3.6.27 \(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [527]

Optimal. Leaf size=277 \[ -\frac {(75 A-163 B+283 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(465 A-985 B+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(45 A-85 B+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A-475 B+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d} \]

[Out]

-1/32*(75*A-163*B+283*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(
A-B+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(5*A-13*B+21*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*
sec(d*x+c))^(3/2)+1/120*(465*A-985*B+1729*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1/80*(45*A-85*B+157*C)*se
c(d*x+c)^2*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)-1/240*(195*A-475*B+787*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)
/a^3/d

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Rubi [A]
time = 0.61, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4169, 4104, 4106, 4095, 4086, 3880, 209} \begin {gather*} -\frac {(75 A-163 B+283 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(195 A-475 B+787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{240 a^3 d}+\frac {(45 A-85 B+157 C) \tan (c+d x) \sec ^2(c+d x)}{80 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(465 A-985 B+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {(5 A-13 B+21 C) \tan (c+d x) \sec ^3(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((75*A - 163*B + 283*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[2]*a^(5
/2)*d) - ((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((5*A - 13*B + 21*C)*Sec
[c + d*x]^3*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + ((465*A - 985*B + 1729*C)*Tan[c + d*x])/(120*a
^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((45*A - 85*B + 157*C)*Sec[c + d*x]^2*Tan[c + d*x])/(80*a^2*d*Sqrt[a + a*Sec[
c + d*x]]) - ((195*A - 475*B + 787*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(240*a^3*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 4104

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

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 4169

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*C
sc[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)]

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {\sec ^4(c+d x) \left (4 a (B-C)+\frac {1}{2} a (5 A-5 B+13 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^3(c+d x) \left (-\frac {3}{2} a^2 (5 A-13 B+21 C)+\frac {1}{4} a^2 (45 A-85 B+157 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(45 A-85 B+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\sec ^2(c+d x) \left (\frac {1}{2} a^3 (45 A-85 B+157 C)-\frac {1}{8} a^3 (195 A-475 B+787 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{20 a^5}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(45 A-85 B+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A-475 B+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}+\frac {\int \frac {\sec (c+d x) \left (-\frac {1}{16} a^4 (195 A-475 B+787 C)+\frac {1}{8} a^4 (465 A-985 B+1729 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{30 a^6}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(465 A-985 B+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(45 A-85 B+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A-475 B+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}-\frac {(75 A-163 B+283 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(465 A-985 B+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(45 A-85 B+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A-475 B+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}+\frac {(75 A-163 B+283 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 )}{16 a^2 d}\\ &=-\frac {(75 A-163 B+283 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B+21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(465 A-985 B+1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(45 A-85 B+157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(195 A-475 B+787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 25.73, size = 7237, normalized size = 26.13 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1438\) vs. \(2(246)=492\).
time = 0.23, size = 1439, normalized size = 5.19

method	result	size
default	\(\text {Expression too large to display}\)	\(1439\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/1920/d*(-1+cos(d*x+c))^2*(-3840*A*cos(d*x+c)^2-1280*B*cos(d*x+c)-768*C-6360*A*cos(d*x+c)^3+13720*B*cos(d*x+
c)^3-23896*C*cos(d*x+c)^3+7680*B*cos(d*x+c)^2-13824*C*cos(d*x+c)^2+2048*C*cos(d*x+c)+1125*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)))^(5/2)*cos(d*x+c)^
4*sin(d*x+c)-2445*B*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)))^(5/2)*cos(d*x+c)^4*sin(d*x+c)+4245*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)))^(5/2)*cos(d*x+c)^4*sin(d*x+c)+21368*C*cos(d*x+c)^5+
5880*A*cos(d*x+c)^5+4320*A*cos(d*x+c)^4+4500*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)))^(5/2)*cos(d*x+c)^3*sin(d*x+c)-9780*B*ln(-(-(-2*cos(d*x+c)/(1+c
os(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)))^(5/2)*cos(d*x+c)^3*sin(d
*x+c)+16980*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)))^(5/2)*cos(d*x+c)^3*sin(d*x+c)-11960*B*cos(d*x+c)^5+15072*C*cos(d*x+c)^4+6750*A*cos(d*x+c)^2*sin
(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)))^(5/2)-14670*B*cos(d*x+c)^2*sin(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)))^(5/2)+25470*C*cos(d*x+c)^2*sin(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)))^(5/2)+4500*A*cos(d*x+c)
*sin(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)))^(5/2)-9780*B*cos(d*x+c)*sin(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)))^(5/2)+16980*C*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+c
os(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)))^(5/2)+1125*A*ln(-(-(-2*c
os(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)))^(5/2)*sin(
d*x+c)-2445*B*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)))^(5/2)*sin(d*x+c)+4245*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)))^(5/2)*sin(d*x+c)-8160*B*cos(d*x+c)^4)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1
/2)/cos(d*x+c)^2/sin(d*x+c)^5/a^3

