3.250 \(\int (a+b \tan ^2(c+d x))^4 \, dx\)

Optimal. Leaf size=115 \[ \frac{b^2 \left (6 a^2-4 a b+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{b (2 a-b) \left (2 a^2-2 a b+b^2\right ) \tan (c+d x)}{d}+\frac{b^3 (4 a-b) \tan ^5(c+d x)}{5 d}+x (a-b)^4+\frac{b^4 \tan ^7(c+d x)}{7 d} \]

[Out]

(a - b)^4*x + ((2*a - b)*b*(2*a^2 - 2*a*b + b^2)*Tan[c + d*x])/d + (b^2*(6*a^2 - 4*a*b + b^2)*Tan[c + d*x]^3)/
(3*d) + ((4*a - b)*b^3*Tan[c + d*x]^5)/(5*d) + (b^4*Tan[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.0732737, antiderivative size = 115, 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} \[ \frac{b^2 \left (6 a^2-4 a b+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{b (2 a-b) \left (2 a^2-2 a b+b^2\right ) \tan (c+d x)}{d}+\frac{b^3 (4 a-b) \tan ^5(c+d x)}{5 d}+x (a-b)^4+\frac{b^4 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^2)^4,x]

[Out]

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

Mathematica [A]  time = 1.88495, size = 137, normalized size = 1.19 \[ \frac{\tan (c+d x) \left (b \left (35 b \left (6 a^2-4 a b+b^2\right ) \tan ^2(c+d x)+105 \left (-6 a^2 b+4 a^3+4 a b^2-b^3\right )+21 b^2 (4 a-b) \tan ^4(c+d x)+15 b^3 \tan ^6(c+d x)\right )+\frac{105 (a-b)^4 \tanh ^{-1}\left (\sqrt{-\tan ^2(c+d x)}\right )}{\sqrt{-\tan ^2(c+d x)}}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^2)^4,x]

[Out]

(Tan[c + d*x]*((105*(a - b)^4*ArcTanh[Sqrt[-Tan[c + d*x]^2]])/Sqrt[-Tan[c + d*x]^2] + b*(105*(4*a^3 - 6*a^2*b
+ 4*a*b^2 - b^3) + 35*b*(6*a^2 - 4*a*b + b^2)*Tan[c + d*x]^2 + 21*(4*a - b)*b^2*Tan[c + d*x]^4 + 15*b^3*Tan[c
+ d*x]^6)))/(105*d)

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Maple [B]  time = 0.006, size = 242, normalized size = 2.1 \begin{align*}{\frac{{b}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{7}}{7\,d}}+{\frac{4\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}a{b}^{3}}{5\,d}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}{b}^{4}}{5\,d}}+2\,{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{a}^{2}{b}^{2}}{d}}-{\frac{4\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}a{b}^{3}}{3\,d}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{4}}{3\,d}}+4\,{\frac{\tan \left ( dx+c \right ){a}^{3}b}{d}}-6\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{a{b}^{3}\tan \left ( dx+c \right ) }{d}}-{\frac{{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{4}}{d}}-4\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}b}{d}}+6\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}{b}^{2}}{d}}-4\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{3}}{d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{4}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c)^2)^4,x)

[Out]

1/7*b^4*tan(d*x+c)^7/d+4/5/d*tan(d*x+c)^5*a*b^3-1/5/d*tan(d*x+c)^5*b^4+2/d*tan(d*x+c)^3*a^2*b^2-4/3/d*tan(d*x+
c)^3*a*b^3+1/3/d*tan(d*x+c)^3*b^4+4/d*tan(d*x+c)*a^3*b-6/d*a^2*b^2*tan(d*x+c)+4/d*a*b^3*tan(d*x+c)-1/d*b^4*tan
(d*x+c)+1/d*arctan(tan(d*x+c))*a^4-4/d*arctan(tan(d*x+c))*a^3*b+6/d*arctan(tan(d*x+c))*a^2*b^2-4/d*arctan(tan(
d*x+c))*a*b^3+1/d*arctan(tan(d*x+c))*b^4

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Maxima [A]  time = 1.69935, size = 219, normalized size = 1.9 \begin{align*} a^{4} x - \frac{4 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} b}{d} + \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} b^{2}}{d} + \frac{4 \,{\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^{3}}{15 \, d} + \frac{{\left (15 \, \tan \left (d x + c\right )^{7} - 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 105 \, d x + 105 \, c - 105 \, \tan \left (d x + c\right )\right )} b^{4}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c)^2)^4,x, algorithm="maxima")

