3.90 \(\int \frac{\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=179 \[ -\frac{4 (2 A-3 B) \tan ^3(c+d x)}{3 a^2 d}-\frac{4 (2 A-3 B) \tan (c+d x)}{a^2 d}+\frac{(7 A-10 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{(7 A-10 B) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(7 A-10 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{(A-B) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

((7*A - 10*B)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (4*(2*A - 3*B)*Tan[c + d*x])/(a^2*d) + ((7*A - 10*B)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) + ((7*A - 10*B)*Sec[c + d*x]^3*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((A -
B)*Sec[c + d*x]^4*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - (4*(2*A - 3*B)*Tan[c + d*x]^3)/(3*a^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.321115, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4019, 3787, 3768, 3770, 3767} \[ -\frac{4 (2 A-3 B) \tan ^3(c+d x)}{3 a^2 d}-\frac{4 (2 A-3 B) \tan (c+d x)}{a^2 d}+\frac{(7 A-10 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{(7 A-10 B) \tan (c+d x) \sec ^3(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(7 A-10 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{(A-B) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

((7*A - 10*B)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (4*(2*A - 3*B)*Tan[c + d*x])/(a^2*d) + ((7*A - 10*B)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) + ((7*A - 10*B)*Sec[c + d*x]^3*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((A -
B)*Sec[c + d*x]^4*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - (4*(2*A - 3*B)*Tan[c + d*x]^3)/(3*a^2*d)

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 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]

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]

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^4(c+d x) (4 a (A-B)-3 a (A-2 B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac{(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \sec ^3(c+d x) \left (3 a^2 (7 A-10 B)-12 a^2 (2 A-3 B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac{(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(7 A-10 B) \int \sec ^3(c+d x) \, dx}{a^2}-\frac{(4 (2 A-3 B)) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=\frac{(7 A-10 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(7 A-10 B) \int \sec (c+d x) \, dx}{2 a^2}+\frac{(4 (2 A-3 B)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=\frac{(7 A-10 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{4 (2 A-3 B) \tan (c+d x)}{a^2 d}+\frac{(7 A-10 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{(7 A-10 B) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac{(A-B) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{4 (2 A-3 B) \tan ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [B]  time = 6.35213, size = 764, normalized size = 4.27 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec (c) \cos \left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^4(c+d x) \left (195 A \sin \left (c-\frac{d x}{2}\right )-51 A \sin \left (c+\frac{d x}{2}\right )+189 A \sin \left (2 c+\frac{d x}{2}\right )-A \sin \left (c+\frac{3 d x}{2}\right )-81 A \sin \left (2 c+\frac{3 d x}{2}\right )+119 A \sin \left (3 c+\frac{3 d x}{2}\right )-129 A \sin \left (c+\frac{5 d x}{2}\right )-9 A \sin \left (2 c+\frac{5 d x}{2}\right )-57 A \sin \left (3 c+\frac{5 d x}{2}\right )+63 A \sin \left (4 c+\frac{5 d x}{2}\right )-75 A \sin \left (2 c+\frac{7 d x}{2}\right )-15 A \sin \left (3 c+\frac{7 d x}{2}\right )-39 A \sin \left (4 c+\frac{7 d x}{2}\right )+21 A \sin \left (5 c+\frac{7 d x}{2}\right )-32 A \sin \left (3 c+\frac{9 d x}{2}\right )-12 A \sin \left (4 c+\frac{9 d x}{2}\right )-20 A \sin \left (5 c+\frac{9 d x}{2}\right )+45 A \sin \left (\frac{d x}{2}\right )-201 A \sin \left (\frac{3 d x}{2}\right )-306 B \sin \left (c-\frac{d x}{2}\right )+42 B \sin \left (c+\frac{d x}{2}\right )-270 B \sin \left (2 c+\frac{d x}{2}\right )+50 B \sin \left (c+\frac{3 d x}{2}\right )+90 B \sin \left (2 c+\frac{3 d x}{2}\right )-170 B \sin \left (3 c+\frac{3 d x}{2}\right )+198 B \sin \left (c+\frac{5 d x}{2}\right )+42 B \sin \left (2 c+\frac{5 d x}{2}\right )+66 B \sin \left (3 c+\frac{5 d x}{2}\right )-90 B \sin \left (4 c+\frac{5 d x}{2}\right )+114 B \sin \left (2 c+\frac{7 d x}{2}\right )+36 B \sin \left (3 c+\frac{7 d x}{2}\right )+48 B \sin \left (4 c+\frac{7 d x}{2}\right )-30 B \sin \left (5 c+\frac{7 d x}{2}\right )+48 B \sin \left (3 c+\frac{9 d x}{2}\right )+22 B \sin \left (4 c+\frac{9 d x}{2}\right )+26 B \sin \left (5 c+\frac{9 d x}{2}\right )-6 B \sin \left (\frac{d x}{2}\right )+310 B \sin \left (\frac{3 d x}{2}\right )\right ) (A+B \sec (c+d x))}{96 d (a \sec (c+d x)+a)^2 (A \cos (c+d x)+B)}+\frac{2 (10 B-7 A) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^2 (A \cos (c+d x)+B)}-\frac{2 (10 B-7 A) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^2 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

