3.171 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=200 \[ \frac {a^{3/2} (75 A+112 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a^2 (75 A+112 C) \sin (c+d x)}{64 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{32 d \sqrt {a \sec (c+d x)+a}}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{8 d} \]
[Out]
1/64*a^(3/2)*(75*A+112*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+
c))^(3/2)*sin(d*x+c)/d+1/64*a^2*(75*A+112*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/32*a^2*(13*A+16*C)*cos(d*x+
c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/8*a*A*cos(d*x+c)^2*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
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Rubi [A] time = 0.57, antiderivative size = 200, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used =
{4087, 4017, 4015, 3805, 3774, 203} \[ \frac {a^2 (75 A+112 C) \sin (c+d x)}{64 d \sqrt {a \sec (c+d x)+a}}+\frac {a^{3/2} (75 A+112 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a^2 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{32 d \sqrt {a \sec (c+d x)+a}}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{8 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
(a^(3/2)*(75*A + 112*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^2*(75*A + 112*C)*
Sin[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(13*A + 16*C)*Cos[c + d*x]*Sin[c + d*x])/(32*d*Sqrt[a + a
*Sec[c + d*x]]) + (a*A*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(8*d) + (A*Cos[c + d*x]^3*(a + a*
Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d)
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 3774
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 3805
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(a*Cot[
e + f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(a*(2*n + 1))/(2*b*d*n), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]
Rule 4015
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 4017
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 4087
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*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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 ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {3 a A}{2}+\frac {1}{2} a (3 A+8 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac {a A \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{4} a^2 (13 A+16 C)+\frac {3}{4} a^2 (9 A+16 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {a^2 (13 A+16 C) \cos (c+d x) \sin (c+d x)}{32 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {1}{64} (a (75 A+112 C)) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (75 A+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+16 C) \cos (c+d x) \sin (c+d x)}{32 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {1}{128} (a (75 A+112 C)) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (75 A+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+16 C) \cos (c+d x) \sin (c+d x)}{32 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}-\frac {\left (a^2 (75 A+112 C)\right ) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}\\ &=\frac {a^{3/2} (75 A+112 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+16 C) \cos (c+d x) \sin (c+d x)}{32 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 1.59, size = 140, normalized size = 0.70 \[ \frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (\sqrt {2} (75 A+112 C) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+\left (\sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) ((62 A+32 C) \cos (c+d x)+20 A \cos (2 (c+d x))+4 A \cos (3 (c+d x))+95 A+112 C)\right )}{128 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
(a*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*(75*A + 112*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[C
os[c + d*x]] + (95*A + 112*C + (62*A + 32*C)*Cos[c + d*x] + 20*A*Cos[2*(c + d*x)] + 4*A*Cos[3*(c + d*x)])*(-Si
n[(c + d*x)/2] + Sin[(3*(c + d*x))/2])))/(128*d)
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fricas [A] time = 0.54, size = 380, normalized size = 1.90 \[ \left [\frac {{\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 112 \, C\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (16 \, A a \cos \left (d x + c\right )^{4} + 40 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (25 \, A + 16 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{128 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {{\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 112 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (16 \, A a \cos \left (d x + c\right )^{4} + 40 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (25 \, A + 16 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/128*(((75*A + 112*C)*a*cos(d*x + c) + (75*A + 112*C)*a)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-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*(16
*A*a*cos(d*x + c)^4 + 40*A*a*cos(d*x + c)^3 + 2*(25*A + 16*C)*a*cos(d*x + c)^2 + (75*A + 112*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), -1/64*(((75*A + 112*C)*a*cos(d*x
+ c) + (75*A + 112*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))) - (16*A*a*cos(d*x + c)^4 + 40*A*a*cos(d*x + c)^3 + 2*(25*A + 16*C)*a*cos(d*x + c)^2 + (75*A + 112*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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giac [B] time = 2.31, size = 1087, normalized size = 5.44 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/128*((75*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 112*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))) - (75*A*sqrt(-a)*a*sgn(cos(d*x + c))
+ 112*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*sqrt(2)*(75*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^14*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 112*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^
2 + a))^14*C*sqrt(-a)*a^2*sgn(cos(d*x + c)) - 2087*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^12*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 2864*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a))^12*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 11975*(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^4*sgn(cos(d*x + c)) + 23344*(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^4*sgn(cos(d*x + c)) - 42483*(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^5*sgn(cos(d*x + c)) - 69360*(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^5*sgn(cos(d*x + c)) + 33889*(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^6*sgn(cos(d*x + c)) + 51536*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 8693*(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^7*sgn(cos(d*x + c)) - 14736*(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^7*sgn(cos(d*x + c)) + 1101*(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^8*sgn(cos(d*x + c)) + 1808*(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^8*sgn(cos(d*x + c)) - 49*A*sqrt(-a)*a^9*sgn(cos(d*x
+ c)) - 80*C*sqrt(-a)*a^9*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)^4)/d
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maple [B] time = 1.89, size = 752, normalized size = 3.76 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x)
[Out]
1/1024/d*(75*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3*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*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+112*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3*arctanh(1/2*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+225*A*2^(1/
2)*sin(d*x+c)*cos(d*x+c)^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
*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+336*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^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*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+225*A*2^(1/2)*sin(d*x+c)*cos(d*
x+c)*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*cos(d*x+c)/(1+cos(d*x
+c)))^(7/2)+336*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/co
s(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+75*A*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))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)+112*C*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))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin
(d*x+c)-256*A*cos(d*x+c)^8-384*A*cos(d*x+c)^7-160*A*cos(d*x+c)^6-512*C*cos(d*x+c)^6-400*A*cos(d*x+c)^5-1280*C*
cos(d*x+c)^5+1200*A*cos(d*x+c)^4+1792*C*cos(d*x+c)^4)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c
)^3*a
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maxima [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)^4*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^4\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2),x)
[Out]
int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2), x)
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sympy [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)**4*(a+a*sec(d*x+c))**(3/2)*(A+C*sec(d*x+c)**2),x)
[Out]
Timed out
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