3.439 \(\int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=209 \[ \frac{2 a^4 (10 A+8 B+7 C) \tan ^3(c+d x)}{15 d}+\frac{4 a^4 (10 A+8 B+7 C) \tan (c+d x)}{5 d}+\frac{7 a^4 (10 A+8 B+7 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (10 A+8 B+7 C) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac{27 a^4 (10 A+8 B+7 C) \tan (c+d x) \sec (c+d x)}{80 d}+\frac{(6 B-C) \tan (c+d x) (a \sec (c+d x)+a)^4}{30 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^5}{6 a d} \]

[Out]

(7*a^4*(10*A + 8*B + 7*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (4*a^4*(10*A + 8*B + 7*C)*Tan[c + d*x])/(5*d) + (27*
a^4*(10*A + 8*B + 7*C)*Sec[c + d*x]*Tan[c + d*x])/(80*d) + (a^4*(10*A + 8*B + 7*C)*Sec[c + d*x]^3*Tan[c + d*x]
)/(40*d) + ((6*B - C)*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(30*d) + (C*(a + a*Sec[c + d*x])^5*Tan[c + d*x])/(6
*a*d) + (2*a^4*(10*A + 8*B + 7*C)*Tan[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.328311, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.18, Rules used = {4082, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac{2 a^4 (10 A+8 B+7 C) \tan ^3(c+d x)}{15 d}+\frac{4 a^4 (10 A+8 B+7 C) \tan (c+d x)}{5 d}+\frac{7 a^4 (10 A+8 B+7 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (10 A+8 B+7 C) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac{27 a^4 (10 A+8 B+7 C) \tan (c+d x) \sec (c+d x)}{80 d}+\frac{(6 B-C) \tan (c+d x) (a \sec (c+d x)+a)^4}{30 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^5}{6 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^4*(10*A + 8*B + 7*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (4*a^4*(10*A + 8*B + 7*C)*Tan[c + d*x])/(5*d) + (27*
a^4*(10*A + 8*B + 7*C)*Sec[c + d*x]*Tan[c + d*x])/(80*d) + (a^4*(10*A + 8*B + 7*C)*Sec[c + d*x]^3*Tan[c + d*x]
)/(40*d) + ((6*B - C)*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(30*d) + (C*(a + a*Sec[c + d*x])^5*Tan[c + d*x])/(6
*a*d) + (2*a^4*(10*A + 8*B + 7*C)*Tan[c + d*x]^3)/(15*d)

Rule 4082

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

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 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 8

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

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]

