3.314 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=169 \[ \frac {a^2 (10 B+9 C) \tan ^3(c+d x)}{15 d}+\frac {a^2 (10 B+9 C) \tan (c+d x)}{5 d}+\frac {a^2 (7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 (7 B+6 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

[Out]

1/8*a^2*(7*B+6*C)*arctanh(sin(d*x+c))/d+1/5*a^2*(10*B+9*C)*tan(d*x+c)/d+1/8*a^2*(7*B+6*C)*sec(d*x+c)*tan(d*x+c
)/d+1/20*a^2*(5*B+6*C)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*C*sec(d*x+c)^3*(a^2+a^2*sec(d*x+c))*tan(d*x+c)/d+1/15*a^2
*(10*B+9*C)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.32, antiderivative size = 169, normalized size of antiderivative = 1.00, 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, 4018, 3997, 3787, 3768, 3770, 3767} \[ \frac {a^2 (10 B+9 C) \tan ^3(c+d x)}{15 d}+\frac {a^2 (10 B+9 C) \tan (c+d x)}{5 d}+\frac {a^2 (7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 (7 B+6 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*B + 6*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(10*B + 9*C)*Tan[c + d*x])/(5*d) + (a^2*(7*B + 6*C)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (a^2*(5*B + 6*C)*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (C*Sec[c + d*x]^3*(a^2 +
a^2*Sec[c + d*x])*Tan[c + d*x])/(5*d) + (a^2*(10*B + 9*C)*Tan[c + d*x]^3)/(15*d)

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 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 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 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 4018

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

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]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac {C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) (a+a \sec (c+d x)) (a (5 B+3 C)+a (5 B+6 C) \sec (c+d x)) \, dx\\ &=\frac {a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{20} \int \sec ^3(c+d x) \left (5 a^2 (7 B+6 C)+4 a^2 (10 B+9 C) \sec (c+d x)\right ) \, dx\\ &=\frac {a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^2 (7 B+6 C)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (a^2 (10 B+9 C)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {a^2 (7 B+6 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^2 (7 B+6 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (10 B+9 C)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {a^2 (7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (10 B+9 C) \tan (c+d x)}{5 d}+\frac {a^2 (7 B+6 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {a^2 (10 B+9 C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [B]  time = 0.83, size = 391, normalized size = 2.31 \[ -\frac {a^2 \sec ^5(c+d x) \left (150 (7 B+6 C) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+75 (7 B+6 C) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-640 B \sin (c+d x)-660 B \sin (2 (c+d x))-800 B \sin (3 (c+d x))-210 B \sin (4 (c+d x))-160 B \sin (5 (c+d x))+105 B \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-105 B \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-960 C \sin (c+d x)-840 C \sin (2 (c+d x))-720 C \sin (3 (c+d x))-180 C \sin (4 (c+d x))-144 C \sin (5 (c+d x))+90 C \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-90 C \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1920*(a^2*Sec[c + d*x]^5*(105*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 90*C*Cos[5*(c +
d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 150*(7*B + 6*C)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 75*(7*B + 6*C)*Cos[3*(c + d*x)]*(Log[Cos[(c + d*x)/2]
- Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 105*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]] - 90*C*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 640*B*Sin[c + d*x] - 9
60*C*Sin[c + d*x] - 660*B*Sin[2*(c + d*x)] - 840*C*Sin[2*(c + d*x)] - 800*B*Sin[3*(c + d*x)] - 720*C*Sin[3*(c
+ d*x)] - 210*B*Sin[4*(c + d*x)] - 180*C*Sin[4*(c + d*x)] - 160*B*Sin[5*(c + d*x)] - 144*C*Sin[5*(c + d*x)]))/
d

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fricas [A]  time = 0.51, size = 165, normalized size = 0.98 \[ \frac {15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 24 \, C a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(7*B + 6*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(7*B + 6*C)*a^2*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(16*(10*B + 9*C)*a^2*cos(d*x + c)^4 + 15*(7*B + 6*C)*a^2*cos(d*x + c)^3 + 8*(10*B + 9*C)*a^2*c
os(d*x + c)^2 + 30*(B + 2*C)*a^2*cos(d*x + c) + 24*C*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 0.32, size = 246, normalized size = 1.46 \[ \frac {15 \, {\left (7 \, B a^{2} + 6 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (7 \, B a^{2} + 6 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 90 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 490 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 420 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 800 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 864 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 790 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 540 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 390 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(7*B*a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(7*B*a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1)) - 2*(105*B*a^2*tan(1/2*d*x + 1/2*c)^9 + 90*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 490*B*a^2*tan(1/2*d*x
+ 1/2*c)^7 - 420*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 800*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 864*C*a^2*tan(1/2*d*x + 1/
2*c)^5 - 790*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 540*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 375*B*a^2*tan(1/2*d*x + 1/2*c)
+ 390*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 1.92, size = 235, normalized size = 1.39 \[ \frac {7 a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {7 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {6 a^{2} C \tan \left (d x +c \right )}{5 d}+\frac {3 a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {4 a^{2} B \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2} B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

7/8*a^2*B*sec(d*x+c)*tan(d*x+c)/d+7/8/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+6/5/d*a^2*C*tan(d*x+c)+3/5/d*a^2*C*tan
(d*x+c)*sec(d*x+c)^2+4/3*a^2*B*tan(d*x+c)/d+2/3*a^2*B*sec(d*x+c)^2*tan(d*x+c)/d+1/2/d*a^2*C*tan(d*x+c)*sec(d*x
+c)^3+3/4/d*a^2*C*sec(d*x+c)*tan(d*x+c)+3/4/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+1/4*a^2*B*sec(d*x+c)^3*tan(d*x+c
)/d+1/5/d*a^2*C*tan(d*x+c)*sec(d*x+c)^4

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maxima [A]  time = 0.48, size = 278, normalized size = 1.64 \[ \frac {160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, C a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*C*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 15*B*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d
*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*C*a^2*(2*(3*sin(d*
x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 60*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 5.64, size = 224, normalized size = 1.33 \[ \frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,B+6\,C\right )}{4\,d}-\frac {\left (\frac {7\,B\,a^2}{4}+\frac {3\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {49\,B\,a^2}{6}-7\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {40\,B\,a^2}{3}+\frac {72\,C\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {79\,B\,a^2}{6}-9\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,B\,a^2}{4}+\frac {13\,C\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/cos(c + d*x)^2,x)

[Out]

(a^2*atanh(tan(c/2 + (d*x)/2))*(7*B + 6*C))/(4*d) - (tan(c/2 + (d*x)/2)*((25*B*a^2)/4 + (13*C*a^2)/2) + tan(c/
2 + (d*x)/2)^9*((7*B*a^2)/4 + (3*C*a^2)/2) - tan(c/2 + (d*x)/2)^7*((49*B*a^2)/6 + 7*C*a^2) - tan(c/2 + (d*x)/2
)^3*((79*B*a^2)/6 + 9*C*a^2) + tan(c/2 + (d*x)/2)^5*((40*B*a^2)/3 + (72*C*a^2)/5))/(d*(5*tan(c/2 + (d*x)/2)^2
- 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int B \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(B*sec(c + d*x)**3, x) + Integral(2*B*sec(c + d*x)**4, x) + Integral(B*sec(c + d*x)**5, x) + Int
egral(C*sec(c + d*x)**4, x) + Integral(2*C*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))

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