3.6.9 \(\int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [509]
Optimal. Leaf size=311 \[ \frac {a^{5/2} (1015 A+1132 B+1304 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{512 d}+\frac {a^3 (1015 A+1132 B+1304 C) \sin (c+d x)}{512 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1015 A+1132 B+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (545 A+628 B+680 C) \cos ^2(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (115 A+156 B+120 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d}+\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d} \]
[Out]
1/512*a^(5/2)*(1015*A+1132*B+1304*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/60*a*(5*A+12*B)*cos
(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/6*A*cos(d*x+c)^5*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/512*a^3
*(1015*A+1132*B+1304*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/768*a^3*(1015*A+1132*B+1304*C)*cos(d*x+c)*sin(d*
x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/960*a^3*(545*A+628*B+680*C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1
/480*a^2*(115*A+156*B+120*C)*cos(d*x+c)^3*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
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Rubi [A]
time = 0.62, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4171, 4102,
4100, 3890, 3859, 209} \begin {gather*} \frac {a^{5/2} (1015 A+1132 B+1304 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{512 d}+\frac {a^3 (1015 A+1132 B+1304 C) \sin (c+d x)}{512 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (545 A+628 B+680 C) \sin (c+d x) \cos ^2(c+d x)}{960 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (1015 A+1132 B+1304 C) \sin (c+d x) \cos (c+d x)}{768 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (115 A+156 B+120 C) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{480 d}+\frac {a (5 A+12 B) \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{60 d}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^{5/2}}{6 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^(5/2)*(1015*A + 1132*B + 1304*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(512*d) + (a^3*(1
015*A + 1132*B + 1304*C)*Sin[c + d*x])/(512*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(1015*A + 1132*B + 1304*C)*Cos[
c + d*x]*Sin[c + d*x])/(768*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(545*A + 628*B + 680*C)*Cos[c + d*x]^2*Sin[c +
d*x])/(960*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(115*A + 156*B + 120*C)*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*
Sin[c + d*x])/(480*d) + (a*(5*A + 12*B)*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(60*d) + (A*Co
s[c + d*x]^5*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(6*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 3890
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 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 ^6(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+12 B)+\frac {1}{2} a (5 A+12 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (115 A+156 B+120 C)+\frac {15}{4} a^2 (5 A+4 B+8 C) \sec (c+d x)\right ) \, dx}{30 a}\\ &=\frac {a^2 (115 A+156 B+120 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d}+\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{8} a^3 (545 A+628 B+680 C)+\frac {5}{8} a^3 (235 A+252 B+312 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^3 (545 A+628 B+680 C) \cos ^2(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (115 A+156 B+120 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d}+\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {1}{384} \left (a^2 (1015 A+1132 B+1304 C)\right ) \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (1015 A+1132 B+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (545 A+628 B+680 C) \cos ^2(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (115 A+156 B+120 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d}+\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {1}{512} \left (a^2 (1015 A+1132 B+1304 C)\right ) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (1015 A+1132 B+1304 C) \sin (c+d x)}{512 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1015 A+1132 B+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (545 A+628 B+680 C) \cos ^2(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (115 A+156 B+120 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d}+\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\left (a^2 (1015 A+1132 B+1304 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx}{1024}\\ &=\frac {a^3 (1015 A+1132 B+1304 C) \sin (c+d x)}{512 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1015 A+1132 B+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (545 A+628 B+680 C) \cos ^2(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (115 A+156 B+120 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d}+\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}-\frac {\left (a^3 (1015 A+1132 B+1304 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 )}{512 d}\\ &=\frac {a^{5/2} (1015 A+1132 B+1304 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{512 d}+\frac {a^3 (1015 A+1132 B+1304 C) \sin (c+d x)}{512 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1015 A+1132 B+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (545 A+628 B+680 C) \cos ^2(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (115 A+156 B+120 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{480 d}+\frac {a (5 A+12 B) \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}\\ \end {align*}
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Mathematica [A]
