3.6.64 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\) [564]

Optimal. Leaf size=237 \[ -\frac {2 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 (5 A b+9 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a^2 (5 a A-b B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3039, 4110, 4159, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 a^2 (5 a A-b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*a^3*A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*
x]])/(5*d) + (2*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Se
c[c + d*x]])/(3*d) + (2*b^2*(5*A*b + 9*a*B)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*a^2*(5*a*A - b*B)*Sqr
t[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*b*B*(b + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[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 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4110

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4159

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx &=\int \frac {(b+a \sec (c+d x))^3 (B+A \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2}{5} \int \frac {(b+a \sec (c+d x)) \left (-\frac {1}{2} b (5 A b+9 a B)-\frac {1}{2} \left (10 a A b+5 a^2 B+3 b^2 B\right ) \sec (c+d x)-\frac {1}{2} a (5 a A-b B) \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 (5 A b+9 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {\frac {3}{4} b \left (15 a A b+14 a^2 B+3 b^2 B\right )+\frac {5}{4} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sec (c+d x)+\frac {3}{4} a^2 (5 a A-b B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b^2 (5 A b+9 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {\frac {3}{4} b \left (15 a A b+14 a^2 B+3 b^2 B\right )+\frac {3}{4} a^2 (5 a A-b B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 b^2 (5 A b+9 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a^2 (5 a A-b B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (-5 a^3 A+15 a A b^2+15 a^2 b B+3 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 (5 A b+9 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a^2 (5 a A-b B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (\left (-5 a^3 A+15 a A b^2+15 a^2 b B+3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 (5 A b+9 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a^2 (5 a A-b B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 1.54, size = 172, normalized size = 0.73 \begin {gather*} \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (12 \left (-5 a^3 A+15 a A b^2+15 a^2 b B+3 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {2 \left (10 b^2 (A b+3 a B) \cos (c+d x)+3 \left (10 a^3 A+b^3 B+b^3 B \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(12*(-5*a^3*A + 15*a*A*b^2 + 15*a^2*b*B + 3*b^3*B)*EllipticE[(c + d*x)/
2, 2] + 20*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B)*EllipticF[(c + d*x)/2, 2] + (2*(10*b^2*(A*b + 3*a*B)*Cos[
c + d*x] + 3*(10*a^3*A + b^3*B + b^3*B*Cos[2*(c + d*x)]))*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(30*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(640\) vs. \(2(265)=530\).
time = 0.50, size = 641, normalized size = 2.70

method	result	size
default	\(-\frac {2 \left (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+20 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+60 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}-30 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}-10 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+45 A \,a^{2} 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 )+5 A \,b^{3} \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 )+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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}-45 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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-30 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+15 a^{3} 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 )+15 B a \,b^{2} \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 )-45 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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b -9 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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{3}\right )}{15 \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}\)	\(641\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*(-24*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b^3+20*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3+60*B
*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+24*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3-30*A*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3-10*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^3+45*A*a^2*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))+5*A*b^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)*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-45*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))*a*b^2-30*B*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^2*a*b^2-6*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^3+15*a^3*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*a*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-45*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-9*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.15, size = 270, normalized size = 1.14 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (3 i \, B a^{3} + 9 i \, A a^{2} b + 3 i \, B a b^{2} + i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, B a^{3} - 9 i \, A a^{2} b - 3 i \, B a b^{2} - i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, A a^{3} - 15 i \, B a^{2} b - 15 i \, A a b^{2} - 3 i \, B b^{3}\right )} {\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 \, \sqrt {2} {\left (-5 i \, A a^{3} + 15 i \, B a^{2} b + 15 i \, A a b^{2} + 3 i \, B b^{3}\right )} {\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 ) - \frac {2 \, {\left (3 \, B b^{3} \cos \left (d x + c\right )^{2} + 15 \, A a^{3} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(5*sqrt(2)*(3*I*B*a^3 + 9*I*A*a^2*b + 3*I*B*a*b^2 + I*A*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) + I
*sin(d*x + c)) + 5*sqrt(2)*(-3*I*B*a^3 - 9*I*A*a^2*b - 3*I*B*a*b^2 - I*A*b^3)*weierstrassPInverse(-4, 0, cos(d
*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(5*I*A*a^3 - 15*I*B*a^2*b - 15*I*A*a*b^2 - 3*I*B*b^3)*weierstrassZeta(-4
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-5*I*A*a^3 + 15*I*B*a^2*b + 15*I*
A*a*b^2 + 3*I*B*b^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*
B*b^3*cos(d*x + c)^2 + 15*A*a^3 + 5*(3*B*a*b^2 + A*b^3)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8010 deep

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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+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^3,x)

[Out]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^3, x)

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