3.5.40 \(\int (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [440]

Optimal. Leaf size=195 \[ a^4 A x+\frac {a^4 (48 A+35 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 (40 A+35 B+28 C) \tan (c+d x)}{8 d}+\frac {a (5 B+4 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {(20 A+35 B+28 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{60 d}+\frac {(32 A+35 B+28 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d} \]

[Out]

a^4*A*x+1/8*a^4*(48*A+35*B+28*C)*arctanh(sin(d*x+c))/d+1/8*a^4*(40*A+35*B+28*C)*tan(d*x+c)/d+1/20*a*(5*B+4*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)/d+1/60*(20*A+35*B+28*C)*(a^2+a^2*sec(d*x+c
))^2*tan(d*x+c)/d+1/24*(32*A+35*B+28*C)*(a^4+a^4*sec(d*x+c))*tan(d*x+c)/d

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Rubi [A]
time = 0.23, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4139, 4002, 3999, 3852, 8, 3855} \begin {gather*} \frac {a^4 (40 A+35 B+28 C) \tan (c+d x)}{8 d}+\frac {a^4 (48 A+35 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+a^4 A x+\frac {(20 A+35 B+28 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{60 d}+\frac {a (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

a^4*A*x + (a^4*(48*A + 35*B + 28*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^4*(40*A + 35*B + 28*C)*Tan[c + d*x])/(8*
d) + (a*(5*B + 4*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*d
) + ((20*A + 35*B + 28*C)*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(60*d) + ((32*A + 35*B + 28*C)*(a^4 + a^4*S
ec[c + d*x])*Tan[c + d*x])/(24*d)

Rule 8

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

Rule 3852

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 3855

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

Rule 3999

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 4002

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] &&
GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4139

