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

Optimal. Leaf size=151 \[ \frac {5 a^4 (8 A+7 B) \tan (c+d x)}{8 d}+\frac {a^4 (48 A+35 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(32 A+35 B) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+a^4 A x+\frac {(4 A+7 B) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[Out]

a^4*A*x+1/8*a^4*(48*A+35*B)*arctanh(sin(d*x+c))/d+5/8*a^4*(8*A+7*B)*tan(d*x+c)/d+1/4*a*B*(a+a*sec(d*x+c))^3*ta
n(d*x+c)/d+1/12*(4*A+7*B)*(a^2+a^2*sec(d*x+c))^2*tan(d*x+c)/d+1/24*(32*A+35*B)*(a^4+a^4*sec(d*x+c))*tan(d*x+c)
/d

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Rubi [A]  time = 0.21, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3917, 3914, 3767, 8, 3770} \[ \frac {5 a^4 (8 A+7 B) \tan (c+d x)}{8 d}+\frac {a^4 (48 A+35 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 A+7 B) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}+\frac {(32 A+35 B) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+a^4 A x+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*A*x + (a^4*(48*A + 35*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^4*(8*A + 7*B)*Tan[c + d*x])/(8*d) + (a*B*(a +
a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + ((4*A + 7*B)*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + ((32*A
+ 35*B)*(a^4 + a^4*Sec[c + d*x])*Tan[c + d*x])/(24*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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {a B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+a \sec (c+d x))^3 (4 a A+a (4 A+7 B) \sec (c+d x)) \, dx\\ &=\frac {a B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}+\frac {1}{12} \int (a+a \sec (c+d x))^2 \left (12 a^2 A+a^2 (32 A+35 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}+\frac {(32 A+35 B) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {1}{24} \int (a+a \sec (c+d x)) \left (24 a^3 A+15 a^3 (8 A+7 B) \sec (c+d x)\right ) \, dx\\ &=a^4 A x+\frac {a B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}+\frac {(32 A+35 B) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {1}{8} \left (5 a^4 (8 A+7 B)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (a^4 (48 A+35 B)\right ) \int \sec (c+d x) \, dx\\ &=a^4 A x+\frac {a^4 (48 A+35 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}+\frac {(32 A+35 B) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}-\frac {\left (5 a^4 (8 A+7 B)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{8 d}\\ &=a^4 A x+\frac {a^4 (48 A+35 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^4 (8 A+7 B) \tan (c+d x)}{8 d}+\frac {a B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 A+7 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}+\frac {(32 A+35 B) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}\\ \end {align*}

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Mathematica [B]  time = 2.08, size = 326, normalized size = 2.16 \[ \frac {a^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^4 \left (\sec (c) (48 A \sin (2 c+d x)+496 A \sin (c+2 d x)-144 A \sin (3 c+2 d x)+48 A \sin (2 c+3 d x)+48 A \sin (4 c+3 d x)+160 A \sin (3 c+4 d x)+72 A d x \cos (c)+48 A d x \cos (c+2 d x)+48 A d x \cos (3 c+2 d x)+12 A d x \cos (3 c+4 d x)+12 A d x \cos (5 c+4 d x)-480 A \sin (c)+48 A \sin (d x)+105 B \sin (2 c+d x)+544 B \sin (c+2 d x)-96 B \sin (3 c+2 d x)+81 B \sin (2 c+3 d x)+81 B \sin (4 c+3 d x)+160 B \sin (3 c+4 d x)-480 B \sin (c)+105 B \sin (d x))-24 (48 A+35 B) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3072 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(-24*(48*A + 35*B)*Cos[c + d*x]^4*(Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(72*A*d*x*Cos[c] + 48*A*d*x*Cos[c + 2*d*x] + 4
8*A*d*x*Cos[3*c + 2*d*x] + 12*A*d*x*Cos[3*c + 4*d*x] + 12*A*d*x*Cos[5*c + 4*d*x] - 480*A*Sin[c] - 480*B*Sin[c]
+ 48*A*Sin[d*x] + 105*B*Sin[d*x] + 48*A*Sin[2*c + d*x] + 105*B*Sin[2*c + d*x] + 496*A*Sin[c + 2*d*x] + 544*B*
Sin[c + 2*d*x] - 144*A*Sin[3*c + 2*d*x] - 96*B*Sin[3*c + 2*d*x] + 48*A*Sin[2*c + 3*d*x] + 81*B*Sin[2*c + 3*d*x
] + 48*A*Sin[4*c + 3*d*x] + 81*B*Sin[4*c + 3*d*x] + 160*A*Sin[3*c + 4*d*x] + 160*B*Sin[3*c + 4*d*x])))/(3072*d
)

