3.476 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=244 \[ -\frac {\left (4 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}-\frac {a \left (4 a^2-53 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}-\frac {a \left (4 a^4-121 a^2 b^2-128 b^4\right ) \tan (c+d x)}{60 b d}+\frac {\left (8 a^4+36 a^2 b^2+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {\left (8 a^4-178 a^2 b^2-75 b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {\tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}-\frac {a \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d} \]

[Out]

1/16*(8*a^4+36*a^2*b^2+5*b^4)*arctanh(sin(d*x+c))/d-1/60*a*(4*a^4-121*a^2*b^2-128*b^4)*tan(d*x+c)/b/d-1/240*(8
*a^4-178*a^2*b^2-75*b^4)*sec(d*x+c)*tan(d*x+c)/d-1/120*a*(4*a^2-53*b^2)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d-1/12
0*(4*a^2-25*b^2)*(a+b*sec(d*x+c))^3*tan(d*x+c)/b/d-1/30*a*(a+b*sec(d*x+c))^4*tan(d*x+c)/b/d+1/6*(a+b*sec(d*x+c
))^5*tan(d*x+c)/b/d

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Rubi [A]  time = 0.45, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3840, 4002, 3997, 3787, 3770, 3767, 8} \[ -\frac {a \left (-121 a^2 b^2+4 a^4-128 b^4\right ) \tan (c+d x)}{60 b d}+\frac {\left (36 a^2 b^2+8 a^4+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {\left (4 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}-\frac {a \left (4 a^2-53 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}-\frac {\left (-178 a^2 b^2+8 a^4-75 b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {\tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}-\frac {a \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^4,x]

[Out]

((8*a^4 + 36*a^2*b^2 + 5*b^4)*ArcTanh[Sin[c + d*x]])/(16*d) - (a*(4*a^4 - 121*a^2*b^2 - 128*b^4)*Tan[c + d*x])
/(60*b*d) - ((8*a^4 - 178*a^2*b^2 - 75*b^4)*Sec[c + d*x]*Tan[c + d*x])/(240*d) - (a*(4*a^2 - 53*b^2)*(a + b*Se
c[c + d*x])^2*Tan[c + d*x])/(120*b*d) - ((4*a^2 - 25*b^2)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(120*b*d) - (a*
(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(30*b*d) + ((a + b*Sec[c + d*x])^5*Tan[c + d*x])/(6*b*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 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 3840

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(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*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

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 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]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (5 b-a \sec (c+d x)) (a+b \sec (c+d x))^4 \, dx}{6 b}\\ &=-\frac {a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (21 a b-\left (4 a^2-25 b^2\right ) \sec (c+d x)\right ) \, dx}{30 b}\\ &=-\frac {\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac {a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (24 a^2+25 b^2\right )-3 a \left (4 a^2-53 b^2\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=-\frac {a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac {\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac {a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (3 a b \left (64 a^2+181 b^2\right )-3 \left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x)\right ) \, dx}{360 b}\\ &=-\frac {\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac {a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac {\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac {a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) \left (45 b \left (8 a^4+36 a^2 b^2+5 b^4\right )-12 a \left (4 a^4-121 a^2 b^2-128 b^4\right ) \sec (c+d x)\right ) \, dx}{720 b}\\ &=-\frac {\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac {a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac {\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac {a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}-\frac {\left (a \left (4 a^4-121 a^2 b^2-128 b^4\right )\right ) \int \sec ^2(c+d x) \, dx}{60 b}+\frac {1}{16} \left (8 a^4+36 a^2 b^2+5 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (8 a^4+36 a^2 b^2+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac {a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac {\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac {a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\left (a \left (4 a^4-121 a^2 b^2-128 b^4\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b d}\\ &=\frac {\left (8 a^4+36 a^2 b^2+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {a \left (4 a^4-121 a^2 b^2-128 b^4\right ) \tan (c+d x)}{60 b d}-\frac {\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac {a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac {\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac {a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}\\ \end {align*}

