3.379 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=143 \[ -\frac{a^3 (3 B+14 C) \sin (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (B+2 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{d}+\frac{a^{5/2} (5 B+2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a C \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]
[Out]
(a^(5/2)*(5*B + 2*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (a^3*(3*B + 14*C)*Sin[c + d*
x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(B + 2*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + (2*a*C*(a + a
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)
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Rubi [A] time = 0.515374, antiderivative size = 143, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.119, Rules used =
{4072, 4018, 4015, 3774, 203} \[ -\frac{a^3 (3 B+14 C) \sin (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (B+2 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{d}+\frac{a^{5/2} (5 B+2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a C \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^(5/2)*(5*B + 2*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (a^3*(3*B + 14*C)*Sin[c + d*
x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(B + 2*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + (2*a*C*(a + a
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)
Rule 4072
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rule 4018
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d*n
) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
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 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 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])
Rubi steps
\begin{align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 a C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2}{3} \int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (3 B-2 C)+\frac{3}{2} a (B+2 C) \sec (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (B+2 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{4}{3} \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (-\frac{1}{4} a^2 (3 B+14 C)+\frac{1}{4} a^2 (9 B+10 C) \sec (c+d x)\right ) \, dx\\ &=-\frac{a^3 (3 B+14 C) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (B+2 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{1}{2} \left (a^2 (5 B+2 C)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=-\frac{a^3 (3 B+14 C) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (B+2 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac{\left (a^3 (5 B+2 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 )}{d}\\ &=\frac{a^{5/2} (5 B+2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}-\frac{a^3 (3 B+14 C) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (B+2 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{d}+\frac{2 a C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.806314, size = 126, normalized size = 0.88 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (3 \sqrt{2} (5 B+2 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{3}{2}}(c+d x)+\sin \left (\frac{1}{2} (c+d x)\right ) (4 (3 B+8 C) \cos (c+d x)+3 B \cos (2 (c+d x))+3 B+4 C)\right )}{6 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(5*B + 2*C)*ArcSin[Sqrt[2]*Sin[(c + d
*x)/2]]*Cos[c + d*x]^(3/2) + (3*B + 4*C + 4*(3*B + 8*C)*Cos[c + d*x] + 3*B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2])
)/(6*d)
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Maple [B] time = 0.364, size = 256, normalized size = 1.8 \begin{align*} -{\frac{{a}^{2}}{6\,d\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) } \left ( 15\,B\cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +6\,C\cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +6\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+6\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+32\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-12\,B\cos \left ( dx+c \right ) -28\,C\cos \left ( dx+c \right ) -4\,C \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
-1/6/d*a^2*(15*B*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+6*C*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+6*B*cos(d*x+c)^
3+6*B*cos(d*x+c)^2+32*C*cos(d*x+c)^2-12*B*cos(d*x+c)-28*C*cos(d*x+c)-4*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/
cos(d*x+c)/sin(d*x+c)
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Maxima [B] time = 2.37032, size = 3753, normalized size = 26.24 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
1/12*(3*(18*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)
^(1/4)*((4*a^2*sin(3*d*x + 3*c) + 5*a^2*sin(2*d*x + 2*c) + 4*a^2*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) + 1)) + (a^2*cos(2*d*x + 2*c)^2*sin(d*x + c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2
*cos(2*d*x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) -
(4*a^2*cos(3*d*x + 3*c) + 5*a^2*cos(2*d*x + 2*c) + 4*a^2*cos(d*x + c) + 5*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) + 1)) - ((a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c)^2 + a^2*cos(d*x + c) + (a^2*cos(d*x + c
) - a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2
*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 5*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x +
2*c) + a^2)*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos
(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c) + 1))) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arcta
n2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))
, (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) -
1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*
c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) + 1)) + 1) + (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a
^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*sqrt(a))*B/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2
+ 2*cos(2*d*x + 2*c) + 1) + 2*(30*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^
(5/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)^(1/4)*((12*a^2*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) -
3*a^2*sin(2*d*x + 2*c) - 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
))))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (12*a^2*sin(2*d*x + 2*c)*sin(3/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c))) + 3*a^2*cos(2*d*x + 2*c) - a^2 + 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*cos(3/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a)
+ 3*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
+ 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos
(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c)))) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c)
+ a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)
^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
+ 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) - 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2
*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + (a^2*cos(2*d*x +
2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c
)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)) - 1))*sqrt(a))*C/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1))/d
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Fricas [A] time = 0.659229, size = 968, normalized size = 6.77 \begin{align*} \left [\frac{3 \,{\left ({\left (5 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (5 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )\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 (3 \, B a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac{3 \,{\left ({\left (5 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (5 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )\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 (3 \, B a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/6*(3*((5*B + 2*C)*a^2*cos(d*x + c)^2 + (5*B + 2*C)*a^2*cos(d*x + c))*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*(3*B*a^2*cos(d*x + c)^2 + 2*(3*B + 8*C)*a^2*cos(d*x + c) + 2*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*
x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), -1/3*(3*((5*B + 2*C)*a^2*cos(d*x + c)^2 + (5*B + 2*
C)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)
)) - (3*B*a^2*cos(d*x + c)^2 + 2*(3*B + 8*C)*a^2*cos(d*x + c) + 2*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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Giac [B] time = 6.68909, size = 644, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/6*(3*(5*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 2*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))) - 3*(5*B*sqrt(-a)*a^2*sgn(cos(d*x +
c)) + 2*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))) + 12*sqrt(2)*(3*(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^3*sgn(cos(d*x + c)) - B*sqrt(-a)*a^4*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*(3*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 9*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (3*sqrt(2)*B*a
^4*sgn(cos(d*x + c)) + 7*sqrt(2)*C*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan
(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d