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

Optimal. Leaf size=509 \[ \frac {\left (16 a^2 A+6 a b B-3 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{24 a^2 d}+\frac {(a-b) \sqrt {a+b} \left (16 a^2 A+6 a b B-3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^2 b d}+\frac {\sqrt {a+b} (2 a+b) (8 a A+6 a B-3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^2 d}-\frac {\sqrt {a+b} \left (8 a^3 B+4 a^2 A b-2 a b^2 B+A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{8 a^3 d}+\frac {(6 a B+A b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{12 a d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 d} \]

[Out]

1/24*(a-b)*(16*A*a^2-3*A*b^2+6*B*a*b)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1
/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/b/d+1/24*(2*a+b)*(8*A*a-3*
A*b+6*B*a)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(
d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d-1/8*(4*A*a^2*b+A*b^3+8*B*a^3-2*B*a*b^2)*cot(d*x+c)*
EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b)
)^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/d+1/24*(16*A*a^2-3*A*b^2+6*B*a*b)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2
)/a^2/d+1/12*(A*b+6*B*a)*cos(d*x+c)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d+1/3*A*cos(d*x+c)^2*sin(d*x+c)*(a+b*s
ec(d*x+c))^(1/2)/d

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Rubi [A]  time = 1.13, antiderivative size = 509, 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 = {4032, 4104, 4058, 3921, 3784, 3832, 4004} \[ \frac {\left (16 a^2 A+6 a b B-3 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{24 a^2 d}+\frac {(a-b) \sqrt {a+b} \left (16 a^2 A+6 a b B-3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^2 b d}-\frac {\sqrt {a+b} \left (4 a^2 A b+8 a^3 B-2 a b^2 B+A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{8 a^3 d}+\frac {\sqrt {a+b} (2 a+b) (8 a A+6 a B-3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^2 d}+\frac {(6 a B+A b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{12 a d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)*Sqrt[a + b]*(16*a^2*A - 3*A*b^2 + 6*a*b*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqr
t[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(24*
a^2*b*d) + (Sqrt[a + b]*(2*a + b)*(8*a*A - 3*A*b + 6*a*B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x
]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))]
)/(24*a^2*d) - (Sqrt[a + b]*(4*a^2*A*b + A*b^3 + 8*a^3*B - 2*a*b^2*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSi
n[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 +
Sec[c + d*x]))/(a - b))])/(8*a^3*d) + ((16*a^2*A - 3*A*b^2 + 6*a*b*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(
24*a^2*d) + ((A*b + 6*a*B)*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(12*a*d) + (A*Cos[c + d*x]^2*Sq
rt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4032

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

Rule 4058

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[(Csc[e + f*
x]*(1 + Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]