[Out]

a^4*x - 4*(d*x + c - tan(d*x + c))*a^3*b/d + 2*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2*b^2/d + 4/1
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^3/d + 1/105*(15*tan(d*x + c)^7 -
21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 105*d*x + 105*c - 105*tan(d*x + c))*b^4/d

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Fricas [A]  time = 1.41045, size = 308, normalized size = 2.68 \begin{align*} \frac{15 \, b^{4} \tan \left (d x + c\right )^{7} + 21 \,{\left (4 \, a b^{3} - b^{4}\right )} \tan \left (d x + c\right )^{5} + 35 \,{\left (6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{3} + 105 \,{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} d x + 105 \,{\left (4 \, a^{3} b - 6 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c)^2)^4,x, algorithm="fricas")

[Out]

1/105*(15*b^4*tan(d*x + c)^7 + 21*(4*a*b^3 - b^4)*tan(d*x + c)^5 + 35*(6*a^2*b^2 - 4*a*b^3 + b^4)*tan(d*x + c)
^3 + 105*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x + 105*(4*a^3*b - 6*a^2*b^2 + 4*a*b^3 - b^4)*tan(d*x +
c))/d

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Sympy [A]  time = 1.54244, size = 209, normalized size = 1.82 \begin{align*} \begin{cases} a^{4} x - 4 a^{3} b x + \frac{4 a^{3} b \tan{\left (c + d x \right )}}{d} + 6 a^{2} b^{2} x + \frac{2 a^{2} b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac{6 a^{2} b^{2} \tan{\left (c + d x \right )}}{d} - 4 a b^{3} x + \frac{4 a b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac{4 a b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac{4 a b^{3} \tan{\left (c + d x \right )}}{d} + b^{4} x + \frac{b^{4} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac{b^{4} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac{b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{4} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan ^{2}{\left (c \right )}\right )^{4} & \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)**4,x)

[Out]

Piecewise((a**4*x - 4*a**3*b*x + 4*a**3*b*tan(c + d*x)/d + 6*a**2*b**2*x + 2*a**2*b**2*tan(c + d*x)**3/d - 6*a
**2*b**2*tan(c + d*x)/d - 4*a*b**3*x + 4*a*b**3*tan(c + d*x)**5/(5*d) - 4*a*b**3*tan(c + d*x)**3/(3*d) + 4*a*b
**3*tan(c + d*x)/d + b**4*x + b**4*tan(c + d*x)**7/(7*d) - b**4*tan(c + d*x)**5/(5*d) + b**4*tan(c + d*x)**3/(
3*d) - b**4*tan(c + d*x)/d, Ne(d, 0)), (x*(a + b*tan(c)**2)**4, True))

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Giac [B]  time = 4.9932, size = 2982, normalized size = 25.93 \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)^4,x, algorithm="giac")

[Out]

