3.179 \(\int (a+b \sec ^2(c+d x))^4 \, dx\)

Optimal. Leaf size=111 \[ \frac{b^2 \left (6 a^2+8 a b+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan (c+d x)}{d}+a^4 x+\frac{b^3 (4 a+3 b) \tan ^5(c+d x)}{5 d}+\frac{b^4 \tan ^7(c+d x)}{7 d} \]

[Out]

a^4*x + (b*(2*a + b)*(2*a^2 + 2*a*b + b^2)*Tan[c + d*x])/d + (b^2*(6*a^2 + 8*a*b + 3*b^2)*Tan[c + d*x]^3)/(3*d
) + (b^3*(4*a + 3*b)*Tan[c + d*x]^5)/(5*d) + (b^4*Tan[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.0648342, antiderivative size = 111, 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 = {4128, 390, 203} \[ \frac{b^2 \left (6 a^2+8 a b+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tan (c+d x)}{d}+a^4 x+\frac{b^3 (4 a+3 b) \tan ^5(c+d x)}{5 d}+\frac{b^4 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x]^2)^4,x]

[Out]

a^4*x + (b*(2*a + b)*(2*a^2 + 2*a*b + b^2)*Tan[c + d*x])/d + (b^2*(6*a^2 + 8*a*b + 3*b^2)*Tan[c + d*x]^3)/(3*d
) + (b^3*(4*a + 3*b)*Tan[c + d*x]^5)/(5*d) + (b^4*Tan[c + d*x]^7)/(7*d)

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

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

Mathematica [B]  time = 1.61481, size = 455, normalized size = 4.1 \[ \frac{\sec (c) \sec ^7(c+d x) \left (-10920 a^2 b^2 \sin (2 c+d x)+15120 a^2 b^2 \sin (2 c+3 d x)-2520 a^2 b^2 \sin (4 c+3 d x)+5880 a^2 b^2 \sin (4 c+5 d x)+840 a^2 b^2 \sin (6 c+7 d x)+18480 a^2 b^2 \sin (d x)-12600 a^3 b \sin (2 c+d x)+12600 a^3 b \sin (2 c+3 d x)-5040 a^3 b \sin (4 c+3 d x)+5040 a^3 b \sin (4 c+5 d x)-840 a^3 b \sin (6 c+5 d x)+840 a^3 b \sin (6 c+7 d x)+16800 a^3 b \sin (d x)+3675 a^4 d x \cos (2 c+d x)+2205 a^4 d x \cos (2 c+3 d x)+2205 a^4 d x \cos (4 c+3 d x)+735 a^4 d x \cos (4 c+5 d x)+735 a^4 d x \cos (6 c+5 d x)+105 a^4 d x \cos (6 c+7 d x)+105 a^4 d x \cos (8 c+7 d x)+3675 a^4 d x \cos (d x)-4480 a b^3 \sin (2 c+d x)+9408 a b^3 \sin (2 c+3 d x)+3136 a b^3 \sin (4 c+5 d x)+448 a b^3 \sin (6 c+7 d x)+11200 a b^3 \sin (d x)+2016 b^4 \sin (2 c+3 d x)+672 b^4 \sin (4 c+5 d x)+96 b^4 \sin (6 c+7 d x)+3360 b^4 \sin (d x)\right )}{13440 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x]^2)^4,x]

[Out]

(Sec[c]*Sec[c + d*x]^7*(3675*a^4*d*x*Cos[d*x] + 3675*a^4*d*x*Cos[2*c + d*x] + 2205*a^4*d*x*Cos[2*c + 3*d*x] +
2205*a^4*d*x*Cos[4*c + 3*d*x] + 735*a^4*d*x*Cos[4*c + 5*d*x] + 735*a^4*d*x*Cos[6*c + 5*d*x] + 105*a^4*d*x*Cos[
6*c + 7*d*x] + 105*a^4*d*x*Cos[8*c + 7*d*x] + 16800*a^3*b*Sin[d*x] + 18480*a^2*b^2*Sin[d*x] + 11200*a*b^3*Sin[
d*x] + 3360*b^4*Sin[d*x] - 12600*a^3*b*Sin[2*c + d*x] - 10920*a^2*b^2*Sin[2*c + d*x] - 4480*a*b^3*Sin[2*c + d*
x] + 12600*a^3*b*Sin[2*c + 3*d*x] + 15120*a^2*b^2*Sin[2*c + 3*d*x] + 9408*a*b^3*Sin[2*c + 3*d*x] + 2016*b^4*Si
n[2*c + 3*d*x] - 5040*a^3*b*Sin[4*c + 3*d*x] - 2520*a^2*b^2*Sin[4*c + 3*d*x] + 5040*a^3*b*Sin[4*c + 5*d*x] + 5
880*a^2*b^2*Sin[4*c + 5*d*x] + 3136*a*b^3*Sin[4*c + 5*d*x] + 672*b^4*Sin[4*c + 5*d*x] - 840*a^3*b*Sin[6*c + 5*
d*x] + 840*a^3*b*Sin[6*c + 7*d*x] + 840*a^2*b^2*Sin[6*c + 7*d*x] + 448*a*b^3*Sin[6*c + 7*d*x] + 96*b^4*Sin[6*c
+ 7*d*x]))/(13440*d)

