3.1243 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=234 \[ \frac{(3 A+4 B+17 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{42 a^4 d}+\frac{(15 A-B-83 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{210 a^4 d (\cos (c+d x)+1)^2}+\frac{(B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^4 d}-\frac{(B+8 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{10 a^4 d (\cos (c+d x)+1)}+\frac{(5 A+2 B-9 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{35 a d (a \cos (c+d x)+a)^3}-\frac{(A-B+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

((B + 8*C)*EllipticE[(c + d*x)/2, 2])/(10*a^4*d) + ((3*A + 4*B + 17*C)*EllipticF[(c + d*x)/2, 2])/(42*a^4*d) +
((15*A - B - 83*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(210*a^4*d*(1 + Cos[c + d*x])^2) - ((B + 8*C)*Sqrt[Cos[c
+ d*x]]*Sin[c + d*x])/(10*a^4*d*(1 + Cos[c + d*x])) - ((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(7*d*(a +
a*Cos[c + d*x])^4) + ((5*A + 2*B - 9*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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Rubi [A]  time = 0.763188, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4112, 3041, 2978, 2748, 2641, 2639} \[ \frac{(3 A+4 B+17 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{42 a^4 d}+\frac{(15 A-B-83 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{210 a^4 d (\cos (c+d x)+1)^2}+\frac{(B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^4 d}-\frac{(B+8 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{10 a^4 d (\cos (c+d x)+1)}+\frac{(5 A+2 B-9 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{35 a d (a \cos (c+d x)+a)^3}-\frac{(A-B+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^4),x]

[Out]

((B + 8*C)*EllipticE[(c + d*x)/2, 2])/(10*a^4*d) + ((3*A + 4*B + 17*C)*EllipticF[(c + d*x)/2, 2])/(42*a^4*d) +
((15*A - B - 83*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(210*a^4*d*(1 + Cos[c + d*x])^2) - ((B + 8*C)*Sqrt[Cos[c
+ d*x]]*Sin[c + d*x])/(10*a^4*d*(1 + Cos[c + d*x])) - ((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(7*d*(a +
a*Cos[c + d*x])^4) + ((5*A + 2*B - 9*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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 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{5}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx &=\int \frac{C+B \cos (c+d x)+A \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^4} \, dx\\ &=-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{-\frac{1}{2} a (A-B-13 C)+\frac{1}{2} a (9 A+5 B-5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(5 A+2 B-9 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{\frac{7}{2} a^2 (B+8 C)+\frac{3}{2} a^2 (5 A+2 B-9 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=\frac{(15 A-B-83 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(5 A+2 B-9 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{\frac{1}{4} a^3 (15 A+41 B+253 C)+\frac{1}{4} a^3 (15 A-B-83 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{105 a^6}\\ &=\frac{(15 A-B-83 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(5 A+2 B-9 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{(B+8 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 (3 A+4 B+17 C)+\frac{21}{4} a^4 (B+8 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{105 a^8}\\ &=\frac{(15 A-B-83 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(5 A+2 B-9 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{(B+8 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac{(B+8 C) \int \sqrt{\cos (c+d x)} \, dx}{20 a^4}+\frac{(3 A+4 B+17 C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{84 a^4}\\ &=\frac{(B+8 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac{(3 A+4 B+17 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{42 a^4 d}+\frac{(15 A-B-83 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{(5 A+2 B-9 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{(B+8 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 7.09044, size = 1862, normalized size = 7.96 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^4),x]

[Out]

(((2*I)/5)*B*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) + (((16*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) - (8*A*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - A
rcTan[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]]]*Sq
rt[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]]]])/(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*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[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) - (
136*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[Co
t[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*(B + 8*C)*Csc[c])/(5*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2
]^3*(15*A*Sin[(d*x)/2] - B*Sin[(d*x)/2] - 83*C*Sin[(d*x)/2]))/(105*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(5*A*
Sin[(d*x)/2] + 2*B*Sin[(d*x)/2] - 9*C*Sin[(d*x)/2]))/(35*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) - (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*(B*Sin[(d*x)/2] + 8*C*Sin[(d*x)/2
]))/(5*d) + (8*(15*A - B - 83*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(105*d) + (8*(5*A + 2*B - 9*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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Maple [B]  time = 3.019, size = 595, normalized size = 2.5 \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)^(5/2)/(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)*(60*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)^7-168*B*cos(1/2*d*x+1/2*c)^10+8
0*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)^7-84*B*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)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))-1344*C*cos(1/2*d*x+1/2*c)^10+340*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)^7-672*C*cos(1/2*d*x+1/2*c)^7*(si
n(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*A*cos(1/2
*d*x+1/2*c)^8+248*B*cos(1/2*d*x+1/2*c)^8+1684*C*cos(1/2*d*x+1/2*c)^8-30*A*cos(1/2*d*x+1/2*c)^6-54*B*cos(1/2*d*
x+1/2*c)^6-282*C*cos(1/2*d*x+1/2*c)^6-90*A*cos(1/2*d*x+1/2*c)^4-8*B*cos(1/2*d*x+1/2*c)^4-34*C*cos(1/2*d*x+1/2*
c)^4+75*A*cos(1/2*d*x+1/2*c)^2-33*B*cos(1/2*d*x+1/2*c)^2-9*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., 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)^(5/2)/(a+a*sec(d*x+c))^4,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^{4} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a^{4} \cos \left (d x + c\right )^{3}}, 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)^(5/2)/(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)^3*sec(d*x + c)^4 + 4*a^4
*cos(d*x + c)^3*sec(d*x + c)^3 + 6*a^4*cos(d*x + c)^3*sec(d*x + c)^2 + 4*a^4*cos(d*x + c)^3*sec(d*x + c) + a^4
*cos(d*x + c)^3), 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)**(5/2)/(a+a*sec(d*x+c))**4,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 )}^{4} \cos \left (d x + c\right )^{\frac{5}{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)^(5/2)/(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)^(5/2)), x)