3.5.97 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [497]
Optimal. Leaf size=165 \[ \frac {a^{3/2} (11 A+14 B+24 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a (A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \]
[Out]
1/8*a^(3/2)*(11*A+14*B+24*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/3*A*cos(d*x+c)^2*(a+a*sec(d
*x+c))^(3/2)*sin(d*x+c)/d+1/24*a^2*(19*A+30*B+24*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a*(A+2*B)*cos(d*x+
c)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
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Rubi [A]
time = 0.33, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {4171, 4102,
4100, 3859, 209} \begin {gather*} \frac {a^{3/2} (11 A+14 B+24 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}+\frac {a (A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^(3/2)*(11*A + 14*B + 24*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^2*(19*A + 30
*B + 24*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(A + 2*B)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*
Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 3859
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 4100
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]
Rule 4102
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]
Rule 4171
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*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Rubi steps
\begin {align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {3}{2} a (A+2 B)+\frac {1}{2} a (A+6 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac {a (A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {\int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (19 A+30 B+24 C)+\frac {1}{4} a^2 (7 A+6 B+24 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a (A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{16} (a (11 A+14 B+24 C)) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a (A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {\left (a^2 (11 A+14 B+24 C)\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}\\ &=\frac {a^{3/2} (11 A+14 B+24 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a (A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A]
time = 1.73, size = 124, normalized size = 0.75 \begin {gather*} \frac {a \left ((33 A+42 B+72 C) \text {ArcTan}\left (\sqrt {-1+\sec (c+d x)}\right )+\cos (c+d x) (37 A+42 B+24 C+2 (11 A+6 B) \cos (c+d x)+4 A \cos (2 (c+d x))) \sqrt {-1+\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{24 d \sqrt {-1+\sec (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a*((33*A + 42*B + 72*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]] + Cos[c + d*x]*(37*A + 42*B + 24*C + 2*(11*A + 6*B)*C
os[c + d*x] + 4*A*Cos[2*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(24*
d*Sqrt[-1 + Sec[c + d*x]])
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(832\) vs.
\(2(145)=290\).
time = 28.26, size = 833, normalized size = 5.05
| | |
| method |
result |
size |
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| default |
\(\text {Expression too large to display}\) |
\(833\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
[Out]
-1/192/d*(33*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+42*B*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+72*C*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*
cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+66*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+84*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^
(1/2)+144*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/c
os(d*x+c)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+33*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)*arctanh(1/2
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+42*B*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin(d*x+c)
+72*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)
/cos(d*x+c)*2^(1/2))*sin(d*x+c)+64*A*cos(d*x+c)^6+112*A*cos(d*x+c)^5+96*B*cos(d*x+c)^5+88*A*cos(d*x+c)^4+240*B
*cos(d*x+c)^4+192*C*cos(d*x+c)^4-264*A*cos(d*x+c)^3-336*B*cos(d*x+c)^3-192*C*cos(d*x+c)^3)*(a*(1+cos(d*x+c))/c
os(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)*a
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [A]
time = 2.02, size = 378, normalized size = 2.29 \begin {gather*} \left [\frac {3 \, {\left ({\left (11 \, A + 14 \, B + 24 \, C\right )} a \cos \left (d x + c\right ) + {\left (11 \, A + 14 \, B + 24 \, C\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, A + 6 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (11 \, A + 14 \, B + 8 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (11 \, A + 14 \, B + 24 \, C\right )} a \cos \left (d x + c\right ) + {\left (11 \, A + 14 \, B + 24 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, A + 6 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (11 \, A + 14 \, B + 8 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/48*(3*((11*A + 14*B + 24*C)*a*cos(d*x + c) + (11*A + 14*B + 24*C)*a)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*s
qrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c)
+ 1)) + 2*(8*A*a*cos(d*x + c)^3 + 2*(11*A + 6*B)*a*cos(d*x + c)^2 + 3*(11*A + 14*B + 8*C)*a*cos(d*x + c))*sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*(3*((11*A + 14*B + 24*C)*a*cos(d
*x + c) + (11*A + 14*B + 24*C)*a)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)
*sin(d*x + c))) - (8*A*a*cos(d*x + c)^3 + 2*(11*A + 6*B)*a*cos(d*x + c)^2 + 3*(11*A + 14*B + 8*C)*a*cos(d*x +
c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1245 vs.
\(2 (145) = 290\).
time = 2.24, size = 1245, normalized size = 7.55 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/48*(3*(11*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 14*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 24*C*sqrt(-a)*a*sgn(cos(d*x
+ c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) -
3*(11*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 14*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 24*C*sqrt(-a)*a*sgn(cos(d*x + c)))
*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(33
*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^2*sgn(cos(d*x +
c)) + 42*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a)*a^2*sgn(
cos(d*x + c)) + 72*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C*sqrt(-a)
*a^2*sgn(cos(d*x + c)) - 303*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A
*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 822*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^8*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 888*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a))^8*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 2394*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 3780*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 3024*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1
/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 1806*sqrt(2)*(sqrt(-a)*tan(1
/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 2508*sqrt(2)*(sqrt
(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 1776*sqr
t(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a^5*sgn(cos(d*x + c))
+ 309*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^6*sgn(cos(d
*x + c)) + 498*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^6*
sgn(cos(d*x + c)) + 360*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*sqrt
(-a)*a^6*sgn(cos(d*x + c)) - 19*sqrt(2)*A*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 30*sqrt(2)*B*sqrt(-a)*a^7*sgn(cos(d
*x + c)) - 24*sqrt(2)*C*sqrt(-a)*a^7*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x
+ 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^3)/d
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
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