3.73 \(\int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=159 \[ \frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {7 a^4 (5 A+4 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 (5 A+4 B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {27 a^4 (5 A+4 B) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

[Out]

7/8*a^4*(5*A+4*B)*arctanh(sin(d*x+c))/d+8/5*a^4*(5*A+4*B)*tan(d*x+c)/d+27/40*a^4*(5*A+4*B)*sec(d*x+c)*tan(d*x+
c)/d+1/20*a^4*(5*A+4*B)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*B*(a+a*sec(d*x+c))^4*tan(d*x+c)/d+4/15*a^4*(5*A+4*B)*tan
(d*x+c)^3/d

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Rubi [A]  time = 0.18, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4001, 3791, 3770, 3767, 8, 3768} \[ \frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {7 a^4 (5 A+4 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 (5 A+4 B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {27 a^4 (5 A+4 B) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^4*(5*A + 4*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (8*a^4*(5*A + 4*B)*Tan[c + d*x])/(5*d) + (27*a^4*(5*A + 4*B)
*Sec[c + d*x]*Tan[c + d*x])/(40*d) + (a^4*(5*A + 4*B)*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (B*(a + a*Sec[c +
d*x])^4*Tan[c + d*x])/(5*d) + (4*a^4*(5*A + 4*B)*Tan[c + d*x]^3)/(15*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)]

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{5} (5 A+4 B) \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx\\ &=\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{5} (5 A+4 B) \int \left (a^4 \sec (c+d x)+4 a^4 \sec ^2(c+d x)+6 a^4 \sec ^3(c+d x)+4 a^4 \sec ^4(c+d x)+a^4 \sec ^5(c+d x)\right ) \, dx\\ &=\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{5} \left (a^4 (5 A+4 B)\right ) \int \sec (c+d x) \, dx+\frac {1}{5} \left (a^4 (5 A+4 B)\right ) \int \sec ^5(c+d x) \, dx+\frac {1}{5} \left (4 a^4 (5 A+4 B)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{5} \left (4 a^4 (5 A+4 B)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{5} \left (6 a^4 (5 A+4 B)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^4 (5 A+4 B) \tanh ^{-1}(\sin (c+d x))}{5 d}+\frac {3 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{5 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{20} \left (3 a^4 (5 A+4 B)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (3 a^4 (5 A+4 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (4 a^4 (5 A+4 B)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 d}-\frac {\left (4 a^4 (5 A+4 B)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {4 a^4 (5 A+4 B) \tanh ^{-1}(\sin (c+d x))}{5 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac {1}{40} \left (3 a^4 (5 A+4 B)\right ) \int \sec (c+d x) \, dx\\ &=\frac {7 a^4 (5 A+4 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.75, size = 306, normalized size = 1.92 \[ -\frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (1680 (5 A+4 B) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-960 (3 A+2 B) \sin (2 c+d x)+80 (64 A+59 B) \sin (d x)+930 A \sin (c+2 d x)+930 A \sin (3 c+2 d x)+3520 A \sin (2 c+3 d x)-480 A \sin (4 c+3 d x)+405 A \sin (3 c+4 d x)+405 A \sin (5 c+4 d x)+800 A \sin (4 c+5 d x)+1320 B \sin (c+2 d x)+1320 B \sin (3 c+2 d x)+3200 B \sin (2 c+3 d x)-120 B \sin (4 c+3 d x)+420 B \sin (3 c+4 d x)+420 B \sin (5 c+4 d x)+664 B \sin (4 c+5 d x))\right )}{30720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30720*(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*Sec[c + d*x]^5*(1680*(5*A + 4*B)*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*(64*A + 59*B)*Sin[d
*x] - 960*(3*A + 2*B)*Sin[2*c + d*x] + 930*A*Sin[c + 2*d*x] + 1320*B*Sin[c + 2*d*x] + 930*A*Sin[3*c + 2*d*x] +
1320*B*Sin[3*c + 2*d*x] + 3520*A*Sin[2*c + 3*d*x] + 3200*B*Sin[2*c + 3*d*x] - 480*A*Sin[4*c + 3*d*x] - 120*B*
Sin[4*c + 3*d*x] + 405*A*Sin[3*c + 4*d*x] + 420*B*Sin[3*c + 4*d*x] + 405*A*Sin[5*c + 4*d*x] + 420*B*Sin[5*c +
4*d*x] + 800*A*Sin[4*c + 5*d*x] + 664*B*Sin[4*c + 5*d*x])))/d

