3.103 \(\int \sec (c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=157 \[ \frac {a^3 (20 A+13 C) \tan ^3(c+d x)}{60 d}+\frac {a^3 (20 A+13 C) \tan (c+d x)}{5 d}+\frac {a^3 (20 A+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (20 A+13 C) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 a d}-\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d} \]

[Out]

1/8*a^3*(20*A+13*C)*arctanh(sin(d*x+c))/d+1/5*a^3*(20*A+13*C)*tan(d*x+c)/d+3/40*a^3*(20*A+13*C)*sec(d*x+c)*tan
(d*x+c)/d-1/20*C*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1/5*C*(a+a*sec(d*x+c))^4*tan(d*x+c)/a/d+1/60*a^3*(20*A+13*C)*
tan(d*x+c)^3/d

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Rubi [A]  time = 0.26, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4083, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac {a^3 (20 A+13 C) \tan ^3(c+d x)}{60 d}+\frac {a^3 (20 A+13 C) \tan (c+d x)}{5 d}+\frac {a^3 (20 A+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (20 A+13 C) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 a d}-\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(20*A + 13*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(20*A + 13*C)*Tan[c + d*x])/(5*d) + (3*a^3*(20*A + 13*C
)*Sec[c + d*x]*Tan[c + d*x])/(40*d) - (C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(20*d) + (C*(a + a*Sec[c + d*x])
^4*Tan[c + d*x])/(5*a*d) + (a^3*(20*A + 13*C)*Tan[c + d*x]^3)/(60*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 4001

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

Rule 4083

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

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {\int \sec (c+d x) (a+a \sec (c+d x))^3 (a (5 A+4 C)-a C \sec (c+d x)) \, dx}{5 a}\\ &=-\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} (20 A+13 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=-\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} (20 A+13 C) \int \left (a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^4(c+d x)\right ) \, dx\\ &=-\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} \left (a^3 (20 A+13 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{20} \left (a^3 (20 A+13 C)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (20 A+13 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (20 A+13 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (20 A+13 C) \tanh ^{-1}(\sin (c+d x))}{20 d}+\frac {3 a^3 (20 A+13 C) \sec (c+d x) \tan (c+d x)}{40 d}-\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{40} \left (3 a^3 (20 A+13 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (20 A+13 C)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{20 d}-\frac {\left (3 a^3 (20 A+13 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{20 d}\\ &=\frac {a^3 (20 A+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (20 A+13 C) \tan (c+d x)}{5 d}+\frac {3 a^3 (20 A+13 C) \sec (c+d x) \tan (c+d x)}{40 d}-\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {a^3 (20 A+13 C) \tan ^3(c+d x)}{60 d}\\ \end {align*}

