3.2.92 \(\int \frac {\sec ^4(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [192]

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

[Out]

1/4*(11*A+19*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A+C)*sec(
d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/105*(455*A+799*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/70*(3
5*A+67*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/14*(7*A+11*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*
sec(d*x+c))^(1/2)+1/210*(245*A+397*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d

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Rubi [A]
time = 0.56, antiderivative size = 259, normalized size of antiderivative = 1.00, 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 = {4170, 4106, 4095, 4086, 3880, 209} \begin {gather*} \frac {(11 A+19 C) \text {ArcTan}\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}} \end {gather*}

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 209

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

Rule 3880

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 4086

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 4095

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)),
Int[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] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4106

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 4170

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)]

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) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(527\) vs. \(2(259)=518\).
time = 6.89, size = 527, normalized size = 2.03 \begin {gather*} \frac {\cos ^2(c+d x) \sqrt {(1+\cos (c+d x)) \sec (c+d x)} (1+\sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {2 (-140 A-448 C+665 A \cos (c)+1201 C \cos (c)) \sin \left (\frac {c}{2}\right )}{105 d \left (\cos \left (\frac {c}{2}\right )+\cos \left (\frac {3 c}{2}\right )\right )}+\frac {\sec \left (\frac {c}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {c}{2}\right )+C \sin \left (\frac {c}{2}\right )\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 {\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 {4 C \sec (c) \sec ^3(c+d x) \sin (d x)}{7 d}-\frac {4 \sec (c) \sec (c+d x) (39 C \sin (c)-35 A \sin (d x)-112 C \sin (d x))}{105 d}+\frac {4 \sec (c) \sec ^2(c+d x) (5 C \sin (c)-13 C \sin (d x))}{35 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) (a (1+\sec (c+d x)))^{3/2}}+\frac {(11 A+19 C) \text {ArcTan}\left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4(c+d x) \sqrt {-1+\sec (c+d x)} (1+\sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \sin (c+d x)}{\sqrt {2} d (1+\cos (c+d x)) \sqrt {1-\cos ^2(c+d x)} (A+2 C+A \cos (2 c+2 d x)) (a (1+\sec (c+d x)))^{3/2} \sqrt {\cos ^2(c+d x) (-1+\sec (c+d x)) (1+\sec (c+d x))}} \end {gather*}

Warning: Unable to verify antiderivative.

[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] Leaf count of result is larger than twice the leaf count of optimal. \(973\) vs. \(2(228)=456\).
time = 13.33, size = 974, normalized size = 3.76

method	result	size
default	\(\text {Expression too large to display}\)	\(974\)

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,method=_RETURNVERBOSE)

[Out]

1/3360/d*(-1+cos(d*x+c))*(1155*A*sin(d*x+c)*cos(d*x+c)^4*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-c
os(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+1995*C*sin(d*x+c)*cos(d*x+c)^4*ln(((-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+4620*A*sin(d
*x+c)*cos(d*x+c)^3*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(7/2)+7980*C*sin(d*x+c)*cos(d*x+c)^3*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos
(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+6930*A*sin(d*x+c)*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+11970*C*sin(d*
x+c)*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(7/2)+4620*A*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*
x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+7980*C*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+1155*A*ln(((-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x
+c)+1995*C*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(7/2)*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+8
960*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*(1
+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)^3/a^2

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 2.06, size = 527, normalized size = 2.03 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 1.51, size = 362, normalized size = 1.40 \begin {gather*} -\frac {\frac {105 \, {\left (11 \, \sqrt {2} A + 19 \, \sqrt {2} 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 (\cos \left (d x + c\right )\right )} - \frac {{\left ({\left ({\left ({\left (\frac {105 \, {\left (\sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} - \frac {4 \, {\left (455 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 877 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {14 \, {\left (305 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 517 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {140 \, {\left (25 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {105 \, {\left (9 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 17 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{3}}\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 )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{420 \, d} \end {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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