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

Optimal. Leaf size=261 \[ -\frac {(15 B-19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{210 a^2 d}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{14 a d \sqrt {a \sec (c+d x)+a}}+\frac {(63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{70 a d \sqrt {a \sec (c+d x)+a}}+\frac {(651 B-799 C) \tan (c+d x)}{105 a d \sqrt {a \sec (c+d x)+a}} \]

[Out]

-1/4*(15*B-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*(B-C)*sec
(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+1/105*(651*B-799*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/70*(
63*B-67*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/14*(7*B-11*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a
*sec(d*x+c))^(1/2)-1/210*(273*B-397*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d

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Rubi [A]  time = 0.92, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4072, 4019, 4021, 4010, 4001, 3795, 203} \[ -\frac {(15 B-19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(273 B-397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{210 a^2 d}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(7 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{14 a d \sqrt {a \sec (c+d x)+a}}+\frac {(63 B-67 C) \tan (c+d x) \sec ^2(c+d x)}{70 a d \sqrt {a \sec (c+d x)+a}}+\frac {(651 B-799 C) \tan (c+d x)}{105 a d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*B - 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) + ((B
- C)*Sec[c + d*x]^4*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((651*B - 799*C)*Tan[c + d*x])/(105*a*d*
Sqrt[a + a*Sec[c + d*x]]) + ((63*B - 67*C)*Sec[c + d*x]^2*Tan[c + d*x])/(70*a*d*Sqrt[a + a*Sec[c + d*x]]) - ((
7*B - 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(14*a*d*Sqrt[a + a*Sec[c + d*x]]) - ((273*B - 397*C)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(210*a^2*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 4019

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 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 4072

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

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=\int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=\frac {(B-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 (4 a (B-C)-\frac {1}{2} a (7 B-11 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(7 B-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 B-11 C)+\frac {1}{4} a^2 (63 B-67 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{7 a^3}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(63 B-67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}-\frac {(7 B-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 (63 B-67 C)-\frac {1}{8} a^3 (273 B-397 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{35 a^4}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(63 B-67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}-\frac {(7 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}-\frac {(273 B-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 (273 B-397 C)+\frac {1}{8} a^4 (651 B-799 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{105 a^5}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(651 B-799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}+\frac {(63 B-67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}-\frac {(7 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}-\frac {(273 B-397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}-\frac {(15 B-19 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(651 B-799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}+\frac {(63 B-67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}-\frac {(7 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}-\frac {(273 B-397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}+\frac {(15 B-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 {(15 B-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 {(B-C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(651 B-799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}+\frac {(63 B-67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}-\frac {(7 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}-\frac {(273 B-397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.43, size = 204, normalized size = 0.78 \[ \frac {\tan (c+d x) \left (\frac {1}{4} \sqrt {1-\sec (c+d x)} \sec ^4(c+d x) (24 (217 B-213 C) \cos (c+d x)+60 (63 B-67 C) \cos (2 (c+d x))+1512 B \cos (3 (c+d x))+1029 B \cos (4 (c+d x))+2751 B-1608 C \cos (3 (c+d x))-1201 C \cos (4 (c+d x))-2339 C)-210 \sqrt {2} (15 B-19 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )\right )}{420 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-210*Sqrt[2]*(15*B - 19*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sec[c + d*x] + ((2751*
B - 2339*C + 24*(217*B - 213*C)*Cos[c + d*x] + 60*(63*B - 67*C)*Cos[2*(c + d*x)] + 1512*B*Cos[3*(c + d*x)] - 1
608*C*Cos[3*(c + d*x)] + 1029*B*Cos[4*(c + d*x)] - 1201*C*Cos[4*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x
]^4)/4)*Tan[c + d*x])/(420*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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fricas [A]  time = 0.55, size = 539, normalized size = 2.07 \[ \left [\frac {105 \, \sqrt {2} {\left ({\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (15 \, B - 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 (1029 \, B - 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (63 \, B - 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (7 \, B - 3 \, C\right )} \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 (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (15 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (15 \, B - 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 (1029 \, B - 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (63 \, B - 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (7 \, B - 3 \, C\right )} \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 ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/840*(105*sqrt(2)*((15*B - 19*C)*cos(d*x + c)^5 + 2*(15*B - 19*C)*cos(d*x + c)^4 + (15*B - 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*((1029*B - 1201*C)*cos(d*x +
c)^4 + 12*(63*B - 67*C)*cos(d*x + c)^3 - 28*(3*B - 7*C)*cos(d*x + c)^2 + 12*(7*B - 3*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*cos(d*x + c)^4 + a^2*d*c
os(d*x + c)^3), 1/420*(105*sqrt(2)*((15*B - 19*C)*cos(d*x + c)^5 + 2*(15*B - 19*C)*cos(d*x + c)^4 + (15*B - 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*((1029*B - 1201*C)*cos(d*x + c)^4 + 12*(63*B - 67*C)*cos(d*x + c)^3 - 28*(3*B - 7*C)*cos(d*x + c
)^2 + 12*(7*B - 3*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*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)]

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giac [A]  time = 4.77, size = 439, normalized size = 1.68 \[ -\frac {\frac {105 \, {\left (15 \, \sqrt {2} B - 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 (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {{\left ({\left ({\left ({\left (\frac {105 \, {\left (\sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - \sqrt {2} C a^{5} \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}}{a^{3}} - \frac {4 \, {\left (693 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 877 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {14 \, {\left (453 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 517 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {140 \, {\left (39 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 47 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1785 \, {\left (\sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(105*(15*sqrt(2)*B - 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)*B*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*(693*sqrt(2)*B*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*
(453*sqrt(2)*B*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)*t
an(1/2*d*x + 1/2*c)^2 - 140*(39*sqrt(2)*B*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 + 1785*(sqrt(2)*B*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))/a^3)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a)))/d

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maple [B]  time = 1.92, size = 983, normalized size = 3.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3360/d*(-1+cos(d*x+c))*(1575*B*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)-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)+6300*B*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)-co
s(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+9450*B*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)+6300*B*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+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)))^(7/2)+1575*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)))^(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+co
s(d*x+c)))^(7/2)*sin(d*x+c)-16464*B*cos(d*x+c)^5+19216*C*cos(d*x+c)^5+4368*B*cos(d*x+c)^4-6352*C*cos(d*x+c)^4+
13440*B*cos(d*x+c)^3-16000*C*cos(d*x+c)^3-2688*B*cos(d*x+c)^2+3712*C*cos(d*x+c)^2+1344*B*cos(d*x+c)-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)]  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*(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)**5/(a*(sec(c + d*x) + 1))**(3/2), x)

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