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

Optimal. Leaf size=131 \[ -\frac{(3 B-4 C) \tan ^3(c+d x)}{3 a d}-\frac{(3 B-4 C) \tan (c+d x)}{a d}+\frac{3 (B-C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac{3 (B-C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.253334, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4072, 4019, 3787, 3768, 3770, 3767} \[ -\frac{(3 B-4 C) \tan ^3(c+d x)}{3 a d}-\frac{(3 B-4 C) \tan (c+d x)}{a d}+\frac{3 (B-C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac{3 (B-C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

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

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

Mathematica [B]  time = 1.11916, size = 550, normalized size = 4.2 \[ -\frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (27 (B-C) \cos \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 )+27 (B-C) \cos \left (\frac{3}{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 )-12 B \sin \left (\frac{1}{2} (c+d x)\right )+18 B \sin \left (\frac{3}{2} (c+d x)\right )-6 B \sin \left (\frac{5}{2} (c+d x)\right )+12 B \sin \left (\frac{7}{2} (c+d x)\right )+9 B \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+9 B \cos \left (\frac{7}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-9 B \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-9 B \cos \left (\frac{7}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-30 C \sin \left (\frac{3}{2} (c+d x)\right )+2 C \sin \left (\frac{5}{2} (c+d x)\right )-16 C \sin \left (\frac{7}{2} (c+d x)\right )-9 C \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-9 C \cos \left (\frac{7}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+9 C \cos \left (\frac{5}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+9 C \cos \left (\frac{7}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{24 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

-(Cos[(c + d*x)/2]*Sec[c + d*x]^3*(9*B*Cos[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 9*C*Cos
[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*B*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] - 9*C*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 27*(B - C)*Cos[(c + d*x
)/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 27*(B - C)*Cos[(3
*(c + d*x))/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 9*B*Cos
[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*C*Cos[(5*(c + d*x))/2]*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] - 9*B*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*C*Cos[(7*(c + d*x))/2
]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 12*B*Sin[(c + d*x)/2] + 18*B*Sin[(3*(c + d*x))/2] - 30*C*Sin[(3*(
c + d*x))/2] - 6*B*Sin[(5*(c + d*x))/2] + 2*C*Sin[(5*(c + d*x))/2] + 12*B*Sin[(7*(c + d*x))/2] - 16*C*Sin[(7*(
c + d*x))/2]))/(24*a*d*(1 + Cos[c + d*x]))

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Maple [B]  time = 0.062, size = 340, normalized size = 2.6 \begin{align*} -{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{3\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{B}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,B}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{5\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,B}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{C}{3\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{B}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{3\,B}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{5\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3\,B}{2\,ad} \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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)

[Out]

-1/a/d*B*tan(1/2*d*x+1/2*c)+1/a/d*C*tan(1/2*d*x+1/2*c)-1/3/a/d*C/(tan(1/2*d*x+1/2*c)+1)^3-1/2/a/d/(tan(1/2*d*x
+1/2*c)+1)^2*B+1/a/d/(tan(1/2*d*x+1/2*c)+1)^2*C-3/2/a/d*ln(tan(1/2*d*x+1/2*c)+1)*C+3/2/a/d*ln(tan(1/2*d*x+1/2*
c)+1)*B-5/2/a/d/(tan(1/2*d*x+1/2*c)+1)*C+3/2/a/d/(tan(1/2*d*x+1/2*c)+1)*B-1/3/a/d*C/(tan(1/2*d*x+1/2*c)-1)^3-1
/a/d/(tan(1/2*d*x+1/2*c)-1)^2*C+1/2/a/d/(tan(1/2*d*x+1/2*c)-1)^2*B+3/2/a/d*ln(tan(1/2*d*x+1/2*c)-1)*C-3/2/a/d*
ln(tan(1/2*d*x+1/2*c)-1)*B-5/2/a/d/(tan(1/2*d*x+1/2*c)-1)*C+3/2/a/d/(tan(1/2*d*x+1/2*c)-1)*B

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Maxima [B]  time = 0.954362, size = 497, normalized size = 3.79 \begin{align*} \frac{C{\left (\frac{2 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{16 \, \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 - \frac{3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{6 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B{\left (\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{2 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(C*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*B*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c
) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*si
n(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 0.519179, size = 417, normalized size = 3.18 \begin{align*} \frac{9 \,{\left ({\left (B - C\right )} \cos \left (d x + c\right )^{4} +{\left (B - C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \,{\left ({\left (B - C\right )} \cos \left (d x + c\right )^{4} +{\left (B - C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (3 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (3 \, B - C\right )} \cos \left (d x + c\right ) - 2 \, C\right )} \sin \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{4} + a 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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(9*((B - C)*cos(d*x + c)^4 + (B - C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 9*((B - C)*cos(d*x + c)^4 +
(B - C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(3*B - 4*C)*cos(d*x + c)^3 + (3*B - 7*C)*cos(d*x + c)^2
- (3*B - C)*cos(d*x + c) - 2*C)*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{B \sec ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{5}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

(Integral(B*sec(c + d*x)**4/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**5/(sec(c + d*x) + 1), x))/a

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Giac [A]  time = 1.18516, size = 246, normalized size = 1.88 \begin{align*} \frac{\frac{9 \,{\left (B - C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{9 \,{\left (B - C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{6 \,{\left (B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} + \frac{2 \,{\left (9 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 16 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, 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 )}^{3} a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(9*(B - C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*(B - C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 6*(B*ta
n(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a + 2*(9*B*tan(1/2*d*x + 1/2*c)^5 - 15*C*tan(1/2*d*x + 1/2*c)^5 -
12*B*tan(1/2*d*x + 1/2*c)^3 + 16*C*tan(1/2*d*x + 1/2*c)^3 + 3*B*tan(1/2*d*x + 1/2*c) - 9*C*tan(1/2*d*x + 1/2*
c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d