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

Optimal. Leaf size=259 \[ \frac{(11 A+19 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(245 A+397 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{210 a^2 d}-\frac{(A+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(7 A+11 C) \tan (c+d x) \sec ^3(c+d x)}{14 a d \sqrt{a \sec (c+d x)+a}}-\frac{(35 A+67 C) \tan (c+d x) \sec ^2(c+d x)}{70 a d \sqrt{a \sec (c+d x)+a}}-\frac{(455 A+799 C) \tan (c+d x)}{105 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

((11*A + 19*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
+ C)*Sec[c + d*x]^4*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((455*A + 799*C)*Tan[c + d*x])/(105*a*d*S
qrt[a + a*Sec[c + d*x]]) - ((35*A + 67*C)*Sec[c + d*x]^2*Tan[c + d*x])/(70*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((7
*A + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(14*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((245*A + 397*C)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(210*a^2*d)

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Rubi [A]  time = 0.845068, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4085, 4021, 4010, 4001, 3795, 203} \[ \frac{(11 A+19 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(245 A+397 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{210 a^2 d}-\frac{(A+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(7 A+11 C) \tan (c+d x) \sec ^3(c+d x)}{14 a d \sqrt{a \sec (c+d x)+a}}-\frac{(35 A+67 C) \tan (c+d x) \sec ^2(c+d x)}{70 a d \sqrt{a \sec (c+d x)+a}}-\frac{(455 A+799 C) \tan (c+d x)}{105 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*A + 19*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
+ C)*Sec[c + d*x]^4*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((455*A + 799*C)*Tan[c + d*x])/(105*a*d*S
qrt[a + a*Sec[c + d*x]]) - ((35*A + 67*C)*Sec[c + d*x]^2*Tan[c + d*x])/(70*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((7
*A + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(14*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((245*A + 397*C)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(210*a^2*d)

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[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[b*C*n + A*b*(
2*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, 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 ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\sec ^4(c+d x) \left (2 a (A+2 C)-\frac{1}{2} a (7 A+11 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(7 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt{a+a \sec (c+d x)}}-\frac{\int \frac{\sec ^3(c+d x) \left (-\frac{3}{2} a^2 (7 A+11 C)+\frac{1}{4} a^2 (35 A+67 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{7 a^3}\\ &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(35 A+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt{a+a \sec (c+d x)}}+\frac{(7 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt{a+a \sec (c+d x)}}-\frac{2 \int \frac{\sec ^2(c+d x) \left (\frac{1}{2} a^3 (35 A+67 C)-\frac{1}{8} a^3 (245 A+397 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{35 a^4}\\ &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(35 A+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt{a+a \sec (c+d x)}}+\frac{(7 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt{a+a \sec (c+d x)}}+\frac{(245 A+397 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}-\frac{4 \int \frac{\sec (c+d x) \left (-\frac{1}{16} a^4 (245 A+397 C)+\frac{1}{8} a^4 (455 A+799 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{105 a^5}\\ &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(455 A+799 C) \tan (c+d x)}{105 a d \sqrt{a+a \sec (c+d x)}}-\frac{(35 A+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt{a+a \sec (c+d x)}}+\frac{(7 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt{a+a \sec (c+d x)}}+\frac{(245 A+397 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}+\frac{(11 A+19 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(455 A+799 C) \tan (c+d x)}{105 a d \sqrt{a+a \sec (c+d x)}}-\frac{(35 A+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt{a+a \sec (c+d x)}}+\frac{(7 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt{a+a \sec (c+d x)}}+\frac{(245 A+397 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}-\frac{(11 A+19 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{(11 A+19 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+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(455 A+799 C) \tan (c+d x)}{105 a d \sqrt{a+a \sec (c+d x)}}-\frac{(35 A+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt{a+a \sec (c+d x)}}+\frac{(7 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt{a+a \sec (c+d x)}}+\frac{(245 A+397 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}\\ \end{align*}

