3.4.84 \(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [384]

Optimal. Leaf size=630 \[ \frac {\left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{24 a^4 b \sqrt {a+b} d}+\frac {\left (105 A b^3+5 a b^2 (7 A-18 B)+4 a^3 (4 A+3 B)-6 a^2 b (A+5 B)\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{24 a^4 \sqrt {a+b} d}+\frac {\sqrt {a+b} \left (12 a^2 A b+35 A b^3-8 a^3 B-30 a b^2 B\right ) \cot (c+d x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{8 a^5 d}+\frac {\left (16 a^2 A+35 A b^2-30 a b B\right ) \sin (c+d x)}{24 a^3 d \sqrt {a+b \sec (c+d x)}}-\frac {(7 A b-6 a B) \cos (c+d x) \sin (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right ) \tan (c+d x)}{24 a^4 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[Out]

1/24*(16*A*a^4+41*A*a^2*b^2-105*A*b^4-42*B*a^3*b+90*B*a*b^3)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)
^(1/2),((a+b)/(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^4/b/d/(a+b)^(1/2)
+1/24*(105*A*b^3+5*a*b^2*(7*A-18*B)+4*a^3*(4*A+3*B)-6*a^2*b*(A+5*B))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/
2)/(a+b)^(1/2),((a+b)/(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^4/d/(a+b)
^(1/2)+1/8*(12*A*a^2*b+35*A*b^3-8*B*a^3-30*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^5/d+1
/24*(16*A*a^2+35*A*b^2-30*B*a*b)*sin(d*x+c)/a^3/d/(a+b*sec(d*x+c))^(1/2)-1/12*(7*A*b-6*B*a)*cos(d*x+c)*sin(d*x
+c)/a^2/d/(a+b*sec(d*x+c))^(1/2)+1/3*A*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(1/2)+1/24*b*(16*A*a^4+41*
A*a^2*b^2-105*A*b^4-42*B*a^3*b+90*B*a*b^3)*tan(d*x+c)/a^4/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.08, antiderivative size = 630, 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 = {4119, 4189, 4145, 4143, 4006, 3869, 3917, 4089} \begin {gather*} -\frac {(7 A b-6 a B) \sin (c+d x) \cos (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (16 a^2 A-30 a b B+35 A b^2\right ) \sin (c+d x)}{24 a^3 d \sqrt {a+b \sec (c+d x)}}+\frac {\sqrt {a+b} \left (-8 a^3 B+12 a^2 A b-30 a b^2 B+35 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};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{8 a^5 d}+\frac {\left (4 a^3 (4 A+3 B)-6 a^2 b (A+5 B)+5 a b^2 (7 A-18 B)+105 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}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^4 d \sqrt {a+b}}+\frac {\left (16 a^4 A-42 a^3 b B+41 a^2 A b^2+90 a b^3 B-105 A b^4\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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^4 b d \sqrt {a+b}}+\frac {b \left (16 a^4 A-42 a^3 b B+41 a^2 A b^2+90 a b^3 B-105 A b^4\right ) \tan (c+d x)}{24 a^4 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((16*a^4*A + 41*a^2*A*b^2 - 105*A*b^4 - 42*a^3*b*B + 90*a*b^3*B)*Cot[c + d*x]*EllipticE[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^4*b*Sqrt[a + b]*d) + ((105*A*b^3 + 5*a*b^2*(7*A - 18*B) + 4*a^3*(4*A + 3*B) - 6*a^2*b*(A + 5*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^4*Sqrt[a + b]*d) + (Sqrt[a + b]*(12*a^2*A*b + 35
*A*b^3 - 8*a^3*B - 30*a*b^2*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/(8*a^5*d) + (
(16*a^2*A + 35*A*b^2 - 30*a*b*B)*Sin[c + d*x])/(24*a^3*d*Sqrt[a + b*Sec[c + d*x]]) - ((7*A*b - 6*a*B)*Cos[c +
d*x]*Sin[c + d*x])/(12*a^2*d*Sqrt[a + b*Sec[c + d*x]]) + (A*Cos[c + d*x]^2*Sin[c + d*x])/(3*a*d*Sqrt[a + b*Sec
[c + d*x]]) + (b*(16*a^4*A + 41*a^2*A*b^2 - 105*A*b^4 - 42*a^3*b*B + 90*a*b^3*B)*Tan[c + d*x])/(24*a^4*(a^2 -
b^2)*d*Sqrt[a + b*Sec[c + d*x]])