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fricas [A]  time = 0.46, size = 165, normalized size = 1.04 \[ \frac {105 \, {\left (5 \, A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (5 \, A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (100 \, A + 83 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (27 \, A + 28 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \, {\left (10 \, A + 17 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 24 \, B a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(105*(5*A + 4*B)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 105*(5*A + 4*B)*a^4*cos(d*x + c)^5*log(-sin(
d*x + c) + 1) + 2*(8*(100*A + 83*B)*a^4*cos(d*x + c)^4 + 15*(27*A + 28*B)*a^4*cos(d*x + c)^3 + 16*(10*A + 17*B
)*a^4*cos(d*x + c)^2 + 30*(A + 4*B)*a^4*cos(d*x + c) + 24*B*a^4)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 2.22, size = 246, normalized size = 1.55 \[ \frac {105 \, {\left (5 \, A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (5 \, A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (525 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2450 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1960 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4480 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3584 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3950 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3160 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1395 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1500 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(105*(5*A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(5*A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*
x + 1/2*c) - 1)) - 2*(525*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 420*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 2450*A*a^4*tan(1/2
*d*x + 1/2*c)^7 - 1960*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 4480*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3584*B*a^4*tan(1/2*d
*x + 1/2*c)^5 - 3950*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 3160*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 1395*A*a^4*tan(1/2*d*x
+ 1/2*c) + 1500*B*a^4*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 = 1.70, size = 234, normalized size = 1.47 \[ \frac {35 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {83 a^{4} B \tan \left (d x +c \right )}{15 d}+\frac {20 A \,a^{4} \tan \left (d x +c \right )}{3 d}+\frac {7 a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {7 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {27 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {34 a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

35/8/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+83/15/d*a^4*B*tan(d*x+c)+20/3/d*A*a^4*tan(d*x+c)+7/2/d*a^4*B*sec(d*x+c)
*tan(d*x+c)+7/2/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+27/8/d*A*a^4*sec(d*x+c)*tan(d*x+c)+34/15/d*a^4*B*tan(d*x+c)*
sec(d*x+c)^2+4/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/d*a^4*B*tan(d*x+c)*sec(d*x+c)^3+1/4/d*A*a^4*tan(d*x+c)*sec(
d*x+c)^3+1/5/d*a^4*B*tan(d*x+c)*sec(d*x+c)^4

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maxima [B]  time = 0.35, size = 369, normalized size = 2.32 \[ \frac {320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{4} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} - 15 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 960 \, A a^{4} \tan \left (d x + c\right ) + 240 \, B a^{4} \tan \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(320*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*B*a^4 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 - 15*A*a^4*(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^4*(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)) - 360*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) -
240*B*a^4*(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^4*l
og(sec(d*x + c) + tan(d*x + c)) + 960*A*a^4*tan(d*x + c) + 240*B*a^4*tan(d*x + c))/d

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mupad [B]  time = 4.54, size = 224, normalized size = 1.41 \[ \frac {7\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (5\,A+4\,B\right )}{4\,d}-\frac {\left (\frac {35\,A\,a^4}{4}+7\,B\,a^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {245\,A\,a^4}{6}-\frac {98\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {224\,A\,a^4}{3}+\frac {896\,B\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {395\,A\,a^4}{6}-\frac {158\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {93\,A\,a^4}{4}+25\,B\,a^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4)/cos(c + d*x),x)

[Out]

(7*a^4*atanh(tan(c/2 + (d*x)/2))*(5*A + 4*B))/(4*d) - (tan(c/2 + (d*x)/2)*((93*A*a^4)/4 + 25*B*a^4) + tan(c/2
+ (d*x)/2)^9*((35*A*a^4)/4 + 7*B*a^4) - tan(c/2 + (d*x)/2)^7*((245*A*a^4)/6 + (98*B*a^4)/3) - tan(c/2 + (d*x)/
2)^3*((395*A*a^4)/6 + (158*B*a^4)/3) + tan(c/2 + (d*x)/2)^5*((224*A*a^4)/3 + (896*B*a^4)/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)^1
0 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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