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

Optimal. Leaf size=194 \[ \frac{2 a^4 (8 A+7 B) \tan ^3(c+d x)}{15 d}+\frac{4 a^4 (8 A+7 B) \tan (c+d x)}{5 d}+\frac{7 a^4 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (8 A+7 B) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac{27 a^4 (8 A+7 B) \tan (c+d x) \sec (c+d x)}{80 d}+\frac{(6 A-B) \tan (c+d x) (a \sec (c+d x)+a)^4}{30 d}+\frac{B \tan (c+d x) (a \sec (c+d x)+a)^5}{6 a d} \]

[Out]

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

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Rubi [A]  time = 0.318361, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4010, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac{2 a^4 (8 A+7 B) \tan ^3(c+d x)}{15 d}+\frac{4 a^4 (8 A+7 B) \tan (c+d x)}{5 d}+\frac{7 a^4 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (8 A+7 B) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac{27 a^4 (8 A+7 B) \tan (c+d x) \sec (c+d x)}{80 d}+\frac{(6 A-B) \tan (c+d x) (a \sec (c+d x)+a)^4}{30 d}+\frac{B \tan (c+d x) (a \sec (c+d x)+a)^5}{6 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !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 ^2(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))^5 \tan (c+d x)}{6 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^4 (5 a B+a (6 A-B) \sec (c+d x)) \, dx}{6 a}\\ &=\frac{(6 A-B) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{B (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{10} (8 A+7 B) \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx\\ &=\frac{(6 A-B) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{B (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{10} (8 A+7 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{(6 A-B) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{B (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{10} \left (a^4 (8 A+7 B)\right ) \int \sec (c+d x) \, dx+\frac{1}{10} \left (a^4 (8 A+7 B)\right ) \int \sec ^5(c+d x) \, dx+\frac{1}{5} \left (2 a^4 (8 A+7 B)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{5} \left (2 a^4 (8 A+7 B)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{5} \left (3 a^4 (8 A+7 B)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^4 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{10 d}+\frac{3 a^4 (8 A+7 B) \sec (c+d x) \tan (c+d x)}{10 d}+\frac{a^4 (8 A+7 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{(6 A-B) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{B (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{1}{40} \left (3 a^4 (8 A+7 B)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{10} \left (3 a^4 (8 A+7 B)\right ) \int \sec (c+d x) \, dx-\frac{\left (2 a^4 (8 A+7 B)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 d}-\frac{\left (2 a^4 (8 A+7 B)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{2 a^4 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{5 d}+\frac{4 a^4 (8 A+7 B) \tan (c+d x)}{5 d}+\frac{27 a^4 (8 A+7 B) \sec (c+d x) \tan (c+d x)}{80 d}+\frac{a^4 (8 A+7 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{(6 A-B) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{B (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{2 a^4 (8 A+7 B) \tan ^3(c+d x)}{15 d}+\frac{1}{80} \left (3 a^4 (8 A+7 B)\right ) \int \sec (c+d x) \, dx\\ &=\frac{7 a^4 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{4 a^4 (8 A+7 B) \tan (c+d x)}{5 d}+\frac{27 a^4 (8 A+7 B) \sec (c+d x) \tan (c+d x)}{80 d}+\frac{a^4 (8 A+7 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{(6 A-B) (a+a \sec (c+d x))^4 \tan (c+d x)}{30 d}+\frac{B (a+a \sec (c+d x))^5 \tan (c+d x)}{6 a d}+\frac{2 a^4 (8 A+7 B) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 2.23417, size = 358, normalized size = 1.85 \[ -\frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (3360 (8 A+7 B) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-160 (83 A+72 B) \sin (c)+30 (88 A+125 B) \sin (d x)+2640 A \sin (2 c+d x)+15840 A \sin (c+2 d x)-4080 A \sin (3 c+2 d x)+3480 A \sin (2 c+3 d x)+3480 A \sin (4 c+3 d x)+7728 A \sin (3 c+4 d x)-240 A \sin (5 c+4 d x)+840 A \sin (4 c+5 d x)+840 A \sin (6 c+5 d x)+1328 A \sin (5 c+6 d x)+3750 B \sin (2 c+d x)+15360 B \sin (c+2 d x)-1920 B \sin (3 c+2 d x)+3845 B \sin (2 c+3 d x)+3845 B \sin (4 c+3 d x)+6912 B \sin (3 c+4 d x)+735 B \sin (4 c+5 d x)+735 B \sin (6 c+5 d x)+1152 B \sin (5 c+6 d x))\right )}{122880 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*Sec[c + d*x]^6*(3360*(8*A + 7*B)*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]]) - Sec[c]*(-160*(83*A + 72*B)*Sin[c] + 30
*(88*A + 125*B)*Sin[d*x] + 2640*A*Sin[2*c + d*x] + 3750*B*Sin[2*c + d*x] + 15840*A*Sin[c + 2*d*x] + 15360*B*Si
n[c + 2*d*x] - 4080*A*Sin[3*c + 2*d*x] - 1920*B*Sin[3*c + 2*d*x] + 3480*A*Sin[2*c + 3*d*x] + 3845*B*Sin[2*c +
3*d*x] + 3480*A*Sin[4*c + 3*d*x] + 3845*B*Sin[4*c + 3*d*x] + 7728*A*Sin[3*c + 4*d*x] + 6912*B*Sin[3*c + 4*d*x]
- 240*A*Sin[5*c + 4*d*x] + 840*A*Sin[4*c + 5*d*x] + 735*B*Sin[4*c + 5*d*x] + 840*A*Sin[6*c + 5*d*x] + 735*B*S
in[6*c + 5*d*x] + 1328*A*Sin[5*c + 6*d*x] + 1152*B*Sin[5*c + 6*d*x])))/(122880*d)

