3.138 \(\int \frac {(a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=156 \[ \frac {8 a^2 (A+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 a^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {8 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-4*a^2*(A-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+8/3*a^2*(
A+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+2/3*A*(a+a*cos(d*
x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)+8/3*A*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(1/2)-2/3*a^2*(5*A-C)
*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.42, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3044, 2975, 2968, 3023, 2748, 2641, 2639} \[ \frac {8 a^2 (A+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 a^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {8 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

(-4*a^2*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (8*a^2*(A + C)*EllipticF[(c + d*x)/2, 2])/(3*d) - (2*a^2*(5*A -
C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)
) + (8*A*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(3*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 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))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^2 \left (2 a A-\frac {3}{2} a (A-C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {4 \int \frac {(a+a \cos (c+d x)) \left (\frac {3}{4} a^2 (3 A+C)-\frac {3}{4} a^2 (5 A-C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{3 a}\\ &=\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {4 \int \frac {\frac {3}{4} a^3 (3 A+C)+\left (-\frac {3}{4} a^3 (5 A-C)+\frac {3}{4} a^3 (3 A+C)\right ) \cos (c+d x)-\frac {3}{4} a^3 (5 A-C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 a^2 (5 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {8 \int \frac {\frac {3}{2} a^3 (A+C)-\frac {9}{4} a^3 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{9 a}\\ &=-\frac {2 a^2 (5 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\left (2 a^2 (A-C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (4 a^2 (A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {8 a^2 (A+C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 a^2 (5 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 6.48, size = 865, normalized size = 5.54 \[ \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^2 \left (\frac {A \sec (c) \sin (d x) \sec ^2(c+d x)}{6 d}+\frac {\sec (c) (A \sin (c)+6 A \sin (d x)) \sec (c+d x)}{6 d}-\frac {(-2 A+C+C \cos (2 c)) \csc (c) \sec (c)}{2 d}+\frac {C \cos (d x) \sin (c)}{6 d}+\frac {C \cos (c) \sin (d x)}{6 d}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+\frac {A (\cos (c+d x) a+a)^2 \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}-\frac {C (\cos (c+d x) a+a)^2 \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}-\frac {2 A (\cos (c+d x) a+a)^2 \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \sqrt {\cot ^2(c)+1}}-\frac {2 C (\cos (c+d x) a+a)^2 \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \sqrt {\cot ^2(c)+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/2*((-2*A + C + C*Cos[2*c])*Csc[c]*Sec[c])/d
+ (C*Cos[d*x]*Sin[c])/(6*d) + (C*Cos[c]*Sin[d*x])/(6*d) + (A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(6*d) + (Sec[c]*
Sec[c + d*x]*(A*Sin[c] + 6*A*Sin[d*x]))/(6*d)) - (2*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 - ArcT
an[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]) - (2*C*(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[-(S
qrt[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]) + (A*(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]) - ((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]]))/(2*d) - (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 + Ar
cTan[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]]))/(2*d)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{2} \cos \left (d x + c\right )^{4} + 2 \, C a^{2} \cos \left (d x + c\right )^{3} + {\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^2/cos(d*x + c)^(5/2), x)

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maple [B]  time = 4.39, size = 651, normalized size = 4.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)

[Out]

4/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)
^2+1)/sin(1/2*d*x+1/2*c)^3*(4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+4*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticF(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+6*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)^2-12*A*cos
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4*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)^2-6*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)^2-4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4
-2*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))-3*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))+7*A*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*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(c
os(1/2*d*x+1/2*c),2^(1/2))+3*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))+C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)*(-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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^2/cos(d*x + c)^(5/2), x)

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mupad [B]  time = 2.02, size = 161, normalized size = 1.03 \[ \frac {2\,C\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+6\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

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))^2)/cos(c + d*x)^(5/2),x)

[Out]

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

[Out]

Timed out

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