time = 3.56, size = 217, normalized size = 0.70 \begin {gather*} \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \left (15 \sqrt {2} (1015 A+1132 B+1304 C) \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+(20965 A+22084 B+23240 C+2 (8085 A+7748 B+7240 C) \cos (c+d x)+4 (1575 A+1324 B+920 C) \cos (2 (c+d x))+2140 A \cos (3 (c+d x))+1392 B \cos (3 (c+d x))+480 C \cos (3 (c+d x))+560 A \cos (4 (c+d x))+192 B \cos (4 (c+d x))+80 A \cos (5 (c+d x))) \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{15360 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(15*Sqrt[2]*(1015*A + 1132*B + 1304*C)*ArcSin[Sqrt[2]*Sin[(c
+ d*x)/2]]*Sqrt[Cos[c + d*x]] + (20965*A + 22084*B + 23240*C + 2*(8085*A + 7748*B + 7240*C)*Cos[c + d*x] + 4*(
1575*A + 1324*B + 920*C)*Cos[2*(c + d*x)] + 2140*A*Cos[3*(c + d*x)] + 1392*B*Cos[3*(c + d*x)] + 480*C*Cos[3*(c
+ d*x)] + 560*A*Cos[4*(c + d*x)] + 192*B*Cos[4*(c + d*x)] + 80*A*Cos[5*(c + d*x)])*(-Sin[(c + d*x)/2] + Sin[(
3*(c + d*x))/2])))/(15360*d)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1653\) vs.
\(2(279)=558\).
time = 29.42, size = 1654, normalized size = 5.32
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
[Out]
1/491520/d*(-144896*B*cos(d*x+c)^8-362240*B*cos(d*x+c)^7-417280*C*cos(d*x+c)^7+1086720*B*cos(d*x+c)^6+974400*A
*cos(d*x+c)^6+169800*B*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+195600*C*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(11/2)*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
)+76125*A*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+84900*B*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
11/2)*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)+97800*C*sin(d*x+
c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+15225*A*sin(d*x+c)*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+16980*B*sin(d*x+c)*cos(d*x+c)^5*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+19560*C*sin(d*x+c)*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+76125*A*sin(d*x+c)*cos(d*x+c)^4*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(11/2)*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
)+84900*B*sin(d*x+c)*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+97800*C*sin(d*x+c)*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(11/2)*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)+152250*A*sin
(d*x+c)*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*si
n(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+169800*B*sin(d*x+c)*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*ar
ctanh(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)+195600*C*sin(d*x+c)*cos(
d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+152250*A*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)-81920*A*cos(d*x+c)^12-204800*A*cos(d
*x+c)^11-98304*B*cos(d*x+c)^11-258048*B*cos(d*x+c)^10-122880*C*cos(d*x+c)^10-348160*C*cos(d*x+c)^9-158720*A*co
s(d*x+c)^10-74240*A*cos(d*x+c)^9-223232*B*cos(d*x+c)^9-363520*C*cos(d*x+c)^8-129920*A*cos(d*x+c)^8-324800*A*co
s(d*x+c)^7+1251840*C*cos(d*x+c)^6+15225*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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)+16980*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
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^(1/2)*sin(d*x+c)+19560*
C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*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))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^5*a^2
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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)^6*(a+a*sec(d*x+c))^(5/2)*(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 = 2.93, size = 536, normalized size = 1.72 \begin {gather*} \left [\frac {15 \, {\left ({\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2}\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 (1280 \, A a^{2} \cos \left (d x + c\right )^{6} + 128 \, {\left (35 \, A + 12 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (145 \, A + 116 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (1015 \, A + 1132 \, B + 920 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \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 )}{15360 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {15 \, {\left ({\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2}\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 (1280 \, A a^{2} \cos \left (d x + c\right )^{6} + 128 \, {\left (35 \, A + 12 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (145 \, A + 116 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (1015 \, A + 1132 \, B + 920 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \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 )}{7680 \, {\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)^6*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/15360*(15*((1015*A + 1132*B + 1304*C)*a^2*cos(d*x + c) + (1015*A + 1132*B + 1304*C)*a^2)*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*(1280*A*a^2*cos(d*x + c)^6 + 128*(35*A + 12*B)*a^2*cos(d*x + c)^5 + 48*(145*A +
116*B + 40*C)*a^2*cos(d*x + c)^4 + 8*(1015*A + 1132*B + 920*C)*a^2*cos(d*x + c)^3 + 10*(1015*A + 1132*B + 1304
*C)*a^2*cos(d*x + c)^2 + 15*(1015*A + 1132*B + 1304*C)*a^2*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), -1/7680*(15*((1015*A + 1132*B + 1304*C)*a^2*cos(d*x + c) + (1015*A + 11
32*B + 1304*C)*a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))
) - (1280*A*a^2*cos(d*x + c)^6 + 128*(35*A + 12*B)*a^2*cos(d*x + c)^5 + 48*(145*A + 116*B + 40*C)*a^2*cos(d*x
+ c)^4 + 8*(1015*A + 1132*B + 920*C)*a^2*cos(d*x + c)^3 + 10*(1015*A + 1132*B + 1304*C)*a^2*cos(d*x + c)^2 + 1
5*(1015*A + 1132*B + 1304*C)*a^2*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)**6*(a+a*sec(d*x+c))**(5/2)*(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. 2319 vs.
