3.148 \(\int \frac {(a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=213 \[ \frac {4 a^3 (13 A+35 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (7 A+5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (53 A+70 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

-4/5*a^3*(7*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+4/2
1*a^3*(13*A+35*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+2/7*
A*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(7/2)+12/35*A*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d/cos(d*x+c)^(5
/2)+2/15*(7*A+5*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(3/2)+8/105*a^3*(53*A+70*C)*sin(d*x+c)/d/cos(d
*x+c)^(1/2)

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Rubi [A]  time = 0.61, antiderivative size = 213, 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 = {3044, 2975, 2968, 3021, 2748, 2641, 2639} \[ \frac {4 a^3 (13 A+35 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (7 A+5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (53 A+70 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(-4*a^3*(7*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(13*A + 35*C)*EllipticF[(c + d*x)/2, 2])/(21*d)
+ (8*a^3*(53*A + 70*C)*Sin[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]) + (2*A*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(7
*d*Cos[c + d*x]^(7/2)) + (12*A*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(35*a*d*Cos[c + d*x]^(5/2)) + (2*(7*A
+ 5*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2))

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 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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3044

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/(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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^3 \left (3 a A-\frac {1}{2} a (A-7 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {12 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {7}{4} a^2 (7 A+5 C)-\frac {1}{4} a^2 (11 A-35 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {12 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{2} a^3 (53 A+70 C)-\frac {1}{4} a^3 (41 A-35 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{105 a}\\ &=\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {12 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 \int \frac {\frac {1}{2} a^4 (53 A+70 C)+\left (-\frac {1}{4} a^4 (41 A-35 C)+\frac {1}{2} a^4 (53 A+70 C)\right ) \cos (c+d x)-\frac {1}{4} a^4 (41 A-35 C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{105 a}\\ &=\frac {8 a^3 (53 A+70 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {12 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 \int \frac {\frac {5}{8} a^4 (13 A+35 C)-\frac {21}{8} a^4 (7 A+5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{105 a}\\ &=\frac {8 a^3 (53 A+70 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {12 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {1}{5} \left (2 a^3 (7 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (2 a^3 (13 A+35 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a^3 (7 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (13 A+35 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {8 a^3 (53 A+70 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {12 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{35 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A+5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C]  time = 6.73, size = 920, normalized size = 4.32 \[ \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^3 \left (\frac {A \sec (c) \sin (d x) \sec ^4(c+d x)}{28 d}+\frac {\sec (c) (5 A \sin (c)+21 A \sin (d x)) \sec ^3(c+d x)}{140 d}+\frac {\sec (c) (63 A \sin (c)+130 A \sin (d x)+35 C \sin (d x)) \sec ^2(c+d x)}{420 d}+\frac {\sec (c) (130 A \sin (c)+35 C \sin (c)+294 A \sin (d x)+315 C \sin (d x)) \sec (c+d x)}{420 d}-\frac {(-28 A-25 C+5 C \cos (2 c)) \csc (c) \sec (c)}{40 d}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )+\frac {7 A (\cos (c+d x) a+a)^3 \csc (c) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt {\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}} \sqrt {\tan ^2(c)+1}}-\frac {\frac {2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac {\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {\tan ^2(c)+1}}}{\sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}}}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{20 d}+\frac {C (\cos (c+d x) a+a)^3 \csc (c) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt {\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}} \sqrt {\tan ^2(c)+1}}-\frac {\frac {2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac {\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {\tan ^2(c)+1}}}{\sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}}}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d}-\frac {13 A (\cos (c+d x) a+a)^3 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {-\sqrt {\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{42 d \sqrt {\cot ^2(c)+1}}-\frac {5 C (\cos (c+d x) a+a)^3 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {-\sqrt {\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{6 d \sqrt {\cot ^2(c)+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/40*((-28*A - 25*C + 5*C*Cos[2*c])*Csc[c]*Se
c[c])/d + (A*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(28*d) + (Sec[c]*Sec[c + d*x]^3*(5*A*Sin[c] + 21*A*Sin[d*x]))/(14
0*d) + (Sec[c]*Sec[c + d*x]^2*(63*A*Sin[c] + 130*A*Sin[d*x] + 35*C*Sin[d*x]))/(420*d) + (Sec[c]*Sec[c + d*x]*(
130*A*Sin[c] + 35*C*Sin[c] + 294*A*Sin[d*x] + 315*C*Sin[d*x]))/(420*d)) - (13*A*(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]]]])/(42*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]) + (7*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
) + (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 + A
rcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + Ar
cTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + Arc
Tan[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)

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{3} \cos \left (d x + c\right )^{5} + 3 \, C a^{3} \cos \left (d x + c\right )^{4} + {\left (A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (3 \, A + C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, A a^{3} \cos \left (d x + c\right ) + A a^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

integral((C*a^3*cos(d*x + c)^5 + 3*C*a^3*cos(d*x + c)^4 + (A + 3*C)*a^3*cos(d*x + c)^3 + (3*A + C)*a^3*cos(d*x
+ c)^2 + 3*A*a^3*cos(d*x + c) + A*a^3)/cos(d*x + c)^(9/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(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)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)

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maple [B]  time = 6.25, size = 1012, normalized size = 4.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x)

[Out]

-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(1/8*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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+1/4*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))+1/8*A*(-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)))+(1/8*A+3/8*C)*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(s
in(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*si
n(1/2*d*x+1/2*c)^2-1)-3/40*A/(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*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+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)+(3/
8*A+1/8*C)*(-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))))/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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(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)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)

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mupad [B]  time = 3.35, size = 279, normalized size = 1.31 \[ \frac {2\,\left (C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7}+\frac {6\,A\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5}+2\,A\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+2\,A\,a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {6\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^3)/cos(c + d*x)^(9/2),x)

[Out]

(2*(C*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*C*a^3*ellipticF(c/2 + (d*x)/2, 2)))/d + ((2*A*a^3*sin(c + d*x)*hyper
geom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + (6*A*a^3*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, c
os(c + d*x)^2))/5 + 2*A*a^3*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 2*A*a^3*
cos(c + d*x)^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(7/2)*(1 - cos(c + d*
x)^2)^(1/2)) + (6*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)) + (2*C*a^3*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))

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

[Out]

Timed out

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