3.251 \(\int (a+b \tan ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=77 \[ \frac{b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)}{d}+\frac{b^2 (3 a-b) \tan ^3(c+d x)}{3 d}+x (a-b)^3+\frac{b^3 \tan ^5(c+d x)}{5 d} \]

[Out]

(a - b)^3*x + (b*(3*a^2 - 3*a*b + b^2)*Tan[c + d*x])/d + ((3*a - b)*b^2*Tan[c + d*x]^3)/(3*d) + (b^3*Tan[c + d
*x]^5)/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.048896, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3661, 390, 203} \[ \frac{b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)}{d}+\frac{b^2 (3 a-b) \tan ^3(c+d x)}{3 d}+x (a-b)^3+\frac{b^3 \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^2)^3,x]

[Out]

(a - b)^3*x + (b*(3*a^2 - 3*a*b + b^2)*Tan[c + d*x])/d + ((3*a - b)*b^2*Tan[c + d*x]^3)/(3*d) + (b^3*Tan[c + d
*x]^5)/(5*d)

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])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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])

Rubi steps

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

Mathematica [A]  time = 0.921627, size = 102, normalized size = 1.32 \[ \frac{\tan (c+d x) \left (b \left (45 a^2-15 a b \left (3-\tan ^2(c+d x)\right )+b^2 \left (3 \tan ^4(c+d x)-5 \tan ^2(c+d x)+15\right )\right )+\frac{15 (a-b)^3 \tanh ^{-1}\left (\sqrt{-\tan ^2(c+d x)}\right )}{\sqrt{-\tan ^2(c+d x)}}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^2)^3,x]

[Out]

(Tan[c + d*x]*((15*(a - b)^3*ArcTanh[Sqrt[-Tan[c + d*x]^2]])/Sqrt[-Tan[c + d*x]^2] + b*(45*a^2 - 15*a*b*(3 - T
an[c + d*x]^2) + b^2*(15 - 5*Tan[c + d*x]^2 + 3*Tan[c + d*x]^4))))/(15*d)

________________________________________________________________________________________

Maple [B]  time = 0.003, size = 154, normalized size = 2. \begin{align*}{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{3\,d}}+3\,{\frac{\tan \left ( dx+c \right ){a}^{2}b}{d}}-3\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d}}-3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}b}{d}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{3}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^3*tan(d*x+c)^5/d+a*b^2*tan(d*x+c)^3/d-1/3/d*tan(d*x+c)^3*b^3+3/d*tan(d*x+c)*a^2*b-3*a*b^2*tan(d*x+c)/d+1
/d*b^3*tan(d*x+c)+1/d*arctan(tan(d*x+c))*a^3-3/d*arctan(tan(d*x+c))*a^2*b+3/d*arctan(tan(d*x+c))*a*b^2-1/d*arc
tan(tan(d*x+c))*b^3

________________________________________________________________________________________

Maxima [A]  time = 1.6906, size = 140, normalized size = 1.82 \begin{align*} a^{3} x - \frac{3 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b}{d} + \frac{{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a b^{2}}{d} + \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 )} b^{3}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 3*(d*x + c - tan(d*x + c))*a^2*b/d + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a*b^2/d + 1/15*(3
*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*b^3/d

________________________________________________________________________________________

Fricas [A]  time = 1.40855, size = 204, normalized size = 2.65 \begin{align*} \frac{3 \, b^{3} \tan \left (d x + c\right )^{5} + 5 \,{\left (3 \, a b^{2} - b^{3}\right )} \tan \left (d x + c\right )^{3} + 15 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x + 15 \,{\left (3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^3*tan(d*x + c)^5 + 5*(3*a*b^2 - b^3)*tan(d*x + c)^3 + 15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*x + 15*(3
*a^2*b - 3*a*b^2 + b^3)*tan(d*x + c))/d

________________________________________________________________________________________

Sympy [A]  time = 0.838256, size = 126, normalized size = 1.64 \begin{align*} \begin{cases} a^{3} x - 3 a^{2} b x + \frac{3 a^{2} b \tan{\left (c + d x \right )}}{d} + 3 a b^{2} x + \frac{a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac{3 a b^{2} \tan{\left (c + d x \right )}}{d} - b^{3} x + \frac{b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac{b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac{b^{3} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan ^{2}{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x - 3*a**2*b*x + 3*a**2*b*tan(c + d*x)/d + 3*a*b**2*x + a*b**2*tan(c + d*x)**3/d - 3*a*b**2*ta
n(c + d*x)/d - b**3*x + b**3*tan(c + d*x)**5/(5*d) - b**3*tan(c + d*x)**3/(3*d) + b**3*tan(c + d*x)/d, Ne(d, 0
)), (x*(a + b*tan(c)**2)**3, True))