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

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {\int (a+a \sec (c+d x))^4 (5 a A+a (5 B+4 C) \sec (c+d x)) \, dx}{5 a}\\ &=\frac {a (5 B+4 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 d}+\frac {\int (a+a \sec (c+d x))^3 \left (20 a^2 A+a^2 (20 A+35 B+28 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac {a (5 B+4 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 d}+\frac {(20 A+35 B+28 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{60 d}+\frac {\int (a+a \sec (c+d x))^2 \left (60 a^3 A+5 a^3 (32 A+35 B+28 C) \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac {a (5 B+4 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 d}+\frac {(20 A+35 B+28 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{60 d}+\frac {(32 A+35 B+28 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {\int (a+a \sec (c+d x)) \left (120 a^4 A+15 a^4 (40 A+35 B+28 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=a^4 A x+\frac {a (5 B+4 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 d}+\frac {(20 A+35 B+28 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{60 d}+\frac {(32 A+35 B+28 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac {1}{8} \left (a^4 (40 A+35 B+28 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (a^4 (48 A+35 B+28 C)\right ) \int \sec (c+d x) \, dx\\ &=a^4 A x+\frac {a^4 (48 A+35 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (5 B+4 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 d}+\frac {(20 A+35 B+28 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{60 d}+\frac {(32 A+35 B+28 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}-\frac {\left (a^4 (40 A+35 B+28 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{8 d}\\ &=a^4 A x+\frac {a^4 (48 A+35 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 (40 A+35 B+28 C) \tan (c+d x)}{8 d}+\frac {a (5 B+4 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 d}+\frac {(20 A+35 B+28 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{60 d}+\frac {(32 A+35 B+28 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(538\) vs. \(2(195)=390\).
time = 5.25, size = 538, normalized size = 2.76 \begin {gather*} \frac {a^4 (1+\cos (c+d x))^4 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (-240 (48 A+35 B+28 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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec (c) (600 A d x \cos (d x)+600 A d x \cos (2 c+d x)+300 A d x \cos (2 c+3 d x)+300 A d x \cos (4 c+3 d x)+60 A d x \cos (4 c+5 d x)+60 A d x \cos (6 c+5 d x)+4880 A \sin (d x)+5120 B \sin (d x)+4720 C \sin (d x)-3120 A \sin (2 c+d x)-2880 B \sin (2 c+d x)-1920 C \sin (2 c+d x)+480 A \sin (c+2 d x)+930 B \sin (c+2 d x)+1320 C \sin (c+2 d x)+480 A \sin (3 c+2 d x)+930 B \sin (3 c+2 d x)+1320 C \sin (3 c+2 d x)+3280 A \sin (2 c+3 d x)+3520 B \sin (2 c+3 d x)+3200 C \sin (2 c+3 d x)-720 A \sin (4 c+3 d x)-480 B \sin (4 c+3 d x)-120 C \sin (4 c+3 d x)+240 A \sin (3 c+4 d x)+405 B \sin (3 c+4 d x)+420 C \sin (3 c+4 d x)+240 A \sin (5 c+4 d x)+405 B \sin (5 c+4 d x)+420 C \sin (5 c+4 d x)+800 A \sin (4 c+5 d x)+800 B \sin (4 c+5 d x)+664 C \sin (4 c+5 d x))\right )}{15360 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^8*Sec[c + d*x]^5*(-240*(48*
A + 35*B + 28*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]*(600*A*d*x*Cos[d*x] + 600*A*d*x*Cos[2*c + d*x] + 300*A*d*x*Cos[2*c + 3*d*x] + 300*A*d*x*Cos[4
*c + 3*d*x] + 60*A*d*x*Cos[4*c + 5*d*x] + 60*A*d*x*Cos[6*c + 5*d*x] + 4880*A*Sin[d*x] + 5120*B*Sin[d*x] + 4720
*C*Sin[d*x] - 3120*A*Sin[2*c + d*x] - 2880*B*Sin[2*c + d*x] - 1920*C*Sin[2*c + d*x] + 480*A*Sin[c + 2*d*x] + 9
30*B*Sin[c + 2*d*x] + 1320*C*Sin[c + 2*d*x] + 480*A*Sin[3*c + 2*d*x] + 930*B*Sin[3*c + 2*d*x] + 1320*C*Sin[3*c
+ 2*d*x] + 3280*A*Sin[2*c + 3*d*x] + 3520*B*Sin[2*c + 3*d*x] + 3200*C*Sin[2*c + 3*d*x] - 720*A*Sin[4*c + 3*d*
x] - 480*B*Sin[4*c + 3*d*x] - 120*C*Sin[4*c + 3*d*x] + 240*A*Sin[3*c + 4*d*x] + 405*B*Sin[3*c + 4*d*x] + 420*C
*Sin[3*c + 4*d*x] + 240*A*Sin[5*c + 4*d*x] + 405*B*Sin[5*c + 4*d*x] + 420*C*Sin[5*c + 4*d*x] + 800*A*Sin[4*c +
5*d*x] + 800*B*Sin[4*c + 5*d*x] + 664*C*Sin[4*c + 5*d*x])))/(15360*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c
+ d*x)]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(405\) vs. \(2(183)=366\).
time = 0.98, size = 406, normalized size = 2.08 Too large to display

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)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-A*a^4*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^4*B*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(se
c(d*x+c)+tan(d*x+c)))-a^4*C*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+4*A*a^4*(1/2*sec(d*x+c)*tan(
d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-4*a^4*B*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+4*a^4*C*(-(-1/4*sec(d*x+c)^3-
3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+6*A*a^4*tan(d*x+c)+6*a^4*B*(1/2*sec(d*x+c)*tan(d*x+c
)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-6*a^4*C*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+4*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+
4*a^4*B*tan(d*x+c)+4*a^4*C*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+A*a^4*(d*x+c)+a^4*B*ln(se
c(d*x+c)+tan(d*x+c))+a^4*C*tan(d*x+c))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 482 vs. \(2 (183) = 366\).
time = 0.28, size = 482, normalized size = 2.47 \begin {gather*} \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 240 \, {\left (d x + c\right )} A a^{4} + 320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B 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 )} C a^{4} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 15 \, 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 )} - 60 \, C 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 )} - 240 \, 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 )} - 360 \, 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 \, C 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 )} + 960 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 240 \, B a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 1440 \, A a^{4} \tan \left (d x + c\right ) + 960 \, B a^{4} \tan \left (d x + c\right ) + 240 \, C a^{4} \tan \left (d x + c\right )}{240 \, d} \end {gather*}

