3.100 \(\int \frac{\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=156 \[ -\frac{(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac{(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-3 B) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

((A - 3*B)*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((7*A - 27*B)*Tan[c + d*x])/(15*a^3*d) + ((A - B)*Sec[c + d*x]^3*T
an[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((4*A - 9*B)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d
*x])^2) - ((A - 3*B)*Tan[c + d*x])/(d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.428503, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4019, 4008, 3787, 3770, 3767, 8} \[ -\frac{(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac{(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-3 B) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(A-B) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(4 A-9 B) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - 3*B)*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((7*A - 27*B)*Tan[c + d*x])/(15*a^3*d) + ((A - B)*Sec[c + d*x]^3*T
an[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((4*A - 9*B)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d
*x])^2) - ((A - 3*B)*Tan[c + d*x])/(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 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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 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]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^3(c+d x) (3 a (A-B)-a (A-6 B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^2(c+d x) \left (2 a^2 (4 A-9 B)-a^2 (7 A-27 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\int \sec (c+d x) \left (-15 a^3 (A-3 B)+a^3 (7 A-27 B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(7 A-27 B) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac{(A-3 B) \int \sec (c+d x) \, dx}{a^3}\\ &=\frac{(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(7 A-27 B) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac{(A-3 B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(7 A-27 B) \tan (c+d x)}{15 a^3 d}+\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(4 A-9 B) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(A-3 B) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 3.97864, size = 480, normalized size = 3.08 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec (c) \cos \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (5 (32 A-51 B) \sin \left (\frac{d x}{2}\right )+(567 B-167 A) \sin \left (\frac{3 d x}{2}\right )+170 A \sin \left (c-\frac{d x}{2}\right )-170 A \sin \left (c+\frac{d x}{2}\right )+160 A \sin \left (2 c+\frac{d x}{2}\right )+75 A \sin \left (c+\frac{3 d x}{2}\right )-167 A \sin \left (2 c+\frac{3 d x}{2}\right )+75 A \sin \left (3 c+\frac{3 d x}{2}\right )-95 A \sin \left (c+\frac{5 d x}{2}\right )+15 A \sin \left (2 c+\frac{5 d x}{2}\right )-95 A \sin \left (3 c+\frac{5 d x}{2}\right )+15 A \sin \left (4 c+\frac{5 d x}{2}\right )-22 A \sin \left (2 c+\frac{7 d x}{2}\right )-22 A \sin \left (4 c+\frac{7 d x}{2}\right )-600 B \sin \left (c-\frac{d x}{2}\right )+375 B \sin \left (c+\frac{d x}{2}\right )-480 B \sin \left (2 c+\frac{d x}{2}\right )-60 B \sin \left (c+\frac{3 d x}{2}\right )+402 B \sin \left (2 c+\frac{3 d x}{2}\right )-225 B \sin \left (3 c+\frac{3 d x}{2}\right )+315 B \sin \left (c+\frac{5 d x}{2}\right )+30 B \sin \left (2 c+\frac{5 d x}{2}\right )+240 B \sin \left (3 c+\frac{5 d x}{2}\right )-45 B \sin \left (4 c+\frac{5 d x}{2}\right )+72 B \sin \left (2 c+\frac{7 d x}{2}\right )+15 B \sin \left (3 c+\frac{7 d x}{2}\right )+57 B \sin \left (4 c+\frac{7 d x}{2}\right )\right )-960 (A-3 B) \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{120 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-960*(A - 3*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]*(5*(32*A - 51*B)*Sin[(d*x)/2] + (-167*A + 567*B)*Si
n[(3*d*x)/2] + 170*A*Sin[c - (d*x)/2] - 600*B*Sin[c - (d*x)/2] - 170*A*Sin[c + (d*x)/2] + 375*B*Sin[c + (d*x)/
2] + 160*A*Sin[2*c + (d*x)/2] - 480*B*Sin[2*c + (d*x)/2] + 75*A*Sin[c + (3*d*x)/2] - 60*B*Sin[c + (3*d*x)/2] -
167*A*Sin[2*c + (3*d*x)/2] + 402*B*Sin[2*c + (3*d*x)/2] + 75*A*Sin[3*c + (3*d*x)/2] - 225*B*Sin[3*c + (3*d*x)
/2] - 95*A*Sin[c + (5*d*x)/2] + 315*B*Sin[c + (5*d*x)/2] + 15*A*Sin[2*c + (5*d*x)/2] + 30*B*Sin[2*c + (5*d*x)/
2] - 95*A*Sin[3*c + (5*d*x)/2] + 240*B*Sin[3*c + (5*d*x)/2] + 15*A*Sin[4*c + (5*d*x)/2] - 45*B*Sin[4*c + (5*d*
x)/2] - 22*A*Sin[2*c + (7*d*x)/2] + 72*B*Sin[2*c + (7*d*x)/2] + 15*B*Sin[3*c + (7*d*x)/2] - 22*A*Sin[4*c + (7*
d*x)/2] + 57*B*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.057, size = 245, normalized size = 1.6 \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}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{A}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) B}{d{a}^{3}}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) B}{d{a}^{3}}}-{\frac{B}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(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-1/3/d/a^3*A*tan(1/2*d*x+1/2*c)^3+1/2/d/a^
3*B*tan(1/2*d*x+1/2*c)^3-7/4/d/a^3*A*tan(1/2*d*x+1/2*c)+17/4/d/a^3*B*tan(1/2*d*x+1/2*c)+1/d/a^3*ln(tan(1/2*d*x
+1/2*c)+1)*A-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*B-1/d/a^3*ln(tan(1/2*d*x+1/2*c)
-1)*A+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*B

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Maxima [A]  time = 1.02667, size = 386, normalized size = 2.47 \begin{align*} \frac{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 )} - A{\left (\frac{\frac{105 \, \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{3 \, \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*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(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) - A*((105*s
in(d*x + c)/(cos(d*x + c) + 1) + 20*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 - 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 = 0.498247, size = 668, normalized size = 4.28 \begin{align*} \frac{15 \,{\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (A - 3 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (A - 3 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (11 \, A - 36 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (17 \, A - 57 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (32 \, A - 117 \, B\right )} \cos \left (d x + c\right ) - 15 \, B\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*((A - 3*B)*cos(d*x + c)^4 + 3*(A - 3*B)*cos(d*x + c)^3 + 3*(A - 3*B)*cos(d*x + c)^2 + (A - 3*B)*cos(d
*x + c))*log(sin(d*x + c) + 1) - 15*((A - 3*B)*cos(d*x + c)^4 + 3*(A - 3*B)*cos(d*x + c)^3 + 3*(A - 3*B)*cos(d
*x + c)^2 + (A - 3*B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(11*A - 36*B)*cos(d*x + c)^3 + 3*(17*A - 57*
B)*cos(d*x + c)^2 + (32*A - 117*B)*cos(d*x + c) - 15*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x
+ c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{4}{\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 ^{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}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + 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))/a**3

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Giac [A]  time = 1.36327, size = 251, normalized size = 1.61 \begin{align*} \frac{\frac{60 \,{\left (A - 3 \, 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 (A - 3 \, B\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac{120 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} 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} + 20 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, 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)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(A - 3*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(A - 3*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^
3 - 120*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^1
2*tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*ta
n(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d