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Mathematica [A]  time = 0.96, size = 154, normalized size = 0.63 \[ \frac {15 \left (8 a^4+36 a^2 b^2+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (64 a b \left (5 \left (a^2+2 b^2\right ) \tan ^2(c+d x)+15 \left (a^2+b^2\right )+3 b^2 \tan ^4(c+d x)\right )+10 b^2 \left (36 a^2+5 b^2\right ) \sec ^3(c+d x)+15 \left (8 a^4+36 a^2 b^2+5 b^4\right ) \sec (c+d x)+40 b^4 \sec ^5(c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^4,x]

[Out]

(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*Sec[c +
d*x] + 10*b^2*(36*a^2 + 5*b^2)*Sec[c + d*x]^3 + 40*b^4*Sec[c + d*x]^5 + 64*a*b*(15*(a^2 + b^2) + 5*(a^2 + 2*b
^2)*Tan[c + d*x]^2 + 3*b^2*Tan[c + d*x]^4)))/(240*d)

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fricas [A]  time = 0.51, size = 217, normalized size = 0.89 \[ \frac {15 \, {\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (128 \, {\left (5 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + 192 \, a b^{3} \cos \left (d x + c\right ) + 15 \, {\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 40 \, b^{4} + 64 \, {\left (5 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/480*(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*
cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(128*(5*a^3*b + 4*a*b^3)*cos(d*x + c)^5 + 192*a*b^3*cos(d*x + c) + 1
5*(8*a^4 + 36*a^2*b^2 + 5*b^4)*cos(d*x + c)^4 + 40*b^4 + 64*(5*a^3*b + 4*a*b^3)*cos(d*x + c)^3 + 10*(36*a^2*b^
2 + 5*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [B]  time = 0.31, size = 592, normalized size = 2.43 \[ \frac {15 \, {\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{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^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 165 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 360 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 3520 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2240 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 25 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5760 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4992 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 450 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5760 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4992 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 450 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 360 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3520 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 25 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 165 \, b^{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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/240*(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*lo
g(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(120*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 90
0*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 165*b^4*tan(1/2*d*x + 1/2*c)^11 - 360*
a^4*tan(1/2*d*x + 1/2*c)^9 + 3520*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 2240*a*
b^3*tan(1/2*d*x + 1/2*c)^9 + 25*b^4*tan(1/2*d*x + 1/2*c)^9 + 240*a^4*tan(1/2*d*x + 1/2*c)^7 - 5760*a^3*b*tan(1
/2*d*x + 1/2*c)^7 + 360*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 4992*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 450*b^4*tan(1/2*d
*x + 1/2*c)^7 + 240*a^4*tan(1/2*d*x + 1/2*c)^5 + 5760*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 360*a^2*b^2*tan(1/2*d*x +
1/2*c)^5 + 4992*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 450*b^4*tan(1/2*d*x + 1/2*c)^5 - 360*a^4*tan(1/2*d*x + 1/2*c)^
3 - 3520*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 2240*a*b^3*tan(1/2*d*x + 1/2*c)^
3 + 25*b^4*tan(1/2*d*x + 1/2*c)^3 + 120*a^4*tan(1/2*d*x + 1/2*c) + 960*a^3*b*tan(1/2*d*x + 1/2*c) + 900*a^2*b^
2*tan(1/2*d*x + 1/2*c) + 960*a*b^3*tan(1/2*d*x + 1/2*c) + 165*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
2 - 1)^6)/d

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maple [A]  time = 1.24, size = 302, normalized size = 1.24 \[ \frac {a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {8 a^{3} b \tan \left (d x +c \right )}{3 d}+\frac {4 a^{3} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {3 a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {9 a^{2} b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {9 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {32 a \,b^{3} \tan \left (d x +c \right )}{15 d}+\frac {4 a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {16 a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {b^{4} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 b^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sec(d*x+c))^4,x)

