3.294 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=252 \[ \frac{\left (15 a^3 A b+52 a^2 b^2 B-3 a^4 B+60 a A b^3+16 b^4 B\right ) \tan (c+d x)}{30 b d}+\frac{\left (12 a^2 A b+4 a^3 B+9 a b^2 B+3 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac{\left (30 a^2 A b-6 a^3 B+71 a b^2 B+45 A b^3\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(5 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]

[Out]

((12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((15*a^3*A*b + 60*a*A*b^3 - 3*a^4
*B + 52*a^2*b^2*B + 16*b^4*B)*Tan[c + d*x])/(30*b*d) + ((30*a^2*A*b + 45*A*b^3 - 6*a^3*B + 71*a*b^2*B)*Sec[c +
d*x]*Tan[c + d*x])/(120*d) + ((15*a*A*b - 3*a^2*B + 16*b^2*B)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(60*b*d) +
((5*A*b - a*B)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(20*b*d) + (B*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*b*d
)

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Rubi [A]  time = 0.47939, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4010, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{\left (15 a^3 A b+52 a^2 b^2 B-3 a^4 B+60 a A b^3+16 b^4 B\right ) \tan (c+d x)}{30 b d}+\frac{\left (12 a^2 A b+4 a^3 B+9 a b^2 B+3 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac{\left (30 a^2 A b-6 a^3 B+71 a b^2 B+45 A b^3\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(5 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

((12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((15*a^3*A*b + 60*a*A*b^3 - 3*a^4
*B + 52*a^2*b^2*B + 16*b^4*B)*Tan[c + d*x])/(30*b*d) + ((30*a^2*A*b + 45*A*b^3 - 6*a^3*B + 71*a*b^2*B)*Sec[c +
d*x]*Tan[c + d*x])/(120*d) + ((15*a*A*b - 3*a^2*B + 16*b^2*B)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(60*b*d) +
((5*A*b - a*B)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(20*b*d) + (B*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*b*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 4002

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[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 (4 b B+(5 A b-a B) \sec (c+d x)) \, dx}{5 b}\\ &=\frac{(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (b (15 A b+13 a B)+\left (15 a A b-3 a^2 B+16 b^2 B\right ) \sec (c+d x)\right ) \, dx}{20 b}\\ &=\frac{\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (75 a A b+33 a^2 B+32 b^2 B\right )+\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x)\right ) \, dx}{60 b}\\ &=\frac{\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) \left (15 b \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right )+4 \left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=\frac{\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{1}{8} \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \int \sec (c+d x) \, dx+\frac{\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \int \sec ^2(c+d x) \, dx}{30 b}\\ &=\frac{\left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}-\frac{\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b d}\\ &=\frac{\left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \tan (c+d x)}{30 b d}+\frac{\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac{(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 3.34059, size = 181, normalized size = 0.72 \[ \frac{15 \left (12 a^2 A b+4 a^3 B+9 a b^2 B+3 A b^3\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 b \left (3 a^2 B+3 a A b+2 b^2 B\right ) \tan ^2(c+d x)+15 \left (a^3 A+3 a^2 b B+3 a A b^2+b^3 B\right )+3 b^3 B \tan ^4(c+d x)\right )+15 \left (12 a^2 A b+4 a^3 B+9 a b^2 B+3 A b^3\right ) \sec (c+d x)+30 b^2 (3 a B+A b) \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(15*(12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(12*a^2*A*b + 3*A*b^
3 + 4*a^3*B + 9*a*b^2*B)*Sec[c + d*x] + 30*b^2*(A*b + 3*a*B)*Sec[c + d*x]^3 + 8*(15*(a^3*A + 3*a*A*b^2 + 3*a^2
*b*B + b^3*B) + 5*b*(3*a*A*b + 3*a^2*B + 2*b^2*B)*Tan[c + d*x]^2 + 3*b^3*B*Tan[c + d*x]^4)))/(120*d)

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Maple [A]  time = 0.042, size = 382, normalized size = 1.5 \begin{align*}{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{3\,A{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{B{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{Aa{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{Aa{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,Ba{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,Ba{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,Ba{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{A{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,B{b}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{B{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,B{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x)

