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

Optimal. Leaf size=236 \[ \frac {2 (21 A+19 C) \tan (c+d x) \sec ^2(c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} (A+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 (21 A+29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 a d}+\frac {4 (147 A+143 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}-\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

-(A+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+4/315*(147*A+143*C)*tan
(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/105*(21*A+19*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/63*C*sec
(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9*C*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/315*(21
*A+29*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a/d

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Rubi [A]  time = 0.80, antiderivative size = 236, 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 = {4089, 4021, 4010, 4001, 3795, 203} \[ \frac {2 (21 A+19 C) \tan (c+d x) \sec ^2(c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} (A+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 (21 A+29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 a d}+\frac {4 (147 A+143 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}-\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (4*(147*A
+ 143*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(21*A + 19*C)*Sec[c + d*x]^2*Tan[c + d*x])/(105*
d*Sqrt[a + a*Sec[c + d*x]]) - (2*C*Sec[c + d*x]^3*Tan[c + d*x])/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c +
d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (2*(21*A + 29*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(315*a*d)

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])

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 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 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 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 4089

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

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx &=\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\sec ^4(c+d x) \left (\frac {1}{2} a (9 A+8 C)-\frac {1}{2} a C \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{9 a}\\ &=-\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {4 \int \frac {\sec ^3(c+d x) \left (-\frac {3 a^2 C}{2}+\frac {3}{4} a^2 (21 A+19 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{63 a^2}\\ &=\frac {2 (21 A+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {8 \int \frac {\sec ^2(c+d x) \left (\frac {3}{2} a^3 (21 A+19 C)-\frac {3}{8} a^3 (21 A+29 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{315 a^3}\\ &=\frac {2 (21 A+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 a d}+\frac {16 \int \frac {\sec (c+d x) \left (-\frac {3}{16} a^4 (21 A+29 C)+\frac {3}{8} a^4 (147 A+143 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{945 a^4}\\ &=\frac {4 (147 A+143 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (21 A+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 a d}+(-A-C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {4 (147 A+143 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (21 A+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 a d}+\frac {(2 (A+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 )}{d}\\ &=-\frac {\sqrt {2} (A+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {4 (147 A+143 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (21 A+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (21 A+29 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 a d}\\ \end {align*}

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Mathematica [B]  time = 6.73, size = 474, normalized size = 2.01 \[ \frac {\cos ^2(c+d x) \sqrt {\sec (c+d x)+1} \sqrt {(\cos (c+d x)+1) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {4 \sec (c) \sec ^2(c+d x) (-63 A \sin (d x)+40 C \sin (c)-97 C \sin (d x))}{315 d}+\frac {4 \sec (c) \sec (c+d x) (63 A \sin (c)-84 A \sin (d x)+97 C \sin (c)-126 C \sin (d x))}{315 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (357 A \sin \left (\frac {d x}{2}\right )+383 C \sin \left (\frac {d x}{2}\right )\right )}{315 d}+\frac {8 \sin \left (\frac {c}{2}\right ) (273 A \cos (c)-84 A+257 C \cos (c)-126 C)}{315 d \left (\cos \left (\frac {c}{2}\right )+\cos \left (\frac {3 c}{2}\right )\right )}+\frac {4 C \sec (c) \sin (d x) \sec ^4(c+d x)}{9 d}+\frac {4 \sec (c) \sec ^3(c+d x) (7 C \sin (c)-8 C \sin (d x))}{63 d}\right )}{\sqrt {a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)}-\frac {2 \sqrt {2} (A+C) \sin (c+d x) \cos ^4(c+d x) \sqrt {\sec (c+d x)-1} (\sec (c+d x)+1)^2 \tan ^{-1}\left (\frac {\sqrt {\sec (c+d x)-1}}{\sqrt {2}}\right ) \left (A+C \sec ^2(c+d x)\right )}{d (\cos (c+d x)+1) \sqrt {1-\cos ^2(c+d x)} \sqrt {a (\sec (c+d x)+1)} \sqrt {\cos ^2(c+d x) (\sec (c+d x)-1) (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^2*Sqrt[(1 + Cos[c + d*x])*Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((8*(-84*A
- 126*C + 273*A*Cos[c] + 257*C*Cos[c])*Sin[c/2])/(315*d*(Cos[c/2] + Cos[(3*c)/2])) + (4*Sec[c/2]*Sec[c/2 + (d
*x)/2]*(357*A*Sin[(d*x)/2] + 383*C*Sin[(d*x)/2]))/(315*d) + (4*C*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(9*d) + (4*Se
c[c]*Sec[c + d*x]*(63*A*Sin[c] + 97*C*Sin[c] - 84*A*Sin[d*x] - 126*C*Sin[d*x]))/(315*d) - (4*Sec[c]*Sec[c + d*
x]^2*(40*C*Sin[c] - 63*A*Sin[d*x] - 97*C*Sin[d*x]))/(315*d) + (4*Sec[c]*Sec[c + d*x]^3*(7*C*Sin[c] - 8*C*Sin[d
*x]))/(63*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[a*(1 + Sec[c + d*x])]) - (2*Sqrt[2]*(A + 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])^2*(A + C*Sec[c + d*x]^2)*S
in[c + d*x])/(d*(1 + Cos[c + d*x])*Sqrt[1 - Cos[c + d*x]^2]*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[a*(1 + Sec[c +
d*x])]*Sqrt[Cos[c + d*x]^2*(-1 + Sec[c + d*x])*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.50, size = 445, normalized size = 1.89 \[ \left [\frac {315 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{5} + {\left (A + C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (273 \, A + 257 \, C\right )} \cos \left (d x + c\right )^{4} - {\left (21 \, A + 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, C \cos \left (d x + c\right ) + 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{630 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}, \frac {2 \, {\left ({\left (273 \, A + 257 \, C\right )} \cos \left (d x + c\right )^{4} - {\left (21 \, A + 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, C \cos \left (d x + c\right ) + 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {315 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{5} + {\left (A + C\right )} a \cos \left (d x + c\right )^{4}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}\right ] \]

