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

Optimal. Leaf size=229 \[ -\frac{(7 A-11 B+15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(15 A-35 B+39 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{30 a^2 d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}+\frac{(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((7*A - 11*B + 15*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d)
- ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((45*A - 65*B + 93*C)*Tan[c +
d*x])/(15*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((5*A - 5*B + 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*S
ec[c + d*x]]) - ((15*A - 35*B + 39*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(30*a^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.682523, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4084, 4021, 4010, 4001, 3795, 203} \[ -\frac{(7 A-11 B+15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(15 A-35 B+39 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{30 a^2 d}-\frac{(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}+\frac{(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*A - 11*B + 15*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d)
- ((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((45*A - 65*B + 93*C)*Tan[c +
d*x])/(15*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((5*A - 5*B + 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*S
ec[c + d*x]]) - ((15*A - 35*B + 39*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(30*a^2*d)

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 4021

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

Rule 4010

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

Rule 4001

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

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec ^3(c+d x) \left (-a (A-3 B+3 C)+\frac{1}{2} a (5 A-5 B+9 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\sec ^2(c+d x) \left (a^2 (5 A-5 B+9 C)-\frac{1}{4} a^2 (15 A-35 B+39 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac{2 \int \frac{\sec (c+d x) \left (-\frac{1}{8} a^3 (15 A-35 B+39 C)+\frac{1}{4} a^3 (45 A-65 B+93 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}-\frac{(7 A-11 B+15 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac{(7 A-11 B+15 C) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{2 a d}\\ &=-\frac{(7 A-11 B+15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}\\ \end{align*}

Mathematica [C]  time = 8.84682, size = 2025, normalized size = 8.84 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - 2
*Sin[(c + d*x)/2]^2]*((4*A*Sin[(c + d*x)/2])/(5*(1 - 2*Sin[(c + d*x)/2]^2)^(5/2)) - ((A - B + C)*(1 - 2*Sin[(c
+ d*x)/2]))/(20*(1 + Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x)/2]^2)^(5/2)) + ((A - B + C)*(1 + 2*Sin[(c + d*x)/
2]))/(20*(1 - Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x)/2]^2)^(5/2)) + (16*A*(Sin[(c + d*x)/2]/(1 - 2*Sin[(c + d*
x)/2]^2)^(3/2) + (2*Sin[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2]))/15 - ((A - B + C)*(-105*ArcTan[(1 - 2*S
in[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2]] + (4 + 3*Sin[(c + d*x)/2])/((1 - Sin[(c + d*x)/2])*(1 - 2*Sin
[(c + d*x)/2]^2)^(3/2)) - (19 + 29*Sin[(c + d*x)/2])/((1 - Sin[(c + d*x)/2])*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]) -
(67*Sqrt[1 - 2*Sin[(c + d*x)/2]^2])/(1 - Sin[(c + d*x)/2])))/30 + ((A - B + C)*(-105*ArcTan[(1 + 2*Sin[(c + d
*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2]] + (4 - 3*Sin[(c + d*x)/2])/((1 + Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x
)/2]^2)^(3/2)) - (19 - 29*Sin[(c + d*x)/2])/((1 + Sin[(c + d*x)/2])*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]) - (67*Sqrt
[1 - 2*Sin[(c + d*x)/2]^2])/(1 + Sin[(c + d*x)/2])))/30 + ((7*A - 3*B - C)*Csc[(c + d*x)/2]^7*(4725*Sin[(c + d
*x)/2]^2 - 48825*Sin[(c + d*x)/2]^4 + 210105*Sin[(c + d*x)/2]^6 - 486630*Sin[(c + d*x)/2]^8 + 655812*Sin[(c +
d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 11/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)
/2]^10 - 40*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[
(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^10 - 518760*Sin[(c + d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2, 11/2, Sin[(
c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^12 + 226656*Sin[(c + d*x)/2]^14 - 1500*Hypergeomet
ric2F1[2, 9/2, 11/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^14 - 42048*Sin[(c + d*x)
/2]^16 + 440*Hypergeometric2F1[2, 9/2, 11/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^
16 + 4725*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c
+ d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^2*Sqrt[S
in[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 291060*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/
2]^2)]]*Sin[(c + d*x)/2]^4*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 833760*ArcTanh[Sqrt[Sin[(c +
d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^6*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2
)] + 1458000*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^8*Sqrt[Sin[(c + d*
x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 1598400*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*S
in[(c + d*x)/2]^10*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 1080000*ArcTanh[Sqrt[Sin[(c + d*x)/2
]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^12*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 4
14720*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^14*Sqrt[Sin[(c + d*x)/2]^
2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 69120*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c +
d*x)/2]^16*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 60*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2,
2, 9/2}, {1, 11/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^10*(-5 + 4*Sin[(c + d*x)/
2]^2)))/(1350*(1 - 2*Sin[(c + d*x)/2]^2)^(7/2)*(-1 + 2*Sin[(c + d*x)/2]^2))))/(d*(A + 2*C + 2*B*Cos[c + d*x] +
A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.376, size = 1152, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240/d/a^2*(-1+cos(d*x+c))*(105*A*cos(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+
c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)-165*B*cos(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)+225*C*c
os(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(5/2)+315*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*
x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*sin(d*x+c)-495*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*sin(d*x+c)+675*C*
ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(5/2)*cos(d*x+c)^2*sin(d*x+c)+315*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d
*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*sin(d*x+c)-495*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*sin(d*x+c)+675*C*ln(
-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5
/2)*cos(d*x+c)*sin(d*x+c)+105*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c)
)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)-165*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+
cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+225*C*ln(-(-(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+600*A*cos(d*
x+c)^4-760*B*cos(d*x+c)^4+1176*C*cos(d*x+c)^4-120*A*cos(d*x+c)^3+280*B*cos(d*x+c)^3-312*C*cos(d*x+c)^3-480*A*c
os(d*x+c)^2+640*B*cos(d*x+c)^2-960*C*cos(d*x+c)^2-160*B*cos(d*x+c)+192*C*cos(d*x+c)-96*C)*(a*(cos(d*x+c)+1)/co
s(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)^3