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)+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^4 + 240*(d*x + c)*A*a^4 + 320*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*B*a^4 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^4 + 480*(tan(d*x + c)^3 + 3*tan(d*x
+ c))*C*a^4 - 15*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)) - 60*C*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)) - 240*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)) - 360*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*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(s
in(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 960*A*a^4*log(sec(d*x + c) + tan(d*x + c)) + 240*B*a^4*log(sec(d*x
+ c) + tan(d*x + c)) + 1440*A*a^4*tan(d*x + c) + 960*B*a^4*tan(d*x + c) + 240*C*a^4*tan(d*x + c))/d

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Fricas [A]
time = 2.38, size = 196, normalized size = 1.01 \begin {gather*} \frac {240 \, A a^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left (48 \, A + 35 \, B + 28 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (48 \, A + 35 \, B + 28 \, C\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 + 100 \, B + 83 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (16 \, A + 27 \, B + 28 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 20 \, B + 34 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, C a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

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

[Out]

1/240*(240*A*a^4*d*x*cos(d*x + c)^5 + 15*(48*A + 35*B + 28*C)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4
8*A + 35*B + 28*C)*a^4*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(100*A + 100*B + 83*C)*a^4*cos(d*x + c)^4
+ 15*(16*A + 27*B + 28*C)*a^4*cos(d*x + c)^3 + 8*(5*A + 20*B + 34*C)*a^4*cos(d*x + c)^2 + 30*(B + 4*C)*a^4*cos
(d*x + c) + 24*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 C \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \end {gather*}

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

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Giac [A]
time = 0.57, size = 352, normalized size = 1.81 \begin {gather*} \frac {120 \, {\left (d x + c\right )} A a^{4} + 15 \, {\left (48 \, A a^{4} + 35 \, B a^{4} + 28 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (48 \, A a^{4} + 35 \, B a^{4} + 28 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (600 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2720 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2450 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1960 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4720 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3584 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3680 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3950 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3160 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1080 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1500 \, C 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} \end {gather*}

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

[Out]

