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

Optimal. Leaf size=202 \[ \frac{8 (9 B-19 C) \tan (c+d x)}{15 a^3 d}-\frac{(6 B-13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{4 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(6 B-13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}+\frac{(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

-((6*B - 13*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) + (8*(9*B - 19*C)*Tan[c + d*x])/(15*a^3*d) - ((6*B - 13*C)*Sec
[c + d*x]*Tan[c + d*x])/(2*a^3*d) + ((B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((6*B
- 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) + (4*(9*B - 19*C)*Sec[c + d*x]^2*Tan[c +
d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.552848, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4019, 3787, 3767, 8, 3768, 3770} \[ \frac{8 (9 B-19 C) \tan (c+d x)}{15 a^3 d}-\frac{(6 B-13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{4 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(6 B-13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}+\frac{(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

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])^3,x]

[Out]

-((6*B - 13*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) + (8*(9*B - 19*C)*Tan[c + d*x])/(15*a^3*d) - ((6*B - 13*C)*Sec
[c + d*x]*Tan[c + d*x])/(2*a^3*d) + ((B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((6*B
- 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) + (4*(9*B - 19*C)*Sec[c + d*x]^2*Tan[c +
d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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]

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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} \, dx &=\int \frac{\sec ^5(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^4(c+d x) (4 a (B-C)-a (2 B-7 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^3(c+d x) \left (3 a^2 (6 B-11 C)-a^2 (18 B-43 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 B-19 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \sec ^2(c+d x) \left (8 a^3 (9 B-19 C)-15 a^3 (6 B-13 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 B-19 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(8 (9 B-19 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}-\frac{(6 B-13 C) \int \sec ^3(c+d x) \, dx}{a^3}\\ &=-\frac{(6 B-13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 B-19 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(6 B-13 C) \int \sec (c+d x) \, dx}{2 a^3}-\frac{(8 (9 B-19 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac{(6 B-13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{8 (9 B-19 C) \tan (c+d x)}{15 a^3 d}-\frac{(6 B-13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 B-19 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.97206, size = 428, normalized size = 2.12 \[ \frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-64 (9 B-19 C) \tan ^3\left (\frac{1}{2} (c+d x)\right )+4 (87 B-197 C) \tan \left (\frac{1}{2} (c+d x)\right )+16 (12 B+13 C) \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+\tan ^5\left (\frac{1}{2} (c+d x)\right ) \left (3 (B-C) \sec ^4\left (\frac{1}{2} (c+d x)\right )+228 B-428 C\right )+30 (6 B-13 C) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\tan ^3\left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) ((12 B-17 C) \cos (c+d x)-6 B+11 C)+\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) ((24 B-34 C) \cos (c+d x)-21 B+31 C)+30 (6 B-13 C) \tan ^4\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-60 (6 B-13 C) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{60 a^3 d} \]

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,x]

[Out]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(30*(6*B - 13*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d
*x)/2] + Sin[(c + d*x)/2]]) + 16*(12*B + 13*C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 4*(87*B - 197*C)*Tan[(c + d
*x)/2] + (-21*B + 31*C + (24*B - 34*C)*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2] - 60*(6*B - 13*C)*(Lo
g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Tan[(c + d*x)/2]^2 - 64*(9*
B - 19*C)*Tan[(c + d*x)/2]^3 - (-6*B + 11*C + (12*B - 17*C)*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2]^
3 + 30*(6*B - 13*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Tan[
(c + d*x)/2]^4 + (228*B - 428*C + 3*(B - C)*Sec[(c + d*x)/2]^4)*Tan[(c + d*x)/2]^5))/(60*a^3*d)

________________________________________________________________________________________

Maple [A]  time = 0.069, size = 334, normalized size = 1.7 \begin{align*}{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{2\,C}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{31\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{7\,C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) B}{d{a}^{3}}}+{\frac{13\,C}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) B}{d{a}^{3}}}-{\frac{13\,C}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{7\,C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}} \end{align*}

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,x)

[Out]

1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*B-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*B-2/3/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*B*tan(1/2*d*x+1/2*c)-31/4/d/a^3*C*tan(1/2*d*x+1/2*c)+7/2/d/a^3/(tan(1/2*d*x
+1/2*c)+1)*C-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*B-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B+13/2/d/a^3*ln(tan(1/2*d*x+1/2
*c)+1)*C-1/2/d/a^3*C/(tan(1/2*d*x+1/2*c)+1)^2+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B-13/2/d/a^3*ln(tan(1/2*d*x+1/2
*c)-1)*C+7/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)*C-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*B+1/2/d/a^3*C/(tan(1/2*d*x+1/2*c)-1
)^2

________________________________________________________________________________________

Maxima [A]  time = 0.976019, size = 509, normalized size = 2.52 \begin{align*} -\frac{C{\left (\frac{60 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac{2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{390 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{390 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, B{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \end{align*}

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,x, algorithm="maxima")

[Out]

-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(40*sin(d*x + c)/((a^3 -
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

________________________________________________________________________________________

Fricas [A]  time = 0.525794, size = 757, normalized size = 3.75 \begin{align*} -\frac{15 \,{\left ({\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (9 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (114 \, B - 239 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (234 \, B - 479 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sin \left (d x + c\right )}{60 \,{\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 )}} \end{align*}

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,x, algorithm="fricas")

[Out]

-1/60*(15*((6*B - 13*C)*cos(d*x + c)^5 + 3*(6*B - 13*C)*cos(d*x + c)^4 + 3*(6*B - 13*C)*cos(d*x + c)^3 + (6*B
- 13*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((6*B - 13*C)*cos(d*x + c)^5 + 3*(6*B - 13*C)*cos(d*x + c)^
4 + 3*(6*B - 13*C)*cos(d*x + c)^3 + (6*B - 13*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(9*B - 19*C)*c
os(d*x + c)^4 + 3*(114*B - 239*C)*cos(d*x + c)^3 + (234*B - 479*C)*cos(d*x + c)^2 + 15*(2*B - 3*C)*cos(d*x + c
) + 15*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)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

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,x)

[Out]

(Integral(B*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c
+ d*x)**6/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

________________________________________________________________________________________

Giac [A]  time = 1.19174, size = 315, normalized size = 1.56 \begin{align*} -\frac{\frac{30 \,{\left (6 \, B - 13 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{30 \,{\left (6 \, B - 13 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{60 \,{\left (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac{3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 255 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 465 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

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,x, algorithm="giac")

[Out]

-1/60*(30*(6*B - 13*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(6*B - 13*C)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(2*B*tan(1/2*d*x + 1/2*c)^3 - 7*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 5*C*tan(1/
2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/2*d*x
+ 1/2*c)^5 + 30*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*B*a^12*tan(1/2*d*x + 1
/2*c) - 465*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d