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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(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.652874, size = 1399, normalized size = 6.11 \begin{align*} \left [-\frac{15 \, \sqrt{2}{\left ({\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{-a} \log \left (-\frac{2 \, \sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left ({\left (75 \, A - 95 \, B + 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (5 \, A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 12 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{120 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac{15 \, \sqrt{2}{\left ({\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) + 2 \,{\left ({\left (75 \, A - 95 \, B + 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (5 \, A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 12 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{60 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/120*(15*sqrt(2)*((7*A - 11*B + 15*C)*cos(d*x + c)^4 + 2*(7*A - 11*B + 15*C)*cos(d*x + c)^3 + (7*A - 11*B +
15*C)*cos(d*x + c)^2)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*
sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((75*A -
95*B + 147*C)*cos(d*x + c)^3 + 12*(5*A - 5*B + 9*C)*cos(d*x + c)^2 + 4*(5*B - 3*C)*cos(d*x + c) + 12*C)*sqrt((
a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x
+ c)^2), 1/60*(15*sqrt(2)*((7*A - 11*B + 15*C)*cos(d*x + c)^4 + 2*(7*A - 11*B + 15*C)*cos(d*x + c)^3 + (7*A -
11*B + 15*C)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqr
t(a)*sin(d*x + c))) + 2*((75*A - 95*B + 147*C)*cos(d*x + c)^3 + 12*(5*A - 5*B + 9*C)*cos(d*x + c)^2 + 4*(5*B -
3*C)*cos(d*x + c) + 12*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2
*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [A]  time = 9.61757, size = 456, normalized size = 1.99 \begin{align*} -\frac{\frac{15 \, \sqrt{2}{\left (7 \, A - 11 \, B + 15 \, C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{{\left ({\left ({\left (\frac{15 \, \sqrt{2}{\left (A a^{3} - B a^{3} + C a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (165 \, A a^{3} - 245 \, B a^{3} + 381 \, C a^{3}\right )}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{5 \, \sqrt{2}{\left (57 \, A a^{3} - 73 \, B a^{3} + 105 \, C a^{3}\right )}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{15 \, \sqrt{2}{\left (9 \, A a^{3} - 9 \, B a^{3} + 17 \, C a^{3}\right )}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/60*(15*sqrt(2)*(7*A - 11*B + 15*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a)))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - (((15*sqrt(2)*(A*a^3 - B*a^3 + C*a^3)*tan(1/2*d*x + 1/2*
c)^2/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(165*A*a^3 - 245*B*a^3 + 381*C*a^3)/(a^2*sgn(tan(1/2*d*x
+ 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 + 5*sqrt(2)*(57*A*a^3 - 73*B*a^3 + 105*C*a^3)/(a^2*sgn(tan(1/2*d*x +
1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 - 15*sqrt(2)*(9*A*a^3 - 9*B*a^3 + 17*C*a^3)/(a^2*sgn(tan(1/2*d*x + 1/2*
c)^2 - 1)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d