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

Optimal. Leaf size=156 \[ \frac{2 (5 B-8 C) \tan (c+d x)}{3 a^2 d}-\frac{(4 B-7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{(5 B-8 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac{(4 B-7 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-((4*B - 7*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + (2*(5*B - 8*C)*Tan[c + d*x])/(3*a^2*d) - ((4*B - 7*C)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) + ((5*B - 8*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((B -
C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.38068, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{2 (5 B-8 C) \tan (c+d x)}{3 a^2 d}-\frac{(4 B-7 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{(5 B-8 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac{(4 B-7 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

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])^2,x]

[Out]

-((4*B - 7*C)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + (2*(5*B - 8*C)*Tan[c + d*x])/(3*a^2*d) - ((4*B - 7*C)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) + ((5*B - 8*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((B -
C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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

Mathematica [B]  time = 1.48753, size = 379, normalized size = 2.43 \[ \frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-8 (5 B-8 C) \tan ^3\left (\frac{1}{2} (c+d x)\right )+(26 B-44 C) \tan \left (\frac{1}{2} (c+d x)\right )-64 (B-C) \sin ^8\left (\frac{1}{2} (c+d x)\right ) \csc ^5(c+d x)+8 (B+5 C) \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+\tan ^5\left (\frac{1}{2} (c+d x)\right ) \left (B \sec ^2\left (\frac{1}{2} (c+d x)\right )+14 B-20 C\right )+3 (4 B-7 C) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+3 (4 B-7 C) \tan ^4\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-6 (4 B-7 C) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-128 C \sin ^{12}\left (\frac{1}{2} (c+d x)\right ) \csc ^7(c+d x)\right )}{6 a^2 d} \]

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])^2,x]

[Out]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(3*(4*B - 7*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]]) + 8*(B + 5*C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 64*(B - C)*Csc[c + d*x]^5*Sin[(c +
d*x)/2]^8 - 128*C*Csc[c + d*x]^7*Sin[(c + d*x)/2]^12 + (26*B - 44*C)*Tan[(c + d*x)/2] - 6*(4*B - 7*C)*(Log[Co
s[(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 - 8*(5*B - 8
*C)*Tan[(c + d*x)/2]^3 + 3*(4*B - 7*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 + (14*B - 20*C + B*Sec[(c + d*x)/2]^2)*Tan[(c + d*x)/2]^5))/(6*a^2*d)

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Maple [B]  time = 0.066, size = 294, normalized size = 1.9 \begin{align*}{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{5\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{B}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{5\,C}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) B}{d{a}^{2}}}+{\frac{7\,C}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{B}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{5\,C}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) B}{d{a}^{2}}}-{\frac{7\,C}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{C}{2\,d{a}^{2}} \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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 0.963206, size = 454, normalized size = 2.91 \begin{align*} -\frac{C{\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 )} - B{\left (\frac{\frac{15 \, \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{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac{a^{2} \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 )}}\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))^2,x, algorithm="maxima")

[Out]

-1/6*(C*(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) + s
in(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) - B*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)
/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*
sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 0.518364, size = 566, normalized size = 3.63 \begin{align*} -\frac{3 \,{\left ({\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (4 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (5 \, B - 8 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (28 \, B - 43 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left (B - C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} 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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(3*((4*B - 7*C)*cos(d*x + c)^4 + 2*(4*B - 7*C)*cos(d*x + c)^3 + (4*B - 7*C)*cos(d*x + c)^2)*log(sin(d*x
+ c) + 1) - 3*((4*B - 7*C)*cos(d*x + c)^4 + 2*(4*B - 7*C)*cos(d*x + c)^3 + (4*B - 7*C)*cos(d*x + c)^2)*log(-si
n(d*x + c) + 1) - 2*(4*(5*B - 8*C)*cos(d*x + c)^3 + (28*B - 43*C)*cos(d*x + c)^2 + 6*(B - C)*cos(d*x + c) + 3*
C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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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 ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{5}{\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)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(B*sec(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**5/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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Giac [A]  time = 1.22009, size = 267, normalized size = 1.71 \begin{align*} -\frac{\frac{3 \,{\left (4 \, B - 7 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \,{\left (4 \, B - 7 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{6 \,{\left (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, 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 ) + 3 \, 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^{2}} - \frac{B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 21 \, C 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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(3*(4*B - 7*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(4*B - 7*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
a^2 + 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x
+ 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3
+ 15*B*a^4*tan(1/2*d*x + 1/2*c) - 21*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d