3.1238 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=268 \[ \frac{(3 A-13 B+33 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}+\frac{(9 A-49 B+119 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(9 A-49 B+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(3 A-13 B+33 C) \sin (c+d x)}{6 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A-49 B+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}+\frac{(B-2 C) \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[Out]

((9*A - 49*B + 119*C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((3*A - 13*B + 33*C)*EllipticF[(c + d*x)/2, 2])/
(6*a^3*d) + ((3*A - 13*B + 33*C)*Sin[c + d*x])/(6*a^3*d*Cos[c + d*x]^(3/2)) - ((9*A - 49*B + 119*C)*Sin[c + d*
x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) - ((A - B + C)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3)
+ ((B - 2*C)*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) - ((9*A - 49*B + 119*C)*Sin[c +
d*x])/(30*d*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.664676, antiderivative size = 268, 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 = {4112, 3041, 2978, 2748, 2636, 2641, 2639} \[ \frac{(3 A-13 B+33 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(9 A-49 B+119 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(9 A-49 B+119 C) \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(3 A-13 B+33 C) \sin (c+d x)}{6 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A-49 B+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}+\frac{(B-2 C) \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

((9*A - 49*B + 119*C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((3*A - 13*B + 33*C)*EllipticF[(c + d*x)/2, 2])/
(6*a^3*d) + ((3*A - 13*B + 33*C)*Sin[c + d*x])/(6*a^3*d*Cos[c + d*x]^(3/2)) - ((9*A - 49*B + 119*C)*Sin[c + d*
x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) - ((A - B + C)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3)
+ ((B - 2*C)*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) - ((9*A - 49*B + 119*C)*Sin[c +
d*x])/(30*d*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c + d*x]))

Rule 4112

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (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 + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

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 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+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=\int \frac{C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\\ &=-\frac{(A-B+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-3 B+13 C)+\frac{1}{2} a (3 A+7 B-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-B+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac{(B-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-8 B+23 C)+\frac{25}{2} a^2 (B-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-B+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac{(B-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-49 B+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{15}{4} a^3 (3 A-13 B+33 C)-\frac{3}{4} a^3 (9 A-49 B+119 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac{(B-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-49 B+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-13 B+33 C) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{4 a^3}-\frac{(9 A-49 B+119 C) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{20 a^3}\\ &=\frac{(3 A-13 B+33 C) \sin (c+d x)}{6 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A-49 B+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac{(B-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-49 B+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-13 B+33 C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}+\frac{(9 A-49 B+119 C) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{(9 A-49 B+119 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(3 A-13 B+33 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(3 A-13 B+33 C) \sin (c+d x)}{6 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(9 A-49 B+119 C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac{(B-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-49 B+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 = 8.02092, size = 2248, normalized size = 8.39 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + 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 + B*Sec[c + d*x] + 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)*S
in[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 + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c +
d*x])^3) - (((49*I)/5)*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + 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))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometr
ic2F1[-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 + 2*B*Cos[c + d*x] + 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 + B*Sec[c + d*x]
+ 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))*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 + 2*B*Cos[c + d*x] + 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]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Si
n[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + 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 + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Se
c[c + d*x])^3) + (52*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot
[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + 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]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + 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]*Sec[c/2
]*Sec[c + d*x]*(A + B*Sec[c + d*x] + 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 + 2*B*Cos[c + d*x] + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(-20*B + 60*C + 9*A*Cos[c] - 29*B*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] - B*Sin[(d*x)/2] + C*Sin[(d*x)/
2]))/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(6*A*Sin[(d*x)/2] - 11*B*Sin[(d*x)/2] + 16*C*Sin[(d*x)/2]))/(15*
d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(9*A*Sin[(d*x)/2] - 29*B*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] + 3*B*Sin[d*x] - 9*C*Sin[d*x]))/(3*d
) - (8*(6*A - 11*B + 16*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (4*(A - B + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2
])/(5*d)))/(Sqrt[Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)

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Maple [B]  time = 10.748, size = 1040, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^3*(1/3*(-2*B+4*C)*(2*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*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)^2-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*si
n(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(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)^2-1)+8*C*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)
^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(-4*B+
12*C)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x
+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d
*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(A-B+C)*(1/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d
*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^5+4/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1
/2*c)^3+18/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)-8/5*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),2^(1/2))+18/5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
))+(8*B-24*C)*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(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)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*
c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/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+B*sec(d*x+c)+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} + B \sec \left (d x + c\right ) + 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+B*sec(d*x+c)+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 + B*sec(d*x + c) + 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*sec(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+B*sec(d*x+c)+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} + B \sec \left (d x + c\right ) + 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+B*sec(d*x+c)+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 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^3*cos(d*x + c)^(7/2)), x)