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

Optimal. Leaf size=244 \[ \frac {4 a^3 (24 A+23 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 A+13 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (13 A+11 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {4 a^3 (13 A+11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (21 A+17 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

4/105*a^3*(24*A+23*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/9*a*B*(a+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(7/2)+2
/63*(9*A+13*B)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(5/2)+4/21*a^3*(13*A+11*B)*sin(d*x+c)/d/sec(d*x+c)
^(1/2)+4/15*a^3*(21*A+17*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/21*a^3*(13*A+11*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticF(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 = 0.54, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2960, 4017, 3996, 3787, 3769, 3771, 2641, 2639} \[ \frac {4 a^3 (24 A+23 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 A+13 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (13 A+11 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {4 a^3 (13 A+11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^3 (21 A+17 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^2}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*(21*A + 17*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^3*(13*A +
11*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^3*(24*A + 23*B)*Sin[c + d
*x])/(105*d*Sec[c + d*x]^(3/2)) + (4*a^3*(13*A + 11*B)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*a*B*(a + a
*Sec[c + d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (2*(9*A + 13*B)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x
])/(63*d*Sec[c + d*x]^(5/2))

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(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) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4017

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[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx &=\int \frac {(a+a \sec (c+d x))^3 (B+A \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2}{9} \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{2} a (9 A+13 B)+\frac {3}{2} a (3 A+B) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (9 A+13 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4}{63} \int \frac {(a+a \sec (c+d x)) \left (\frac {3}{2} a^2 (24 A+23 B)+\frac {15}{2} a^2 (3 A+2 B) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {4 a^3 (24 A+23 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (9 A+13 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {8}{315} \int \frac {-\frac {45}{4} a^3 (13 A+11 B)-\frac {21}{4} a^3 (21 A+17 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 a^3 (24 A+23 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (9 A+13 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (2 a^3 (13 A+11 B)\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{15} \left (2 a^3 (21 A+17 B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {4 a^3 (24 A+23 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (13 A+11 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (9 A+13 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{21} \left (2 a^3 (13 A+11 B)\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{15} \left (2 a^3 (21 A+17 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {4 a^3 (21 A+17 B) \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 {4 a^3 (24 A+23 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (13 A+11 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (9 A+13 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{21} \left (2 a^3 (13 A+11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^3 (21 A+17 B) \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 {4 a^3 (13 A+11 B) \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 {4 a^3 (24 A+23 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (13 A+11 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 a B (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 (9 A+13 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C]  time = 2.75, size = 196, normalized size = 0.80 \[ \frac {a^3 \sqrt {\sec (c+d x)} \left (-112 i (21 A+17 B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+\cos (c+d x) (30 (107 A+97 B) \sin (c+d x)+14 (54 A+73 B) \sin (2 (c+d x))+90 A \sin (3 (c+d x))+7056 i A+270 B \sin (3 (c+d x))+35 B \sin (4 (c+d x))+5712 i B)+240 (13 A+11 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sqrt[Sec[c + d*x]]*(240*(13*A + 11*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (112*I)*(21*A + 17*B
)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + Cos[c
+ d*x]*((7056*I)*A + (5712*I)*B + 30*(107*A + 97*B)*Sin[c + d*x] + 14*(54*A + 73*B)*Sin[2*(c + d*x)] + 90*A*S
in[3*(c + d*x)] + 270*B*Sin[3*(c + d*x)] + 35*B*Sin[4*(c + d*x)])))/(1260*d)

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fricas [F]  time = 2.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B a^{3} \cos \left (d x + c\right )^{4} + {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + A a^{3}}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]

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))/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 1.39, size = 413, normalized size = 1.69 \[ -\frac {4 \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 )}\, a^{3} \left (-560 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (360 A +2200 B \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-1296 A -3412 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (1806 A +2702 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-624 A -738 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+195 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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-441 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 )+165 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 )-357 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 )\right )}{315 \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 )}\, \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} \]

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))/sec(d*x+c)^(1/2),x)

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-560*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^10+(360*A+2200*B)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1296*A-3412*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x
+1/2*c)+(1806*A+2702*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-624*A-738*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)+195*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))-441*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))
+165*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))-357
*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*si
n(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 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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)*(a*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 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((A + B*cos(c + d*x))*(a + a*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 \[ a^{3} \left (\int \frac {A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {3 A \cos {\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {3 A \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {A \cos ^{3}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \cos {\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {3 B \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {3 B \cos ^{3}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \cos ^{4}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx\right ) \]

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))/sec(d*x+c)**(1/2),x)

[Out]

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

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