________________________________________________________________________________________

Giac [B]  time = 2.33848, size = 1386, normalized size = 18. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a^3*d*x*tan(d*x)^5*tan(c)^5 - 45*a^2*b*d*x*tan(d*x)^5*tan(c)^5 + 45*a*b^2*d*x*tan(d*x)^5*tan(c)^5 - 1
5*b^3*d*x*tan(d*x)^5*tan(c)^5 - 75*a^3*d*x*tan(d*x)^4*tan(c)^4 + 225*a^2*b*d*x*tan(d*x)^4*tan(c)^4 - 225*a*b^2
*d*x*tan(d*x)^4*tan(c)^4 + 75*b^3*d*x*tan(d*x)^4*tan(c)^4 - 45*a^2*b*tan(d*x)^5*tan(c)^4 + 45*a*b^2*tan(d*x)^5
*tan(c)^4 - 15*b^3*tan(d*x)^5*tan(c)^4 - 45*a^2*b*tan(d*x)^4*tan(c)^5 + 45*a*b^2*tan(d*x)^4*tan(c)^5 - 15*b^3*
tan(d*x)^4*tan(c)^5 + 150*a^3*d*x*tan(d*x)^3*tan(c)^3 - 450*a^2*b*d*x*tan(d*x)^3*tan(c)^3 + 450*a*b^2*d*x*tan(
d*x)^3*tan(c)^3 - 150*b^3*d*x*tan(d*x)^3*tan(c)^3 - 15*a*b^2*tan(d*x)^5*tan(c)^2 + 5*b^3*tan(d*x)^5*tan(c)^2 +
180*a^2*b*tan(d*x)^4*tan(c)^3 - 225*a*b^2*tan(d*x)^4*tan(c)^3 + 75*b^3*tan(d*x)^4*tan(c)^3 + 180*a^2*b*tan(d*
x)^3*tan(c)^4 - 225*a*b^2*tan(d*x)^3*tan(c)^4 + 75*b^3*tan(d*x)^3*tan(c)^4 - 15*a*b^2*tan(d*x)^2*tan(c)^5 + 5*
b^3*tan(d*x)^2*tan(c)^5 - 150*a^3*d*x*tan(d*x)^2*tan(c)^2 + 450*a^2*b*d*x*tan(d*x)^2*tan(c)^2 - 450*a*b^2*d*x*
tan(d*x)^2*tan(c)^2 + 150*b^3*d*x*tan(d*x)^2*tan(c)^2 - 3*b^3*tan(d*x)^5 + 30*a*b^2*tan(d*x)^4*tan(c) - 25*b^3
*tan(d*x)^4*tan(c) - 270*a^2*b*tan(d*x)^3*tan(c)^2 + 360*a*b^2*tan(d*x)^3*tan(c)^2 - 150*b^3*tan(d*x)^3*tan(c)
^2 - 270*a^2*b*tan(d*x)^2*tan(c)^3 + 360*a*b^2*tan(d*x)^2*tan(c)^3 - 150*b^3*tan(d*x)^2*tan(c)^3 + 30*a*b^2*ta
n(d*x)*tan(c)^4 - 25*b^3*tan(d*x)*tan(c)^4 - 3*b^3*tan(c)^5 + 75*a^3*d*x*tan(d*x)*tan(c) - 225*a^2*b*d*x*tan(d
*x)*tan(c) + 225*a*b^2*d*x*tan(d*x)*tan(c) - 75*b^3*d*x*tan(d*x)*tan(c) - 15*a*b^2*tan(d*x)^3 + 5*b^3*tan(d*x)
^3 + 180*a^2*b*tan(d*x)^2*tan(c) - 225*a*b^2*tan(d*x)^2*tan(c) + 75*b^3*tan(d*x)^2*tan(c) + 180*a^2*b*tan(d*x)
*tan(c)^2 - 225*a*b^2*tan(d*x)*tan(c)^2 + 75*b^3*tan(d*x)*tan(c)^2 - 15*a*b^2*tan(c)^3 + 5*b^3*tan(c)^3 - 15*a
^3*d*x + 45*a^2*b*d*x - 45*a*b^2*d*x + 15*b^3*d*x - 45*a^2*b*tan(d*x) + 45*a*b^2*tan(d*x) - 15*b^3*tan(d*x) -
45*a^2*b*tan(c) + 45*a*b^2*tan(c) - 15*b^3*tan(c))/(d*tan(d*x)^5*tan(c)^5 - 5*d*tan(d*x)^4*tan(c)^4 + 10*d*tan
(d*x)^3*tan(c)^3 - 10*d*tan(d*x)^2*tan(c)^2 + 5*d*tan(d*x)*tan(c) - d)