Rule 3869

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

Rule 3917

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

Rule 4006

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 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4119

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 + 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*(n +
1)*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, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4143

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 4145

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

Rule 4189

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 \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx &=\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\cos ^2(c+d x) \left (\frac {1}{2} (7 A b-6 a B)-2 a A \sec (c+d x)-\frac {5}{2} A b \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a}\\ &=-\frac {(7 A b-6 a B) \cos (c+d x) \sin (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (\frac {1}{4} \left (16 a^2 A+5 b (7 A b-6 a B)\right )+\frac {3}{2} a (A b+2 a B) \sec (c+d x)-\frac {3}{4} b (7 A b-6 a B) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{6 a^2}\\ &=\frac {\left (16 a^2 A+35 A b^2-30 a b B\right ) \sin (c+d x)}{24 a^3 d \sqrt {a+b \sec (c+d x)}}-\frac {(7 A b-6 a B) \cos (c+d x) \sin (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\frac {3}{8} \left (12 a^2 A b+35 A b^3-8 a^3 B-30 a b^2 B\right )+\frac {3}{4} a b (7 A b-6 a B) \sec (c+d x)-\frac {1}{8} b \left (16 a^2 A+35 A b^2-30 a b B\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{6 a^3}\\ &=\frac {\left (16 a^2 A+35 A b^2-30 a b B\right ) \sin (c+d x)}{24 a^3 d \sqrt {a+b \sec (c+d x)}}-\frac {(7 A b-6 a B) \cos (c+d x) \sin (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right ) \tan (c+d x)}{24 a^4 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {-\frac {3}{16} \left (a^2-b^2\right ) \left (12 a^2 A b+35 A b^3-8 a^3 B-30 a b^2 B\right )-\frac {1}{8} a b \left (11 a^2 A b-35 A b^3-6 a^3 B+30 a b^2 B\right ) \sec (c+d x)-\frac {1}{16} b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )}\\ &=\frac {\left (16 a^2 A+35 A b^2-30 a b B\right ) \sin (c+d x)}{24 a^3 d \sqrt {a+b \sec (c+d x)}}-\frac {(7 A b-6 a B) \cos (c+d x) \sin (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right ) \tan (c+d x)}{24 a^4 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {-\frac {3}{16} \left (a^2-b^2\right ) \left (12 a^2 A b+35 A b^3-8 a^3 B-30 a b^2 B\right )+\left (-\frac {1}{8} a b \left (11 a^2 A b-35 A b^3-6 a^3 B+30 a b^2 B\right )+\frac {1}{16} b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )}-\frac {\left (b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{48 a^4 \left (a^2-b^2\right )}\\ &=\frac {\left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 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^4 b \sqrt {a+b} d}+\frac {\left (16 a^2 A+35 A b^2-30 a b B\right ) \sin (c+d x)}{24 a^3 d \sqrt {a+b \sec (c+d x)}}-\frac {(7 A b-6 a B) \cos (c+d x) \sin (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right ) \tan (c+d x)}{24 a^4 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {\left (12 a^2 A b+35 A b^3-8 a^3 B-30 a b^2 B\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{16 a^4}+\frac {\left (b \left (105 A b^3+5 a b^2 (7 A-18 B)+4 a^3 (4 A+3 B)-6 a^2 b (A+5 B)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{48 a^4 (a+b)}\\ &=\frac {\left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 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^4 b \sqrt {a+b} d}+\frac {\left (105 A b^3+5 a b^2 (7 A-18 B)+4 a^3 (4 A+3 B)-6 a^2 b (A+5 B)\right ) \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^4 \sqrt {a+b} d}+\frac {\sqrt {a+b} \left (12 a^2 A b+35 A b^3-8 a^3 B-30 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^5 d}+\frac {\left (16 a^2 A+35 A b^2-30 a b B\right ) \sin (c+d x)}{24 a^3 d \sqrt {a+b \sec (c+d x)}}-\frac {(7 A b-6 a B) \cos (c+d x) \sin (c+d x)}{12 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (16 a^4 A+41 a^2 A b^2-105 A b^4-42 a^3 b B+90 a b^3 B\right ) \tan (c+d x)}{24 a^4 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(2319\) vs. \(2(630)=1260\).