1/105*(105*a^4*d*x*tan(d*x)^7*tan(c)^7 - 420*a^3*b*d*x*tan(d*x)^7*tan(c)^7 + 630*a^2*b^2*d*x*tan(d*x)^7*tan(c)
^7 - 420*a*b^3*d*x*tan(d*x)^7*tan(c)^7 + 105*b^4*d*x*tan(d*x)^7*tan(c)^7 - 735*a^4*d*x*tan(d*x)^6*tan(c)^6 + 2
940*a^3*b*d*x*tan(d*x)^6*tan(c)^6 - 4410*a^2*b^2*d*x*tan(d*x)^6*tan(c)^6 + 2940*a*b^3*d*x*tan(d*x)^6*tan(c)^6
- 735*b^4*d*x*tan(d*x)^6*tan(c)^6 - 420*a^3*b*tan(d*x)^7*tan(c)^6 + 630*a^2*b^2*tan(d*x)^7*tan(c)^6 - 420*a*b^
3*tan(d*x)^7*tan(c)^6 + 105*b^4*tan(d*x)^7*tan(c)^6 - 420*a^3*b*tan(d*x)^6*tan(c)^7 + 630*a^2*b^2*tan(d*x)^6*t
an(c)^7 - 420*a*b^3*tan(d*x)^6*tan(c)^7 + 105*b^4*tan(d*x)^6*tan(c)^7 + 2205*a^4*d*x*tan(d*x)^5*tan(c)^5 - 882
0*a^3*b*d*x*tan(d*x)^5*tan(c)^5 + 13230*a^2*b^2*d*x*tan(d*x)^5*tan(c)^5 - 8820*a*b^3*d*x*tan(d*x)^5*tan(c)^5 +
2205*b^4*d*x*tan(d*x)^5*tan(c)^5 - 210*a^2*b^2*tan(d*x)^7*tan(c)^4 + 140*a*b^3*tan(d*x)^7*tan(c)^4 - 35*b^4*t
an(d*x)^7*tan(c)^4 + 2520*a^3*b*tan(d*x)^6*tan(c)^5 - 4410*a^2*b^2*tan(d*x)^6*tan(c)^5 + 2940*a*b^3*tan(d*x)^6
*tan(c)^5 - 735*b^4*tan(d*x)^6*tan(c)^5 + 2520*a^3*b*tan(d*x)^5*tan(c)^6 - 4410*a^2*b^2*tan(d*x)^5*tan(c)^6 +
2940*a*b^3*tan(d*x)^5*tan(c)^6 - 735*b^4*tan(d*x)^5*tan(c)^6 - 210*a^2*b^2*tan(d*x)^4*tan(c)^7 + 140*a*b^3*tan
(d*x)^4*tan(c)^7 - 35*b^4*tan(d*x)^4*tan(c)^7 - 3675*a^4*d*x*tan(d*x)^4*tan(c)^4 + 14700*a^3*b*d*x*tan(d*x)^4*
tan(c)^4 - 22050*a^2*b^2*d*x*tan(d*x)^4*tan(c)^4 + 14700*a*b^3*d*x*tan(d*x)^4*tan(c)^4 - 3675*b^4*d*x*tan(d*x)
^4*tan(c)^4 - 84*a*b^3*tan(d*x)^7*tan(c)^2 + 21*b^4*tan(d*x)^7*tan(c)^2 + 840*a^2*b^2*tan(d*x)^6*tan(c)^3 - 98
0*a*b^3*tan(d*x)^6*tan(c)^3 + 245*b^4*tan(d*x)^6*tan(c)^3 - 6300*a^3*b*tan(d*x)^5*tan(c)^4 + 11970*a^2*b^2*tan
(d*x)^5*tan(c)^4 - 8820*a*b^3*tan(d*x)^5*tan(c)^4 + 2205*b^4*tan(d*x)^5*tan(c)^4 - 6300*a^3*b*tan(d*x)^4*tan(c
)^5 + 11970*a^2*b^2*tan(d*x)^4*tan(c)^5 - 8820*a*b^3*tan(d*x)^4*tan(c)^5 + 2205*b^4*tan(d*x)^4*tan(c)^5 + 840*
a^2*b^2*tan(d*x)^3*tan(c)^6 - 980*a*b^3*tan(d*x)^3*tan(c)^6 + 245*b^4*tan(d*x)^3*tan(c)^6 - 84*a*b^3*tan(d*x)^
2*tan(c)^7 + 21*b^4*tan(d*x)^2*tan(c)^7 + 3675*a^4*d*x*tan(d*x)^3*tan(c)^3 - 14700*a^3*b*d*x*tan(d*x)^3*tan(c)
^3 + 22050*a^2*b^2*d*x*tan(d*x)^3*tan(c)^3 - 14700*a*b^3*d*x*tan(d*x)^3*tan(c)^3 + 3675*b^4*d*x*tan(d*x)^3*tan