Rubi steps

\begin{align*} \int \sec (c+d x) (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))^5 \tan (c+d x)}{6 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^4 (a (6 A+5 C)+a (6 B-C) \sec (c+d x)) \, dx}{6 a}\\ &=\frac{(6 B-C) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{C (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{10} (10 A+8 B+7 C) \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx\\ &=\frac{(6 B-C) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{C (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{10} (10 A+8 B+7 C) \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{(6 B-C) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{C (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{10} \left (a^4 (10 A+8 B+7 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{10} \left (a^4 (10 A+8 B+7 C)\right ) \int \sec ^5(c+d x) \, dx+\frac{1}{5} \left (2 a^4 (10 A+8 B+7 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{5} \left (2 a^4 (10 A+8 B+7 C)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{5} \left (3 a^4 (10 A+8 B+7 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^4 (10 A+8 B+7 C) \tanh ^{-1}(\sin (c+d x))}{10 d}+\frac{3 a^4 (10 A+8 B+7 C) \sec (c+d x) \tan (c+d x)}{10 d}+\frac{a^4 (10 A+8 B+7 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{(6 B-C) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{C (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{40} \left (3 a^4 (10 A+8 B+7 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{10} \left (3 a^4 (10 A+8 B+7 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (2 a^4 (10 A+8 B+7 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 d}-\frac{\left (2 a^4 (10 A+8 B+7 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{2 a^4 (10 A+8 B+7 C) \tanh ^{-1}(\sin (c+d x))}{5 d}+\frac{4 a^4 (10 A+8 B+7 C) \tan (c+d x)}{5 d}+\frac{27 a^4 (10 A+8 B+7 C) \sec (c+d x) \tan (c+d x)}{80 d}+\frac{a^4 (10 A+8 B+7 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{(6 B-C) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{C (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{2 a^4 (10 A+8 B+7 C) \tan ^3(c+d x)}{15 d}+\frac{1}{80} \left (3 a^4 (10 A+8 B+7 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{7 a^4 (10 A+8 B+7 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{4 a^4 (10 A+8 B+7 C) \tan (c+d x)}{5 d}+\frac{27 a^4 (10 A+8 B+7 C) \sec (c+d x) \tan (c+d x)}{80 d}+\frac{a^4 (10 A+8 B+7 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{(6 B-C) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{C (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{2 a^4 (10 A+8 B+7 C) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 5.92255, size = 359, normalized size = 1.72 \[ -\frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (105 (10 A+8 B+7 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) \cos ^5(c+d x) (15 \sin (c) (54 A+56 B+49 C)+16 (100 A+83 B+72 C) \sin (d x))-\sec (c) \cos ^4(c+d x) (32 \sin (c) (10 A+17 B+18 C)+15 (54 A+56 B+49 C) \sin (d x))-2 \sec (c) \cos ^3(c+d x) (5 \sin (c) (6 A+24 B+41 C)+16 (10 A+17 B+18 C) \sin (d x))-2 \sec (c) \cos ^2(c+d x) (5 (6 A+24 B+41 C) \sin (d x)+24 (B+4 C) \sin (c))-8 \sec (c) \cos (c+d x) (6 (B+4 C) \sin (d x)+5 C \sin (c))-40 C \sec (c) \sin (d x)\right )}{1920 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(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]^6*(105*(10*
A + 8*B + 7*C)*Cos[c + d*x]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]]) - 40*C*Sec[c]*Sin[d*x] - 8*Cos[c + d*x]*Sec[c]*(5*C*Sin[c] + 6*(B + 4*C)*Sin[d*x]) - 2*Cos[c + d*x]^3*Se
c[c]*(5*(6*A + 24*B + 41*C)*Sin[c] + 16*(10*A + 17*B + 18*C)*Sin[d*x]) - 2*Cos[c + d*x]^2*Sec[c]*(24*(B + 4*C)
*Sin[c] + 5*(6*A + 24*B + 41*C)*Sin[d*x]) - Cos[c + d*x]^4*Sec[c]*(32*(10*A + 17*B + 18*C)*Sin[c] + 15*(54*A +
56*B + 49*C)*Sin[d*x]) - Cos[c + d*x]^5*Sec[c]*(15*(54*A + 56*B + 49*C)*Sin[c] + 16*(100*A + 83*B + 72*C)*Sin
[d*x])))/(1920*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.08, size = 385, normalized size = 1.8 \begin{align*}{\frac{49\,{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{7\,B{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{27\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{41\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{24\,{a}^{4}C\tan \left ( dx+c \right ) }{5\,d}}+{\frac{12\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{83\,B{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{34\,B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{20\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{4\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{49\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{7\,B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{35\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

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

[Out]

49/16/d*a^4*C*sec(d*x+c)*tan(d*x+c)+1/d*B*a^4*tan(d*x+c)*sec(d*x+c)^3+1/4/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+1/6/
d*a^4*C*tan(d*x+c)*sec(d*x+c)^5+7/2/d*B*a^4*sec(d*x+c)*tan(d*x+c)+27/8/d*A*a^4*sec(d*x+c)*tan(d*x+c)+41/24/d*a
^4*C*tan(d*x+c)*sec(d*x+c)^3+24/5/d*a^4*C*tan(d*x+c)+12/5/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+83/15/d*B*a^4*tan(d*
x+c)+34/15/d*B*a^4*tan(d*x+c)*sec(d*x+c)^2+20/3/d*A*a^4*tan(d*x+c)+4/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+4/5/d*a
^4*C*tan(d*x+c)*sec(d*x+c)^4+1/5/d*B*a^4*tan(d*x+c)*sec(d*x+c)^4+49/16/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+7/2/d
*B*a^4*ln(sec(d*x+c)+tan(d*x+c))+35/8/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [B]  time = 0.98382, size = 861, normalized size = 4.12 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

