3.1126 \(\int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\)
Optimal. Leaf size=242 \[ \frac{(A+11 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{2 a^3 d}+\frac{(9 A+119 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(A+11 C) \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}-\frac{2 C \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]
[Out]
((9*A + 119*C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((A + 11*C)*EllipticF[(c + d*x)/2, 2])/(2*a^3*d) + ((A
+ 11*C)*Sin[c + d*x])/(2*a^3*d*Cos[c + d*x]^(3/2)) - ((9*A + 119*C)*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]
) - ((A + C)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3) - (2*C*Sin[c + d*x])/(3*a*d*Cos[c +
d*x]^(3/2)*(a + a*Cos[c + d*x])^2) - ((9*A + 119*C)*Sin[c + d*x])/(30*d*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c +
d*x]))
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Rubi [A] time = 0.566726, antiderivative size = 242, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{4114, 3042, 2978, 2748, 2636, 2641, 2639} \[ \frac{(A+11 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{(9 A+119 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(A+11 C) \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}-\frac{2 C \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]
[Out]
((9*A + 119*C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((A + 11*C)*EllipticF[(c + d*x)/2, 2])/(2*a^3*d) + ((A
+ 11*C)*Sin[c + d*x])/(2*a^3*d*Cos[c + d*x]^(3/2)) - ((9*A + 119*C)*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]
) - ((A + C)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3) - (2*C*Sin[c + d*x])/(3*a*d*Cos[c +
d*x]^(3/2)*(a + a*Cos[c + d*x])^2) - ((9*A + 119*C)*Sin[c + d*x])/(30*d*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c +
d*x]))
Rule 4114
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sec[(e_.)
+ (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Rule 3042
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2636
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=\int \frac{C+A \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac{\int \frac{\frac{1}{2} a (3 A+13 C)+\frac{1}{2} a (3 A-7 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 C \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a^2 (3 A+23 C)-25 a^2 C \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 C \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \frac{\frac{45}{4} a^3 (A+11 C)-\frac{3}{4} a^3 (9 A+119 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 C \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(3 (A+11 C)) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{4 a^3}-\frac{(9 A+119 C) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{20 a^3}\\ &=\frac{(A+11 C) \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 C \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(A+11 C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{4 a^3}+\frac{(9 A+119 C) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{(9 A+119 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A+11 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{(A+11 C) \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 C \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{(9 A+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 7.5439, size = 1505, normalized size = 6.22 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]
[Out]
(((9*I)/5)*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hype
rgeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I
)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*
d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2
*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*
d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1
+ E^((2*I)*d*x))*Sin[c])))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (((119*I)/5)*C*Cos[c/2 +
(d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4,
7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*
Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*C
os[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Si
n[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)
*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c]
)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) - (4*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hypergeometri
cPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*Sec[d*x - Ar
cTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]
*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x]
)^3) - (44*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*S
ec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt
[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*
Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*((
-4*(60*C + 9*A*Cos[c] + 59*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin
[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) - (16*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(3*A*Sin[(d*x)/2] + 8*C*Sin[(d*x)/2]))/
(15*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(9*A*Sin[(d*x)/2] + 59*C*Sin[(d*x)/2]))/(5*d) + (32*C*Sec[c]*Sec[c + d
*x]^2*Sin[d*x])/(3*d) + (32*Sec[c]*Sec[c + d*x]*(C*Sin[c] - 9*C*Sin[d*x]))/(3*d) - (16*(3*A + 8*C)*Sec[c/2 + (
d*x)/2]^2*Tan[c/2])/(15*d) - (4*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(Sqrt[Cos[c + d*x]]*(A + 2*C +
A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)
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Maple [B] time = 3.187, size = 876, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x)
[Out]
1/60*(12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+55*C*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-119*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^6-30*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+55*C*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-119*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x
+1/2*c)+24*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+55*C*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-119*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)-6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+55*C*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-119*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-24*(-2*sin(1
/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(9*A+119*C)*sin(1/2*d*x+1/2*c)^10+24*(-2*sin(1/2*d*x+1/2*c)^4+sin(
1/2*d*x+1/2*c)^2)^(1/2)*(29*A+389*C)*sin(1/2*d*x+1/2*c)^8-10*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1
/2)*(81*A+1111*C)*sin(1/2*d*x+1/2*c)^6+4*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(99*A+1414*C)*si
n(1/2*d*x+1/2*c)^4-3*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(23*A+343*C)*sin(1/2*d*x+1/2*c)^2)/a
^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/
sin(1/2*d*x+1/2*c)/d
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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+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{\cos \left (d x + c\right )}}{a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
integral((C*sec(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a^3*cos(d*x + c)^4*sec(d*x + c)^3 + 3*a^3*cos(d*x + c)^4*s
ec(d*x + c)^2 + 3*a^3*cos(d*x + c)^4*sec(d*x + c) + a^3*cos(d*x + c)^4), x)
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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+C*sec(d*x+c)**2)/cos(d*x+c)**(7/2)/(a+a*sec(d*x+c))**3,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)^3*cos(d*x + c)^(7/2)), x)