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Maple [A]  time = 0.041, size = 130, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ( dx+c \right ) +4\,{a}^{3}b\tan \left ( dx+c \right ) -6\,{a}^{2}{b}^{2} \left ( -2/3-1/3\, \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \tan \left ( dx+c \right ) -4\,a{b}^{3} \left ( -{\frac{8}{15}}-1/5\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) -{b}^{4} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^4*(d*x+c)+4*a^3*b*tan(d*x+c)-6*a^2*b^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)-4*a*b^3*(-8/15-1/5*sec(d*x+c)
^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-b^4*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.00227, size = 181, normalized size = 1.63 \begin{align*} a^{4} x + \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b^{2}}{d} + \frac{4 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a b^{3}}{15 \, d} + \frac{{\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} b^{4}}{35 \, d} + \frac{4 \, a^{3} b \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*x + 2*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2*b^2/d + 4/15*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*
x + c))*a*b^3/d + 1/35*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*b^4/d + 4*
a^3*b*tan(d*x + c)/d

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Fricas [A]  time = 0.52788, size = 317, normalized size = 2.86 \begin{align*} \frac{105 \, a^{4} d x \cos \left (d x + c\right )^{7} +{\left (4 \,{\left (105 \, a^{3} b + 105 \, a^{2} b^{2} + 56 \, a b^{3} + 12 \, b^{4}\right )} \cos \left (d x + c\right )^{6} + 2 \,{\left (105 \, a^{2} b^{2} + 56 \, a b^{3} + 12 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, b^{4} + 6 \,{\left (14 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(105*a^4*d*x*cos(d*x + c)^7 + (4*(105*a^3*b + 105*a^2*b^2 + 56*a*b^3 + 12*b^4)*cos(d*x + c)^6 + 2*(105*a
^2*b^2 + 56*a*b^3 + 12*b^4)*cos(d*x + c)^4 + 15*b^4 + 6*(14*a*b^3 + 3*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*co
s(d*x + c)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec ^{2}{\left (c + d x \right )}\right )^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(c + d*x)**2)**4, x)

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Giac [A]  time = 1.23412, size = 200, normalized size = 1.8 \begin{align*} \frac{15 \, b^{4} \tan \left (d x + c\right )^{7} + 84 \, a b^{3} \tan \left (d x + c\right )^{5} + 63 \, b^{4} \tan \left (d x + c\right )^{5} + 210 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 280 \, a b^{3} \tan \left (d x + c\right )^{3} + 105 \, b^{4} \tan \left (d x + c\right )^{3} + 105 \,{\left (d x + c\right )} a^{4} + 420 \, a^{3} b \tan \left (d x + c\right ) + 630 \, a^{2} b^{2} \tan \left (d x + c\right ) + 420 \, a b^{3} \tan \left (d x + c\right ) + 105 \, b^{4} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*b^4*tan(d*x + c)^7 + 84*a*b^3*tan(d*x + c)^5 + 63*b^4*tan(d*x + c)^5 + 210*a^2*b^2*tan(d*x + c)^3 +
280*a*b^3*tan(d*x + c)^3 + 105*b^4*tan(d*x + c)^3 + 105*(d*x + c)*a^4 + 420*a^3*b*tan(d*x + c) + 630*a^2*b^2*t
an(d*x + c) + 420*a*b^3*tan(d*x + c) + 105*b^4*tan(d*x + c))/d