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

[Out]

Timed out

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Fricas [A]
time = 3.80, size = 638, normalized size = 2.30 \begin {gather*} \left [-\frac {15 \, \sqrt {2} {\left ({\left (75 \, A - 163 \, B + 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (75 \, A - 163 \, B + 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (75 \, A - 163 \, B + 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (75 \, A - 163 \, B + 283 \, 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 (735 \, A - 1495 \, B + 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (255 \, A - 503 \, B + 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (15 \, A - 25 \, B + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 160 \, {\left (B - C\right )} \cos \left (d x + c\right ) + 96 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{960 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}, \frac {15 \, \sqrt {2} {\left ({\left (75 \, A - 163 \, B + 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (75 \, A - 163 \, B + 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (75 \, A - 163 \, B + 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (75 \, A - 163 \, B + 283 \, 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 (735 \, A - 1495 \, B + 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (255 \, A - 503 \, B + 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (15 \, A - 25 \, B + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 160 \, {\left (B - C\right )} \cos \left (d x + c\right ) + 96 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{480 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/960*(15*sqrt(2)*((75*A - 163*B + 283*C)*cos(d*x + c)^5 + 3*(75*A - 163*B + 283*C)*cos(d*x + c)^4 + 3*(75*A
- 163*B + 283*C)*cos(d*x + c)^3 + (75*A - 163*B + 283*C)*cos(d*x + c)^2)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sq
rt((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*((735*A - 1495*B + 2671*C)*cos(d*x + c)^4 + 5*(255*A - 503*B + 911*C
)*cos(d*x + c)^3 + 32*(15*A - 25*B + 49*C)*cos(d*x + c)^2 + 160*(B - C)*cos(d*x + c) + 96*C)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 +
a^3*d*cos(d*x + c)^2), 1/480*(15*sqrt(2)*((75*A - 163*B + 283*C)*cos(d*x + c)^5 + 3*(75*A - 163*B + 283*C)*co
s(d*x + c)^4 + 3*(75*A - 163*B + 283*C)*cos(d*x + c)^3 + (75*A - 163*B + 283*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)/(sqrt(a)*sin(d*x + c))) + 2*((735*A - 1495*B + 2
671*C)*cos(d*x + c)^4 + 5*(255*A - 503*B + 911*C)*cos(d*x + c)^3 + 32*(15*A - 25*B + 49*C)*cos(d*x + c)^2 + 16
0*(B - C)*cos(d*x + c) + 96*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3
*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )} + 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 {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [A]
time = 1.94, size = 447, normalized size = 1.61 \begin {gather*} -\frac {\frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{2} \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^{2}} + \frac {13 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 21 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 29 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {1725 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 3685 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 6733 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, {\left (549 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 1133 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 1973 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (83 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 155 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 291 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\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}} - \frac {15 \, {\left (75 \, \sqrt {2} A - 163 \, \sqrt {2} B + 283 \, \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^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/480*((((15*(2*(sqrt(2)*A*a^2*sgn(cos(d*x + c)) - sqrt(2)*B*a^2*sgn(cos(d*x + c)) + sqrt(2)*C*a^2*sgn(cos(d*
x + c)))*tan(1/2*d*x + 1/2*c)^2/a^2 + (13*sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 21*sqrt(2)*B*a^2*sgn(cos(d*x + c))
+ 29*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (1725*sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 3
685*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 6733*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 + 5*(5
49*sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 1133*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 1973*sqrt(2)*C*a^2*sgn(cos(d*x + c
)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(83*sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 155*sqrt(2)*B*a^2*sgn(cos(d*x + c))
+ 291*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a)) - 15*(75*sqrt(2)*A - 163*sqrt(2)*B + 283*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^2*sgn(cos(d*x + c))))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\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 )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(5/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(5/2)), x)

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