3.99 \(\int \frac{\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.474776, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4019, 3787, 3767, 8, 3768, 3770} \[ \frac{8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac{(6 A-13 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{4 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(6 A-13 B) \tan (c+d x) \sec (c+d x)}{2 a^3 d}+\frac{(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

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

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 ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac{(A-B) \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 (A-B)-a (2 A-7 B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 A-11 B) \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 A-11 B)-a^2 (18 A-43 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 A-19 B) \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 A-19 B)-15 a^3 (6 A-13 B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(8 (9 A-19 B)) \int \sec ^2(c+d x) \, dx}{15 a^3}-\frac{(6 A-13 B) \int \sec ^3(c+d x) \, dx}{a^3}\\ &=-\frac{(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(6 A-13 B) \int \sec (c+d x) \, dx}{2 a^3}-\frac{(8 (9 A-19 B)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac{(6 A-13 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac{(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{4 (9 A-19 B) \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 = 6.16131, size = 610, normalized size = 3.02 \[ \frac{1920 (6 A-13 B) \cos ^6\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 )+\sec \left (\frac{c}{2}\right ) \sec (c) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left ((1235 B-870 A) \sin \left (\frac{d x}{2}\right )+5 (366 A-761 B) \sin \left (\frac{3 d x}{2}\right )-2094 A \sin \left (c-\frac{d x}{2}\right )+1314 A \sin \left (c+\frac{d x}{2}\right )-1650 A \sin \left (2 c+\frac{d x}{2}\right )-450 A \sin \left (c+\frac{3 d x}{2}\right )+1230 A \sin \left (2 c+\frac{3 d x}{2}\right )-1050 A \sin \left (3 c+\frac{3 d x}{2}\right )+1278 A \sin \left (c+\frac{5 d x}{2}\right )-90 A \sin \left (2 c+\frac{5 d x}{2}\right )+918 A \sin \left (3 c+\frac{5 d x}{2}\right )-450 A \sin \left (4 c+\frac{5 d x}{2}\right )+630 A \sin \left (2 c+\frac{7 d x}{2}\right )+60 A \sin \left (3 c+\frac{7 d x}{2}\right )+480 A \sin \left (4 c+\frac{7 d x}{2}\right )-90 A \sin \left (5 c+\frac{7 d x}{2}\right )+144 A \sin \left (3 c+\frac{9 d x}{2}\right )+30 A \sin \left (4 c+\frac{9 d x}{2}\right )+114 A \sin \left (5 c+\frac{9 d x}{2}\right )+4329 B \sin \left (c-\frac{d x}{2}\right )-1989 B \sin \left (c+\frac{d x}{2}\right )+3575 B \sin \left (2 c+\frac{d x}{2}\right )+475 B \sin \left (c+\frac{3 d x}{2}\right )-2005 B \sin \left (2 c+\frac{3 d x}{2}\right )+2275 B \sin \left (3 c+\frac{3 d x}{2}\right )-2673 B \sin \left (c+\frac{5 d x}{2}\right )-105 B \sin \left (2 c+\frac{5 d x}{2}\right )-1593 B \sin \left (3 c+\frac{5 d x}{2}\right )+975 B \sin \left (4 c+\frac{5 d x}{2}\right )-1325 B \sin \left (2 c+\frac{7 d x}{2}\right )-255 B \sin \left (3 c+\frac{7 d x}{2}\right )-875 B \sin \left (4 c+\frac{7 d x}{2}\right )+195 B \sin \left (5 c+\frac{7 d x}{2}\right )-304 B \sin \left (3 c+\frac{9 d x}{2}\right )-90 B \sin \left (4 c+\frac{9 d x}{2}\right )-214 B \sin \left (5 c+\frac{9 d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(1920*(6*A - 13*B)*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*((-870*A + 1235*B)*Sin[(d*x)/2] + 5*(366*A - 7
61*B)*Sin[(3*d*x)/2] - 2094*A*Sin[c - (d*x)/2] + 4329*B*Sin[c - (d*x)/2] + 1314*A*Sin[c + (d*x)/2] - 1989*B*Si
n[c + (d*x)/2] - 1650*A*Sin[2*c + (d*x)/2] + 3575*B*Sin[2*c + (d*x)/2] - 450*A*Sin[c + (3*d*x)/2] + 475*B*Sin[
c + (3*d*x)/2] + 1230*A*Sin[2*c + (3*d*x)/2] - 2005*B*Sin[2*c + (3*d*x)/2] - 1050*A*Sin[3*c + (3*d*x)/2] + 227
5*B*Sin[3*c + (3*d*x)/2] + 1278*A*Sin[c + (5*d*x)/2] - 2673*B*Sin[c + (5*d*x)/2] - 90*A*Sin[2*c + (5*d*x)/2] -
105*B*Sin[2*c + (5*d*x)/2] + 918*A*Sin[3*c + (5*d*x)/2] - 1593*B*Sin[3*c + (5*d*x)/2] - 450*A*Sin[4*c + (5*d*
x)/2] + 975*B*Sin[4*c + (5*d*x)/2] + 630*A*Sin[2*c + (7*d*x)/2] - 1325*B*Sin[2*c + (7*d*x)/2] + 60*A*Sin[3*c +
(7*d*x)/2] - 255*B*Sin[3*c + (7*d*x)/2] + 480*A*Sin[4*c + (7*d*x)/2] - 875*B*Sin[4*c + (7*d*x)/2] - 90*A*Sin[
5*c + (7*d*x)/2] + 195*B*Sin[5*c + (7*d*x)/2] + 144*A*Sin[3*c + (9*d*x)/2] - 304*B*Sin[3*c + (9*d*x)/2] + 30*A
*Sin[4*c + (9*d*x)/2] - 90*B*Sin[4*c + (9*d*x)/2] + 114*A*Sin[5*c + (9*d*x)/2] - 214*B*Sin[5*c + (9*d*x)/2]))/
(480*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.075, size = 334, normalized size = 1.7 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{2\,B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{31\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{7\,B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{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 ) A}{d{a}^{3}}}+{\frac{13\,B}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{B}{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 ) A}{d{a}^{3}}}-{\frac{13\,B}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{7\,B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{B}{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)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.07032, size = 509, normalized size = 2.52 \begin{align*} -\frac{B{\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 \, A{\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)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(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*A*(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

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Fricas [A]  time = 0.505824, size = 757, normalized size = 3.75 \begin{align*} -\frac{15 \,{\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (9 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (114 \, A - 239 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (234 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 15 \, B\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)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^4 + 3*(6*A - 13*B)*cos(d*x + c)^3 + (6*A
- 13*B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^
4 + 3*(6*A - 13*B)*cos(d*x + c)^3 + (6*A - 13*B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(9*A - 19*B)*c
os(d*x + c)^4 + 3*(114*A - 239*B)*cos(d*x + c)^3 + (234*A - 479*B)*cos(d*x + c)^2 + 15*(2*A - 3*B)*cos(d*x + c
) + 15*B)*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., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \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{B \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)**5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.34511, size = 315, normalized size = 1.56 \begin{align*} -\frac{\frac{30 \,{\left (6 \, A - 13 \, B\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 \, A - 13 \, B\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 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, B \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 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 465 \, B 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)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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