1/480*(640*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*B*a^4 + 960*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*
x + c))*C*a^4 + 640*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 5*C*a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^
3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 1
5*log(sin(d*x + c) - 1)) - 30*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)) - 120*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)) - 180*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)) - 720*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)) - 480*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)) - 120*C*
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)) + 480*A*a^4*log(sec(
d*x + c) + tan(d*x + c)) + 1920*A*a^4*tan(d*x + c) + 480*B*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.543912, size = 539, normalized size = 2.58 \begin{align*} \frac{105 \,{\left (10 \, A + 8 \, B + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (10 \, A + 8 \, B + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (100 \, A + 83 \, B + 72 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 15 \,{\left (54 \, A + 56 \, B + 49 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 32 \,{\left (10 \, A + 17 \, B + 18 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 10 \,{\left (6 \, A + 24 \, B + 41 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 48 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 40 \, C a^{4}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \end{align*}

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

[Out]

1/480*(105*(10*A + 8*B + 7*C)*a^4*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 105*(10*A + 8*B + 7*C)*a^4*cos(d*x +
c)^6*log(-sin(d*x + c) + 1) + 2*(16*(100*A + 83*B + 72*C)*a^4*cos(d*x + c)^5 + 15*(54*A + 56*B + 49*C)*a^4*cos
(d*x + c)^4 + 32*(10*A + 17*B + 18*C)*a^4*cos(d*x + c)^3 + 10*(6*A + 24*B + 41*C)*a^4*cos(d*x + c)^2 + 48*(B +
4*C)*a^4*cos(d*x + c) + 40*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{4}{\left (c + d x \right )}\, dx + \int 6 C \sec ^{5}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{7}{\left (c + d x \right )}\, dx\right ) \end{align*}

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)+C*sec(d*x+c)**2),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) + Integral(C*sec(c + d*x)**3, x) + Integral(4*C*sec(c + d*x)**4, x) + Integral(6*C*sec(c + d*x)**
5, x) + Integral(4*C*sec(c + d*x)**6, x) + Integral(C*sec(c + d*x)**7, x))

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Giac [B]  time = 1.33016, size = 529, normalized size = 2.53 \begin{align*} \frac{105 \,{\left (10 \, A a^{4} + 8 \, B a^{4} + 7 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 105 \,{\left (10 \, A a^{4} + 8 \, B a^{4} + 7 \, 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 (1050 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 840 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 735 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 5950 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 4760 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 4165 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 13860 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 11088 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9702 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 16860 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 13488 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 11802 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10690 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9320 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7355 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2790 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3000 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3105 \, 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 )}^{6}}}{240 \, d} \end{align*}

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

[Out]

1/240*(105*(10*A*a^4 + 8*B*a^4 + 7*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(10*A*a^4 + 8*B*a^4 + 7*C*a
^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(1050*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 840*B*a^4*tan(1/2*d*x + 1/2*c
)^11 + 735*C*a^4*tan(1/2*d*x + 1/2*c)^11 - 5950*A*a^4*tan(1/2*d*x + 1/2*c)^9 - 4760*B*a^4*tan(1/2*d*x + 1/2*c)
^9 - 4165*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 13860*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 11088*B*a^4*tan(1/2*d*x + 1/2*c)
^7 + 9702*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 16860*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 13488*B*a^4*tan(1/2*d*x + 1/2*c)
^5 - 11802*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 10690*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9320*B*a^4*tan(1/2*d*x + 1/2*c)
^3 + 7355*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 2790*A*a^4*tan(1/2*d*x + 1/2*c) - 3000*B*a^4*tan(1/2*d*x + 1/2*c) - 3
105*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d