1/120*(120*(d*x + c)*A*a^4 + 15*(48*A*a^4 + 35*B*a^4 + 28*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(48*A
*a^4 + 35*B*a^4 + 28*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(600*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a
^4*tan(1/2*d*x + 1/2*c)^9 + 420*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 2720*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 2450*B*a^4*
tan(1/2*d*x + 1/2*c)^7 - 1960*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 4720*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*ta
n(1/2*d*x + 1/2*c)^5 + 3584*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 3680*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 3950*B*a^4*tan(
1/2*d*x + 1/2*c)^3 - 3160*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 1080*A*a^4*tan(1/2*d*x + 1/2*c) + 1395*B*a^4*tan(1/2*
d*x + 1/2*c) + 1500*C*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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Mupad [B]
time = 4.75, size = 996, normalized size = 5.11 \begin {gather*} \frac {30\,A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+80\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,a^4\,\sin \left (4\,c+4\,d\,x\right )+25\,A\,a^4\,\sin \left (5\,c+5\,d\,x\right )+\frac {465\,B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{8}+95\,B\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {405\,B\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{16}+25\,B\,a^4\,\sin \left (5\,c+5\,d\,x\right )+\frac {165\,C\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {385\,C\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {105\,C\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{4}+\frac {83\,C\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{4}+55\,A\,a^4\,\sin \left (c+d\,x\right )+70\,B\,a^4\,\sin \left (c+d\,x\right )+\frac {175\,C\,a^4\,\sin \left (c+d\,x\right )}{2}+\frac {75\,A\,a^4\,\mathrm {atan}\left (\frac {2368\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A^2+3360\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A\,B+2688\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A\,C+1225\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B^2+1960\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B\,C+784\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,C^2}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2368\,A^2+3360\,A\,B+2688\,A\,C+1225\,B^2+1960\,B\,C+784\,C^2\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}+\frac {15\,A\,a^4\,\mathrm {atan}\left (\frac {2368\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A^2+3360\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A\,B+2688\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A\,C+1225\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B^2+1960\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B\,C+784\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,C^2}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2368\,A^2+3360\,A\,B+2688\,A\,C+1225\,B^2+1960\,B\,C+784\,C^2\right )}\right )\,\cos \left (5\,c+5\,d\,x\right )}{2}+450\,A\,a^4\,\cos \left (c+d\,x\right )\,\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 )+\frac {2625\,B\,a^4\,\cos \left (c+d\,x\right )\,\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 )}{8}+\frac {525\,C\,a^4\,\cos \left (c+d\,x\right )\,\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 )}{2}+225\,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 )\,\cos \left (3\,c+3\,d\,x\right )+45\,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 )\,\cos \left (5\,c+5\,d\,x\right )+\frac {2625\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{16}+\frac {525\,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 )\,\cos \left (5\,c+5\,d\,x\right )}{16}+\frac {525\,C\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{4}+\frac {105\,C\,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 )\,\cos \left (5\,c+5\,d\,x\right )}{4}+75\,A\,a^4\,\mathrm {atan}\left (\frac {2368\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A^2+3360\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A\,B+2688\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,A\,C+1225\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B^2+1960\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B\,C+784\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,C^2}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2368\,A^2+3360\,A\,B+2688\,A\,C+1225\,B^2+1960\,B\,C+784\,C^2\right )}\right )\,\cos \left (c+d\,x\right )}{60\,d\,\left (\frac {5\,\cos \left (c+d\,x\right )}{8}+\frac {5\,\cos \left (3\,c+3\,d\,x\right )}{16}+\frac {\cos \left (5\,c+5\,d\,x\right )}{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(30*A*a^4*sin(2*c + 2*d*x) + 80*A*a^4*sin(3*c + 3*d*x) + 15*A*a^4*sin(4*c + 4*d*x) + 25*A*a^4*sin(5*c + 5*d*x)
+ (465*B*a^4*sin(2*c + 2*d*x))/8 + 95*B*a^4*sin(3*c + 3*d*x) + (405*B*a^4*sin(4*c + 4*d*x))/16 + 25*B*a^4*sin
(5*c + 5*d*x) + (165*C*a^4*sin(2*c + 2*d*x))/2 + (385*C*a^4*sin(3*c + 3*d*x))/4 + (105*C*a^4*sin(4*c + 4*d*x))
/4 + (83*C*a^4*sin(5*c + 5*d*x))/4 + 55*A*a^4*sin(c + d*x) + 70*B*a^4*sin(c + d*x) + (175*C*a^4*sin(c + d*x))/
2 + (75*A*a^4*atan((2368*A^2*sin(c/2 + (d*x)/2) + 1225*B^2*sin(c/2 + (d*x)/2) + 784*C^2*sin(c/2 + (d*x)/2) + 3
360*A*B*sin(c/2 + (d*x)/2) + 2688*A*C*sin(c/2 + (d*x)/2) + 1960*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(2
368*A^2 + 1225*B^2 + 784*C^2 + 3360*A*B + 2688*A*C + 1960*B*C)))*cos(3*c + 3*d*x))/2 + (15*A*a^4*atan((2368*A^
2*sin(c/2 + (d*x)/2) + 1225*B^2*sin(c/2 + (d*x)/2) + 784*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2)
+ 2688*A*C*sin(c/2 + (d*x)/2) + 1960*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(2368*A^2 + 1225*B^2 + 784*C^
2 + 3360*A*B + 2688*A*C + 1960*B*C)))*cos(5*c + 5*d*x))/2 + 450*A*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/co
s(c/2 + (d*x)/2)) + (2625*B*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + (525*C*a^4*cos(
c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + 225*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/
2))*cos(3*c + 3*d*x) + 45*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x) + (2625*B*a^4*at
anh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/16 + (525*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2))*cos(5*c + 5*d*x))/16 + (525*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4
+ (105*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/4 + 75*A*a^4*atan((2368*A^2*sin(c
/2 + (d*x)/2) + 1225*B^2*sin(c/2 + (d*x)/2) + 784*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2) + 2688*
A*C*sin(c/2 + (d*x)/2) + 1960*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(2368*A^2 + 1225*B^2 + 784*C^2 + 336
0*A*B + 2688*A*C + 1960*B*C)))*cos(c + d*x))/(60*d*((5*cos(c + d*x))/8 + (5*cos(3*c + 3*d*x))/16 + cos(5*c + 5
*d*x)/16))

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