3.2.42 \(\int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [142]

Optimal. Leaf size=171 \[ -\frac {4 a^3 (9 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (3 A+5 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a^3 (21 A+20 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (9 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-4/5*a^3*(9*A+5*B)*(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)*(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+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/15*(9*A+5*B)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^
(3/2)+4/15*a^3*(21*A+20*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.30, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3054, 3047, 3100, 2827, 2720, 2719} \begin {gather*} \frac {4 a^3 (3 A+5 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^3 (9 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (21 A+20 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(7/2),x]

[Out]

(-4*a^3*(9*A + 5*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(3*A + 5*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (
4*a^3*(21*A + 20*B)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*a*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(5*d
*Cos[c + d*x]^(5/2)) + (2*(9*A + 5*B)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

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 3047

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 3054

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
*Sin[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 3100

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]

Rubi steps

\begin {align*} \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{2} a (9 A+5 B)-\frac {1}{2} a (A-5 B) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (9 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{2} a^2 (21 A+20 B)-\frac {1}{2} a^2 (6 A-5 B) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (9 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {\frac {1}{2} a^3 (21 A+20 B)+\left (-\frac {1}{2} a^3 (6 A-5 B)+\frac {1}{2} a^3 (21 A+20 B)\right ) \cos (c+d x)-\frac {1}{2} a^3 (6 A-5 B) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 a^3 (21 A+20 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (9 A+5 B) \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}{15} \int \frac {\frac {5}{4} a^3 (3 A+5 B)-\frac {3}{4} a^3 (9 A+5 B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^3 (21 A+20 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (9 A+5 B) \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}{3} \left (2 a^3 (3 A+5 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (2 a^3 (9 A+5 B)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {4 a^3 (9 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (3 A+5 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {4 a^3 (21 A+20 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (9 A+5 B) \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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.62, size = 890, normalized size = 5.20 \begin {gather*} \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {(-36 A-25 B+5 B \cos (2 c)) \csc (c) \sec (c)}{40 d}+\frac {A \sec (c) \sec ^3(c+d x) \sin (d x)}{20 d}+\frac {\sec (c) \sec ^2(c+d x) (3 A \sin (c)+15 A \sin (d x)+5 B \sin (d x))}{60 d}+\frac {\sec (c) \sec (c+d x) (15 A \sin (c)+5 B \sin (c)+54 A \sin (d x)+45 B \sin (d x))}{60 d}\right )-\frac {A (a+a \cos (c+d x))^3 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{2 d \sqrt {1+\cot ^2(c)}}-\frac {5 B (a+a \cos (c+d x))^3 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{6 d \sqrt {1+\cot ^2(c)}}+\frac {9 A (a+a \cos (c+d x))^3 \csc (c) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{20 d}+\frac {B (a+a \cos (c+d x))^3 \csc (c) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{4 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/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*B*Cos[2*c])*Csc[c]*Se
c[c])/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]*Sec[c + d*x]*(15*A*Sin[c] + 5*B*Sin[c] + 54*A*Sin[d*x] + 45*B*Sin[d*x]))/(60*d))
- (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 + 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 + 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[Ta
n[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*((HypergeometricPF
Q[{-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 + ArcT
an[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqr
t[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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(915\) vs. \(2(207)=414\).
time = 0.69, size = 916, normalized size = 5.36

method	result	size
default	\(\text {Expression too large to display}\)	\(916\)

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))/cos(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[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*(216*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*A*(2*si
n(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)^4-108*A*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)^4+180*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-100*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)^4-60*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)^4-246*A*co
s(1/2*d*x+1/2*c)*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)^2+108*A*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)^2-190*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^4+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)^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)^2+72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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))-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))+50*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^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*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)))
*(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 243, normalized size = 1.42 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (9 \, {\left (6 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 3 \, A a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

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))/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(5*I*sqrt(2)*(3*A + 5*B)*a^3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) -
5*I*sqrt(2)*(3*A + 5*B)*a^3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqr
t(2)*(9*A + 5*B)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c))) - 3*I*sqrt(2)*(9*A + 5*B)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c))) - (9*(6*A + 5*B)*a^3*cos(d*x + c)^2 + 5*(3*A + B)*a^3*cos(d*x + c) + 3*A*a^3)*sqrt(co
s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)

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Mupad [B]
time = 2.50, size = 287, normalized size = 1.68 \begin {gather*} \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 {2\,A\,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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^3)/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 + (2*A*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)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(s
in(c + d*x)^2)^(1/2)) + (2*A*a^3*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)) + (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*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*c
os(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))

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