3.424 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=280 \[ -\frac{(63 A-35 B+11 C) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(39 A-20 B+8 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac{(219 A-115 B+43 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(31 A-15 B+7 C) \tan (c+d x) \sec (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(19 A-11 B+3 C) \tan (c+d x) \sec (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
[Out]
((39*A - 20*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B
+ 43*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((63*A -
35*B + 11*C)*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(4*d
*(a + a*Cos[c + d*x])^(5/2)) - ((19*A - 11*B + 3*C)*Sec[c + d*x]*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3
/2)) + ((31*A - 15*B + 7*C)*Sec[c + d*x]*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])
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Rubi [A] time = 1.01811, antiderivative size = 280, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used =
{3041, 2978, 2984, 2985, 2649, 206, 2773} \[ -\frac{(63 A-35 B+11 C) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(39 A-20 B+8 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac{(219 A-115 B+43 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(31 A-15 B+7 C) \tan (c+d x) \sec (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(19 A-11 B+3 C) \tan (c+d x) \sec (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
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])^(5/2),x]
[Out]
((39*A - 20*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B
+ 43*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((63*A -
35*B + 11*C)*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(4*d
*(a + a*Cos[c + d*x])^(5/2)) - ((19*A - 11*B + 3*C)*Sec[c + d*x]*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3
/2)) + ((31*A - 15*B + 7*C)*Sec[c + d*x]*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])
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)]
Rule 2978
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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 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 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 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]
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))^{5/2}} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\left (2 a (3 A-B+C)-\frac{1}{2} a (7 A-7 B-C) \cos (c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(19 A-11 B+3 C) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\left (a^2 (31 A-15 B+7 C)-\frac{5}{4} a^2 (19 A-11 B+3 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(19 A-11 B+3 C) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(31 A-15 B+7 C) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (-a^3 (63 A-35 B+11 C)+\frac{3}{2} a^3 (31 A-15 B+7 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{16 a^5}\\ &=-\frac{(63 A-35 B+11 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(19 A-11 B+3 C) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(31 A-15 B+7 C) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (2 a^4 (39 A-20 B+8 C)-\frac{1}{2} a^4 (63 A-35 B+11 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{16 a^6}\\ &=-\frac{(63 A-35 B+11 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(19 A-11 B+3 C) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(31 A-15 B+7 C) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(39 A-20 B+8 C) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a^3}-\frac{(219 A-115 B+43 C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(63 A-35 B+11 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(19 A-11 B+3 C) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(31 A-15 B+7 C) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(39 A-20 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^2 d}+\frac{(219 A-115 B+43 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 )}{16 a^2 d}\\ &=\frac{(39 A-20 B+8 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 a^{5/2} d}-\frac{(219 A-115 B+43 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(63 A-35 B+11 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(19 A-11 B+3 C) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(31 A-15 B+7 C) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 5.2668, size = 248, normalized size = 0.89 \[ \frac{\sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-\frac{1}{8} \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5\left (\frac{1}{2} (c+d x)\right ) ((269 A-169 B+33 C) \cos (c+d x)+10 (19 A-11 B+3 C) \cos (2 (c+d x))+63 A \cos (3 (c+d x))+158 A-35 B \cos (3 (c+d x))-110 B+11 C \cos (3 (c+d x))+30 C)-(219 A-115 B+43 C) \cos ^2(c+d x) \cos ^9\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \sqrt{2} (39 A-20 B+8 C) \cos ^2(c+d x) \cos ^9\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{8 a d (a (\cos (c+d x)+1))^{3/2}} \]
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])^(5/2),x]
[Out]
(Sec[(c + d*x)/2]^6*Sec[c + d*x]^2*(-((219*A - 115*B + 43*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^9*Cos[
c + d*x]^2) + 4*Sqrt[2]*(39*A - 20*B + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^9*Cos[c + d*x]^
2 - (Cos[(c + d*x)/2]^5*(158*A - 110*B + 30*C + (269*A - 169*B + 33*C)*Cos[c + d*x] + 10*(19*A - 11*B + 3*C)*C
os[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] - 35*B*Cos[3*(c + d*x)] + 11*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])/8))
/(8*a*d*(a*(1 + Cos[c + d*x]))^(3/2))
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Maple [B] time = 0.343, size = 2366, normalized size = 8.4 \begin{align*} \text{result too large to display} \end{align*}
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))^(5/2),x)
[Out]
-1/8*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-128*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+
a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^8*a-624*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2
^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^
8*a-624*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d
