3.969 \(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=552 \[ -\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^4 d}-\frac {\cot (c+d x) \left (-2 a^2 (A+2 B-4 C)+a b (5 A-12 B)+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^3 d \sqrt {a+b}}+\frac {b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\cot (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^3 b d \sqrt {a+b}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}} \]
[Out]
1/4*(15*A*b^3+4*a^3*B-12*a*b^2*B-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*a*b*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*a^3*B-12*a*b^2*B
-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)
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Rubi [A] time = 1.25, antiderivative size = 552, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used =
{4104, 4060, 4058, 3921, 3784, 3832, 4004} \[ \frac {b \tan (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right )}{4 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\cot (c+d x) \left (-2 a^2 (A+2 B-4 C)+a b (5 A-12 B)+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^3 d \sqrt {a+b}}+\frac {\cot (c+d x) \left (-a^2 (7 A b-8 b C)+4 a^3 B-12 a b^2 B+15 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^3 b d \sqrt {a+b}}-\frac {\sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{4 a^4 d}-\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[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 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 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 4060
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 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 \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 {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 (\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}}}{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 (\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}}}{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) 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}}}{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) \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}}}{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*}
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Mathematica [C] time = 25.12, size = 3404, normalized size = 6.17 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[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*Sqrt[(-a + b)
/(a + b)]*Tan[(c + d*x)/2] - 7*a^2*A*b^2*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + 15*a*A*b^3*Sqrt[(-a + b)/(a
+ b)]*Tan[(c + d*x)/2] + 15*A*b^4*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + 4*a^4*Sqrt[(-a + b)/(a + b)]*B*Ta
n[(c + d*x)/2] + 4*a^3*b*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2] - 12*a^2*b^2*Sqrt[(-a + b)/(a + b)]*B*Tan[(
c + d*x)/2] - 12*a*b^3*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2] + 8*a^3*b*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d
*x)/2] + 8*a^2*b^2*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/2] + 14*a^3*A*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)
/2]^3 - 30*a*A*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3 - 8*a^4*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]
^3 + 24*a^2*b^2*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^3 - 16*a^3*b*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/
2]^3 - 7*a^3*A*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 + 7*a^2*A*b^2*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/
2]^5 + 15*a*A*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - 15*A*b^4*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]
^5 + 4*a^4*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5 - 4*a^3*b*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5 -
12*a^2*b^2*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5 + 12*a*b^3*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5
+ 8*a^3*b*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/2]^5 - 8*a^2*b^2*Sqrt[(-a + b)/(a + b)]*C*Tan[(c + d*x)/2]^5
- (8*I)*a^4*A*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)
*a^2*A*b^2*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*A*b
^4*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*a^3*b*B*Ell
ipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*a*b^3*B*EllipticP
i[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*a^4*C*EllipticPi[-((a +
b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*a^2*b^2*C*EllipticPi[-((a + b)/
(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*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)] - (8*I)*a^4*A*EllipticPi[-((a + b)/(a - b)),
I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*a^2*A*b^2*EllipticPi[-((a
+ b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*A*b^4*Elli
pticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)
*a^3*b*B*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*T
an[(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*I)*a*b^3*B*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*a^4*C*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*a^2*b^2*C*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)
]*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)] - I*(a - b)*(15*A*b^3 + 4*a^3*B - 12*a*b^2*B + a^2*(-7*A*b + 8*b*C
))*EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*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*I)*(a - b)*
(15*A*b^3 + 2*a*b^2*(5*A - 6*B) + 2*a^3*(A + 2*C) + a^2*b*(A - 8*B + 8*C))*EllipticF[I*ArcSinh[Sqrt[(-a + b)/(
a + b)]*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*Sqrt[(-a + b)/(a + b)]*(a^2 - b^2)*d*(a + b*S
ec[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
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fricas [F] time = 2.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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maple [B] time = 2.05, size = 5176, normalized size = 9.38 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
[Out]
result too large to display
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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