\(2 (279) = 558\).
time = 3.09, size = 2319, normalized size = 7.46 \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)^6*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/15360*(15*(1015*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 1132*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 1304*C*sqrt(-a)*
a^2*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))) - 15*(1015*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 1132*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 1304*C*sq
rt(-a)*a^2*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*(15225*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))
^22*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 16980*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/
2*c)^2 + a))^22*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 19560*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^22*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 502425*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^20*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 560340*sqrt(2)*(sqrt(-a)*tan(1/2*d
*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^20*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 645480*sqrt(2)*(sqrt(
-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^20*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 6518495*
sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*A*sqrt(-a)*a^5*sgn(cos(d*x +
c)) + 7963020*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*B*sqrt(-a)*a^5*
sgn(cos(d*x + c)) + 8467800*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*C
*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 49683495*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^16*A*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 56336940*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a))^16*B*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 59757720*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c
) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^16*C*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 191286330*sqrt(2)*(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^7*sgn(cos(d*x + c)) + 219014472*sqrt(
2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*sqrt(-a)*a^7*sgn(cos(d*x + c)) +
244004880*sqrt(2)*(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^7*sgn
(cos(d*x + c)) - 418895130*sqrt(2)*(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^8*sgn(cos(d*x + c)) - 474348232*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^12*B*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 531000080*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a))^12*C*sqrt(-a)*a^8*sgn(cos(d*x + c)) + 374587230*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^9*sgn(cos(d*x + c)) + 421769112*sqrt(2)*(sqrt(-a)*t
an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a)*a^9*sgn(cos(d*x + c)) + 473308080*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))^10*C*sqrt(-a)*a^9*sgn(cos(d*x + c))
- 154254030*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^10*s
gn(cos(d*x + c)) - 174597720*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^10*sgn(cos(d*x + c)) - 198757680*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^10*sgn(cos(d*x + c)) + 35939005*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^11*sgn(cos(d*x + c)) + 40114980*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*B*sqrt(-a)*a^11*sgn(cos(d*x + c)) + 45352200*sqrt(2)*(sqrt(-a)*ta
n(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^11*sgn(cos(d*x + c)) - 4649085*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^12*sgn(cos(d*x + c)) -
5273124*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^12*sgn(co
s(d*x + c)) - 5884680*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*C*sqrt(-
a)*a^12*sgn(cos(d*x + c)) + 324435*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^13*sgn(cos(d*x + c)) + 367644*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^13*sgn(cos(d*x + c)) + 411000*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^13*sgn(cos(d*x + c)) - 9435*sqrt(2)*A*sqrt(-a)*a^14*sgn(cos(d*x +
c)) - 10684*sqrt(2)*B*sqrt(-a)*a^14*sgn(cos(d*x + c)) - 11960*sqrt(2)*C*sqrt(-a)*a^14*sgn(cos(d*x + c)))/((sqr
t(-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)^6)/d
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\cos \left (c+d\,x\right )}^6\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\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)^6*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int(cos(c + d*x)^6*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
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