Mathematica [B]  time = 6.80202, size = 527, normalized size = 2.03 \[ \frac{\cos ^2(c+d x) (\sec (c+d x)+1)^{3/2} \sqrt{(\cos (c+d x)+1) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (\frac{\sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+C \sin \left (\frac{d x}{2}\right )\right )}{2 d}+\frac{\sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{c}{2}\right )+C \sin \left (\frac{c}{2}\right )\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (-805 A \sin \left (\frac{d x}{2}\right )-1649 C \sin \left (\frac{d x}{2}\right )\right )}{105 d}-\frac{4 \sec (c) \sec (c+d x) (-35 A \sin (d x)+39 C \sin (c)-112 C \sin (d x))}{105 d}-\frac{2 \sin \left (\frac{c}{2}\right ) (665 A \cos (c)-140 A+1201 C \cos (c)-448 C)}{105 d \left (\cos \left (\frac{c}{2}\right )+\cos \left (\frac{3 c}{2}\right )\right )}+\frac{4 C \sec (c) \sin (d x) \sec ^3(c+d x)}{7 d}+\frac{4 \sec (c) \sec ^2(c+d x) (5 C \sin (c)-13 C \sin (d x))}{35 d}\right )}{(a (\sec (c+d x)+1))^{3/2} (A \cos (2 c+2 d x)+A+2 C)}+\frac{(11 A+19 C) \sin (c+d x) \cos ^4(c+d x) \sqrt{\sec (c+d x)-1} (\sec (c+d x)+1)^3 \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right ) \left (A+C \sec ^2(c+d x)\right )}{\sqrt{2} d (\cos (c+d x)+1) \sqrt{1-\cos ^2(c+d x)} (a (\sec (c+d x)+1))^{3/2} \sqrt{\cos ^2(c+d x) (\sec (c+d x)-1) (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^2*Sqrt[(1 + Cos[c + d*x])*Sec[c + d*x]]*(1 + Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*((-2*(-1
40*A - 448*C + 665*A*Cos[c] + 1201*C*Cos[c])*Sin[c/2])/(105*d*(Cos[c/2] + Cos[(3*c)/2])) + (Sec[c/2]*Sec[c/2 +
(d*x)/2]^2*(A*Sin[c/2] + C*Sin[c/2]))/(2*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]*(-805*A*Sin[(d*x)/2] - 1649*C*Sin[
(d*x)/2]))/(105*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(2*d) + (4*C*Sec[c]*Sec
[c + d*x]^3*Sin[d*x])/(7*d) - (4*Sec[c]*Sec[c + d*x]*(39*C*Sin[c] - 35*A*Sin[d*x] - 112*C*Sin[d*x]))/(105*d) +
(4*Sec[c]*Sec[c + d*x]^2*(5*C*Sin[c] - 13*C*Sin[d*x]))/(35*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c
+ d*x]))^(3/2)) + ((11*A + 19*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[c + d*x]^4*Sqrt[-1 + Sec[c + d*x
]]*(1 + Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*Sin[c + d*x])/(Sqrt[2]*d*(1 + Cos[c + d*x])*Sqrt[1 - Cos[c + d*
x]^2]*(A + 2*C + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(3/2)*Sqrt[Cos[c + d*x]^2*(-1 + Sec[c + d*x])*(1 +
Sec[c + d*x])])

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Maple [B]  time = 0.438, size = 974, normalized size = 3.8 \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)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/3360/d/a^2*(-1+cos(d*x+c))*(1155*A*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))^(7/2)*cos(d*x+c)^4+1995*C*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(co
s(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))^(7/2)*cos(d*x+c)^4+4620
*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/si
n(d*x+c))*sin(d*x+c)*cos(d*x+c)^3+7980*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)^3+6930*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(7/2)*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))*sin(d*x+c)*cos(d*x+c)^2+
11970*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)^2+4620*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)+7980*C*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)
+1155*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)+1995*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)-10640*A*cos(d*x+c)^5-19216*C*cos(d*x+c)^5+3920*A*cos(d*x+c)
^4+6352*C*cos(d*x+c)^4+8960*A*cos(d*x+c)^3+16000*C*cos(d*x+c)^3-2240*A*cos(d*x+c)^2-3712*C*cos(d*x+c)^2+1536*C
*cos(d*x+c)-960*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(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)^4*(A+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 = 0.649925, size = 1392, normalized size = 5.37 \begin{align*} \left [-\frac{105 \, \sqrt{2}{\left ({\left (11 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (11 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (11 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3}\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 (665 \, A + 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \,{\left (35 \, A + 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \,{\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 36 \, C \cos \left (d x + c\right ) - 60 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{840 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}, -\frac{105 \, \sqrt{2}{\left ({\left (11 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (11 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (11 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3}\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 (665 \, A + 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \,{\left (35 \, A + 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \,{\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 36 \, C \cos \left (d x + c\right ) - 60 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{420 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/840*(105*sqrt(2)*((11*A + 19*C)*cos(d*x + c)^5 + 2*(11*A + 19*C)*cos(d*x + c)^4 + (11*A + 19*C)*cos(d*x +
c)^3)*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*((665*A + 1201*C)*cos(d*x +
c)^4 + 12*(35*A + 67*C)*cos(d*x + c)^3 - 28*(5*A + 7*C)*cos(d*x + c)^2 + 36*C*cos(d*x + c) - 60*C)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c
)^3), -1/420*(105*sqrt(2)*((11*A + 19*C)*cos(d*x + c)^5 + 2*(11*A + 19*C)*cos(d*x + c)^4 + (11*A + 19*C)*cos(d
*x + c)^3)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))
+ 2*((665*A + 1201*C)*cos(d*x + c)^4 + 12*(35*A + 67*C)*cos(d*x + c)^3 - 28*(5*A + 7*C)*cos(d*x + c)^2 + 36*C
*cos(d*x + c) - 60*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*co
s(d*x + c)^4 + a^2*d*cos(d*x + c)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\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)**4*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [A]  time = 9.30967, size = 593, normalized size = 2.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(105*(11*sqrt(2)*A + 19*sqrt(2)*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)) - ((((105*(sqrt(2)*A*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
) + sqrt(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2/a^3 - 4*(455*sqrt(2)*A*a^5*sgn(tan(1
/2*d*x + 1/2*c)^2 - 1) + 877*sqrt(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^3)*tan(1/2*d*x + 1/2*c)^2 + 14*(
305*sqrt(2)*A*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 517*sqrt(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^3)*ta
n(1/2*d*x + 1/2*c)^2 - 140*(25*sqrt(2)*A*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 47*sqrt(2)*C*a^5*sgn(tan(1/2*d*
x + 1/2*c)^2 - 1))/a^3)*tan(1/2*d*x + 1/2*c)^2 + 105*(9*sqrt(2)*A*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 17*sqr
t(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^3)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-
a*tan(1/2*d*x + 1/2*c)^2 + a)))/d