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

Optimal. Leaf size=313 \[ \frac{(157 A-85 B+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(2671 A-1495 B+735 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{240 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(787 A-475 B+195 C) \sin (c+d x)}{240 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{(283 A-163 B+75 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-((283*A - 163*B + 75*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]
)/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sin[c + d*x])/(4*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)) - ((
21*A - 13*B + 5*C)*Sin[c + d*x])/(16*a*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((157*A - 85*B + 45*
C)*Sin[c + d*x])/(80*a^2*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((787*A - 475*B + 195*C)*Sin[c + d*x
])/(240*a^2*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + ((2671*A - 1495*B + 735*C)*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(240*a^2*d*Sqrt[a + a*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.990029, antiderivative size = 313, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4084, 4020, 4022, 4013, 3808, 206} \[ \frac{(157 A-85 B+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(2671 A-1495 B+735 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{240 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(787 A-475 B+195 C) \sin (c+d x)}{240 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{(283 A-163 B+75 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((283*A - 163*B + 75*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]
)/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sin[c + d*x])/(4*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)) - ((
21*A - 13*B + 5*C)*Sin[c + d*x])/(16*a*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((157*A - 85*B + 45*
C)*Sin[c + d*x])/(80*a^2*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((787*A - 475*B + 195*C)*Sin[c + d*x
])/(240*a^2*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + ((2671*A - 1495*B + 735*C)*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(240*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 3808

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/(a*f), Subst[Int[1/(2*b - d*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])], x
] /; FreeQ[{a, b, d, e, f}, 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])

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (13 A-5 B+5 C)-4 a (A-B) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 (157 A-85 B+45 C)-\frac{3}{2} a^2 (21 A-13 B+5 C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(157 A-85 B+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{-\frac{1}{8} a^3 (787 A-475 B+195 C)+\frac{1}{2} a^3 (157 A-85 B+45 C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{20 a^5}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(157 A-85 B+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(787 A-475 B+195 C) \sin (c+d x)}{240 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\frac{1}{16} a^4 (2671 A-1495 B+735 C)-\frac{1}{8} a^4 (787 A-475 B+195 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{30 a^6}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(157 A-85 B+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(787 A-475 B+195 C) \sin (c+d x)}{240 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(2671 A-1495 B+735 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(283 A-163 B+75 C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(157 A-85 B+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(787 A-475 B+195 C) \sin (c+d x)}{240 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(2671 A-1495 B+735 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(283 A-163 B+75 C) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=-\frac{(283 A-163 B+75 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac{(21 A-13 B+5 C) \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(157 A-85 B+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(787 A-475 B+195 C) \sin (c+d x)}{240 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(2671 A-1495 B+735 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.29578, size = 221, normalized size = 0.71 \[ -\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (15 (283 A-163 B+75 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\frac{1}{2} \sin \left (\frac{1}{2} (c+d x)\right ) (5 (887 A-479 B+255 C) \cos (c+d x)+16 (52 A-25 B+15 C) \cos (2 (c+d x))-40 A \cos (3 (c+d x))+12 A \cos (4 (c+d x))+3491 A+40 B \cos (3 (c+d x))-1895 B+975 C)\right )}{60 a d \sqrt{\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[(c + d*x)/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(15*(283*A - 163*B + 75*C)*ArcTanh[Sin[(c + d*x)/2]
]*Cos[(c + d*x)/2]^4 - ((3491*A - 1895*B + 975*C + 5*(887*A - 479*B + 255*C)*Cos[c + d*x] + 16*(52*A - 25*B +
15*C)*Cos[2*(c + d*x)] - 40*A*Cos[3*(c + d*x)] + 40*B*Cos[3*(c + d*x)] + 12*A*Cos[4*(c + d*x)])*Sin[(c + d*x)/
2])/2))/(60*a*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sqrt[Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3
/2))

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Maple [B]  time = 0.423, size = 657, normalized size = 2.1 \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)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/480/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(4245*A*sin(d*x+c)*cos(d*x+c)^2*arctan(1/2*s
in(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-192*A*cos(d*x+c)^5-2445*B*sin(d*x+c)*cos(d*x+c)
^2*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)+1125*C*arctan(1/2*sin(d*x+c)*(-2
/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*cos(d*x+c)^2+8490*A*sin(d*x+c)*cos(d*x+c)*arctan(
1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)+512*A*cos(d*x+c)^4-4890*B*sin(d*x+c)*cos(d
*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-320*B*cos(d*x+c)^4+2250*C*sin
(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+4245*arctan(1/2*
sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*A*sin(d*x+c)-3456*A*cos(d*x+c)^3-2445*arctan(1
/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*B*sin(d*x+c)+1920*B*cos(d*x+c)^3+1125*C*(-2
/(cos(d*x+c)+1))^(1/2)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-960*C*cos(d*x+c)^3-5974*A*c
os(d*x+c)^2+3430*B*cos(d*x+c)^2-1590*C*cos(d*x+c)^2+3768*A*cos(d*x+c)-2040*B*cos(d*x+c)+1080*C*cos(d*x+c)+5342
*A-2990*B+1470*C)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)^5

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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)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.586198, size = 1747, normalized size = 5.58 \begin{align*} \text{result too large to display} \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)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/960*(15*sqrt(2)*((283*A - 163*B + 75*C)*cos(d*x + c)^3 + 3*(283*A - 163*B + 75*C)*cos(d*x + c)^2 + 3*(283*A
- 163*B + 75*C)*cos(d*x + c) + 283*A - 163*B + 75*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 +
2*cos(d*x + c) + 1)) + 4*(96*A*cos(d*x + c)^5 - 160*(A - B)*cos(d*x + c)^4 + 32*(49*A - 25*B + 15*C)*cos(d*x
+ c)^3 + 5*(911*A - 503*B + 255*C)*cos(d*x + c)^2 + (2671*A - 1495*B + 735*C)*cos(d*x + c))*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*
d*cos(d*x + c) + a^3*d), 1/480*(15*sqrt(2)*((283*A - 163*B + 75*C)*cos(d*x + c)^3 + 3*(283*A - 163*B + 75*C)*c
os(d*x + c)^2 + 3*(283*A - 163*B + 75*C)*cos(d*x + c) + 283*A - 163*B + 75*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*(96*A*cos(d*x + c)^5 - 160*(
A - B)*cos(d*x + c)^4 + 32*(49*A - 25*B + 15*C)*cos(d*x + c)^3 + 5*(911*A - 503*B + 255*C)*cos(d*x + c)^2 + (2
671*A - 1495*B + 735*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))
/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]

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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)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(5/2),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 )}^{\frac{5}{2}} \sec \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)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^(5/2)), x)