*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^8*a-128*C*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/
2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^8*a+320*B*ln(-4*(a*2^(1/2)*cos(1
/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+
1/2*c)^8*a+320*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*
d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^8*a-320*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*
sin(1/2*d*x+1/2*c)^2)^(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-320*B*ln(4/(2*cos(1/2*
d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2
*d*x+1/2*c)^6*a+80*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+80*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*
(a*sin(1/2*d*x+1/2*c)^2)^(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*(a*2^(1/
2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+128*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*c
os(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a+128*C*ln(-4*(a*2
^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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+624*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)
*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a+624*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)
*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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-156*A*ln(4
/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2
*a))*cos(1/2*d*x+1/2*c)^4*a-156*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(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-2*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)+2*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-
460*B*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)^8*a-1
40*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)^6+460*B*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+100*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+172*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)^8*a+876*A*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)^8*a+252*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)^6-87
6*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-172
*C*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-188*
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)^4-115*B*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-11*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+219*A*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+43*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+44*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)^6-36*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+19*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+3*C*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)/a^(7/2)/c
os(1/2*d*x+1/2*c)^3/(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*co
s(1/2*d*x+1/2*c)^2)^(1/2)/d
________________________________________________________________________________________
Maxima [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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A] time = 4.9397, size = 1218, normalized size = 4.35 \begin{align*} \frac{\sqrt{2}{\left ({\left (219 \, A - 115 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (219 \, A - 115 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (219 \, A - 115 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (219 \, A - 115 \, B + 43 \, 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 ) + 4 \,{\left ({\left (39 \, A - 20 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (39 \, A - 20 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (39 \, A - 20 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (39 \, A - 20 \, 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 (63 \, A - 35 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \,{\left (19 \, A - 11 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (5 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 8 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}
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))^(5/2),x, algorithm="fricas")
[Out]
1/64*(sqrt(2)*((219*A - 115*B + 43*C)*cos(d*x + c)^5 + 3*(219*A - 115*B + 43*C)*cos(d*x + c)^4 + 3*(219*A - 11
5*B + 43*C)*cos(d*x + c)^3 + (219*A - 115*B + 43*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)
) + 4*((39*A - 20*B + 8*C)*cos(d*x + c)^5 + 3*(39*A - 20*B + 8*C)*cos(d*x + c)^4 + 3*(39*A - 20*B + 8*C)*cos(d
*x + c)^3 + (39*A - 20*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*((63*A
- 35*B + 11*C)*cos(d*x + c)^3 + 5*(19*A - 11*B + 3*C)*cos(d*x + c)^2 + 4*(5*A - 4*B)*cos(d*x + c) - 8*A)*sqrt
(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a
^3*d*cos(d*x + c)^2)
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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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 6.92407, size = 876, normalized size = 3.13 \begin{align*} \text{result too large to display} \end{align*}
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))^(5/2),x, algorithm="giac")
[Out]
-1/64*(2*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/a^8 + sq
rt(2)*(29*A*a^5 - 21*B*a^5 + 13*C*a^5)/a^8)*tan(1/2*d*x + 1/2*c) - sqrt(2)*(219*A*sqrt(a) - 115*B*sqrt(a) + 43
*C*sqrt(a))*log((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/a^3 - 8*(39*A*sqrt(a) -
20*B*sqrt(a) + 8*C*sqrt(a))*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)))/a^3 + 8*(39*A*sqrt(a) - 20*B*sqrt(a) + 8*C*sqrt(a))*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)))/a^3 + 32*sqrt(2)*(41*(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) - 12*(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) - 209*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a^(3/
2) + 76*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(3/2) + 91*(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^(5/2) - 36*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan
(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(5/2) - 11*A*a^(7/2) + 4*B*a^(7/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(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)^2*a^2))/d