Rule 4104

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_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)) \, dx &=\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} \int \frac {\cos ^2(c+d x) \left (\frac {1}{2} (A b+6 a B)+(2 a A+3 b B) \sec (c+d x)+\frac {3}{2} A b \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {(A b+6 a B) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}-\frac {\int \frac {\cos (c+d x) \left (\frac {1}{4} \left (-16 a^2 A+3 A b^2-6 a b B\right )-\frac {1}{2} a (7 A b+6 a B) \sec (c+d x)-\frac {1}{4} b (A b+6 a B) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a}\\ &=\frac {\left (16 a^2 A-3 A b^2+6 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^2 d}+\frac {(A b+6 a B) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {\int \frac {\frac {3}{8} \left (4 a^2 A b+A b^3+8 a^3 B-2 a b^2 B\right )+\frac {1}{4} a b (A b+6 a B) \sec (c+d x)-\frac {1}{8} b \left (16 a^2 A-3 A b^2+6 a b B\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^2}\\ &=\frac {\left (16 a^2 A-3 A b^2+6 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^2 d}+\frac {(A b+6 a B) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {\int \frac {\frac {3}{8} \left (4 a^2 A b+A b^3+8 a^3 B-2 a b^2 B\right )+\left (\frac {1}{4} a b (A b+6 a B)+\frac {1}{8} b \left (16 a^2 A-3 A b^2+6 a b B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^2}-\frac {\left (b \left (16 a^2 A-3 A b^2+6 a b B\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{48 a^2}\\ &=\frac {(a-b) \sqrt {a+b} \left (16 a^2 A-3 A b^2+6 a b B\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{24 a^2 b d}+\frac {\left (16 a^2 A-3 A b^2+6 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^2 d}+\frac {(A b+6 a B) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {(b (2 a+b) (8 a A-3 A b+6 a B)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{48 a^2}+\frac {\left (4 a^2 A b+A b^3+8 a^3 B-2 a b^2 B\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{16 a^2}\\ &=\frac {(a-b) \sqrt {a+b} \left (16 a^2 A-3 A b^2+6 a b B\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{24 a^2 b d}+\frac {\sqrt {a+b} (2 a+b) (8 a A-3 A b+6 a B) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{24 a^2 d}-\frac {\sqrt {a+b} \left (4 a^2 A b+A b^3+8 a^3 B-2 a b^2 B\right ) \cot (c+d x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{8 a^3 d}+\frac {\left (16 a^2 A-3 A b^2+6 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^2 d}+\frac {(A b+6 a B) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 20.04, size = 1548, normalized size = 3.04 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[a + b*Sec[c + d*x]]*((A*Sin[c + d*x])/12 + ((A*b + 6*a*B)*Sin[2*(c + d*x)])/(24*a) + (A*Sin[3*(c + d*x)]
)/12))/d - (Sqrt[a + b*Sec[c + d*x]]*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 +
b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(16*a^3*A*Tan[(c + d*x)/2] + 16*a^2*A*b*Tan[(c + d*x)/2] - 3*a
*A*b^2*Tan[(c + d*x)/2] - 3*A*b^3*Tan[(c + d*x)/2] + 6*a^2*b*B*Tan[(c + d*x)/2] + 6*a*b^2*B*Tan[(c + d*x)/2] -
32*a^3*A*Tan[(c + d*x)/2]^3 + 6*a*A*b^2*Tan[(c + d*x)/2]^3 - 12*a^2*b*B*Tan[(c + d*x)/2]^3 + 16*a^3*A*Tan[(c
+ d*x)/2]^5 - 16*a^2*A*b*Tan[(c + d*x)/2]^5 - 3*a*A*b^2*Tan[(c + d*x)/2]^5 + 3*A*b^3*Tan[(c + d*x)/2]^5 + 6*a^
2*b*B*Tan[(c + d*x)/2]^5 - 6*a*b^2*B*Tan[(c + d*x)/2]^5 + 24*a^2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]],
(a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a +
b)] + 6*A*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a +
b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 48*a^3*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]],
(a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a +
b)] - 12*a*b^2*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[
(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 24*a^2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x
)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b
*Tan[(c + d*x)/2]^2)/(a + b)] + 6*A*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x
)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 48*a
^3*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]
*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 12*a*b^2*B*EllipticPi[-1, ArcSin[Tan[(c
+ d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]
^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*(16*a^2*A - 3*A*b^2 + 6*a*b*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]
], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 +
b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*(-(A*b^2) + 2*a*b*(7*A - 3*B) + 12*a^2*B)*EllipticF[ArcSin[Tan[(c + d*x)
/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^
2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(24*a^2*d*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Tan[(c + d
*x)/2]^2]*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2)))

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fricas [F]  time = 59.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{3}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.49, size = 2954, normalized size = 5.80 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/d*(-1+cos(d*x+c))^2*(16*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*sin(d*x+c)+8*A*cos(d*x+c)^3*a^3-16*A*cos(d*x+c
)^2*a^3-3*A*cos(d*x+c)^2*b^3-12*B*cos(d*x+c)^2*a^3-A*cos(d*x+c)^3*a*b^2+6*A*cos(d*x+c)^2*a^2*b+3*A*cos(d*x+c)^
2*a*b^2-16*A*cos(d*x+c)*a^2*b-2*A*cos(d*x+c)*a*b^2+18*B*cos(d*x+c)^3*a^2*b-6*B*cos(d*x+c)^2*a^2*b+6*B*cos(d*x+
c)^2*a*b^2-12*B*cos(d*x+c)*a^2*b-6*B*cos(d*x+c)*a*b^2+3*A*cos(d*x+c)*b^3+10*A*cos(d*x+c)^4*a^2*b-3*A*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((
a-b)/(a+b))^(1/2))*b^3*sin(d*x+c)+12*B*cos(d*x+c)^4*a^3-24*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)
*a^3+24*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d
*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)-28*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*
x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)+2
*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)+16*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+co
s(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)-3*A*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(
(a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)-12*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(
a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)+12*B*EllipticF((-1+
cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*sin(d*x+c)*b+6*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(
1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)+6*B*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*a*b^2*sin(d*x+c)+6*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipti
cPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*b^3+16*A*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/
2))*sin(d*x+c)*cos(d*x+c)*a^3-3*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*b^3+48*B*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)
/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^3+48*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c
))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^3*sin(d*x+c)+6*A*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((
a-b)/(a+b))^(1/2))*b^3*sin(d*x+c)-24*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b
))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*sin(d*x+c)+8*A*cos(d*x+c)^5*a^3+6*B*Ell
ipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*b^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a+24*A*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b
))^(1/2))*cos(d*x+c)*a^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d
*x+c)*b-28*A*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*a^2*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*b+2*A*EllipticF((-1+cos(d*x+c))/sin(d*x+c),
((a-b)/(a+b))^(1/2))*cos(d*x+c)*b^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^
(1/2)*sin(d*x+c)*a+16*A*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*a^2*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*b-3*A*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*sin(d*x+c)*cos(d*x+c)*a*b^2-12*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(
1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a*b^2+12*B*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a
-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^2*b+6*B*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*co
s(d*x+c)*a^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*b)*(1+
cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)^5/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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