3.741 \(\int \frac {\cos ^3(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\)
Optimal. Leaf size=506 \[ -\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{12 a^2 d}+\frac {b \sqrt {a+b} \left (4 a^2 (A+2 C)+5 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{8 a^4 d}+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{24 a^3 d}-\frac {\sqrt {a+b} \left (-8 a^2 (2 A+3 C)+10 a A b-15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^3 d}+\frac {(a-b) \sqrt {a+b} \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^3 b d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d} \]
[Out]
1/24*(a-b)*(15*A*b^2+8*a^2*(2*A+3*C))*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^3/b/d-1/24*(10*a*A*b-15*A*b^
2-8*a^2*(2*A+3*C))*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^3/d+1/8*b*(5*A*b^2+4*a^2*(A+2*C))*cot(d*x+c)*El
lipticPi((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+1/24*(15*A*b^2+8*a^2*(2*A+3*C))*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/
a^3/d-5/12*A*b*cos(d*x+c)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^2/d+1/3*A*cos(d*x+c)^2*sin(d*x+c)*(a+b*sec(d*x+c
))^(1/2)/a/d
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Rubi [A] time = 1.03, antiderivative size = 506, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{4105, 4104, 4058, 3921, 3784, 3832, 4004} \[ \frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{24 a^3 d}-\frac {\sqrt {a+b} \left (-8 a^2 (2 A+3 C)+10 a A b-15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^3 d}+\frac {(a-b) \sqrt {a+b} \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{24 a^3 b d}+\frac {b \sqrt {a+b} \left (4 a^2 (A+2 C)+5 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{8 a^4 d}-\frac {5 A b \sin (c+d x) \cos (c+d x) \sqrt {a+b \sec (c+d x)}}{12 a^2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
((a - b)*Sqrt[a + b]*(15*A*b^2 + 8*a^2*(2*A + 3*C))*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^3*b*d) - (Sqrt[a + b]*(10*a*A*b - 15*A*b^2 - 8*a^2*(2*A + 3*C))*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^3*d) + (b*Sqrt[a + b]*(5*A*b^2 + 4*a^2*(A + 2*C))*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sq
rt[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^4*d) + ((15*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(24*a
^3*d) - (5*A*b*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(12*a^2*d) + (A*Cos[c + d*x]^2*Sqrt[a + b*S
ec[c + d*x]]*Sin[c + d*x])/(3*a*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 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]
Rule 4105
Int[((A_.) + 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*(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, C, m}, x] && NeQ[
a^2 - b^2, 0] && LeQ[n, -1]
Rubi steps
\begin {align*} \int \frac {\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {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 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (\frac {5 A b}{2}-a (2 A+3 C) \sec (c+d x)-\frac {3}{2} A b \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a}\\ &=-\frac {5 A b \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a^2 d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 a d}+\frac {\int \frac {\cos (c+d x) \left (\frac {1}{4} \left (15 A b^2+4 a^2 (4 A+6 C)\right )+\frac {1}{2} a A b \sec (c+d x)-\frac {5}{4} A b^2 \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^2}\\ &=\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^3 d}-\frac {5 A b \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a^2 d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 a d}-\frac {\int \frac {\frac {3}{8} b \left (5 A b^2+4 a^2 (A+2 C)\right )+\frac {5}{4} a A b^2 \sec (c+d x)+\frac {1}{8} b \left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^3}\\ &=\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^3 d}-\frac {5 A b \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a^2 d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 a d}-\frac {\int \frac {\frac {3}{8} b \left (5 A b^2+4 a^2 (A+2 C)\right )+\left (\frac {5}{4} a A b^2-\frac {1}{8} b \left (15 A b^2+8 a^2 (2 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^3}-\frac {\left (b \left (15 A b^2+8 a^2 (2 A+3 C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{48 a^3}\\ &=\frac {(a-b) \sqrt {a+b} \left (15 A b^2+8 a^2 (2 A+3 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}}}{24 a^3 b d}+\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^3 d}-\frac {5 A b \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a^2 d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 a d}-\frac {\left (b \left (5 A b^2+4 a^2 (A+2 C)\right )\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{16 a^3}-\frac {\left (b \left (10 a A b-15 A b^2-8 a^2 (2 A+3 C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{48 a^3}\\ &=\frac {(a-b) \sqrt {a+b} \left (15 A b^2+8 a^2 (2 A+3 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}}}{24 a^3 b d}-\frac {\sqrt {a+b} \left (10 a A b-15 A b^2-8 a^2 (2 A+3 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}}}{24 a^3 d}+\frac {b \sqrt {a+b} \left (5 A b^2+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}}}{8 a^4 d}+\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 a^3 d}-\frac {5 A b \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{12 a^2 d}+\frac {A \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 a d}\\ \end {align*}
