3.375 \(\int (a+b \tan ^3(c+d x))^3 \, dx\)

Optimal. Leaf size=168 \[ \frac {b \left (3 a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac {3 a b^2 \tan ^5(c+d x)}{5 d}-\frac {a b^2 \tan ^3(c+d x)}{d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b^3 \tan ^8(c+d x)}{8 d}-\frac {b^3 \tan ^6(c+d x)}{6 d}+\frac {b^3 \tan ^4(c+d x)}{4 d} \]

[Out]

a*(a^2-3*b^2)*x+b*(3*a^2-b^2)*ln(cos(d*x+c))/d+3*a*b^2*tan(d*x+c)/d+1/2*b*(3*a^2-b^2)*tan(d*x+c)^2/d-a*b^2*tan
(d*x+c)^3/d+1/4*b^3*tan(d*x+c)^4/d+3/5*a*b^2*tan(d*x+c)^5/d-1/6*b^3*tan(d*x+c)^6/d+1/8*b^3*tan(d*x+c)^8/d

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Rubi [A]  time = 0.10, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3661, 1810, 635, 203, 260} \[ \frac {b \left (3 a^2-b^2\right ) \tan ^2(c+d x)}{2 d}+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac {3 a b^2 \tan ^5(c+d x)}{5 d}-\frac {a b^2 \tan ^3(c+d x)}{d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b^3 \tan ^8(c+d x)}{8 d}-\frac {b^3 \tan ^6(c+d x)}{6 d}+\frac {b^3 \tan ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^3)^3,x]

[Out]

a*(a^2 - 3*b^2)*x + (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/d + (3*a*b^2*Tan[c + d*x])/d + (b*(3*a^2 - b^2)*Tan[c
+ d*x]^2)/(2*d) - (a*b^2*Tan[c + d*x]^3)/d + (b^3*Tan[c + d*x]^4)/(4*d) + (3*a*b^2*Tan[c + d*x]^5)/(5*d) - (b^
3*Tan[c + d*x]^6)/(6*d) + (b^3*Tan[c + d*x]^8)/(8*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \left (a+b \tan ^3(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^3\right )^3}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (3 a b^2+b \left (3 a^2-b^2\right ) x-3 a b^2 x^2+b^3 x^3+3 a b^2 x^4-b^3 x^5+b^3 x^7+\frac {a^3-3 a b^2-b \left (3 a^2-b^2\right ) x}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \tan ^2(c+d x)}{2 d}-\frac {a b^2 \tan ^3(c+d x)}{d}+\frac {b^3 \tan ^4(c+d x)}{4 d}+\frac {3 a b^2 \tan ^5(c+d x)}{5 d}-\frac {b^3 \tan ^6(c+d x)}{6 d}+\frac {b^3 \tan ^8(c+d x)}{8 d}+\frac {\operatorname {Subst}\left (\int \frac {a^3-3 a b^2-b \left (3 a^2-b^2\right ) x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \tan ^2(c+d x)}{2 d}-\frac {a b^2 \tan ^3(c+d x)}{d}+\frac {b^3 \tan ^4(c+d x)}{4 d}+\frac {3 a b^2 \tan ^5(c+d x)}{5 d}-\frac {b^3 \tan ^6(c+d x)}{6 d}+\frac {b^3 \tan ^8(c+d x)}{8 d}+\frac {\left (a \left (a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (b \left (3 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=a \left (a^2-3 b^2\right ) x+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \tan ^2(c+d x)}{2 d}-\frac {a b^2 \tan ^3(c+d x)}{d}+\frac {b^3 \tan ^4(c+d x)}{4 d}+\frac {3 a b^2 \tan ^5(c+d x)}{5 d}-\frac {b^3 \tan ^6(c+d x)}{6 d}+\frac {b^3 \tan ^8(c+d x)}{8 d}\\ \end {align*}

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Mathematica [C]  time = 0.51, size = 160, normalized size = 0.95 \[ \frac {-60 b \left (b^2-3 a^2\right ) \tan ^2(c+d x)+72 a b^2 \tan ^5(c+d x)-120 a b^2 \tan ^3(c+d x)+360 a b^2 \tan (c+d x)+60 \left (i (a+i b)^3 \log (\tan (c+d x)+i)-i (a-i b)^3 \log (-\tan (c+d x)+i)\right )+15 b^3 \tan ^8(c+d x)-20 b^3 \tan ^6(c+d x)+30 b^3 \tan ^4(c+d x)}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^3)^3,x]

[Out]

(60*((-I)*(a - I*b)^3*Log[I - Tan[c + d*x]] + I*(a + I*b)^3*Log[I + Tan[c + d*x]]) + 360*a*b^2*Tan[c + d*x] -
60*b*(-3*a^2 + b^2)*Tan[c + d*x]^2 - 120*a*b^2*Tan[c + d*x]^3 + 30*b^3*Tan[c + d*x]^4 + 72*a*b^2*Tan[c + d*x]^
5 - 20*b^3*Tan[c + d*x]^6 + 15*b^3*Tan[c + d*x]^8)/(120*d)

