3.673 \(\int (a+b \sec (c+d x))^3 (a^2-b^2 \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=158 \[ a^5 x+\frac {a b^2 \left (5 a^2-4 b^2\right ) \tan (c+d x)}{2 d}+\frac {b^3 \left (2 a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b \left (24 a^4-8 a^2 b^2-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {a b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}-\frac {b^2 \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

[Out]

a^5*x+1/8*b*(24*a^4-8*a^2*b^2-3*b^4)*arctanh(sin(d*x+c))/d+1/2*a*b^2*(5*a^2-4*b^2)*tan(d*x+c)/d+1/8*b^3*(2*a^2
-3*b^2)*sec(d*x+c)*tan(d*x+c)/d-1/4*a*b^2*(a+b*sec(d*x+c))^2*tan(d*x+c)/d-1/4*b^2*(a+b*sec(d*x+c))^3*tan(d*x+c
)/d

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Rubi [A]  time = 0.29, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {4042, 3918, 4056, 4048, 3770, 3767, 8} \[ \frac {a b^2 \left (5 a^2-4 b^2\right ) \tan (c+d x)}{2 d}+\frac {b \left (-8 a^2 b^2+24 a^4-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^3 \left (2 a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+a^5 x-\frac {a b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}-\frac {b^2 \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^5*x + (b*(24*a^4 - 8*a^2*b^2 - 3*b^4)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*b^2*(5*a^2 - 4*b^2)*Tan[c + d*x])/(2
*d) + (b^3*(2*a^2 - 3*b^2)*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (a*b^2*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(4*d
) - (b^2*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*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 3918

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 - 2)*Simp[a^2*c
*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4042

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Dist[
C/b^2, Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x
] && EqQ[A*b^2 + a^2*C, 0]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 4056