(2*(-7*A + 10*B)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]*(A + B*Sec[c +
d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) - (2*(-7*A + 10*B)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 +
(d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])
^2) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^4*(A + B*Sec[c + d*x])*(45*A*Sin[(d*x)/2] - 6*B*Sin[(d*
x)/2] - 201*A*Sin[(3*d*x)/2] + 310*B*Sin[(3*d*x)/2] + 195*A*Sin[c - (d*x)/2] - 306*B*Sin[c - (d*x)/2] - 51*A*S
in[c + (d*x)/2] + 42*B*Sin[c + (d*x)/2] + 189*A*Sin[2*c + (d*x)/2] - 270*B*Sin[2*c + (d*x)/2] - A*Sin[c + (3*d
*x)/2] + 50*B*Sin[c + (3*d*x)/2] - 81*A*Sin[2*c + (3*d*x)/2] + 90*B*Sin[2*c + (3*d*x)/2] + 119*A*Sin[3*c + (3*
d*x)/2] - 170*B*Sin[3*c + (3*d*x)/2] - 129*A*Sin[c + (5*d*x)/2] + 198*B*Sin[c + (5*d*x)/2] - 9*A*Sin[2*c + (5*
d*x)/2] + 42*B*Sin[2*c + (5*d*x)/2] - 57*A*Sin[3*c + (5*d*x)/2] + 66*B*Sin[3*c + (5*d*x)/2] + 63*A*Sin[4*c + (
5*d*x)/2] - 90*B*Sin[4*c + (5*d*x)/2] - 75*A*Sin[2*c + (7*d*x)/2] + 114*B*Sin[2*c + (7*d*x)/2] - 15*A*Sin[3*c
+ (7*d*x)/2] + 36*B*Sin[3*c + (7*d*x)/2] - 39*A*Sin[4*c + (7*d*x)/2] + 48*B*Sin[4*c + (7*d*x)/2] + 21*A*Sin[5*
c + (7*d*x)/2] - 30*B*Sin[5*c + (7*d*x)/2] - 32*A*Sin[3*c + (9*d*x)/2] + 48*B*Sin[3*c + (9*d*x)/2] - 12*A*Sin[
4*c + (9*d*x)/2] + 22*B*Sin[4*c + (9*d*x)/2] - 20*A*Sin[5*c + (9*d*x)/2] + 26*B*Sin[5*c + (9*d*x)/2]))/(96*d*(
B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.065, size = 382, normalized size = 2.1 \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{9\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{7\,A}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-5\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) B}{d{a}^{2}}}+{\frac{3\,B}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-5\,{\frac{B}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{5\,A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{3\,B}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{7\,A}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+5\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) B}{d{a}^{2}}}-5\,{\frac{B}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}+{\frac{5\,A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}} \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))^2,x)

[Out]

-1/6/d/a^2*A*tan(1/2*d*x+1/2*c)^3+1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3-7/2/d/a^2*A*tan(1/2*d*x+1/2*c)+9/2/d/a^2*B*
tan(1/2*d*x+1/2*c)+7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*A-5/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*B+3/2/d/a^2/(tan(1/2*
d*x+1/2*c)+1)^2*B-1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2*A-5/d/a^2/(tan(1/2*d*x+1/2*c)+1)*B+5/2/d/a^2/(tan(1/2*d*x
+1/2*c)+1)*A-1/3/d/a^2*B/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2*A-3/2/d/a^2/(tan(1/2*d*x+
1/2*c)-1)^2*B-7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*A+5/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B-5/d/a^2/(tan(1/2*d*x+1/2
*c)-1)*B+5/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)*A-1/3/d/a^2*B/(tan(1/2*d*x+1/2*c)-1)^3

________________________________________________________________________________________

Maxima [B]  time = 1.01261, size = 574, normalized size = 3.21 \begin{align*} \frac{B{\left (\frac{4 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - A{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, 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))^2,x, algorithm="maxima")

[Out]

1/6*(B*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)
/a^2) - A*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) +
sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x +
c)/(cos(d*x + c) + 1) - 1)/a^2))/d

________________________________________________________________________________________

Fricas [A]  time = 0.506427, size = 616, normalized size = 3.44 \begin{align*} \frac{3 \,{\left ({\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (7 \, A - 10 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (16 \,{\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{4} +{\left (43 \, A - 66 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} -{\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 2 \, B\right )} \sin \left (d x + c\right )}{12 \,{\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 )}} \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))^2,x, algorithm="fricas")

[Out]

1/12*(3*((7*A - 10*B)*cos(d*x + c)^5 + 2*(7*A - 10*B)*cos(d*x + c)^4 + (7*A - 10*B)*cos(d*x + c)^3)*log(sin(d*
x + c) + 1) - 3*((7*A - 10*B)*cos(d*x + c)^5 + 2*(7*A - 10*B)*cos(d*x + c)^4 + (7*A - 10*B)*cos(d*x + c)^3)*lo
g(-sin(d*x + c) + 1) - 2*(16*(2*A - 3*B)*cos(d*x + c)^4 + (43*A - 66*B)*cos(d*x + c)^3 + 6*(A - 2*B)*cos(d*x +
c)^2 - (3*A - 2*B)*cos(d*x + c) - 2*B)*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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.35096, size = 305, normalized size = 1.7 \begin{align*} \frac{\frac{3 \,{\left (7 \, A - 10 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \,{\left (7 \, A - 10 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 30 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 24 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 18 \, 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 )}^{3} a^{2}} - \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 27 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

1/6*(3*(7*A - 10*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(7*A - 10*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))
/a^2 + 2*(15*A*tan(1/2*d*x + 1/2*c)^5 - 30*B*tan(1/2*d*x + 1/2*c)^5 - 24*A*tan(1/2*d*x + 1/2*c)^3 + 40*B*tan(1
/2*d*x + 1/2*c)^3 + 9*A*tan(1/2*d*x + 1/2*c) - 18*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2)
- (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + 21*A*a^4*tan(1/2*d*x + 1/2*c) - 27*B*a^4*tan
(1/2*d*x + 1/2*c))/a^6)/d