time = 21.88, size = 2319, normalized size = 3.68 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*(-1/12*((a^4*A - a^2*A*b^2 + 24*A*b^4 - 24*a*b^3*B)*Sin[c + d*x])/(a^4*
(-a^2 + b^2)) - (2*(A*b^5*Sin[c + d*x] - a*b^4*B*Sin[c + d*x]))/(a^4*(a^2 - b^2)*(b + a*Cos[c + d*x])) + ((-11
*A*b + 6*a*B)*Sin[2*(c + d*x)])/(24*a^3) + (A*Sin[3*(c + d*x)])/(12*a^2)))/(d*(a + b*Sec[c + d*x])^(3/2)) - ((
b + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*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^5*A*Tan[(c + d*x)/2] + 16*a^4*A*b*Tan[(c + d*x)/
2] + 41*a^3*A*b^2*Tan[(c + d*x)/2] + 41*a^2*A*b^3*Tan[(c + d*x)/2] - 105*a*A*b^4*Tan[(c + d*x)/2] - 105*A*b^5*
Tan[(c + d*x)/2] - 42*a^4*b*B*Tan[(c + d*x)/2] - 42*a^3*b^2*B*Tan[(c + d*x)/2] + 90*a^2*b^3*B*Tan[(c + d*x)/2]
+ 90*a*b^4*B*Tan[(c + d*x)/2] - 32*a^5*A*Tan[(c + d*x)/2]^3 - 82*a^3*A*b^2*Tan[(c + d*x)/2]^3 + 210*a*A*b^4*T
an[(c + d*x)/2]^3 + 84*a^4*b*B*Tan[(c + d*x)/2]^3 - 180*a^2*b^3*B*Tan[(c + d*x)/2]^3 + 16*a^5*A*Tan[(c + d*x)/
2]^5 - 16*a^4*A*b*Tan[(c + d*x)/2]^5 + 41*a^3*A*b^2*Tan[(c + d*x)/2]^5 - 41*a^2*A*b^3*Tan[(c + d*x)/2]^5 - 105
*a*A*b^4*Tan[(c + d*x)/2]^5 + 105*A*b^5*Tan[(c + d*x)/2]^5 - 42*a^4*b*B*Tan[(c + d*x)/2]^5 + 42*a^3*b^2*B*Tan[
(c + d*x)/2]^5 + 90*a^2*b^3*B*Tan[(c + d*x)/2]^5 - 90*a*b^4*B*Tan[(c + d*x)/2]^5 - 72*a^4*A*b*EllipticPi[-1, A
rcSin[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)] - 138*a^2*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)] + 210*A*b^5*Elliptic
Pi[-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^5*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)] + 132*a^3*b^2*B*E
llipticPi[-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)] - 180*a*b^4*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)] - 72*a^
4*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)] - 138*a^2*A*b^3*EllipticPi[-1, ArcSin[Ta
n[(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)] + 210*A*b^5*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^5*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)] + 132*a^3*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*T
an[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 180*a*b^4*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^4*A + 41*a^2*A*b^2 - 105*A*b^4 - 42*a^3*b*B + 90*a*b^3*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)*(-35*A*b^3 + 12*a^3*B - 2*a^2*b*(5*A + 9*B) + 3*
a*b^2*(7*A + 10*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^4*(a^2 - b^2)*d*(
a + b*Sec[c + d*x])^(3/2)*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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(5085\) vs. \(2(581)=1162\).
time = 9.31, size = 5086, normalized size = 8.07

method	result	size
default	\(\text {Expression too large to display}\)	\(5086\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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

[Out]

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

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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))**(3/2),x)

[Out]

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

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

[Out]

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

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

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))^(3/2),x)

[Out]

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

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