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fricas [A]  time = 0.48, size = 148, normalized size = 0.88 \[ \frac {15 \, b^{3} \tan \left (d x + c\right )^{8} - 20 \, b^{3} \tan \left (d x + c\right )^{6} + 72 \, a b^{2} \tan \left (d x + c\right )^{5} + 30 \, b^{3} \tan \left (d x + c\right )^{4} - 120 \, a b^{2} \tan \left (d x + c\right )^{3} + 360 \, a b^{2} \tan \left (d x + c\right ) + 120 \, {\left (a^{3} - 3 \, a b^{2}\right )} d x + 60 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{2} + 60 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*b^3*tan(d*x + c)^8 - 20*b^3*tan(d*x + c)^6 + 72*a*b^2*tan(d*x + c)^5 + 30*b^3*tan(d*x + c)^4 - 120*a
*b^2*tan(d*x + c)^3 + 360*a*b^2*tan(d*x + c) + 120*(a^3 - 3*a*b^2)*d*x + 60*(3*a^2*b - b^3)*tan(d*x + c)^2 + 6
0*(3*a^2*b - b^3)*log(1/(tan(d*x + c)^2 + 1)))/d

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.03, size = 201, normalized size = 1.20 \[ \frac {b^{3} \left (\tan ^{8}\left (d x +c \right )\right )}{8 d}-\frac {b^{3} \left (\tan ^{6}\left (d x +c \right )\right )}{6 d}+\frac {3 a \,b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}+\frac {b^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a \,b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 a \,b^{2} \tan \left (d x +c \right )}{d}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b}{2 d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3}}{2 d}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d}-\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^3*tan(d*x+c)^8/d-1/6*b^3*tan(d*x+c)^6/d+3/5*a*b^2*tan(d*x+c)^5/d+1/4*b^3*tan(d*x+c)^4/d-a*b^2*tan(d*x+c)
^3/d+3/2/d*a^2*b*tan(d*x+c)^2-1/2*b^3*tan(d*x+c)^2/d+3*a*b^2*tan(d*x+c)/d-3/2/d*ln(1+tan(d*x+c)^2)*a^2*b+1/2/d
*ln(1+tan(d*x+c)^2)*b^3+1/d*arctan(tan(d*x+c))*a^3-3/d*arctan(tan(d*x+c))*a*b^2

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maxima [A]  time = 0.66, size = 183, normalized size = 1.09 \[ a^{3} x + \frac {{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a b^{2}}{5 \, d} + \frac {b^{3} {\left (\frac {48 \, \sin \left (d x + c\right )^{6} - 108 \, \sin \left (d x + c\right )^{4} + 88 \, \sin \left (d x + c\right )^{2} - 25}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 12 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{24 \, d} - \frac {3 \, a^{2} b {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + 1/5*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a*b^2/d + 1/24*b^3*((48*si
n(d*x + c)^6 - 108*sin(d*x + c)^4 + 88*sin(d*x + c)^2 - 25)/(sin(d*x + c)^8 - 4*sin(d*x + c)^6 + 6*sin(d*x + c
)^4 - 4*sin(d*x + c)^2 + 1) - 12*log(sin(d*x + c)^2 - 1))/d - 3/2*a^2*b*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x
+ c)^2 - 1))/d

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mupad [B]  time = 11.60, size = 174, normalized size = 1.04 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{2}\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^8}{8}-\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{2}\right )-a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^2-3\,b^2\right )}{3\,a\,b^2-a^3}\right )\,\left (a^2-3\,b^2\right )-a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^3+\frac {3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+3\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x)^3)^3,x)

[Out]

(tan(c + d*x)^2*((3*a^2*b)/2 - b^3/2) + (b^3*tan(c + d*x)^4)/4 - (b^3*tan(c + d*x)^6)/6 + (b^3*tan(c + d*x)^8)
/8 - log(tan(c + d*x)^2 + 1)*((3*a^2*b)/2 - b^3/2) - a*atan((a*tan(c + d*x)*(a^2 - 3*b^2))/(3*a*b^2 - a^3))*(a
^2 - 3*b^2) - a*b^2*tan(c + d*x)^3 + (3*a*b^2*tan(c + d*x)^5)/5 + 3*a*b^2*tan(c + d*x))/d

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sympy [A]  time = 1.65, size = 194, normalized size = 1.15 \[ \begin {cases} a^{3} x - \frac {3 a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 a^{2} b \tan ^{2}{\left (c + d x \right )}}{2 d} - 3 a b^{2} x + \frac {3 a b^{2} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} + \frac {3 a b^{2} \tan {\left (c + d x \right )}}{d} + \frac {b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{3} \tan ^{8}{\left (c + d x \right )}}{8 d} - \frac {b^{3} \tan ^{6}{\left (c + d x \right )}}{6 d} + \frac {b^{3} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan ^{3}{\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x - 3*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) + 3*a**2*b*tan(c + d*x)**2/(2*d) - 3*a*b**2*x + 3*
a*b**2*tan(c + d*x)**5/(5*d) - a*b**2*tan(c + d*x)**3/d + 3*a*b**2*tan(c + d*x)/d + b**3*log(tan(c + d*x)**2 +
1)/(2*d) + b**3*tan(c + d*x)**8/(8*d) - b**3*tan(c + d*x)**6/(6*d) + b**3*tan(c + d*x)**4/(4*d) - b**3*tan(c
+ d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan(c)**3)**3, True))

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