(c)^3 - 15*b^4*tan(d*x)^7 + 168*a*b^3*tan(d*x)^6*tan(c) - 147*b^4*tan(d*x)^6*tan(c) - 1260*a^2*b^2*tan(d*x)^5*
tan(c)^2 + 1680*a*b^3*tan(d*x)^5*tan(c)^2 - 735*b^4*tan(d*x)^5*tan(c)^2 + 8400*a^3*b*tan(d*x)^4*tan(c)^3 - 163
80*a^2*b^2*tan(d*x)^4*tan(c)^3 + 12600*a*b^3*tan(d*x)^4*tan(c)^3 - 3675*b^4*tan(d*x)^4*tan(c)^3 + 8400*a^3*b*t
an(d*x)^3*tan(c)^4 - 16380*a^2*b^2*tan(d*x)^3*tan(c)^4 + 12600*a*b^3*tan(d*x)^3*tan(c)^4 - 3675*b^4*tan(d*x)^3
*tan(c)^4 - 1260*a^2*b^2*tan(d*x)^2*tan(c)^5 + 1680*a*b^3*tan(d*x)^2*tan(c)^5 - 735*b^4*tan(d*x)^2*tan(c)^5 +
168*a*b^3*tan(d*x)*tan(c)^6 - 147*b^4*tan(d*x)*tan(c)^6 - 15*b^4*tan(c)^7 - 2205*a^4*d*x*tan(d*x)^2*tan(c)^2 +
8820*a^3*b*d*x*tan(d*x)^2*tan(c)^2 - 13230*a^2*b^2*d*x*tan(d*x)^2*tan(c)^2 + 8820*a*b^3*d*x*tan(d*x)^2*tan(c)
^2 - 2205*b^4*d*x*tan(d*x)^2*tan(c)^2 - 84*a*b^3*tan(d*x)^5 + 21*b^4*tan(d*x)^5 + 840*a^2*b^2*tan(d*x)^4*tan(c
) - 980*a*b^3*tan(d*x)^4*tan(c) + 245*b^4*tan(d*x)^4*tan(c) - 6300*a^3*b*tan(d*x)^3*tan(c)^2 + 11970*a^2*b^2*t
an(d*x)^3*tan(c)^2 - 8820*a*b^3*tan(d*x)^3*tan(c)^2 + 2205*b^4*tan(d*x)^3*tan(c)^2 - 6300*a^3*b*tan(d*x)^2*tan
(c)^3 + 11970*a^2*b^2*tan(d*x)^2*tan(c)^3 - 8820*a*b^3*tan(d*x)^2*tan(c)^3 + 2205*b^4*tan(d*x)^2*tan(c)^3 + 84
0*a^2*b^2*tan(d*x)*tan(c)^4 - 980*a*b^3*tan(d*x)*tan(c)^4 + 245*b^4*tan(d*x)*tan(c)^4 - 84*a*b^3*tan(c)^5 + 21
*b^4*tan(c)^5 + 735*a^4*d*x*tan(d*x)*tan(c) - 2940*a^3*b*d*x*tan(d*x)*tan(c) + 4410*a^2*b^2*d*x*tan(d*x)*tan(c
) - 2940*a*b^3*d*x*tan(d*x)*tan(c) + 735*b^4*d*x*tan(d*x)*tan(c) - 210*a^2*b^2*tan(d*x)^3 + 140*a*b^3*tan(d*x)
^3 - 35*b^4*tan(d*x)^3 + 2520*a^3*b*tan(d*x)^2*tan(c) - 4410*a^2*b^2*tan(d*x)^2*tan(c) + 2940*a*b^3*tan(d*x)^2
*tan(c) - 735*b^4*tan(d*x)^2*tan(c) + 2520*a^3*b*tan(d*x)*tan(c)^2 - 4410*a^2*b^2*tan(d*x)*tan(c)^2 + 2940*a*b
^3*tan(d*x)*tan(c)^2 - 735*b^4*tan(d*x)*tan(c)^2 - 210*a^2*b^2*tan(c)^3 + 140*a*b^3*tan(c)^3 - 35*b^4*tan(c)^3
- 105*a^4*d*x + 420*a^3*b*d*x - 630*a^2*b^2*d*x + 420*a*b^3*d*x - 105*b^4*d*x - 420*a^3*b*tan(d*x) + 630*a^2*
b^2*tan(d*x) - 420*a*b^3*tan(d*x) + 105*b^4*tan(d*x) - 420*a^3*b*tan(c) + 630*a^2*b^2*tan(c) - 420*a*b^3*tan(c
) + 105*b^4*tan(c))/(d*tan(d*x)^7*tan(c)^7 - 7*d*tan(d*x)^6*tan(c)^6 + 21*d*tan(d*x)^5*tan(c)^5 - 35*d*tan(d*x
)^4*tan(c)^4 + 35*d*tan(d*x)^3*tan(c)^3 - 21*d*tan(d*x)^2*tan(c)^2 + 7*d*tan(d*x)*tan(c) - d)