3.6.57 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [557]

Optimal. Leaf size=177 \[ -\frac {2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^2 A+3 A b^2+6 a b B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a (5 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x)) \sin (c+d x)}{3 d} \]

[Out]

2/3*a*(5*A*b+3*B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/3*a*A*(b+a*sec(d*x+c))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2*(2*
A*a*b+B*a^2-B*b^2)*(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*(A*a^2+3*A*b^2+6*B*a*b)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticF(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 = 0.22, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3039, 4111, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 \left (a^2 A+6 a b B+3 A b^2\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 (a^2 B+2 a A b-b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (3 a B+5 A b) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+b)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*a*A*b + a^2*B - b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*(a^2*A +
3*A*b^2 + 6*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a*(5*A*b + 3*a
*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a*A*Sqrt[Sec[c + d*x]]*(b + a*Sec[c + d*x])*Sin[c + d*x])/(3*d
)

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 4111

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

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]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx &=\int \frac {(b+a \sec (c+d x))^2 (B+A \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x)) \sin (c+d x)}{3 d}+\frac {2}{3} \int \frac {-\frac {1}{2} b (a A-3 b B)+\frac {1}{2} \left (a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x)+\frac {1}{2} a (5 A b+3 a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x)) \sin (c+d x)}{3 d}+\frac {2}{3} \int \frac {-\frac {1}{2} b (a A-3 b B)+\frac {1}{2} a (5 A b+3 a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (a^2 A+3 A b^2+6 a b B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 a (5 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x)) \sin (c+d x)}{3 d}+\left (-2 a A b-a^2 B+b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (a^2 A+3 A b^2+6 a b 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 (a^2 A+3 A b^2+6 a b 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 a (5 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x)) \sin (c+d x)}{3 d}+\left (\left (-2 a A b-a^2 B+b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^2 A+3 A b^2+6 a b 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 a (5 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x)) \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 1.24, size = 125, normalized size = 0.71 \begin {gather*} \frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 \left (2 a A b+a^2 B-b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (a^2 A+3 A b^2+6 a b B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {a (a A+3 (2 A b+a B) \cos (c+d x)) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-3*(2*a*A*b + a^2*B - b^2*B)*EllipticE[(c + d*x)/2, 2] + (a^2*A + 3*
A*b^2 + 6*a*b*B)*EllipticF[(c + d*x)/2, 2] + (a*(a*A + 3*(2*A*b + a*B)*Cos[c + d*x])*Sin[c + d*x])/Cos[c + d*x
]^(3/2)))/(3*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(649\) vs. \(2(211)=422\).
time = 0.68, size = 650, normalized size = 3.67

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {2 B \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\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 )}}+\frac {2 A \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}}+\frac {4 B a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}}-\frac {2 B \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}}+2 a^{2} A \left (-\frac {\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 )}}{6 \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}{3 \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 )}}\right )+\frac {2 a \left (2 A b +a B \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 (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \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 )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\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}\)	\(650\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*b^2*(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)))+2*A*b^2*(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))+4*B*a*b*(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)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*B*b^2*(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))+2*a^2*A*(-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)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*a*(2*A*b+B*a)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)*EllipticE(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 \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))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 247, normalized size = 1.40 \begin {gather*} \frac {\sqrt {2} {\left (-i \, A a^{2} - 6 i \, B a b - 3 i \, A b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A a^{2} + 6 i \, B a b + 3 i \, A b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, A a b - i \, B b^{2}\right )} \cos \left (d x + c\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 (-i \, B a^{2} - 2 i \, A a b + i \, B b^{2}\right )} \cos \left (d x + c\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 (A a^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(2)*(-I*A*a^2 - 6*I*B*a*b - 3*I*A*b^2)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c)) + sqrt(2)*(I*A*a^2 + 6*I*B*a*b + 3*I*A*b^2)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*
sin(d*x + c)) - 3*sqrt(2)*(I*B*a^2 + 2*I*A*a*b - I*B*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInve
rse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(-I*B*a^2 - 2*I*A*a*b + I*B*b^2)*cos(d*x + c)*weierstra
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(A*a^2 + 3*(B*a^2 + 2*A*a*b)*cos(
d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))

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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+b*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**(5/2),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+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,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))^(5/2)*(a + b*cos(c + d*x))^2,x)

[Out]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^2, x)

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