3.401 \(\int \frac {\sec ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=261 \[ \frac {(163 B-283 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{240 a^3 d}-\frac {(85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{80 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac {(13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

1/32*(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*(B-C)*s
ec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)+1/16*(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*(985*B-1729*C)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)-1/80*(85*B-157*C)*sec(d*x+c)^2*tan(d*x+c)/a
^2/d/(a+a*sec(d*x+c))^(1/2)+1/240*(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 = 0.93, 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} \[ -\frac {(85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{80 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(163 B-283 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(475 B-787 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{240 a^3 d}-\frac {(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac {(13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

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])^(5/2),x]

[Out]

((163*B - 283*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (
(B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((13*B - 21*C)*Sec[c + d*x]^3*Tan[c +
d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((985*B - 1729*C)*Tan[c + d*x])/(120*a^2*d*Sqrt[a + a*Sec[c + d*x]
]) - ((85*B - 157*C)*Sec[c + d*x]^2*Tan[c + d*x])/(80*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((475*B - 787*C)*Sqrt[
a + a*Sec[c + d*x]]*Tan[c + d*x])/(240*a^3*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))^{5/2}} \, dx &=\int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=\frac {(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 B-13 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(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 (13 B-21 C)-\frac {1}{4} a^2 (85 B-157 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(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 {(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 (85 B-157 C)+\frac {1}{8} a^3 (475 B-787 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{20 a^5}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(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 {(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 {(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 (475 B-787 C)-\frac {1}{8} a^4 (985 B-1729 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{30 a^6}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(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 {(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(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 {(475 B-787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}+\frac {(163 B-283 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(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 {(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(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 {(475 B-787 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}-\frac {(163 B-283 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 )}{16 a^2 d}\\ &=\frac {(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 {(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(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 {(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(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 {(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 [A]  time = 2.87, size = 177, normalized size = 0.68 \[ \frac {\tan (c+d x) \left (\sqrt {1-\sec (c+d x)} \left (160 (B-C) \sec ^3(c+d x)-32 (25 B-49 C) \sec ^2(c+d x)-5 (503 B-911 C) \sec (c+d x)-1495 B+96 C \sec ^4(c+d x)+2671 C\right )+30 \sqrt {2} (163 B-283 C) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )\right )}{240 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]

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])^(5/2),x]

[Out]

((30*Sqrt[2]*(163*B - 283*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2 + Sqrt[
1 - Sec[c + d*x]]*(-1495*B + 2671*C - 5*(503*B - 911*C)*Sec[c + d*x] - 32*(25*B - 49*C)*Sec[c + d*x]^2 + 160*(
B - C)*Sec[c + d*x]^3 + 96*C*Sec[c + d*x]^4))*Tan[c + d*x])/(240*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]
))^(5/2))

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fricas [A]  time = 0.50, size = 596, normalized size = 2.28 \[ \left [\frac {15 \, \sqrt {2} {\left ({\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (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 (1495 \, B - 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (503 \, B - 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (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 (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (163 \, B - 283 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (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 (1495 \, B - 2671 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (503 \, B - 911 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (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 ] \]

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))^(5/2),x, algorithm="fricas")

[Out]

[1/960*(15*sqrt(2)*((163*B - 283*C)*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos(d*x + c)^4 + 3*(163*B - 283*C)*cos(
d*x + c)^3 + (163*B - 283*C)*cos(d*x + c)^2)*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*((1495*B - 2671*C)*cos(d*x + c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 32*(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)*((163*B -
283*C)*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos(d*x + c)^4 + 3*(163*B - 283*C)*cos(d*x + c)^3 + (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*((1495*B - 2671*C)*cos(d*x + c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 32*(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)]

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giac [A]  time = 3.53, size = 439, normalized size = 1.68 \[ -\frac {\frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} B a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - \sqrt {2} C a^{2} \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^{2}} + \frac {21 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 29 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3685 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 6733 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, {\left (1133 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 1973 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (155 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 291 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\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 (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 (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{480 \, 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))^(5/2),x, algorithm="giac")

[Out]

-1/480*((((15*(2*(sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
))*tan(1/2*d*x + 1/2*c)^2/a^2 + (21*sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 29*sqrt(2)*C*a^2*sgn(tan(1
/2*d*x + 1/2*c)^2 - 1))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (3685*sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 67
33*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^2)*tan(1/2*d*x + 1/2*c)^2 + 5*(1133*sqrt(2)*B*a^2*sgn(tan(
1/2*d*x + 1/2*c)^2 - 1) - 1973*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15
*(155*sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 291*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^2)*
tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 15*(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(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.99, size = 985, 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))^(5/2),x)

[Out]

1/1920/d*(-1+cos(d*x+c))^2*(2445*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)))^(5/2)-4245*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)))^(5/2)+9780*B*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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)
)*cos(d*x+c)^3*sin(d*x+c)-16980*C*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))+14670*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))*cos(d*x+c)^2*sin(d*x+c)-25470*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)*cos(d*x+c)^2+9780*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))*cos(d*x+c)*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)*sin(d*x+c)*cos(d*x+c)+2445*B*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))*si
n(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)+11960*B*cos(d*x+c)^5-21368*C*cos(d*x+c)^5+8160*B*cos(d*x+c)^4-15072*C*cos(d*x+
c)^4-13720*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+1280*B*cos(d*x+c)-2048
*C*cos(d*x+c)+768*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(d*x+c)^2/a^3

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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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

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))^(5/2)),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))^(5/2)), x)

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

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))**(5/2),x)

[Out]

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

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