\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [969]

Optimal result

Integrand size = 43, antiderivative size = 552 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 b \sqrt {a+b} d}-\frac {\left (15 A b^2+a b (5 A-12 B)-2 a^2 (A+2 B-4 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 \sqrt {a+b} d}-\frac {\sqrt {a+b} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^4 d}-\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[Out]

1/4*(15*A*b^3+4*B*a^3-12*B*a*b^2-a^2*(7*A*b-8*C*b))*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^3/b/d/(a+b)^(1/2)-1/4*(15*
A*b^2+a*b*(5*A-12*B)-2*a^2*(A+2*B-4*C))*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^3/d/(a+b)^(1/2)-1/4*(15*A*b^2-12*B*a*b
+4*a^2*(A+2*C))*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^4/d-1/4*(5*A*b-4*B*a)*sin(d*x+c)/a^2/d/(a
+b*sec(d*x+c))^(1/2)+1/2*A*cos(d*x+c)*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(1/2)+1/4*b*(15*A*b^3+4*B*a^3-12*B*a*b^2
-a^2*(7*A*b-8*C*b))*tan(d*x+c)/a^3/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 552, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4189, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{4 a^4 d}-\frac {\cot (c+d x) \left (-2 a^2 (A+2 B-4 C)+a b (5 A-12 B)+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{4 a^3 d \sqrt {a+b}}+\frac {\cot (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^3 b d \sqrt {a+b}}+\frac {b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}} \]

[In]

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

[Out]

((15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*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))])
/(4*a^3*b*Sqrt[a + b]*d) - ((15*A*b^2 + a*b*(5*A - 12*B) - 2*a^2*(A + 2*B - 4*C))*Cot[c + d*x]*EllipticF[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))])/(4*a^3*Sqrt[a + b]*d) - (Sqrt[a + b]*(15*A*b^2 - 12*a*b*B + 4*a^2*(A + 2*C))*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))])/(4*a^4*d) - ((5*A*b - 4*a*B)*Sin[c + d*x])/(4*a^2*d
*Sqrt[a + b*Sec[c + d*x]]) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*Sqrt[a + b*Sec[c + d*x]]) + (b*(15*A*b^3 + 4
*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Tan[c + d*x])/(4*a^3*(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 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*} \text {integral}& = \frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (\frac {1}{2} (5 A b-4 a B)-a (A+2 C) \sec (c+d x)-\frac {3}{2} A b \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{2 a} \\ & = -\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {\frac {1}{4} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac {3}{2} a A b \sec (c+d x)-\frac {1}{4} b (5 A b-4 a B) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{2 a^2} \\ & = -\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-\frac {1}{8} \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\frac {1}{4} a b \left (5 A b^2-4 a b B-a^2 (A-4 C)\right ) \sec (c+d x)+\frac {1}{8} b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )} \\ & = -\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-\frac {1}{8} \left (a^2-b^2\right ) \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right )+\left (\frac {1}{4} a b \left (5 A b^2-4 a b B-a^2 (A-4 C)\right )-\frac {1}{8} b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )}-\frac {\left (b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )} \\ & = \frac {\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cot (c+d x) E\left (\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}}}{4 a^3 b \sqrt {a+b} d}-\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {\left (b \left (15 A b^2+a b (5 A-12 B)-2 a^2 (A+2 B-4 C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 a^3 (a+b)}+\frac {\left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 a^3} \\ & = \frac {\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cot (c+d x) E\left (\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}}}{4 a^3 b \sqrt {a+b} d}-\frac {\left (15 A b^2+a b (5 A-12 B)-2 a^2 (A+2 B-4 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\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}}}{4 a^3 \sqrt {a+b} d}-\frac {\sqrt {a+b} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\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}}}{4 a^4 d}-\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(2489\) vs. \(2(552)=1104\).

Time = 26.53 (sec) , antiderivative size = 2489, normalized size of antiderivative = 4.51 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

(((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((4*b*(A*b^2 - a*b*B + a^2*C)*Sin[c + d*x])/(a^3*(-a^2 + b^2)) + (4*(A
*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x]))/(a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])) + (
A*Sin[2*(c + d*x)])/(2*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[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(-7*a^3*A*b*Tan[(c + d*x)
/2] - 7*a^2*A*b^2*Tan[(c + d*x)/2] + 15*a*A*b^3*Tan[(c + d*x)/2] + 15*A*b^4*Tan[(c + d*x)/2] + 4*a^4*B*Tan[(c
+ d*x)/2] + 4*a^3*b*B*Tan[(c + d*x)/2] - 12*a^2*b^2*B*Tan[(c + d*x)/2] - 12*a*b^3*B*Tan[(c + d*x)/2] + 8*a^3*b
*C*Tan[(c + d*x)/2] + 8*a^2*b^2*C*Tan[(c + d*x)/2] + 14*a^3*A*b*Tan[(c + d*x)/2]^3 - 30*a*A*b^3*Tan[(c + d*x)/
2]^3 - 8*a^4*B*Tan[(c + d*x)/2]^3 + 24*a^2*b^2*B*Tan[(c + d*x)/2]^3 - 16*a^3*b*C*Tan[(c + d*x)/2]^3 - 7*a^3*A*
b*Tan[(c + d*x)/2]^5 + 7*a^2*A*b^2*Tan[(c + d*x)/2]^5 + 15*a*A*b^3*Tan[(c + d*x)/2]^5 - 15*A*b^4*Tan[(c + d*x)
/2]^5 + 4*a^4*B*Tan[(c + d*x)/2]^5 - 4*a^3*b*B*Tan[(c + d*x)/2]^5 - 12*a^2*b^2*B*Tan[(c + d*x)/2]^5 + 12*a*b^3
*B*Tan[(c + d*x)/2]^5 + 8*a^3*b*C*Tan[(c + d*x)/2]^5 - 8*a^2*b^2*C*Tan[(c + d*x)/2]^5 + 8*a^4*A*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)] + 22*a^2*A*b^2*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)] - 30*A*b^4*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)] - 24*a^3*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sq
rt[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*b^3*B*El
lipticPi[-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)] + 16*a^4*C*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)] - 16*a^2*b^
2*C*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)] + 8*a^4*A*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)] + 22*a^2*A*b^2*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)] - 30*A*b^4*Ellipti
cPi[-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)] - 24*a^3*b*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)] + 24*a*b^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)] + 16*a^4*
C*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]*Sq
rt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 16*a^2*b^2*C*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)*(15*A*b^3 + 4*a^3*B - 12*a*b^2*B + a^2*(-7*A*b + 8*b*C))*Elliptic
E[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)*(5*A*b^2 - a*b*(3*A + 4*B) + 2*a^2*(A +
2*C))*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)]))/(2*a^3*(a^2 - b^2)*d*(a + b*Sec[c +
d*x])^(3/2)*(-1 + Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x)/2]^2)]*(a*(-1 + Tan[(c
+ d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2))))/2

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(7019\) vs. \(2(507)=1014\).

Time = 6.12 (sec) , antiderivative size = 7020, normalized size of antiderivative = 12.72

method	result	size
default	\(\text {Expression too large to display}\)	\(7020\)

[In]

int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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