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

Optimal. Leaf size=82 \[ \frac{b (2 a+b) \tan ^3(c+d x)}{3 d}-\frac{b (2 a+b) \tan (c+d x)}{d}+x (a+b)^2+\frac{b^2 \tan ^7(c+d x)}{7 d}-\frac{b^2 \tan ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0547545, antiderivative size = 82, 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, 1154, 203} \[ \frac{b (2 a+b) \tan ^3(c+d x)}{3 d}-\frac{b (2 a+b) \tan (c+d x)}{d}+x (a+b)^2+\frac{b^2 \tan ^7(c+d x)}{7 d}-\frac{b^2 \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^4)^2,x]

[Out]

(a + b)^2*x - (b*(2*a + b)*Tan[c + d*x])/d + (b*(2*a + b)*Tan[c + d*x]^3)/(3*d) - (b^2*Tan[c + d*x]^5)/(5*d) +
(b^2*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 1154

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a
+ c*x^4)^p, x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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

Mathematica [A]  time = 0.526095, size = 75, normalized size = 0.91 \[ \frac{105 (a+b)^2 \tan ^{-1}(\tan (c+d x))+b \tan (c+d x) \left (35 (2 a+b) \tan ^2(c+d x)-105 (2 a+b)+15 b \tan ^6(c+d x)-21 b \tan ^4(c+d x)\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^4)^2,x]

[Out]

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

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Maple [A]  time = 0.006, size = 134, normalized size = 1.6 \begin{align*}{\frac{{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{7}}{7\,d}}-{\frac{{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{2\,ab \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-2\,{\frac{a\tan \left ( dx+c \right ) b}{d}}-{\frac{{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d}}+2\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) ab}{d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.57343, size = 123, normalized size = 1.5 \begin{align*} a^{2} x + \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a b}{3 \, 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^{2}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.43597, size = 208, normalized size = 2.54 \begin{align*} \frac{15 \, b^{2} \tan \left (d x + c\right )^{7} - 21 \, b^{2} \tan \left (d x + c\right )^{5} + 35 \,{\left (2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{3} + 105 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 105 \,{\left (2 \, a b + b^{2}\right )} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.35195, size = 116, normalized size = 1.41 \begin{align*} \begin{cases} a^{2} x + 2 a b x + \frac{2 a b \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 a b \tan{\left (c + d x \right )}}{d} + b^{2} x + \frac{b^{2} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac{b^{2} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac{b^{2} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{2} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan ^{4}{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 6.44607, size = 1594, normalized size = 19.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(105*a^2*d*x*tan(d*x)^7*tan(c)^7 + 210*a*b*d*x*tan(d*x)^7*tan(c)^7 + 105*b^2*d*x*tan(d*x)^7*tan(c)^7 - 7
35*a^2*d*x*tan(d*x)^6*tan(c)^6 - 1470*a*b*d*x*tan(d*x)^6*tan(c)^6 - 735*b^2*d*x*tan(d*x)^6*tan(c)^6 + 210*a*b*
tan(d*x)^7*tan(c)^6 + 105*b^2*tan(d*x)^7*tan(c)^6 + 210*a*b*tan(d*x)^6*tan(c)^7 + 105*b^2*tan(d*x)^6*tan(c)^7
+ 2205*a^2*d*x*tan(d*x)^5*tan(c)^5 + 4410*a*b*d*x*tan(d*x)^5*tan(c)^5 + 2205*b^2*d*x*tan(d*x)^5*tan(c)^5 - 70*
a*b*tan(d*x)^7*tan(c)^4 - 35*b^2*tan(d*x)^7*tan(c)^4 - 1470*a*b*tan(d*x)^6*tan(c)^5 - 735*b^2*tan(d*x)^6*tan(c
)^5 - 1470*a*b*tan(d*x)^5*tan(c)^6 - 735*b^2*tan(d*x)^5*tan(c)^6 - 70*a*b*tan(d*x)^4*tan(c)^7 - 35*b^2*tan(d*x
)^4*tan(c)^7 - 3675*a^2*d*x*tan(d*x)^4*tan(c)^4 - 7350*a*b*d*x*tan(d*x)^4*tan(c)^4 - 3675*b^2*d*x*tan(d*x)^4*t
an(c)^4 + 21*b^2*tan(d*x)^7*tan(c)^2 + 280*a*b*tan(d*x)^6*tan(c)^3 + 245*b^2*tan(d*x)^6*tan(c)^3 + 3990*a*b*ta
n(d*x)^5*tan(c)^4 + 2205*b^2*tan(d*x)^5*tan(c)^4 + 3990*a*b*tan(d*x)^4*tan(c)^5 + 2205*b^2*tan(d*x)^4*tan(c)^5
+ 280*a*b*tan(d*x)^3*tan(c)^6 + 245*b^2*tan(d*x)^3*tan(c)^6 + 21*b^2*tan(d*x)^2*tan(c)^7 + 3675*a^2*d*x*tan(d
*x)^3*tan(c)^3 + 7350*a*b*d*x*tan(d*x)^3*tan(c)^3 + 3675*b^2*d*x*tan(d*x)^3*tan(c)^3 - 15*b^2*tan(d*x)^7 - 147
*b^2*tan(d*x)^6*tan(c) - 420*a*b*tan(d*x)^5*tan(c)^2 - 735*b^2*tan(d*x)^5*tan(c)^2 - 5460*a*b*tan(d*x)^4*tan(c
)^3 - 3675*b^2*tan(d*x)^4*tan(c)^3 - 5460*a*b*tan(d*x)^3*tan(c)^4 - 3675*b^2*tan(d*x)^3*tan(c)^4 - 420*a*b*tan
(d*x)^2*tan(c)^5 - 735*b^2*tan(d*x)^2*tan(c)^5 - 147*b^2*tan(d*x)*tan(c)^6 - 15*b^2*tan(c)^7 - 2205*a^2*d*x*ta
n(d*x)^2*tan(c)^2 - 4410*a*b*d*x*tan(d*x)^2*tan(c)^2 - 2205*b^2*d*x*tan(d*x)^2*tan(c)^2 + 21*b^2*tan(d*x)^5 +
280*a*b*tan(d*x)^4*tan(c) + 245*b^2*tan(d*x)^4*tan(c) + 3990*a*b*tan(d*x)^3*tan(c)^2 + 2205*b^2*tan(d*x)^3*tan
(c)^2 + 3990*a*b*tan(d*x)^2*tan(c)^3 + 2205*b^2*tan(d*x)^2*tan(c)^3 + 280*a*b*tan(d*x)*tan(c)^4 + 245*b^2*tan(
d*x)*tan(c)^4 + 21*b^2*tan(c)^5 + 735*a^2*d*x*tan(d*x)*tan(c) + 1470*a*b*d*x*tan(d*x)*tan(c) + 735*b^2*d*x*tan
(d*x)*tan(c) - 70*a*b*tan(d*x)^3 - 35*b^2*tan(d*x)^3 - 1470*a*b*tan(d*x)^2*tan(c) - 735*b^2*tan(d*x)^2*tan(c)
- 1470*a*b*tan(d*x)*tan(c)^2 - 735*b^2*tan(d*x)*tan(c)^2 - 70*a*b*tan(c)^3 - 35*b^2*tan(c)^3 - 105*a^2*d*x - 2
10*a*b*d*x - 105*b^2*d*x + 210*a*b*tan(d*x) + 105*b^2*tan(d*x) + 210*a*b*tan(c) + 105*b^2*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*t
an(c)^3 - 21*d*tan(d*x)^2*tan(c)^2 + 7*d*tan(d*x)*tan(c) - d)