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fricas [A]  time = 0.45, size = 157, normalized size = 1.04 \[ \frac {48 \, A a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (48 \, A + 35 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (48 \, A + 35 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (160 \, {\left (A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (16 \, A + 27 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, B a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*A*a^4*d*x*cos(d*x + c)^4 + 3*(48*A + 35*B)*a^4*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(48*A + 35*B)
*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(160*(A + B)*a^4*cos(d*x + c)^3 + 3*(16*A + 27*B)*a^4*cos(d*x +
c)^2 + 8*(A + 4*B)*a^4*cos(d*x + c) + 6*B*a^4)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.63, size = 223, normalized size = 1.48 \[ \frac {24 \, {\left (d x + c\right )} A a^{4} + 3 \, {\left (48 \, A a^{4} + 35 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (48 \, A a^{4} + 35 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 424 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 385 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 520 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 511 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 216 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 279 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(d*x + c)*A*a^4 + 3*(48*A*a^4 + 35*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(48*A*a^4 + 35*B*a^4
)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 105*B*a^4*tan(1/2*d*x + 1/2*c)^7
- 424*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 385*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 520*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 511
*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 216*A*a^4*tan(1/2*d*x + 1/2*c) - 279*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 1.42, size = 204, normalized size = 1.35 \[ A \,a^{4} x +\frac {A \,a^{4} c}{d}+\frac {35 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {20 a^{4} B \tan \left (d x +c \right )}{3 d}+\frac {20 A \,a^{4} \tan \left (d x +c \right )}{3 d}+\frac {27 a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {2 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {4 a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a^4*x+1/d*A*a^4*c+35/8/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+6/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+20/3/d*a^4*B*ta
n(d*x+c)+20/3/d*A*a^4*tan(d*x+c)+27/8/d*a^4*B*sec(d*x+c)*tan(d*x+c)+2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+4/3/d*a^4*
B*tan(d*x+c)*sec(d*x+c)^2+1/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^4*B*tan(d*x+c)*sec(d*x+c)^3

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maxima [B]  time = 0.36, size = 293, normalized size = 1.94 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 48 \, {\left (d x + c\right )} A a^{4} + 64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} - 3 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, B a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 288 \, A a^{4} \tan \left (d x + c\right ) + 192 \, B a^{4} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 48*(d*x + c)*A*a^4 + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*B
*a^4 - 3*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)) - 48*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)) - 72*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)) + 192*A*a^4*log(sec(d*x + c) + tan(d*x + c)) + 48*B*a^4*log(sec(d*x + c) + tan(d*x + c)) + 288*A*a^
4*tan(d*x + c) + 192*B*a^4*tan(d*x + c))/d

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mupad [B]  time = 2.10, size = 255, normalized size = 1.69 \[ \frac {2\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {35\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {20\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {27\,B\,a^4\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {4\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4} \]

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,x)

[Out]

(2*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)))/d + (35*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (20*A*a^4*sin(c + d*x))/(3*d*cos(c + d
*x)) + (2*A*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (A*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (20*B*a^4*sin(c
+ d*x))/(3*d*cos(c + d*x)) + (27*B*a^4*sin(c + d*x))/(8*d*cos(c + d*x)^2) + (4*B*a^4*sin(c + d*x))/(3*d*cos(c
+ d*x)^3) + (B*a^4*sin(c + d*x))/(4*d*cos(c + d*x)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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