[Out]

1/2*a^4*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+8/3/d*a^3*b*tan(d*x+c)+4/3/d*a^3*b*tan(d*x
+c)*sec(d*x+c)^2+3/2/d*a^2*b^2*tan(d*x+c)*sec(d*x+c)^3+9/4/d*a^2*b^2*sec(d*x+c)*tan(d*x+c)+9/4/d*a^2*b^2*ln(se
c(d*x+c)+tan(d*x+c))+32/15*a*b^3*tan(d*x+c)/d+4/5/d*a*b^3*tan(d*x+c)*sec(d*x+c)^4+16/15/d*a*b^3*tan(d*x+c)*sec
(d*x+c)^2+1/6/d*b^4*tan(d*x+c)*sec(d*x+c)^5+5/24/d*b^4*tan(d*x+c)*sec(d*x+c)^3+5/16/d*b^4*sec(d*x+c)*tan(d*x+c
)+5/16/d*b^4*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.38, size = 275, normalized size = 1.13 \[ \frac {640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} b + 128 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a b^{3} - 5 \, b^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, a^{2} b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/480*(640*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3*b + 128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x +
c))*a*b^3 - 5*b^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)) - 180*a^2*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)) - 120*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)))/d

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mupad [B]  time = 4.88, size = 370, normalized size = 1.52 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4+\frac {9\,a^2\,b^2}{2}+\frac {5\,b^4}{8}\right )}{d}+\frac {\left (a^4-8\,a^3\,b+\frac {15\,a^2\,b^2}{2}-8\,a\,b^3+\frac {11\,b^4}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-3\,a^4+\frac {88\,a^3\,b}{3}-\frac {21\,a^2\,b^2}{2}+\frac {56\,a\,b^3}{3}+\frac {5\,b^4}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,a^4-48\,a^3\,b+3\,a^2\,b^2-\frac {208\,a\,b^3}{5}+\frac {15\,b^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (2\,a^4+48\,a^3\,b+3\,a^2\,b^2+\frac {208\,a\,b^3}{5}+\frac {15\,b^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-3\,a^4-\frac {88\,a^3\,b}{3}-\frac {21\,a^2\,b^2}{2}-\frac {56\,a\,b^3}{3}+\frac {5\,b^4}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^4+8\,a^3\,b+\frac {15\,a^2\,b^2}{2}+8\,a\,b^3+\frac {11\,b^4}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^4/cos(c + d*x)^3,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*(a^4 + (5*b^4)/8 + (9*a^2*b^2)/2))/d + (tan(c/2 + (d*x)/2)^9*((56*a*b^3)/3 + (88*a^
3*b)/3 - 3*a^4 + (5*b^4)/24 - (21*a^2*b^2)/2) - tan(c/2 + (d*x)/2)^3*((56*a*b^3)/3 + (88*a^3*b)/3 + 3*a^4 - (5
*b^4)/24 + (21*a^2*b^2)/2) + tan(c/2 + (d*x)/2)^5*((208*a*b^3)/5 + 48*a^3*b + 2*a^4 + (15*b^4)/4 + 3*a^2*b^2)
+ tan(c/2 + (d*x)/2)^7*(2*a^4 - 48*a^3*b - (208*a*b^3)/5 + (15*b^4)/4 + 3*a^2*b^2) + tan(c/2 + (d*x)/2)*(8*a*b
^3 + 8*a^3*b + a^4 + (11*b^4)/8 + (15*a^2*b^2)/2) + tan(c/2 + (d*x)/2)^11*(a^4 - 8*a^3*b - 8*a*b^3 + (11*b^4)/
8 + (15*a^2*b^2)/2))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c
/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))**4,x)

[Out]

Integral((a + b*sec(c + d*x))**4*sec(c + d*x)**3, x)

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