[Out]

1/d*A*a^3*tan(d*x+c)+1/2/d*B*a^3*sec(d*x+c)*tan(d*x+c)+1/2/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*A*a^2*b*sec
(d*x+c)*tan(d*x+c)+3/2/d*A*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+2/d*B*a^2*b*tan(d*x+c)+1/d*B*a^2*b*tan(d*x+c)*sec(d
*x+c)^2+2/d*A*a*b^2*tan(d*x+c)+1/d*A*a*b^2*tan(d*x+c)*sec(d*x+c)^2+3/4/d*B*a*b^2*tan(d*x+c)*sec(d*x+c)^3+9/8/d
*B*a*b^2*sec(d*x+c)*tan(d*x+c)+9/8/d*B*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*A*b^3*tan(d*x+c)*sec(d*x+c)^3+3/8
/d*A*b^3*sec(d*x+c)*tan(d*x+c)+3/8/d*A*b^3*ln(sec(d*x+c)+tan(d*x+c))+8/15/d*B*b^3*tan(d*x+c)+1/5/d*B*b^3*tan(d
*x+c)*sec(d*x+c)^4+4/15/d*B*b^3*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.00424, size = 460, normalized size = 1.83 \begin{align*} \frac{240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} + 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 b^{3} - 45 \, B a b^{2}{\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 )} - 15 \, A b^{3}{\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^{3}{\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 )} - 180 \, A a^{2} b{\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^{3} \tan \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2*b + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 + 16*(3*t
an(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*b^3 - 45*B*a*b^2*(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)) - 15*A*b^3*(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(si
n(d*x + c) - 1)) - 60*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) -
1)) - 180*A*a^2*b*(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^3*tan(d*x + c))/d

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Fricas [A]  time = 0.545972, size = 612, normalized size = 2.43 \begin{align*} \frac{15 \,{\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (15 \, A a^{3} + 30 \, B a^{2} b + 30 \, A a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, B b^{3} + 15 \,{\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (15 \, B a^{2} b + 15 \, A a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*B*a^3 + 12
*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(15*A*a^3 + 30*B*a^2*b + 30*A*a*b
^2 + 8*B*b^3)*cos(d*x + c)^4 + 24*B*b^3 + 15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c)^3 + 8*(
15*B*a^2*b + 15*A*a*b^2 + 4*B*b^3)*cos(d*x + c)^2 + 30*(3*B*a*b^2 + A*b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(
d*x + c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**3*sec(c + d*x)**2, x)

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Giac [B]  time = 1.26441, size = 975, normalized size = 3.87 \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)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*B*a^3 + 12*A
*a^2*b + 9*B*a*b^2 + 3*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 60*B*
a^3*tan(1/2*d*x + 1/2*c)^9 - 180*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a
*b^2*tan(1/2*d*x + 1/2*c)^9 - 225*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 75*A*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*B*b^3
*tan(1/2*d*x + 1/2*c)^9 - 480*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 360*A*a^2*b*ta
n(1/2*d*x + 1/2*c)^7 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 90*B*a*b^2*ta
n(1/2*d*x + 1/2*c)^7 + 30*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 160*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 720*A*a^3*tan(1/2*
d*x + 1/2*c)^5 + 1200*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 464*B*b^3*tan(1/2
*d*x + 1/2*c)^5 - 480*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 120*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 360*A*a^2*b*tan(1/2*d*
x + 1/2*c)^3 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 90*B*a*b^2*tan(1/2*d*
x + 1/2*c)^3 - 30*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 160*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/
2*c) + 60*B*a^3*tan(1/2*d*x + 1/2*c) + 180*A*a^2*b*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b*tan(1/2*d*x + 1/2*c) + 3
60*A*a*b^2*tan(1/2*d*x + 1/2*c) + 225*B*a*b^2*tan(1/2*d*x + 1/2*c) + 75*A*b^3*tan(1/2*d*x + 1/2*c) + 120*B*b^3
*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d