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))^(1/2),x, algorithm="fricas")

[Out]

[1/630*(315*sqrt(2)*((A + C)*a*cos(d*x + c)^5 + (A + C)*a*cos(d*x + c)^4)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(c
os(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((273*A + 257*C)*cos(d*x + c)^4 - (21*A + 29*C)*cos(d*x + c)^3 + 3*(2
1*A + 19*C)*cos(d*x + c)^2 - 5*C*cos(d*x + c) + 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a
*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4), 1/315*(2*((273*A + 257*C)*cos(d*x + c)^4 - (21*A + 29*C)*cos(d*x + c)
^3 + 3*(21*A + 19*C)*cos(d*x + c)^2 - 5*C*cos(d*x + c) + 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x
+ c) + 315*sqrt(2)*((A + C)*a*cos(d*x + c)^5 + (A + C)*a*cos(d*x + c)^4)*arctan(sqrt(2)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4)]

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giac [A]  time = 4.72, size = 412, normalized size = 1.75 \[ -\frac {\frac {315 \, {\left (\sqrt {2} A + \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} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {2 \, {\left (315 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 315 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - {\left (1050 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 840 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - {\left (1512 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 1638 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - {\left (1134 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 936 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - {\left (357 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 383 \, \sqrt {2} C a^{4} \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}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{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 )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{315 \, d} \]

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))^(1/2),x, algorithm="giac")

[Out]

-1/315*(315*(sqrt(2)*A + 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)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 2*(315*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 315*s
qrt(2)*C*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (1050*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 840*sqrt(
2)*C*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (1512*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 1638*sqrt(2)*
C*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (1134*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 936*sqrt(2)*C*a^
4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (357*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 383*sqrt(2)*C*a^4*sgn
(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*
d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

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maple [B]  time = 2.22, size = 966, normalized size = 4.09 \[ \text {result too large to display} \]

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))^(1/2),x)

[Out]

-1/5040/d*(315*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*cos(d*x+c)^4*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))+315*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*cos(d*x+c)^4*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))+1260*A*(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(9/2)*cos(d*x+c)^3*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))+1260*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*cos(d*x+c)^3*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))+1890*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*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))+1890*C*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(9/2)*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))+1260*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*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))+1260*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*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))+315*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(9/2)*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))*sin(d*x+c)+315*C*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(9/2)*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))*sin(
d*x+c)+8736*A*cos(d*x+c)^5+8224*C*cos(d*x+c)^5-9408*A*cos(d*x+c)^4-9152*C*cos(d*x+c)^4+2688*A*cos(d*x+c)^3+275
2*C*cos(d*x+c)^3-2016*A*cos(d*x+c)^2-1984*C*cos(d*x+c)^2+1280*C*cos(d*x+c)-1120*C)*(a*(1+cos(d*x+c))/cos(d*x+c
))^(1/2)/cos(d*x+c)^4/sin(d*x+c)/a

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

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))^(1/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

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))**(1/2),x)

[Out]

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

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