3.1242 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=229 \[ \frac {(4 A+3 B+4 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}+\frac {(41 A+15 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{210 a^4 d (\cos (c+d x)+1)^2}-\frac {(A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{10 a^4 d (\cos (c+d x)+1)}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {(A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 a d (a \cos (c+d x)+a)^3} \]

[Out]

-1/10*(A-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d+1/42*(
4*A+3*B+4*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d-1/7*(
A-B+C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+1/210*(41*A+15*B-C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^4/d/
(1+cos(d*x+c))^2+1/10*(A-C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^4/d/(1+cos(d*x+c))-1/5*(A-C)*sin(d*x+c)*cos(d*x+c)^(
1/2)/a/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.77, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4112, 3041, 2977, 2978, 2748, 2641, 2639} \[ \frac {(4 A+3 B+4 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}+\frac {(41 A+15 B-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{210 a^4 d (\cos (c+d x)+1)^2}-\frac {(A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{10 a^4 d (\cos (c+d x)+1)}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {(A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^4),x]

[Out]

-((A - C)*EllipticE[(c + d*x)/2, 2])/(10*a^4*d) + ((4*A + 3*B + 4*C)*EllipticF[(c + d*x)/2, 2])/(42*a^4*d) + (
(41*A + 15*B - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(210*a^4*d*(1 + Cos[c + d*x])^2) + ((A - C)*Sqrt[Cos[c + d*
x]]*Sin[c + d*x])/(10*a^4*d*(1 + Cos[c + d*x])) - ((A - B + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(7*d*(a + a*Co
s[c + d*x])^4) - ((A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*a*d*(a + a*Cos[c + d*x])^3)

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]

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 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 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

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 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 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]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx &=\int \frac {\sqrt {\cos (c+d x)} \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx\\ &=-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\frac {1}{2} a (3 A-3 B-11 C)+\frac {1}{2} a (11 A+3 B-3 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {-\frac {7}{2} a^2 (A-C)+\frac {1}{2} a^2 (34 A+15 B+6 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(41 A+15 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {-\frac {1}{4} a^3 (A-15 B-41 C)+\frac {1}{4} a^3 (41 A+15 B-C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{105 a^6}\\ &=\frac {(41 A+15 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \frac {\frac {5}{4} a^4 (4 A+3 B+4 C)-\frac {21}{4} a^4 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{105 a^8}\\ &=\frac {(41 A+15 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(A-C) \int \sqrt {\cos (c+d x)} \, dx}{20 a^4}+\frac {(4 A+3 B+4 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{84 a^4}\\ &=-\frac {(A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(4 A+3 B+4 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}+\frac {(41 A+15 B-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 7.10, size = 1862, normalized size = 8.13 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^4),x]

[Out]

(((-2*I)/5)*A*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(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])^4) + (((2*I)/5)*C*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(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*Hypergeo
metric2F1[-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])^4) - (32*A*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x -
ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*S
qrt[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]]]])/(21*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])^4) - (8*B*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]
]^2]*Sec[c/2]*Sec[c + d*x]^2*(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[Co
t[c]]]])/(7*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])^4) - (
32*C*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*
Sec[c + d*x]^2*(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]]]])/(21*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])^4) + (Cos[c/2 + (d*
x)/2]^8*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((16*(A - C)*Csc[c])/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*
(12*A*Sin[(d*x)/2] - 5*B*Sin[(d*x)/2] - 2*C*Sin[(d*x)/2]))/(35*d) + (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*
x)/2] - C*Sin[(d*x)/2]))/(5*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(41*A*Sin[(d*x)/2] + 15*B*Sin[(d*x)/2] - C*S
in[(d*x)/2]))/(105*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^7*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(
7*d) + (8*(41*A + 15*B - C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(105*d) - (8*(12*A - 5*B - 2*C)*Sec[c/2 + (d*x)/2]^
4*Tan[c/2])/(35*d) + (4*(A - B + C)*Sec[c/2 + (d*x)/2]^6*Tan[c/2])/(7*d)))/(Cos[c + d*x]^(3/2)*(A + 2*C + 2*B*
Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4)

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{a^{4} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a^{4} \cos \left (d x + c\right )^{2}}, x\right ) \]

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)^(3/2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(a^4*cos(d*x + c)^2*sec(d*x + c)^4 + 4*a^4
*cos(d*x + c)^2*sec(d*x + c)^3 + 6*a^4*cos(d*x + c)^2*sec(d*x + c)^2 + 4*a^4*cos(d*x + c)^2*sec(d*x + c) + a^4
*cos(d*x + c)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

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)^(3/2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^4*cos(d*x + c)^(3/2)), x)

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maple [B]  time = 5.85, size = 595, normalized size = 2.60 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (168 A \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+84 A \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-168 C \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-84 C \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-88 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60 B \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+248 C \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-306 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-30 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-54 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+328 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-90 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-117 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-33 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 A -15 B +15 C \right )}{840 a^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

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)^(3/2)/(a+a*sec(d*x+c))^4,x)

[Out]

-1/840*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(168*A*cos(1/2*d*x+1/2*c)^10+80*A*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^7+8
4*A*cos(1/2*d*x+1/2*c)^7*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))+60*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))*cos(1/2*d*x+1/2*c)^7-168*C*cos(1/2*d*x+1/2*c)^10+80*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^7-84*C*cos(1/2*d*x+1/2*c)^7*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-88*A*cos(1/2*d*
x+1/2*c)^8+60*B*cos(1/2*d*x+1/2*c)^8+248*C*cos(1/2*d*x+1/2*c)^8-306*A*cos(1/2*d*x+1/2*c)^6-30*B*cos(1/2*d*x+1/
2*c)^6-54*C*cos(1/2*d*x+1/2*c)^6+328*A*cos(1/2*d*x+1/2*c)^4-90*B*cos(1/2*d*x+1/2*c)^4-8*C*cos(1/2*d*x+1/2*c)^4
-117*A*cos(1/2*d*x+1/2*c)^2+75*B*cos(1/2*d*x+1/2*c)^2-33*C*cos(1/2*d*x+1/2*c)^2+15*A-15*B+15*C)/a^4/cos(1/2*d*
x+1/2*c)^7/(-2*sin(1/2*d*x+1/2*c)^4+sin(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)^2-1)^
(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*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4} \,d x \]

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)^(3/2)*(a + a/cos(c + d*x))^4),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^4), x)

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sympy [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*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+a*sec(d*x+c))**4,x)

[Out]

Timed out

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