3.442 \(\int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=172 \[ \frac {4 a^2 (3 A+2 B+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 a^2 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}+\frac {4 a^2 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}} \]

[Out]

4/5*a^2*(5*B+4*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^2*(3*A+2*B+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*A*(a
+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)-2/15*a^2*(15*A-5*B-7*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d-2/5*(5*A-
C)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.48, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3043, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac {4 a^2 (3 A+2 B+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 a^2 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{5 d}+\frac {4 a^2 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(5*B + 4*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(3*A + 2*B + C)*EllipticF[(c + d*x)/2, 2])/(3*d)
- (2*a^2*(15*A - 5*B - 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x]
)/(d*Sqrt[Cos[c + d*x]]) - (2*(5*A - C)*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(5*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 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 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{2} a (4 A+B)-\frac {1}{2} a (5 A-C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{a}\\ &=\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 (5 A-C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {4 \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (15 A+5 B+C)-\frac {1}{4} a^2 (15 A-5 B-7 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{5 a}\\ &=\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 (5 A-C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {4 \int \frac {\frac {1}{4} a^3 (15 A+5 B+C)+\left (-\frac {1}{4} a^3 (15 A-5 B-7 C)+\frac {1}{4} a^3 (15 A+5 B+C)\right ) \cos (c+d x)-\frac {1}{4} a^3 (15 A-5 B-7 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{5 a}\\ &=-\frac {2 a^2 (15 A-5 B-7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 (5 A-C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {8 \int \frac {\frac {5}{4} a^3 (3 A+2 B+C)+\frac {3}{4} a^3 (5 B+4 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac {2 a^2 (15 A-5 B-7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 (5 A-C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{3} \left (2 a^2 (3 A+2 B+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (2 a^2 (5 B+4 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {4 a^2 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (3 A+2 B+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 a^2 (15 A-5 B-7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 (5 A-C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C]  time = 6.58, size = 1039, normalized size = 6.04 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/20*((-5*A + 10*B + 8*C + 5*A*Cos[2*c] + 10*
B*Cos[2*c] + 8*C*Cos[2*c])*Csc[c]*Sec[c])/d + ((B + 2*C)*Cos[d*x]*Sin[c])/(6*d) + (C*Cos[2*d*x]*Sin[2*c])/(20*
d) + ((B + 2*C)*Cos[c]*Sin[d*x])/(6*d) + (A*Sec[c]*Sec[c + d*x]*Sin[d*x])/(2*d) + (C*Cos[2*c]*Sin[2*d*x])/(20*
d)) - (A*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c
/2 + (d*x)/2]^4*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]]]])/(d*Sqrt[1 + Cot[c]^2]) - (2*B*(a + a*Cos[c +
d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*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]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (C*(a + a*Cos[c + d*x])^2*Csc[c]*Hyperge
ometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*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]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (B*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((Hyperge
ometricPFQ[{-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 + Arc
Tan[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]]))/(2*d) - (2*C*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((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]) - ((S
in[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]]))/(5*d)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{2} \cos \left (d x + c\right )^{4} + {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.24, size = 595, normalized size = 3.46 \[ -\frac {4 a^{2} \left (-12 C \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (5 B +16 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (15 A +5 B +13 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+10 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-15 B \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*a^2*(-12*C*(-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)^6+
2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*B+16*C)*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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(15*A+5*B+13*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+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))*(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+10*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-15*B*(-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))+5*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)*(-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*(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*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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.51, size = 237, normalized size = 1.38 \[ \frac {2\,C\,a^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+6\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*a^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*B*a^2*(cos(c + d*
x)^(1/2)*sin(c + d*x) + 6*ellipticE(c/2 + (d*x)/2, 2) + 4*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*A*a^2*ellip
ticE(c/2 + (d*x)/2, 2))/d + (4*A*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a^2*ellipticE(c/2 + (d*x)/2, 2))/d
+ (2*A*a^2*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*C*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(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+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2),x)

[Out]

Timed out

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