3.416 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=232 \[ \frac {(19 A-12 B+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B+5 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A-6 B+2 C) \tan (c+d x)}{4 a d \sqrt {a \cos (c+d x)+a}}+\frac {(2 A-B+C) \tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]
[Out]
1/4*(19*A-12*B+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-1/4*(13*A-9*B+5*C)*arctanh(1/
2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A-B+C)*sec(d*x+c)*tan(d*x+c)/d/(a+
a*cos(d*x+c))^(3/2)-1/4*(7*A-6*B+2*C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/2*(2*A-B+C)*sec(d*x+c)*tan(d*x+c
)/a/d/(a+a*cos(d*x+c))^(1/2)
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Rubi [A] time = 0.78, antiderivative size = 232, 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 =
{3041, 2984, 2985, 2649, 206, 2773} \[ \frac {(19 A-12 B+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B+5 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A-6 B+2 C) \tan (c+d x)}{4 a d \sqrt {a \cos (c+d x)+a}}+\frac {(2 A-B+C) \tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(3/2),x]
[Out]
((19*A - 12*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(3/2)*d) - ((13*A - 9*B +
5*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((7*A - 6*B +
2*C)*Tan[c + d*x])/(4*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(2*d*(a + a*Cos
[c + d*x])^(3/2)) + ((2*A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d*x]])
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2649
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 2773
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 2984
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Rule 2985
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3041
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Rubi steps
\begin {align*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (2 a (2 A-B+C)-\frac {1}{2} a (5 A-5 B+C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-a^2 (7 A-6 B+2 C)+3 a^2 (2 A-B+C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^3}\\ &=-\frac {(7 A-6 B+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (\frac {1}{2} a^3 (19 A-12 B+8 C)-\frac {1}{2} a^3 (7 A-6 B+2 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^4}\\ &=-\frac {(7 A-6 B+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A-9 B+5 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}+\frac {(19 A-12 B+8 C) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a^2}\\ &=-\frac {(7 A-6 B+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {(13 A-9 B+5 C) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{2 a d}-\frac {(19 A-12 B+8 C) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a d}\\ &=\frac {(19 A-12 B+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B+5 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A-6 B+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.58, size = 186, normalized size = 0.80 \[ \frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\frac {1}{2} \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) ((7 A-6 B+2 C) \cos (2 (c+d x))+(6 A-8 B) \cos (c+d x)+3 A-6 B+2 C)-\sqrt {2} (19 A-12 B+8 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sin ^2\left (\frac {1}{2} (c+d x)\right )-1}-2 (13 A-9 B+5 C) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{2 d (a (\cos (c+d x)+1))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(3/2),x]
[Out]
(Cos[(c + d*x)/2]^3*(-2*(13*A - 9*B + 5*C)*ArcTanh[Sin[(c + d*x)/2]] + (-(Sqrt[2]*(19*A - 12*B + 8*C)*ArcTanh[
Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^2) + ((3*A - 6*B + 2*C + (6*A - 8*B)*Cos[c + d*x] + (7*A - 6*B + 2*
C)*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Sin[(c + d*x)/2])/2)/(-1 + Sin[(c + d*x)/2]^2)))/(2*d*(a*(1 + Cos[c + d*x]
))^(3/2))
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fricas [A] time = 0.68, size = 382, normalized size = 1.65 \[ \frac {2 \, \sqrt {2} {\left ({\left (13 \, A - 9 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (13 \, A - 9 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 9 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + {\left ({\left (19 \, A - 12 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (19 \, A - 12 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (19 \, A - 12 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (7 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
1/16*(2*sqrt(2)*((13*A - 9*B + 5*C)*cos(d*x + c)^4 + 2*(13*A - 9*B + 5*C)*cos(d*x + c)^3 + (13*A - 9*B + 5*C)*
cos(d*x + c)^2)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a
*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + ((19*A - 12*B + 8*C)*cos(d*x + c)^4 + 2*(19*A -
12*B + 8*C)*cos(d*x + c)^3 + (19*A - 12*B + 8*C)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x +
c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x +
c)^2)) - 4*((7*A - 6*B + 2*C)*cos(d*x + c)^2 + (3*A - 4*B)*cos(d*x + c) - 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*
x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(cos((d*t_nostep+c)/2))]Discontinuities at
zeroes of cos((d*t_nostep+c)/2) were not checkedEvaluation time: 1.21Unable to divide, perhaps due to rounding
error%%%{%%{[%%%{%%{[23574053482485268906770432,0]:[1,0,-2]%%},[16]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0]%%%} /
%%%{%%%{%%{[604462909807314587353088,0]:[1,0,-2]%%},[16]%%%},[0]%%%} Error: Bad Argument Value
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maple [B] time = 3.28, size = 2261, normalized size = 9.75 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(3/2),x)
[Out]
-1/2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-19*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^2*a-76*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+
1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^
6*a-76*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2
*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^6*a+48*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c
)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^6*a+48*B*ln(4/(2*cos(1/2*d*x+1/2*c)
+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c
)^6*a-32*C*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*
d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^6*a-8*C*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^2*a+32*C*ln(-4*(a*2^(1/2)*cos(1/2
*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/
2*c)^4*a-32*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*co
s(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^6*a-8*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+
1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^2*a+8*C*2^(1/2)*(a*sin(1/2*d*x+1
/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+26*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d
*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-2*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-19*A*ln(4/(2*cos(
1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos
(1/2*d*x+1/2*c)^2*a+104*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(
1/2*d*x+1/2*c)^6*a-72*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/
2*d*x+1/2*c)^6*a+28*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+76*A*ln(4/(2*cos(1/2
*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/
2*d*x+1/2*c)^4*a+76*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/
(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^4*a-48*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^4*a-48*B*ln(-4*(a*2^(1
/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos
(1/2*d*x+1/2*c)^4*a-40*C*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1
/2*d*x+1/2*c)^4*a+2*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-18*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c
)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a+10*C*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^2*a-8*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^
(1/2)*cos(1/2*d*x+1/2*c)^2-22*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2+16*B*a^(1/
2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2+40*C*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^6*a+12*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*
(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^2*a+12*B*ln(4/(
2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a
))*cos(1/2*d*x+1/2*c)^2*a-24*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+32*C*ln(4/(
2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a
))*cos(1/2*d*x+1/2*c)^4*a+2*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-104*A*ln(2*(2*a^(1/2)*(a*sin(1/2*
d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a+72*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a)/a^(5/2)/cos(1/2*d*x+1/2*c)/(2*cos(1
/2*d*x+1/2*c)+2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^2/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
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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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(3/2)),x)
[Out]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(3/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**(3/2),x)
[Out]
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**3/(a*(cos(c + d*x) + 1))**(3/2), x)
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