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Mathematica [B]  time = 2.20, size = 323, normalized size = 2.06 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (240 (20 A+13 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (7 A+3 C) \sin (2 c+d x)+360 A \sin (c+2 d x)+360 A \sin (3 c+2 d x)+1840 A \sin (2 c+3 d x)-360 A \sin (4 c+3 d x)+180 A \sin (3 c+4 d x)+180 A \sin (5 c+4 d x)+440 A \sin (4 c+5 d x)+80 (34 A+29 C) \sin (d x)+750 C \sin (c+2 d x)+750 C \sin (3 c+2 d x)+1520 C \sin (2 c+3 d x)+195 C \sin (3 c+4 d x)+195 C \sin (5 c+4 d x)+304 C \sin (4 c+5 d x))\right )}{7680 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7680*(a^3*(1 + Cos[c + d*x])^3*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^6*Sec[c + d*x]^5*(240*(20*A + 13*C)*
Cos[c + d*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*
(80*(34*A + 29*C)*Sin[d*x] - 240*(7*A + 3*C)*Sin[2*c + d*x] + 360*A*Sin[c + 2*d*x] + 750*C*Sin[c + 2*d*x] + 36
0*A*Sin[3*c + 2*d*x] + 750*C*Sin[3*c + 2*d*x] + 1840*A*Sin[2*c + 3*d*x] + 1520*C*Sin[2*c + 3*d*x] - 360*A*Sin[
4*c + 3*d*x] + 180*A*Sin[3*c + 4*d*x] + 195*C*Sin[3*c + 4*d*x] + 180*A*Sin[5*c + 4*d*x] + 195*C*Sin[5*c + 4*d*
x] + 440*A*Sin[4*c + 5*d*x] + 304*C*Sin[4*c + 5*d*x])))/(d*(A + 2*C + A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.42, size = 161, normalized size = 1.03 \[ \frac {15 \, {\left (20 \, A + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (20 \, A + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (55 \, A + 38 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (12 \, A + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 19 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 90 \, C a^{3} \cos \left (d x + c\right ) + 24 \, C a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(20*A + 13*C)*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(20*A + 13*C)*a^3*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(8*(55*A + 38*C)*a^3*cos(d*x + c)^4 + 15*(12*A + 13*C)*a^3*cos(d*x + c)^3 + 8*(5*A + 19*C)
*a^3*cos(d*x + c)^2 + 90*C*a^3*cos(d*x + c) + 24*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 0.38, size = 246, normalized size = 1.57 \[ \frac {15 \, {\left (20 \, A a^{3} + 13 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (20 \, A a^{3} + 13 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (300 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 195 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1400 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 910 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2560 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1664 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1330 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 660 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 765 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(20*A*a^3 + 13*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(20*A*a^3 + 13*C*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(300*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 195*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 1400*A*a^3*tan(1
/2*d*x + 1/2*c)^7 - 910*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 2560*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1664*C*a^3*tan(1/2*
d*x + 1/2*c)^5 - 2120*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 1330*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 660*A*a^3*tan(1/2*d*x
+ 1/2*c) + 765*C*a^3*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 = 1.74, size = 212, normalized size = 1.35 \[ \frac {5 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {13 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {13 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {11 A \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {38 a^{3} C \tan \left (d x +c \right )}{15 d}+\frac {19 C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {3 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)

[Out]

5/2/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+13/8/d*C*a^3*sec(d*x+c)*tan(d*x+c)+13/8/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c)
)+11/3/d*A*a^3*tan(d*x+c)+38/15*a^3*C*tan(d*x+c)/d+19/15/d*C*a^3*tan(d*x+c)*sec(d*x+c)^2+3/2/d*A*a^3*sec(d*x+c
)*tan(d*x+c)+3/4/d*C*a^3*tan(d*x+c)*sec(d*x+c)^3+1/3/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+1/5/d*C*a^3*tan(d*x+c)*se
c(d*x+c)^4

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maxima [A]  time = 0.35, size = 285, normalized size = 1.82 \[ \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 45 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 720 \, A a^{3} \tan \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c)
)*C*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 45*C*a^3*(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)) - 180*A*a^3*(2*sin(d*x
+ c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*C*a^3*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 7
20*A*a^3*tan(d*x + c))/d

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mupad [B]  time = 5.11, size = 224, normalized size = 1.43 \[ \frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (20\,A+13\,C\right )}{4\,d}-\frac {\left (5\,A\,a^3+\frac {13\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {70\,A\,a^3}{3}-\frac {91\,C\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {128\,A\,a^3}{3}+\frac {416\,C\,a^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {106\,A\,a^3}{3}-\frac {133\,C\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,A\,a^3+\frac {51\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3)/cos(c + d*x),x)

[Out]

(a^3*atanh(tan(c/2 + (d*x)/2))*(20*A + 13*C))/(4*d) - (tan(c/2 + (d*x)/2)*(11*A*a^3 + (51*C*a^3)/4) + tan(c/2
+ (d*x)/2)^9*(5*A*a^3 + (13*C*a^3)/4) - tan(c/2 + (d*x)/2)^7*((70*A*a^3)/3 + (91*C*a^3)/6) - tan(c/2 + (d*x)/2
)^3*((106*A*a^3)/3 + (133*C*a^3)/6) + tan(c/2 + (d*x)/2)^5*((128*A*a^3)/3 + (416*C*a^3)/15))/(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 \[ a^{3} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 3 A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(A*sec(c + d*x), x) + Integral(3*A*sec(c + d*x)**2, x) + Integral(3*A*sec(c + d*x)**3, x) + Inte
gral(A*sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**3, x) + Integral(3*C*sec(c + d*x)**4, x) + Integral(3*C*
sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))

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