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Mathematica [B] time = 23.09, size = 1352, normalized size = 2.67 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
((b + a*Cos[c + d*x])*Sec[c + d*x]*((A*Sin[c + d*x])/(12*a) - (5*A*b*Sin[2*(c + d*x)])/(24*a^2) + (A*Sin[3*(c
+ d*x)])/(12*a)))/(d*Sqrt[a + b*Sec[c + d*x]]) + (Sqrt[b + a*Cos[c + d*x]]*Sqrt[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] - 15*a*A*b^2*Tan[(c + d*x)/2] - 15*A*b^3*Tan[(c + d*x)/2]
- 24*a^3*C*Tan[(c + d*x)/2] - 24*a^2*b*C*Tan[(c + d*x)/2] + 32*a^3*A*Tan[(c + d*x)/2]^3 + 30*a*A*b^2*Tan[(c +
d*x)/2]^3 + 48*a^3*C*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 - 15*a*
A*b^2*Tan[(c + d*x)/2]^5 + 15*A*b^3*Tan[(c + d*x)/2]^5 - 24*a^3*C*Tan[(c + d*x)/2]^5 + 24*a^2*b*C*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]*Sqr
t[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 30*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^2*b*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)] + 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)] + 30*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^2*b*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^2 + 8*a^2*(2*A
+ 3*C))*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*a + 5*b)*EllipticF[ArcS
in[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*T
an[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(24*a^3*d*Sqrt[a + b*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(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^3/sqrt(b*sec(d*x + c) + a), x)
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maple [B] time = 2.20, size = 2347, normalized size = 4.64 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x)
[Out]
1/24/d*(-1+cos(d*x+c))^2*(-8*A*cos(d*x+c)^3*a^3+16*A*cos(d*x+c)^2*a^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))*a^3*sin
(d*x+c)-15*A*cos(d*x+c)^2*b^3-5*A*cos(d*x+c)^3*a*b^2-18*A*cos(d*x+c)^2*a^2*b+15*A*cos(d*x+c)^2*a*b^2+16*A*cos(
d*x+c)*a^2*b-10*A*cos(d*x+c)*a*b^2+15*A*cos(d*x+c)*b^3-15*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)+2*A*cos
(d*x+c)^4*a^2*b+24*C*cos(d*x+c)^2*a^3-24*C*cos(d*x+c)^2*a^2*b-24*C*cos(d*x+c)^3*a^3-8*A*cos(d*x+c)^5*a^3+24*C*
cos(d*x+c)*a^2*b+30*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)*Elliptic
Pi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*b^3*sin(d*x+c)-24*C*(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)-24*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)*cos
(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b+48*C*sin(d*x+c)*cos(d*x+c)*(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+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+4*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+10*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*Ellipt
icE((-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-15*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
+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)+4*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)+10*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*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+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)-15*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)+48*C*(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)-24*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)*EllipticE((-1+cos(d*x+c))/sin(d*
x+c),((a-b)/(a+b))^(1/2))*b+30*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))*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^3-15*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^3-24*C*a^3*(c
os(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)*cos(d*x+c)*EllipticE(
(-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)))*(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^3
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^3/sqrt(b*sec(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^3*(A + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(1/2),x)
[Out]
int((cos(c + d*x)^3*(A + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(1/2),x)
[Out]
Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**3/sqrt(a + b*sec(c + d*x)), x)
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