4*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/(7*d*cos(c + d*x)^(7/2)*(sin(c + d*x)^2)^(1/2))
+ (2*C*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^
(1/2)) + (8*A*a*b^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d
*x)^2)^(1/2)) + (8*A*a^3*b*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*
(sin(c + d*x)^2)^(1/2)) + (8*C*a^3*b*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)
^(1/2)*(sin(c + d*x)^2)^(1/2)) + (4*A*a^2*b^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos
(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))

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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)/cos(d*x+c)**(9/2),x)

[Out]

Timed out

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