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

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^3 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx &=-\int (-a+b \sec (c+d x)) (a+b \sec (c+d x))^4 \, dx\\ &=-\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {1}{4} \int (a+b \sec (c+d x))^2 \left (-4 a^3-b \left (4 a^2-3 b^2\right ) \sec (c+d x)+3 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=-\frac {a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {1}{12} \int (a+b \sec (c+d x)) \left (-12 a^4-3 a b \left (8 a^2-5 b^2\right ) \sec (c+d x)-3 b^2 \left (2 a^2-3 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (-24 a^5-3 b \left (24 a^4-8 a^2 b^2-3 b^4\right ) \sec (c+d x)-12 a b^2 \left (5 a^2-4 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^5 x+\frac {b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{2} \left (a b^2 \left (5 a^2-4 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (b \left (24 a^4-8 a^2 b^2-3 b^4\right )\right ) \int \sec (c+d x) \, dx\\ &=a^5 x+\frac {b \left (24 a^4-8 a^2 b^2-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (a b^2 \left (5 a^2-4 b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^5 x+\frac {b \left (24 a^4-8 a^2 b^2-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a b^2 \left (5 a^2-4 b^2\right ) \tan (c+d x)}{2 d}+\frac {b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac {b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [B]  time = 6.41, size = 1299, normalized size = 8.22 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^5*(c + d*x)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2))/(d*(b + a*Cos[c + d*x])^3*(
a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x])) + ((-24*a^4*b + 8*a^2*b^3 + 3*b^5)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2))/(4*d*(b + a*Cos[c + d*x])^3*(a^2 - 2*b^2
+ a^2*Cos[2*c + 2*d*x])) + ((24*a^4*b - 8*a^2*b^3 - 3*b^5)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]]*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2))/(4*d*(b + a*Cos[c + d*x])^3*(a^2 - 2*b^2 + a^2*Cos[2*c
+ 2*d*x])) - (b^5*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2))/(8*d*(b + a*Cos[c + d*x])
^3*(a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + ((-8*a^2*b^3 - 4*a*b^4 - 3*
b^5)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2))/(8*d*(b + a*Cos[c + d*x])^3*(a^2 - 2*b^
2 + a^2*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (a*b^4*Cos[c + d*x]^5*(a + b*Sec[c + d*x]
)^3*(a^2 - b^2*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(d*(b + a*Cos[c + d*x])^3*(a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x]
)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (b^5*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x
]^2))/(8*d*(b + a*Cos[c + d*x])^3*(a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4
) - (a*b^4*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(d*(b + a*Cos[c
+ d*x])^3*(a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((8*a^2*b^3 + 4*a*b^
4 + 3*b^5)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2))/(8*d*(b + a*Cos[c + d*x])^3*(a^2
- 2*b^2 + a^2*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (4*Cos[c + d*x]^5*(a + b*Sec[c + d*
x])^3*(a^2 - b^2*Sec[c + d*x]^2)*(-(a^3*b^2*Sin[(c + d*x)/2]) + a*b^4*Sin[(c + d*x)/2]))/(d*(b + a*Cos[c + d*x
])^3*(a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (4*Cos[c + d*x]^5*(a + b*Se
c[c + d*x])^3*(a^2 - b^2*Sec[c + d*x]^2)*(-(a^3*b^2*Sin[(c + d*x)/2]) + a*b^4*Sin[(c + d*x)/2]))/(d*(b + a*Cos
[c + d*x])^3*(a^2 - 2*b^2 + a^2*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.50, size = 181, normalized size = 1.15 \[ \frac {16 \, a^{5} d x \cos \left (d x + c\right )^{4} + {\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (8 \, a b^{4} \cos \left (d x + c\right ) + 2 \, b^{5} - 16 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a^5*d*x*cos(d*x + c)^4 + (24*a^4*b - 8*a^2*b^3 - 3*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (24*a^
4*b - 8*a^2*b^3 - 3*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 2*(8*a*b^4*cos(d*x + c) + 2*b^5 - 16*(a^3*b^2
- a*b^4)*cos(d*x + c)^3 + (8*a^2*b^3 + 3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [B]  time = 0.31, size = 379, normalized size = 2.40 \[ \frac {8 \, {\left (d x + c\right )} a^{5} + {\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (16 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(a^2-b^2*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/8*(8*(d*x + c)*a^5 + (24*a^4*b - 8*a^2*b^3 - 3*b^5)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (24*a^4*b - 8*a^2*b
^3 - 3*b^5)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(16*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 + 8*a^2*b^3*tan(1/2*d*x
+ 1/2*c)^7 - 24*a*b^4*tan(1/2*d*x + 1/2*c)^7 + 5*b^5*tan(1/2*d*x + 1/2*c)^7 - 48*a^3*b^2*tan(1/2*d*x + 1/2*c)^
5 - 8*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 40*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*b^5*tan(1/2*d*x + 1/2*c)^5 + 48*a^3
*b^2*tan(1/2*d*x + 1/2*c)^3 - 8*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 40*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 3*b^5*tan(1
/2*d*x + 1/2*c)^3 - 16*a^3*b^2*tan(1/2*d*x + 1/2*c) + 8*a^2*b^3*tan(1/2*d*x + 1/2*c) + 24*a*b^4*tan(1/2*d*x +
1/2*c) + 5*b^5*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 1.53, size = 205, normalized size = 1.30 \[ a^{5} x +\frac {a^{5} c}{d}+\frac {2 b^{2} a^{3} \tan \left (d x +c \right )}{d}+\frac {3 b \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {a^{2} b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}-\frac {a^{2} b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {2 a \,b^{4} \tan \left (d x +c \right )}{d}-\frac {a \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}-\frac {b^{5} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 b^{5} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}-\frac {3 b^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 192, normalized size = 1.22 \[ \frac {16 \, {\left (d x + c\right )} a^{5} - 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{4} + b^{5} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, a^{2} b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, a^{4} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 32 \, a^{3} b^{2} \tan \left (d x + c\right )}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*(d*x + c)*a^5 - 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b^4 + b^5*(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)) + 8*a^2*b^3*(2
*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*a^4*b*log(sec(d*x + c
) + tan(d*x + c)) + 32*a^3*b^2*tan(d*x + c))/d

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mupad [B]  time = 4.07, size = 274, normalized size = 1.73 \[ \frac {2\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {3\,b^5\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {b^5\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4}-\frac {2\,a\,b^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {a\,b^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^3}+\frac {2\,a^3\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {a^2\,b^3\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {b^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{4\,d}+\frac {a^2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {a^4\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2/cos(c + d*x)^2)*(a + b/cos(c + d*x))^3,x)

[Out]

(2*a^5*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (b^5*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
3i)/(4*d) - (3*b^5*sin(c + d*x))/(8*d*cos(c + d*x)^2) - (b^5*sin(c + d*x))/(4*d*cos(c + d*x)^4) + (a^2*b^3*ata
n((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (a^4*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
6i)/d - (2*a*b^4*sin(c + d*x))/(d*cos(c + d*x)) - (a*b^4*sin(c + d*x))/(d*cos(c + d*x)^3) + (2*a^3*b^2*sin(c +
d*x))/(d*cos(c + d*x)) - (a^2*b^3*sin(c + d*x))/(d*cos(c + d*x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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