3.452 \(\int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=230 \[ \frac {4 a^3 (3 A+5 (B+C)) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^3 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {4 a^3 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
[Out]
-4/5*a^3*(9*A+5*B-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
+4/3*a^3*(3*A+5*B+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))/d
+2/5*A*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/15*(6*A+5*B)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d/c
os(d*x+c)^(3/2)+2/15*(33*A+35*B+15*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(1/2)-4/15*a^3*(21*A+20*B+5
*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d
________________________________________________________________________________________
Rubi [A] time = 0.68, antiderivative size = 230, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used =
{3043, 2975, 2968, 3023, 2748, 2641, 2639} \[ \frac {4 a^3 (3 A+5 (B+C)) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^3 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {4 a^3 (21 A+20 B+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}+\frac {2 (33 A+35 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
[Out]
(-4*a^3*(9*A + 5*B - 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(3*A + 5*(B + C))*EllipticF[(c + d*x)/2, 2
])/(3*d) - (4*a^3*(21*A + 20*B + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*(a + a*Cos[c + d*x])^3*Si
n[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(6*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(15*a*d*Cos[c +
d*x]^(3/2)) + (2*(33*A + 35*B + 15*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d*Sqrt[Cos[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]
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 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 2975
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])
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 3043
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/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Rubi steps
\begin {align*} \int \frac {(a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^3 \left (\frac {1}{2} a (6 A+5 B)-\frac {1}{2} a (3 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{4} a^2 (33 A+35 B+15 C)-\frac {3}{4} a^2 (9 A+5 B-5 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {8 \int \frac {(a+a \cos (c+d x)) \left (\frac {3}{4} a^3 (12 A+15 B+10 C)-\frac {3}{4} a^3 (21 A+20 B+5 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{15 a}\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {8 \int \frac {\frac {3}{4} a^4 (12 A+15 B+10 C)+\left (-\frac {3}{4} a^4 (21 A+20 B+5 C)+\frac {3}{4} a^4 (12 A+15 B+10 C)\right ) \cos (c+d x)-\frac {3}{4} a^4 (21 A+20 B+5 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac {4 a^3 (21 A+20 B+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {16 \int \frac {\frac {15}{8} a^4 (3 A+5 (B+C))-\frac {9}{8} a^4 (9 A+5 B-5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{45 a}\\ &=-\frac {4 a^3 (21 A+20 B+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}-\frac {1}{5} \left (2 a^3 (9 A+5 B-5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^3 (3 A+5 (B+C))\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a^3 (9 A+5 B-5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (3 A+5 (B+C)) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^3 (21 A+20 B+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (6 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (33 A+35 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] time = 6.87, size = 1298, normalized size = 5.64 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
[Out]
Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/40*((-36*A - 25*B + 5*C + 5*B*Cos[2*c] + 15
*C*Cos[2*c])*Csc[c]*Sec[c])/d + (C*Cos[d*x]*Sin[c])/(12*d) + (C*Cos[c]*Sin[d*x])/(12*d) + (A*Sec[c]*Sec[c + d*
x]^3*Sin[d*x])/(20*d) + (Sec[c]*Sec[c + d*x]^2*(3*A*Sin[c] + 15*A*Sin[d*x] + 5*B*Sin[d*x]))/(60*d) + (Sec[c]*S
ec[c + d*x]*(15*A*Sin[c] + 5*B*Sin[c] + 54*A*Sin[d*x] + 45*B*Sin[d*x] + 15*C*Sin[d*x]))/(60*d)) - (A*(a + a*Co
s[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Se
c[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[
Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(2*d*Sqrt[1 + Cot[c]^2]) - (5*B*(a + a*Cos[c + d*x])^3*Csc[c]*
HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]
]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + S
in[d*x - ArcTan[Cot[c]]]])/(6*d*Sqrt[1 + Cot[c]^2]) - (5*C*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x -
ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c
]]]])/(6*d*Sqrt[1 + Cot[c]^2]) + (9*A*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{
-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[
Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1
+ Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]
*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(20*d)
+ (B*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + Ar
cTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + Arc
Tan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcT
an[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 +
Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(4*d) - (C*(a + a*Cos[c + d*x])^3*Csc[
c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan
[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x
+ ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[
c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x +
ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(4*d)
________________________________________________________________________________________
fricas [F] time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{3} \cos \left (d x + c\right )^{5} + {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + {\left (A + 3 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (3 \, A + 3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + A a^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="fricas")
[Out]
integral((C*a^3*cos(d*x + c)^5 + (B + 3*C)*a^3*cos(d*x + c)^4 + (A + 3*B + 3*C)*a^3*cos(d*x + c)^3 + (3*A + 3*
B + C)*a^3*cos(d*x + c)^2 + (3*A + B)*a^3*cos(d*x + c) + A*a^3)/cos(d*x + c)^(7/2), x)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)
________________________________________________________________________________________
maple [B] time = 6.69, size = 1328, normalized size = 5.77 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
[Out]
4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(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)^3*(190*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-50*B*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+25*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))-120*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^6-180*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+90*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-20*C*cos(
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+40*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-216*A*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^6+246*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-15*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))+27*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))+25*B*(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))+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*(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))-60*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)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+60*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))*sin(1/2*d*x+1/2*c)^2+108*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)*sin(1/2*d*x+1/2*c)^4-108*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)*sin(1/2*d*x+1/2*c)^2-60*A*(
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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x
+1/2*c)^2-60*B*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)^2-100*C*(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))*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+100*B*EllipticF(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)^4-100*B*(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))*sin(1/2*d*x+1/2*c)^2+60*B*(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)^(1/2)*sin(1/2*d*x+1/2*c)^4+100*C*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+
60*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1
/2*d*x+1/2*c)^4)*(-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
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)
________________________________________________________________________________________
mupad [B] time = 3.67, size = 408, normalized size = 1.77 \[ \frac {2\,\left (B\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,B\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {C\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a^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 {2\,A\,a^3\,\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}}+\frac {2\,A\,a^3\,\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 {6\,B\,a^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 {2\,B\,a^3\,\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 {2\,C\,a^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}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + a*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(7/2),x)
[Out]
(2*(B*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*B*a^3*ellipticF(c/2 + (d*x)/2, 2)))/d + (C*a^3*((2*cos(c + d*x)^(1/2
)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*A*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*C*a^3
*ellipticE(c/2 + (d*x)/2, 2))/d + (6*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*A*a^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)) + (2*A*a^3*sin(c + d*x)*hyperge
om([-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)) + (2*A*a^3*sin(c + d*x)*hy
pergeom([-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)) + (6*B*a^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)) + (2*B*a^3*si
n(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)) + (2*C
*a^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))
________________________________________________________________________________________
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+a*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)
[Out]
Timed out
________________________________________________________________________________________