3.566 \(\int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\)

Optimal. Leaf size=295 \[ \frac {2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 b^2 (13 a B+9 A b) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

2/63*b^2*(9*A*b+13*B*a)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/45*b*(27*A*a*b+22*B*a^2+7*B*b^2)*sin(d*x+c)/d/sec(d*x+
c)^(3/2)+2/9*b*B*(b+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(7/2)+2/21*(21*A*a^2*b+5*A*b^3+7*B*a^3+15*B*a*b^2)
*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/15*(15*A*a^3+27*A*a*b^2+27*B*a^2*b+7*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+2/21*(21*A*a^2*b+5*A*
b^3+7*B*a^3+15*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 = 0.58, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2960, 4025, 4074, 4047, 3769, 3771, 2641, 4045, 2639} \[ \frac {2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 b^2 (13 a B+9 A b) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15*a^3*A + 27*a*A*b^2 + 27*a^2*b*B + 7*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*
x]])/(15*d) + (2*(21*a^2*A*b + 5*A*b^3 + 7*a^3*B + 15*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sq
rt[Sec[c + d*x]])/(21*d) + (2*b^2*(9*A*b + 13*a*B)*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)) + (2*b*(27*a*A*b +
22*a^2*B + 7*b^2*B)*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (2*(21*a^2*A*b + 5*A*b^3 + 7*a^3*B + 15*a*b^2*B)
*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*b*B*(b + a*Sec[c + d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)
)

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 2960

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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

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 4025

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 4045

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

Rule 4047

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 4074

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 \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx &=\int \frac {(b+a \sec (c+d x))^3 (B+A \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}-\frac {2}{9} \int \frac {(b+a \sec (c+d x)) \left (-\frac {1}{2} b (9 A b+13 a B)-\frac {1}{2} \left (18 a A b+9 a^2 B+7 b^2 B\right ) \sec (c+d x)-\frac {3}{2} a (3 a A+b B) \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4}{63} \int \frac {\frac {7}{4} b \left (27 a A b+22 a^2 B+7 b^2 B\right )+\frac {9}{4} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sec (c+d x)+\frac {21}{4} a^2 (3 a A+b B) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4}{63} \int \frac {\frac {7}{4} b \left (27 a A b+22 a^2 B+7 b^2 B\right )+\frac {21}{4} a^2 (3 a A+b B) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\frac {1}{7} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{21} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{21} \left (\left (21 a^2 A b+5 A b^3+7 a^3 B+15 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 {1}{15} \left (\left (15 a^3 A+27 a A b^2+27 a^2 b B+7 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 (15 a^3 A+27 a A b^2+27 a^2 b B+7 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)}}{15 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 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)}}{21 d}+\frac {2 b^2 (9 A b+13 a B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 1.83, size = 219, normalized size = 0.74 \[ \frac {\sqrt {\sec (c+d x)} \left (\sin (2 (c+d x)) \left (7 b \left (108 a^2 B+108 a A b+43 b^2 B\right ) \cos (c+d x)+5 \left (84 a^3 B+252 a^2 A b+18 b^2 (3 a B+A b) \cos (2 (c+d x))+234 a b^2 B+78 A b^3+7 b^3 B \cos (3 (c+d x))\right )\right )+120 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+168 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(168*(15*a^3*A + 27*a*A*b^2 + 27*a^2*b*B + 7*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)
/2, 2] + 120*(21*a^2*A*b + 5*A*b^3 + 7*a^3*B + 15*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (7*b
*(108*a*A*b + 108*a^2*B + 43*b^2*B)*Cos[c + d*x] + 5*(252*a^2*A*b + 78*A*b^3 + 84*a^3*B + 234*a*b^2*B + 18*b^2
*(A*b + 3*a*B)*Cos[2*(c + d*x)] + 7*b^3*B*Cos[3*(c + d*x)]))*Sin[2*(c + d*x)]))/(1260*d)

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B b^{3} \cos \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

integral((B*b^3*cos(d*x + c)^4 + A*a^3 + (3*B*a*b^2 + A*b^3)*cos(d*x + c)^3 + 3*(B*a^2*b + A*a*b^2)*cos(d*x +
c)^2 + (B*a^3 + 3*A*a^2*b)*cos(d*x + c))/sqrt(sec(d*x + c)), x)

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

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 1.60, size = 745, normalized size = 2.53 \[ \text {result too large to display} \]

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)^(1/2),x)

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10+(720*A*b^3+2160*B*a*b^2+2240*B*b^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1512*A*a*b^2-1080*A*b^3-1
512*B*a^2*b-3240*B*a*b^2-2072*B*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(1260*A*a^2*b+1512*A*a*b^2+840*A*
b^3+420*B*a^3+1512*B*a^2*b+2520*B*a*b^2+952*B*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-630*A*a^2*b-378*A
*a*b^2-240*A*b^3-210*B*a^3-378*B*a^2*b-720*B*a*b^2-168*B*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-315*A*(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))*a^3-567*A*(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))*a*b^2+315*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))+75*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))-567*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))*a^2*b-147
*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))*b^3+105
*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))+225
*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)))/
(-2*sin(1/2*d*x+1/2*c)^4+sin(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 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

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)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]

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)**(1/2),x)

[Out]

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

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