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Maple [A]  time = 0.054, size = 280, normalized size = 1.4 \begin{align*}{\frac{83\,A{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{49\,B{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{49\,B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{7\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{7\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{24\,B{a}^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{12\,B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{34\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{41\,B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{4\,B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{A{a}^{4}\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 ) ^{5}}{6\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

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

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Maxima [B]  time = 1.00517, size = 626, normalized size = 3.23 \begin{align*} \frac{32 \,{\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 )} A a^{4} + 960 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 128 \,{\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} + 640 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} - 5 \, B a^{4}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, 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 )} - 180 \, 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 )} - 480 \, 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 )} - 120 \, 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 )} + 480 \, A a^{4} \tan \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 960*(tan(d*x + c)^3 + 3*tan(d*x + c
))*A*a^4 + 128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^4 + 640*(tan(d*x + c)^3 + 3*tan(d*
x + c))*B*a^4 - 5*B*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) + 15*log(sin(d*x + c) - 1)) - 120*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)) - 180*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*lo
g(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 480*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)) - 120*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)) + 480*A*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.511613, size = 481, normalized size = 2.48 \begin{align*} \frac{105 \,{\left (8 \, A + 7 \, B\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (8 \, A + 7 \, B\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (83 \, A + 72 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} + 105 \,{\left (8 \, A + 7 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} + 32 \,{\left (17 \, A + 18 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 10 \,{\left (24 \, A + 41 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 48 \,{\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 40 \, B 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)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/480*(105*(8*A + 7*B)*a^4*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 105*(8*A + 7*B)*a^4*cos(d*x + c)^6*log(-sin(
d*x + c) + 1) + 2*(16*(83*A + 72*B)*a^4*cos(d*x + c)^5 + 105*(8*A + 7*B)*a^4*cos(d*x + c)^4 + 32*(17*A + 18*B)
*a^4*cos(d*x + c)^3 + 10*(24*A + 41*B)*a^4*cos(d*x + c)^2 + 48*(A + 4*B)*a^4*cos(d*x + c) + 40*B*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 ^{2}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 A \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \sec ^{6}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 B \sec ^{4}{\left (c + d x \right )}\, dx + \int 6 B \sec ^{5}{\left (c + d x \right )}\, dx + \int 4 B \sec ^{6}{\left (c + d x \right )}\, dx + \int B \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)**2*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

a**4*(Integral(A*sec(c + d*x)**2, x) + Integral(4*A*sec(c + d*x)**3, x) + Integral(6*A*sec(c + d*x)**4, x) + I
ntegral(4*A*sec(c + d*x)**5, x) + Integral(A*sec(c + d*x)**6, x) + Integral(B*sec(c + d*x)**3, x) + Integral(4
*B*sec(c + d*x)**4, x) + Integral(6*B*sec(c + d*x)**5, x) + Integral(4*B*sec(c + d*x)**6, x) + Integral(B*sec(
c + d*x)**7, x))

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Giac [A]  time = 1.3626, size = 378, normalized size = 1.95 \begin{align*} \frac{105 \,{\left (8 \, A a^{4} + 7 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 105 \,{\left (8 \, A a^{4} + 7 \, 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 (840 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 735 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 4760 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 4165 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 11088 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9702 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 13488 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 11802 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9320 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7355 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3000 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3105 \, 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 )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/240*(105*(8*A*a^4 + 7*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(8*A*a^4 + 7*B*a^4)*log(abs(tan(1/2*d*
x + 1/2*c) - 1)) - 2*(840*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 735*B*a^4*tan(1/2*d*x + 1/2*c)^11 - 4760*A*a^4*tan(1
/2*d*x + 1/2*c)^9 - 4165*B*a^4*tan(1/2*d*x + 1/2*c)^9 + 11088*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 9702*B*a^4*tan(1/
2*d*x + 1/2*c)^7 - 13488*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 11802*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 9320*A*a^4*tan(1/
2*d*x + 1/2*c)^3 + 7355*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 3000*A*a^4*tan(1/2*d*x + 1/2*c) - 3105*B*a^4*tan(1/2*d*
x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d